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Unformatted text preview: PHYSICS OLYMPIAD
(ΠΗΨΣΙΧΣ ΟΛΨΜΠΙΑ∆)
1993 MULTIPLE CHOICE SCREENING TEST
30 QUESTIONS—40 MINUTES DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN
This test contains 30 multiple choice questions. Your answer to each question must be
marked on the optical mark answer sheet that accompanies the test. Only the boxes
Select the single answer that provides the best response to each question. Please be sure to
an answer, the previous mark must be completely erased.
programs. Calculators may not be shared.
Your grade on this multiple choice test will equal your number of correct answers. There
is no penalty for guessing. It is to your advantage to answer every question.
The values of some possibly useful constants are given below:
mass of electron me = 9.1 x 1031 kg mass of proton mp = 1.7 x 1027 kg electronic charge e = 1.6 x 1019 C speed of light c = 3.0 x 108 m/s Coulomb's constant k = 9.0 x 109 Nm2/C2 permittivity constant ε0 = 8.9 x 1012 C2/Nm2 permeability constant µ0 = gravitational constant G = 6.7 x 1011 Nm2/kg2 acceleration due to gravity g = 10 m/s2 speed of sound (20 oC) vs = 340 m/s 4π x 107 Tm/A DO NOT OPEN THIS TEST UNTIL YOU ARE INSTRUCTED TO BEGIN
Copyright © 1993, AAPT 1. A ball of mass m is thrown vertically upward. Air resistance is not negligible. Assume
the force of air resistance has magnitude proportional to the velocity, and direction
opposite to the velocity's. At the highest point, the ball's acceleration is:
(A) 0 (B) less than g (C) g (D) greater than g (E) upward 2. A train is moving forward at a velocity of 2.0 m/s. At the instant the train begins to
accelerate at 0.80 m/s2, a passenger drops a quarter which takes 0.50 s to fall to the floor.
Relative to a spot on the floor directly under the quarter at release, it lands:
(A) 1.1 m toward the rear of the train.
(B) 1.0 m toward the rear of the train.
(C) 0.10 m toward the rear of the train.
(D) directly on the spot.
(E) 0.90 m toward the front of the train.
3. The dropped quarter in the preceding question (#2) is viewed by an observer standing
next to the tracks. Relative to this observer, the quarter moves
before landing.
(A) forward 1.1 m
(B) forward 1.0 m
(C) forward 0.10 m
(D) straight down
(E) backward 0.10 m
x 4. The accompanying graph of position x versus time t
represents the motion of a particle. If b and c are both
positive constants, which of the following expressions best
describes the acceleration a of the particle?
(A) a = 0 t (B) a = +b
(C) a = c
(D) a = b + ct
(E) a = b  ct
5. In the system shown to the right, a force F pushes on
block A, giving the system an acceleration a. The
coefficient of static friction between the blocks is µ. The
correct equation for block B not to slip is:
(A) a > µg (B) a < µg (C) a > g (D) a > g/µ
Page 1 of 7 F (E) a < g/µ A B 6. A block of mass m starts at rest at height h on a
frictionless inclined plane. The block slides down the
plane, travels a total distance d across a rough surface
h
with coefficient of kinetic friction µ, and compresses a
spring with force constant k a distance x before
momentarily coming to rest. Then the spring extends and the block travels back across the
rough surface, sliding up the plane. The correct expression for the maximum height h' that
the block reaches on its return is:
(A) mgh' = mgh  2µmgd
(B) mgh' = mgh + 2µmgd
(C) mgh' = mgh + 2µmgd + kx2
(D) mgh' = mgh  2µmgd kx2
(E) mgh' = mgh  2µmgd kx2  1/2 mv2
7. Air track car A has mass m and velocity v. Air track car B has mass 2m and velocity
3v. The same constant force F is applied to each car until it stops. Car A is brought to
rest in time t. The time required to stop car B is:
(A) 2t (B) 3t (C) 6t (D) 9t (E) 18t 8. In the preceding question (#7), car A travels a distance d before coming to rest. The
distance traveled by car B before coming to rest is:
(A) 2d (B) 3d (C) 6d (D) 9d (E) 18d 9. Three air track cars are initially placed as shown in the
A
accompanying figure. Car A has mass m and initial velocity v to
the right. Car B with mass m and car C with mass 4m are both
m
initially at rest. Car A collides elastically with car B, which in
turn collides elastically with car C. After the collision, car C has
a velocity of 0.4v to the right. The final velocities of cars A and B are:
(A) Car A: 0.6v to the left Car B: at rest (B) Car A: 1.4v to the left Car B: at rest (C) Car A: v to the left Car B: 0.6v to the left (D) Car A: 0.4v to the left Car B: v to the left (E) Car A: 1.6v to the left Car B: v to the right Page 2 of 7 v B C m 4m 10. Three cylinders, all of mass M, roll without slipping down an inclined plane of height
H. The cylinders are described as follows:
I. hollow of radius R
II. solid of radius R/2
III. solid of radius R
If all cylinders are released simultaneously from the same height, the cylinder (or cylinders)
reaching the bottom first is (are):
(A) I (B) II (C) III (D) I & II (E) II & III 11. The system shown to the right is free to rotate about a
frictionless axis through its center and perpendicular to the
page. All three forces are exerted tangent to their respective
rims. The magnitude of the net torque acting on the system
is: 37 R/2 F
R (A) 1.5 FR
2F (B) 1.9 FR
(C) 2.3 FR
(D) 2.5 FR
(E) 3.5 FR 12. Two identical disks are positioned on a vertical axis. The bottom
disk is rotating at angular velocity ω0 and has rotational kinetic energy K0.
The top disk is initially at rest. It is allowed to fall, and sticks to the
bottom disk. What is the rotational kinetic energy of the system after the
collision?
(A) 1/4 K0 F O (B) 1/2 K0 (C) K0 (D) 2 K0 (E) 4 K0 13. If the sun were suddenly replaced by a black hole of one solar mass, what would
happen to the earth's orbit immediately after the replacement?
(A) The earth would spiral into the black hole.
(B) The earth would spiral out away from the black hole.
(C) The radius of the earth's orbit would be unchanged, but the period of the earth's
motion would increase.
(D) Neither the radius of the orbit nor the period would change.
(E) The radius of the earth's orbit would be unchanged, but the period of the earth's
motion would decrease. Page 3 of 7 14. A hypothetical planet has density ρ, radius R, and surface gravitational acceleration g.
If the radius of the planet were doubled, but the planetary density stayed the same, the
acceleration due to gravity at the planet's surface would be:
(A) 4g (B) 2g (C) g (D) g/2 (E) g/4 15. An ideal organ pipe resonates at frequencies of 50 Hz, 150 Hz, 250 Hz,.... The pipe
is:
(A) open at both ends and of length 1.7 m.
(B) open at both ends and of length 3.4 m.
(C) open at both ends and of length 6.8 m.
(D) closed at one end, open at the other, and of length 1.7 m.
(E) closed at one end, open at the other, and of length 3.4 m.
16. A porpoise, whistleclicking at a frequency f0, swims toward an underwater vertical
cliff at a velocity that is 1.0% of the velocity of sound in sea water. The reflected
frequency experienced by the swimming porpoise is:
(A) 0.98 f0 (B) 0.99 f0 (C) f0 (D) 1.01 f0 17. Three processes compose a thermodynamic cycle shown
in the accompanying pV diagram. Process 1→2 takes place
at constant temperature. Process 2→3 takes place at constant
volume, and process 3→1 is adiabatic. During the complete
cycle, the total amount of work done is 10 J. During process
2→3, the internal energy decreases by 20 J; and during
process 3→1, 20 J of work is done on the system. How
much heat is added to the system during process 1→2? (E) 1.02 f0 1 p 2
3
V (A) 0
Q (B) 10 J 1¨ 2
2¨ 3
3¨ 1
cycle (C) 20 J
(D) 30 J
(E) 40 J W 20J
20J
+10J 18. The root mean square velocity of oxygen gas is v at room temperature. What is the
root mean square velocity of hydrogen gas at the same temperature?
(A) 16 v (B) 4 v (C) v (D) v/4 Page 4 of 7 (E) v/16 ∆U 19. Monochromatic light of wavelength λ is shone on a grating consisting of six equally
spaced slits. The first order interference maximum occurs at an angle of 0.00100 radians.
If the outer two slits are covered, the first order maximum will occur at ______ radians.
(A) 0.00025 (B) 0.00067 (C) 0.00100 (D) 0.00150 (E) 0.00400 20. You are given two lenses, a converging lens with focal length +10 cm and a diverging
lens with focal length 20 cm. Which of the following would produce a virtual image that
is larger than the object?
(A) Placing the object 5 cm from the converging lens.
(B) Placing the object 15 cm from the converging lens.
(C) Placing the object 25 cm from the converging lens.
(D) Placing the object 15 cm from the diverging lens.
(E) Placing the object 25 cm from the diverging lens.
21. Two small identical conducting spheres are separated by a distance much larger than
their diameter. They are initially given charges of 2.00 x 106 C and +4.00 x 106 C, and
found to exert a force on each other of magnitude 1.000 N. Without changing their
position, they are connected by a conducting wire. When the wire is removed, what is the
magnitude of the force between them?
(A) 0 (B) 0.125 N (C) 0.250 N (D) 1.000 N (E) 1.125 N 22. Charges of ±Q, with Q = 2.0 x 107 C, are placed at three corners of
a square whose sides are 0.10 m. The magnitude of the total field at the
center of the square is: +Q (B) 2.5 x 104 V/m
(C) 7.5 x 104 (D) 3.6 x Q V/m 105 X 0.10m (A) 5.1 x 103 V/m +Q V/m 0.10m (E) 1.08 x 106 V/m
23. Three 60 W light bulbs are mistakenly wired in series and connected to a 120 V power
supply. Assume the light bulbs are rated for single connection to 120 V. With the
mistaken connection, the power dissipated by each bulb is:
(A) 6.7 W (B) 13.3 W (C) 20 W (D) 40 W Page 5 of 7 (E) 60 W 24. Two thin spherical conducting shells are centered on the same point. The inner sphere
has radius b and total charge q. The outer sphere has radius B and total charge Q. At a
point a distance R from the common center, where b< R<B, and assuming the electric
potential is zero an infinite distance away, the total potential due to the two spheres is:
(A) kq/b + kQ/B
(B) kq/b + kQ/R
(C) kq/R
(D) kq/R + kQ/B
(E) kq/R + kQ/R
25. What is the magnitude of the total electric field for the two spheres at the point
described in the preceding question (#24)?
(A) kq/b2 + kQ/B2
(B) kq/b2 + kQ/R2
(C) kq/R2
(D) kq/R2+ kQ/B2
(E) kq/R2 + kQ/R2
26. A current of 0.10 A flows through the 25 ½
resistor represented in the diagram to the right. The
current through the 80 ½ resistor is: 80 Ω
0.1 A 25 Ω
V 20 Ω (A) 0.10 A
(B) 0.20 A 20 Ω (C) 0.30 A 60 Ω (D) 0.40 A
(E) 0.60 A
27. The circuit shown in the accompanying
figure has been connected for a very long time.
The voltage across the capacitor is:
(A) 1.2 V
(B) 2.0 V
(C) 2.4 V
(D) 4.0 V
(E) 6.0 V Page 6 of 7 100 Ω
6V
200 Ω 100 Ω
4 µF 28. The accompanying sketch represents a bubble chamber photograph.
field is directed out of the plane of the sketch. All
particles have the same velocity initially from left to
...
right. The one that is most likely an electron is:
.A . .
...
(A) A
...
(B) B
...
...
(C) C
E. .
.
(D) D The magnetic
.
.
.
.
.
.
. .
.B
.
.
.
.D
. .
.
.
.C
.
.
. (E) E
29. What is the correct expression for the magnetic field at point
X in the diagram to the right? HINT: The magnitude of the
magnetic field at the center of a circular current loop of radius R
is µ 0i/(2R).
(A) (µ 0i/4)(1/a + 1/b) into the page (B) (µ 0i/4)(1/a  1/b) out of the page (C) (µ 0i/4)(1/a  1/b)  µ 0i/(2πa) out of the page (E) (µ 0i/2)(1/a + 1/b) + µ 0i/(2πa) i out of the page (D) (µ 0i/2)(1/a  1/b) b into the page a
X 30. You are given a bar magnet and a looped coil of wire.
Which of the following would induce an emf in the coil?
I. Moving the magnet away from the coil.
II. Moving the coil toward the magnet.
III. Turning the coil about a vertical axis.
(A) I only (B) II only (C) I & II (D) I & III Page 7 of 7 (E) I, II, III .
.
.
.
.
.
. .
.
.
.
.
.
. 1993 MULTIPLE CHOICE SCREENING TEST
ANSWER KEY
1. C 16. E 2. C 17. D 3. B 18. B 4. E 19. C 5. D 20. A 6. A 21. B 7. C 22. D 8. E 23. A 9. A 24. D 10. E 25. C 11. A
26. C 12. B 27. D 13. D 28. A 14. B 29. B 15. D 30. E Physics Olympiad
Entia non multiplicanda sunt praeter necessitatem
1996 MULTIPLE CHOICE SCREENING TEST
30 QUESTIONS—40 MINUTES
INSTRUCTIONS
DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN
This test contains 30 multiple choice questions. Your answer to each question must be marked on
the optical mark answer sheet that accompanies the test. Only the boxes preceded by numbers 1
through 30 are to be used on the answer sheet.
Select the single answer that provides the best response to each question. Please be sure to use a
No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer,
the previous mark must be completely erased.
A handheld calculator may be used. However, any memory must be cleared of data and
programs. Calculators may not be shared.
Your grade on this multiple choice test will be your number of correct answers. There is no
penalty for guessing. It is to your advantage to answer every question.
The values of some possibly useful constants are given below:
mass of electron me = 9.1 x 1031 kg mass of proton mp = 1.7 x 1027 kg electronic charge e = 1.6 x 1019 C speed of light c = 3.0 x 108 m/s Coulomb's constant k = 9.0 x 109 N·m2/C2 permittivity constant ε0 = 8.9 x 1012 C2/N·m2 permeability constant µ0 = gravitational constant G = 6.7 x 1011 N·m2/kg2 acceleration due to gravity g = 10 m/s2 speed of sound (20 oC) vs = 340 m/s 4π x 107 T·m/A DO NOT OPEN THIS TEST UNTIL YOU ARE INSTRUCTED TO BEGIN
Copyright © 1996, AAPT 1. An object is projected straight upward from ground level with a velocity of 50 m/s. Ignoring
air resistance, it will return to ground level in approximately:
A. 2.5 s B. 5.0 s C. 7.5 s D. 10 s E. 15 s 2. A jogger runs with constant speed v through a forest of pine trees. A pine cone starts to fall
from a height h when the jogger is directly below it. How far behind the jogger will the pine cone
land?
2 hv2
g A. B. hv2
2g gh 2
C.
2v2 v2
E.
2g 2 gh2
D.
v2 3. A ball is thrown downward with speed 15 m/s from the roof of a 30 m building. At the same
instant a ball is thrown upward with speed 15 m/s from ground level. Relative to ground level,
the two balls pass each other at a height of:
A. 0 B. 5.0 m C. 10 m D. 15 m E. 20 m 4. A swimmer can swim with a velocity of 1.0 m/s in still water. The swimmer wishes to swim
directly across a river with a current of 0.50 m/s directed from upstream to downstream. To end
up directly across the river the swimmer must head at an angle of:
A.
B.
C.
D.
E. tan1(1/2) upstream.
sin1(1/2) upstream.
directly across the river.
sin1(1/2) downstream
tan1(1/2) downstream 5. What is the tension T in the rope if the 10N weight is moving
upward with constant velocity?
A. 3.5 N
B. 5.0 N
C. 7.1 N
D. 10 N
E. 14 N 1996 Physics Olympiad T 45o 10 N page 1 6. As shown to the right, two
Case I
Case II
blocks with masses m and M
F
(M > m) are pushed by a force F
F
M
M
m
m
in both Case I and Case II. The
surface is horizontal and
frictionless. Let RI be the force
that m exerts on M in case I and RII be the force that m exerts on M in case II. Which of the
following statements is true?
A.
B.
C.
D.
E. RI = RII = 0
RI = RII and is not equal to zero or F.
RI = RII = F
RI < RII
RI > RII 7. Two blocks, with masses 17 kg and 15 kg, are
connected by a light string that passes over a
frictionless pulley of negligible mass as shown to the
right. The surfaces of the planes are frictionless. The
blocks are released from rest. T1 and T2 are the
tensions in the strings. Which of the following
statements is correct?
A.
B.
C.
D.
E. T2 T1 17 kg
15 kg
45o 60o The 15kg block accelerates down the plane.
The 17kg block accelerates down the plane.
Both blocks remain at rest.
T 1 > T2
T 1 < T2 8. A small block of mass m starts from rest at the top of a globe of
radius R. At what angle θ does it slide off the surface of the
globe? Assume the system is frictionless.
A. θ = 0o
B. θ = cos1(1/3)
C. θ = cos1(2/3)
D. θ = 60o
E. θ = 90o R
θ 9. An object with mass m and initial velocity v is brought to rest by a constant force F acting for
a time t and through a distance d. Possible expressions for the magnitude of the force F are:
I. (mv2)/(2d)
II. (2md)/t2
III. (mv)/t
Which of these give(s) the correct expression for the magnitude of the force F?
A. II only B. III only 1996 Physics Olympiad C. I and II only D. II and III only E. I, II, and III
page 2 10. A small sphere is moving at a constant speed in a vertical circle. Below is a list of quantities
that could be used to describe some aspect of the motion of the sphere.
I – kinetic energy
II – potential energy
III – momentum
Which of these quantities will change as this sphere moves around the circle?
A. I and II only B. I and III only C. II only D. III only E. II and III only 11. A roller coaster travels with speed vA at point A. Point B is a height H above point A.
Assuming no frictional losses and no work done by a motor, what is the speed at point B?
vA 2 − 2gH A. B. v A − 2gH C. v A − 2gH D. v A + 2gH E. vA 2 + 2gH 12. Three air track cars are initially placed as shown in the
A
B
C
accompanying figure. Car A has mass m and initial velocity v to
v
the right. Car B with mass m and car C with mass 4m are both
m
m
4m
initially at rest. Car A collides elastically with car B, which in
turn collides elastically with car C. After the collision, car B is at rest. The final velocities of cars
A and C are:
A.
B.
C.
D.
E. Car A: 0.6v to the left
Car A: 2.6v to the left
Car A: at rest
Car A: at rest
Car A: at rest Car C: 0.4v to the right
Car C: 0.4v to the right
Car C: 0.5v to the right
Car C: 0.25v to the right
Car C: v to the right 13. A child with mass m is standing at the edge of a playground merrygoround with moment of inertia I, radius R, and initial angular velocity ω .
See figure to the right. The child jumps off the edge of the merrygoround
with tangential velocity v with respect to the ground. The new angular
velocity of the merrygoround is: v
ω A. ω
Iω 2 − mv 2
I B.
C. (I + mR 2 )ω 2 − mv2 I
Iω − mvR
D.
I
I + mR 2 ω − mvR
E.
I ( ) 1996 Physics Olympiad page 3 14. As shown in the figure to the right, a
spool has outer radius R and axle radius r. A
string is wrapped around the axle of the
spool and can be pulled in any of the
directions labeled by I, II, or III. The spool
will slide to the right without rolling on the
horizontal surface if it is pulled in
direction(s):
A. I only
B. II only
C. III only
D. I and II only
E. II and III only I
II
III r
R 15. A uniform flag pole of length L and mass M is pivoted on
the ground with a frictionless hinge. The flag pole makes an
angle θ with the horizontal. The moment of inertia of the flag
pole about one end is (1/3)ML2. If it starts falling from the
position shown in the accompanying figure, the linear
acceleration of the free end of the flag pole – labeled P – would
be:
A.
Β.
C.
D.
E. P
L θ (2/3) g cos θ
(2/3) g
g
(3/2) g cos θ
(3/2) g 16. The root mean square velocity of oxygen gas (atomic mass 16) is v at room temperature.
What is the root mean square velocity of helium (atomic mass 4) at the same temperature?
A. 4 v B. 2 v C. v D. v/2 E. v/4 17. Which of the accompanying PV diagrams best represents an adiabatic process (process where
no heat enters or leaves the system)?
A.
P C.
P B.
P V V E.
P D.
P V V V 18. String A and String B have the same mass and length. String A is under tension T and string
B is under tension 2T. The speed of waves in B is _____ times the speed of waves in A.
A. 0.50
B. 0.71
C. 1.00
D. 1.4
E. 2.0
1996 Physics Olympiad page 4 19. On a day when the velocity of sound in air is v, a whistle moves
with velocity u toward a stationary wall. The whistle emits sound with
frequency f. What frequency of reflected sound will be heard by an
observer traveling along with the whistle?
v u
A. f v + u B. f v v u +
C. f
v
D. f v u u E. f v + u v u 20. You are given two lenses, a converging lens with focal length + 10 cm and a diverging lens
with focal length  20 cm. Which of the following would produce a real image that is larger than
the object?
A.
B.
C.
D.
E. Placing the object 5 cm from the converging lens.
Placing the object 15 cm from the converging lens.
Placing the object 25 cm from the converging lens.
Placing the object 15 cm from the diverging lens.
Placing the object 25 cm from the diverging lens. 21. Two mirrors, labeled LM for left mirror and RM for right mirror in
the accompanying figure, are parallel to each other and 3.0 m apart. A
person standing 1.0 m from the right mirror (RM) looks into this mirror
and sees a series of images. How far from the person is the second
closest image seen in the right mirror (RM)?
A.
B.
C.
D.
E. 2.0 m
4.0 m
6.0 m
8.0 m
10.0 m LM RM 2.0 m 1.0 m 22. A thin film of thickness t and index of refraction 1.33 coats a glass
with index of refraction 1.50. What is the least thickness t that will
strongly reflect light with wavelength 600 nm? Hint: Light undergoes a
180o phase shift when it is reflected off a material with a higher index of
refraction. t A. 225 nm E. 600 nm B. 300 nm 1996 Physics Olympiad C. 400 nm D. 450 nm n=1.00
n=1.33
n=1.50 page 5 23. The accompanying figure shows two concentric spherical shells
isolated from each other. The smaller shell has radius b and net
charge +Q. The larger shell has radius 2b and an equal net charge
+Q. If R is the distance from the common center, the highest
electric field magnitude E occurs:
A.
B.
C.
D.
E. +Q
b
+Q only at R = 0, where E is infinite.
anywhere R < b, where E is constant
immediately outside the smaller (R = b) shell.
immediately outside the larger (R = 2b) shell.
far away from the shells; E increases with distance. 24. An infinite conducting plate of thickness
0.0200 m is surrounded by a uniform field E =
400 V/m directed left to right. See the figure to
the right. Let the potential Vo = 0 a distance
0.0200 m to the right of the plate. What is V3, the
potential 0.0300 m to the left of the plate?
A.  28 V
B.  20 V
C. + 12 V
D. + 20 V
E. + 28 V 2b E E V V 3 1 0.020m 0.030m 25. A sphere of radius a has uniform charge density ρ. A spherical
cavity of radius c is formed in the sphere. The cavity is centered a
distance b (b>c) from the center of the sphere. What is the magnitude
of the electric field at the center of the cavity? B. ρa3 3ε b 2
o ρ a3 − c3 C. 3ε o b2 c3 ρ b 2 b2 D.
3ε o 26. As shown in the diagram to the right, two fixed charges,
q1 = +1.00 µC and q2 =  4.00 µC, are 0.200 m apart. Where
is the total field zero?
A. 0.40 m to the right of q1
B. 0.13 m to the right of q1
C. 0.1m to the right of q1
D. 0.067 m to the left of q1
E. 0.20 m to the left of q1 1996 Physics Olympiad q a
c c3 ρ b b2 E.
3ε o 0.200 m
1 0 0.020m b ρb
A.
3ε o V V 2 q
2 page 6 27. The switch is closed in the circuit shown to the right.
What is the charge on the capacitor when it is fully charged?
A.
B.
C.
D.
E. 5.0 µC
10 µC
20 µC
40 µC
60 µC 100Ω
200Ω
6.00V 10.0µF 28. Two identical resistors with resistance R are connected in the two circuits drawn below. The
battery in both circuits is a 12 volt battery. Which statement is correct? I II
12 V R 12 V R R R A.
B.
C.
D.
E. More current will flow through each R in circuit I than in circuit II.
More total power will be delivered by the battery in circuit II than in circuit I.
The potential drop across each R will be greater in circuit I than in circuit II.
The equivalent resistance will be greater in circuit II than in circuit I.
The power dissipated in each R will be greater in circuit I than in circuit II. 29. The resistors – R1, R2, and R3 – have been adjusted so that the
current in the ammeter (labeled A in the accompanying circuit diagram)
is zero. What is R?
A. R2
B. R3
C. R1R2/R3
D. R1R3/R2
E. R2R3/R1 30. A long cylindrical conducting wire – shown in cross section to the
right – carries a conventional current out of the page. The wire has
uniform current density J and radius a. What is the magnetic field inside
the wire, a distance R (R<a) from the wire’s center?
A. (1/2) µoJa clockwise
B. (1/2) µoJa2/R clockwise
C. (1/2) µoJR counterclockwise
D. (1/2) µoJa2/R counterclockwise
E. (1/2) µoJa counterclockwise 1996 Physics Olympiad R3 R
I=0
A R1 R2 V a
R page 7 1996 MULTIPLE CHOICE SCREENING TEST
ANSWER KEY 1. D 11. A 21. C 2. A 12. A 22. A 3. C 13. E 23. C 4. B 14. B 24. D 5. B 15. D 25. A 6. D 16. B 26. E 7. A 17. C 27. D 8. C 18. D 28. B 9. E 19. E 29. E 10. E 20. B 30. C Physics Olympiad
(Πηψσιχσ Ολψµπιαδ)
1995 MULTIPLE CHOICE SCREENING TEST
30 QUESTIONS—40 MINUTES
INSTRUCTIONS
DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN
This test contains 30 multiple choice questions. Your answer to each question must be marked on
the optical mark answer sheet that accompanies the test. Only the boxes preceded by numbers 1
through 30 are to be used on the answer sheet.
Select the single answer that provides the best response to each question. Please be sure to use a
No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer,
the previous mark must be completely erased.
A handheld calculator may be used. However, any memory must be cleared of data and
programs. Calculators may not be shared.
Your grade on this multiple choice test will be your number of correct answers. There is no
penalty for guessing. It is to your advantage to answer every question.
The values of some possibly useful constants are given below:
mass of electron me = 9.1 x 1031 kg mass of proton mp = 1.7 x 1027 kg electronic charge e = 1.6 x 1019 C speed of light c = 3.0 x 108 m/s Coulomb's constant k = 9.0 x 109 N·m2/C2 permittivity constant ε0 = 8.9 x 1012 C2/N·m2 permeability constant µ0 = gravitational constant G = 6.7 x 1011 N·m2/kg2 acceleration due to gravity g = 10 m/s2 speed of sound (20 oC) vs = 340 m/s 4¹ x 107 T·m/A DO NOT OPEN THIS TEST UNTIL YOU ARE INSTRUCTED TO BEGIN
Copyright © 1995, AAPT 1. The accompanying figure is a graph of an
object’s position as a function of time. Which
lettered segment corresponds to a time when the
object has a negative acceleration?
A.
B.
C.
D.
E. +x C
B
D A
B
C
D
E A
t E 2. A car is moving at a constant velocity to the right along a straight level highway. Just as the
car passes a cliff, a rock falls straight down in the cliff’s reference system. Which of the
accompanying curves best depicts the path the rock takes in the car’s reference system?
A. B. D. C. E. 1
2
3
3. Three blocks – 1, 2, and 3 – rest on a horizontal
F
frictionless surface, as shown in the accompanying figure.
m
m
m
Each block has a mass m, and the blocks are connected by
massless strings. Block 3 is pulled to the right by a force F. The resultant force on block 2 is:
A. Zero B. (1/3) F C. (1/2) F D. (2/3) F E. F 4. Which solid vector in the accompanying figures best represents the acceleration of a pendulum
mass at an intermediate point in its swing?
A. B. C. D. E. 16 cm 5. A mass attached to a horizontal massless spring is displaced 16 cm
to the right of its equilibrium position (see accompanying figure) and
released from rest. At its 16 cm extension, the springmass system has
1.28 joules of potential energy. Upon release, it slides across a rough
surface and comes momentarily to rest 8 cm to the left of its
equilibrium position. How much mechanical energy was dissipated by friction?
A. 0.16 J B. 0.32 J 1995 Physics Olympiad C. 0.64 J D. 0.96 J 8 cm E. 1.12 J page 1 6. Two identical masses m are connected to a massless string which is hung over two frictionless
pulleys as shown on the right. If everything is at rest,
what is the tension in the cord?
A.
B.
C.
D.
E. Less than mg.
Exactly mg.
More than mg but less than 2mg.
Exactly 2mg.
More than 2mg. m m 7. Car A and car B are both traveling down a straight highway at 25 m/s (about 56 mph). Car A
is only 6.0 m behind car B. The driver of car B brakes, slowing down with a constant
acceleration of 2.0 m/s2. After a time 1.2 s, the driver of car A begins to brake, also at 2.0 m/s2.
What is the relative velocity of the two cars when they collide? Hint: Both cars are still moving
forward when they collide.
A. 2.4 m/s B. 5.0 m/s C. 9.5 m/s D. 21 m/s E. 24 m/s 8. During the collision between car A and car B described in Problem # 7 above, which car
experiences the greater change in momentum?
A.
B.
C.
D.
E. The more massive car.
The less massive car.
Car A because its velocity at the start of the collision is greater.
Car B because its velocity at the start of the collision is less.
Both cars experience the same magnitude of momentum change. 9. A 70 kg hunter ropes a 350 kg polar bear. Both are initially at rest, 30 m apart on a
frictionless and level ice surface. When the hunter pulls the polar bear to him, the polar bear will
move:
A. 5 m B. 6 m C. 15 m D. 24 m E. 25 m 10. A bullet of mass m is fired at a block of mass M initially at rest. The bullet, moving at an
initial speed v, embeds itself in the block. The speed of the block after the collision is:
A. Mv
M +m B. (M + m )v
m m v D. M + m v C. M + m m E. mv
M +m 11. The driver of a 1000 kg car tries to turn through a circle of radius 100 m on an unbanked
curve at a speed of 10 m/s. The maximum frictional force between the tires and the slippery road
is 900 N. The car will:
A.
B.
C.
D.
E. Slide into the inside of the curve.
Make the turn.
Slow down due to the centrifugal force.
Make the turn only if it speeds up.
Slide off to the outside of the curve. 1995 Physics Olympiad page 2 12. A solid cylinder weighing 200 N has a fixed axis and a string wrapped around it. The string
is pulled with a force equal to the weight of the cylinder. The acceleration of the string is
approximately:
(A) 10 m/s2 (B) 20 m/s2 (C) 30 m/s2 (D) 40 m/s2 (E) 50 m/s2 13. A spinning ice skater has an initial kinetic energy (1/2)Iω2. She pulls in her outstretched
arms, decreasing her moment of inertia to (1/4)I. Her new angular speed is:
A. ω/4 B. ω/2 C. ω D. 2ω E. 4ω
1 2
L/4 14. A heavy rod of length L and weight W is suspended
horizontally by two vertical ropes as shown on the right. The
first rope is attached to the left end of the rod while the second
rope is attached a distance L/4 from the right end. The tension
in the second rope is:
A. (1/2)W B. (1/4)W C. (1/3)W L D. (2/3)W E. W 15. A block of ice with mass m falls into a lake. After impact, a mass of ice m/5 melts. Both the
block of ice and the lake have a temperature of 0oC. If L represents the heat of fusion, the
minimum distance the ice fell is:
L
5L
gL
mL
5gL
A.
B.
C.
D.
E.
5g
g
5m
5g
m 16. Which of the accompanying PV diagrams best represents an isothermal (constant
temperature) process?
A.
P B.
P V C.
P V E.
P D.
P V V V 17. If the heat is added at constant volume, 6300 joules of heat are required to raise the
temperature of an ideal gas by 150 K. If instead, the heat is added at constant pressure, 8800
joules are needed for the same temperature change. When the temperature of the gas changes by
150 K, the internal energy of the gas changes by:
A. 2500 J B. 6300 J 1995 Physics Olympiad C. 8800 J D. 11,300 J E. 15,100 J page 3 18. A firecracker exploding on the surface of a lake is heard as two sounds – a time interval t
apart – by the paddler of a nearby canoe. Sound travels with a speed u in water and a speed v in
air. The distance from the exploding firecracker to the canoe is:
A. uvt
u +v B. t ( u + v)
uv C. t ( u − v)
uv D. vt E. uvt
u −v 19. Two interfering waves have the same wavelength, frequency, and amplitude. They are
traveling in the same direction but are 90o out of phase. Compared to the individual waves, the
resultant wave will have the same:
A.
B.
C.
D.
E. amplitude and velocity but different wavelength.
amplitude and wavelength but different velocity.
wavelength and velocity but different amplitude.
amplitude and frequency but different velocity.
frequency and velocity but different wavelength. 20. An organ pipe which is open at both ends resonates with fundamental frequency 300 Hz. If
one end of the pipe is closed, it will resonate with a fundamental frequency of:
A. 75 Hz B. 150 Hz C. 300 Hz D. 600 Hz E. 1200 Hz 21. You are given two lenses, a converging lens with focal length + 10 cm and a diverging lens
with focal length  20 cm. Which of the following would produce a virtual image that is larger
than the object?
A.
B.
C.
D.
E. Placing the object 5 cm from the converging lens.
Placing the object 15 cm from the converging lens.
Placing the object 25 cm from the converging lens.
Placing the object 15 cm from the diverging lens.
Placing the object 25 cm from the diverging lens. 22. Light shining through two very narrow slits
produces an interference maximum at point P when
the entire apparatus is in air (see accompanying
figure). For the interference maximum represented,
light through the bottom slit travels one wavelength
further than light through the top slit before reaching
point P. If the entire apparatus is immersed in water,
the angle θ to the interference maximum:
A.
B.
C.
D.
E. P
x
d θ is unchanged
decreases because the frequency decreases.
decreases because the wavelength decreases.
increases because the frequency increases.
increases because the wavelength increases. 1995 Physics Olympiad page 4 23. The magnitude of the field due to an infinite plate of charge is
σ/(2εo), where σ is the charge per unit area and εo is the vacuum
+
permittivity. The figure to the right depicts three infinite plates of
charge perpendicular to the plane of the page with charge per unit area
+
+ σ1, + 2σ1, and  σ1. The total field at the point labeled X is:
+
A. 4σ1/(2εo) to the left.
+
B. σ1/(2εo) to the left.
C. 0
+σ
1
D. σ1/(2εo) to the right.
E. 4σ1/(2εo) to the right. +
+
+
+
+
+
+
+ X  + 2σ1  σ1 24. There are three conducting spheres of
2
1
identical diameter, shown to the right.
Spheres 1 and 2 are separated by a distance
3
that is large compared to their diameter. They
have equal charges and repel each other with
an electrostatic force F. Sphere 3 is initially
uncharged and has an insulated handle. It is touched first to sphere 1, then to sphere 2, and then
removed. If the distance between spheres 1 and 2 has not changed, the force between these two
spheres is:
A. 0 B. (1/16)F C. (1/4)F D. (3/8)F 25. Two conducting spheres, A of radius a and B of radius b,
are shown in the accompanying figure. Both are positively
charged and isolated from their surroundings. They had been
connected by a conducting wire but the wire has been removed.
Assuming the potential an infinite distance away is zero, which A
of the following statements about the spheres are true:
I. Sphere A is at the higher potential.
II. Sphere B is at the higher potential.
III. Both spheres have the same potential.
A. I and IV B. I and VI C. II and VI D. III and IV only at R = 0, where V is infinite.
anywhere R ≤ b, where V is constant
in the region b < R < B.
immediately outside the larger shell; V is zero everywhere within it.
far away from the shells; V increases with distance. 1995 Physics Olympiad a b
B IV. Sphere A has the larger charge.
V. Sphere B has the larger charge.
VI. Both spheres have the same charge. 26. The accompanying figure shows two concentric spherical shells
isolated from each other. The smaller shell has radius b and net charge
+ q. The larger shell has radius B and net charge + Q. Assume that
the potential is zero at an infinite distance from the shells. If R is the
distance from the common center, the highest electric potential V
occurs:
A.
B.
C.
D.
E. E. (1/2)F E. III and V +Q
b +q
B page 5 27. In the circuit represented to the right, the current I equals:
A.
B.
C.
D.
E. R V/5R
V/4R
2V/5R
2V/4R
2V/R V R R I R V Use the circuit below to answer questions 28 and 29. L1, L2, L3, and L4 are identical light bulbs.
There are six voltmeters connected to the circuit as shown. Assume that the voltmeters do not
effect the circuit.
L1
V1 L3
V2 L2 V3 L4 V0 V4 V5 28. Which of the following combinations are equal to V0?
A. V1 + V3 B. V2 + V3 + V4 C. V1 + V5 D. V3 + V4 E. V1 + V4 29. If L3 were to burn out, opening the circuit, which voltmeter(s) would read zero volts?
A.
B.
C.
D.
E. None would read zero.
only V3
only V4
only V3, V4, and V5
they would all read zero 30. Identical currents flow in two perpendicular wires, as shown in
the accompanying figure. The wires are very close but do not touch.
The magnetic field can be zero: 2 A.
B.
C.
D.
E. 3 at a point in region 1 only
at a point in region 2 only
at points in both regions 1 and 2
at points in both regions 1 and 4
at points in both regions 2 and 4 1995 Physics Olympiad 1
I I
4 page 6 1995 MULTIPLE CHOICE SCREENING TEST
ANSWER KEY 1. C 11. E 21. A 2. D 12. B 22. C 3. B 13. E 23. C 4. D 14. D 24. D 5. D 15. A 25. D 6. B 16. B 26. B 7. A 17. B 27. A 8. E 18. E 28. C 9. A 19. C 29. C 10. E 20. B 30. E Physics Olympiad
(Πηψσιχσ Ολψµπιαδ)
1994 MULTIPLE CHOICE SCREENING TEST
30 QUESTIONS—40 MINUTES
INSTRUCTIONS
DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN
This test contains 30 multiple choice questions. Your answer to each question must be marked on
the optical mark answer sheet that accompanies the test. Only the boxes preceded by numbers 1
through 30 are to be used on the answer sheet.
Select the single answer that provides the best response to each question. Please be sure to use a
No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer,
the previous mark must be completely erased.
A handheld calculator may be used. However, any memory must be cleared of data and
programs. Calculators may not be shared.
Your grade on this multiple choice test will be your number of correct answers. There is no
penalty for guessing. It is to your advantage to answer every question.
The values of some possibly useful constants are given below:
mass of electron me = 9.1 x 1031 kg mass of proton mp = 1.7 x 1027 kg electronic charge e = 1.6 x 1019 C speed of light c = 3.0 x 108 m/s Coulomb's constant k = 9.0 x 109 N·m2/C2 permittivity constant ε0 = 8.9 x 1012 C2/N·m2 permeability constant µ0 = gravitational constant G = 6.7 x 1011 N·m2/kg2 acceleration due to gravity g = 10 m/s2 speed of sound (20 oC) vs = 340 m/s 4π x 107 T·m/A DO NOT OPEN THIS TEST UNTIL YOU ARE INSTRUCTED TO BEGIN
Copyright © 1994, AAPT 1. A motorist travels 320 km at 80 km/h and then 320 km at 100 km/h. What is the average
speed of the motorist for the entire trip?
A. 84 km/h B. 89 km/h C. 90 km/h D. 91 km/h E. 95 km/h 2. A sports car is stopped at a light. At t = 0 the light changes and the sports car accelerates at a
constant 2.0 m/s2. At t = (10/3) s a station wagon traveling at a constant 15 m/s in the same
direction passes the stop light. When does the station wagon catch up to the sports car?
A. never
B. t = 5.0 s
C. t = 7.5 s
D. t = 12 s
E. t = 21 s
3. A ball is thrown from ground level with an initial velocity of vo in the upward direction. It
reaches a maximum height y and returns to ground level at time t seconds after it was thrown. If
its initial velocity is doubled, it will be in the air _____ seconds and reach a maximum height
_____.
A. t, 4y
B. 2t, y
C. 2t, 2y
D. 2t, 4y
E. 4t, 2y
4. The ball in the preceding question (#3) is taken to Mars where the acceleration due to gravity
is approximately 4 m/s2 (0.4 ge). It is thrown from ground level with the same initial velocity as it
originally had on Earth, vo in the upward direction. On Mars, it will be in the air _____ seconds
and reach a maximum height _____.
A. 0.4 t, 2.5 y
B. 0.4 t, 6.25 y
C. 2.5 t, 2.5y
D. 6.25 t, 2.5 y
E. 6.25 t, 6.25 y
5. A top is spinning in the direction shown in the accompanying figure. Its axis of
rotation makes an angle of 15o with the vertical. Assume friction can be neglected.
The magnitude of the top’s angular momentum will _____ while its direction will
_____.
A. increase, precess in the counterclockwise direction when seen from above.
B. increase, not precess
C. remain the same, not precess
D. remain the same, precess in the clockwise direction when seen from above.
E. remain the same, precess in the counterclockwise direction when seen from above.
6. Three air track cars, shown in the accompanying
v
1
figure, all have the same mass m. Cars 2 and 3 are
initially at rest. Car 1 is moving to the right with speed v.
m
Car 1 collides with car 2 and sticks to it. The 12
combination collides elastically with car 3. Which of the
following is most nearly the final speed of the 12 combination?
A. 0.17 v
B. 0.50 v
C. 0.67 v
D. 0.80 v
7. A cube with mass M starts at rest at point 1 at a height 4R, where R is
the radius of the circular part of the track. The cube slides down the
frictionless track and around the loop. The force that the track exerts on
the cube at point 2 is most nearly _____ times the cube’s weight Mg.
A. 1
B. 2
C. 3
D. 4
E. 5 2 3 m m E. 1.0 v
1
4R 2
R Page 1 of 6 8. An astronaut with weight W on Earth lands on a planet with mass 0.1 times the mass of Earth
and radius 0.5 times the radius of Earth. The astronaut’s weight is _____ on the planet.
A. 0.02 W
B. 0.04 W
C. 0.2 W
D. 0.4 W
E. W
9. A horizontal force F, represented by the arrow in the figure to the right, is
used to push a block of weight mg up an inclined plane making an angle of θ
with the horizontal. The coefficient of friction between the plane and the block
is µ. The magnitude of the frictional force acting on the block is:
A. µ mg cosθ
B. µ mg/( cosθ )
C. µ ( mg cosθ + F sinθ )
D. µ ( F cosθ – mg sinθ )
E. µ F cosθ θ 10. A child of mass M stands on the edge of a merrygoround of radius R and moment of inertia
I. Both the merrygoround and child are initially at rest. The child walks around the
circumference with speed v with respect to the ground. What is the magnitude of the angular
velocity of the merrygoround with respect to the ground?
v
MRv
MRv
MRv
C. ω =
D. ω =
E. ω =
A. 0
B. ω =
2
R
I − MR
I + MR 2
I
11. A rigid rod of mass M and length L has moment of
1
inertia 12 ML2 about its center of mass. A sphere of mass m
2
2
5 mR axis
M and radius R has moment of inertia
about its center of
mass. A combined system is formed by centering the sphere
L
at one end of the rod and placing an axis at the other (see
accompanying figure). What is the moment of inertia of the combined system about the axis
shown? m 1
A. I = 12 ML2 + 2 mR2
5
1
B. I = 12 ML2 + 2 mR2 + mL2
5 C. I = 1 ML2 + 2 mR 2 + mL2
3
5
1
D. I = 12 ML2 + mL2
2
2
E. I = 1 ML + mL
3 12. A rocket is launched from the surface of a planet with mass M and radius R. What is the
minimum velocity the rocket must be given to completely escape from the planet’s gravitational
field?
2GM
2GM
GM
GM
A. v =
B. v =
C. v =
D. v =
E. v = GM
2
2
R
R
R
R Page 2 of 6 13. A uniform rod of length L and weight WR is suspended as shown in the
accompanying figure. A weight W is added to the end of the rod. The
support wire is at an angle θ to the rod. What is the tension T in the wire?
W
A. T =
sin θ
B. T = W + W R T
θ
W C. T = W + 1 W R
2
D. T = W + 1 WR
2 sin θ
W + WR
E. T =
sin θ 14. A hollow vertical cylinder of radius R is rotated with angular
velocity ω about an axis through its center. What is the minimum
coefficient of static friction necessary to keep the mass M suspended
on the inside of the cylinder as it rotates? M
top view
side view A. µ = gR
ω2 B. µ = ω 2g
R C. µ = ω 2R
g D. µ = ω2
gR 15. A block of mass M is attached to a relaxed spring with force constant k,
placed on a frictionless inclined plane as shown in the accompanying figure,
and released. What is the maximum extension of the spring?
2 Mg sin θ
Mg sin θ
2 Mg
Mg
A. x =
B. x =
C. x =
D. x =
k
k
k
k
16. A simple pendulum of length L and mass m is attached to a moving
support. In order for the pendulum string to make a constant angle θ with the
vertical, the support must be moving to the:
A. right with constant acceleration a = g tanθ.
B. left with constant acceleration a = g tanθ.
C. right with constant acceleration a = g sinθ.
D. right with constant velocity v = Lg tanθ.
E. left with constant velocity v = Lg tanθ. E. µ = g
ω 2R θ
E. x = 2gM θ L 17. A Carnot cycle takes in 1000 J of heat at a high temperature of 400 K. How much heat is
expelled at the cooler temperature of 300 K?
A. 0
B. 250 J
C. 500 J
D. 750 J
E. 1000 J Page 3 of 6 18. An ideal gas is expanded at constant pressure from initial volume Vi and temperature Ti to
final volume Vf and temperature Tf . The gas has molar heat capacity CP at constant pressure.
The amount of work done by n moles of the gas during the process can be expressed:
A. 0
B. nRTi ln(Vf/Vi)
C. CPn(Τf – Ti) D. nk(Vf – Vi)
E. nR(Τf – Ti)
19. The average translational kinetic energy of any ideal gas depends only on:
A. the absolute (Kelvin) temperature.
B. the mass of the gas.
C. the pressure of the gas.
D. the amount of the gas.
E. whether the gas is monatomic or diatomic.
20. A soap film of thickness t is surrounded by air (see accompanying figure).
It is illuminated at near normal incidence by monochromatic light which has
wavelength λ in the film. A film thickness _____ will produce maximum
brightness of the reflected light.
A. 1/4 λ
B. 1/2 λ
C. 1 λ
D. 2 λ
E. 4 λ air
t air 21. You are given two lenses, a converging lens with focal length +10 cm and
a diverging lens with focal length –20 cm. Which of the following would produce a real image
that is smaller than the object?
A. Placing the object 5 cm from the converging lens.
B. Placing the object 15 cm from the converging lens.
C. Placing the object 25 cm from the converging lens.
D. Placing the object 15 cm from the diverging lens.
E. Placing the object 25 cm from the diverging lens.
22. A source at rest emits waves with wavelength λ in a medium with
velocity v. If the source moves to the right with velocity vs(see
accompanying figure), the distance between adjacent crests λ’ directly
behind the source is: A. λv
v + vs B. λv
v − vs v
C. λ 1 + vs 23. Two sources, in phase and a distance d apart,
each emit a wave of wavelength λ. See
accompanying figure. Which of the choices for the
path difference ∆L = L1 – L2 will always produce
constructive interference at point P?
A. d sinθ
B. x/L1
C. (x/L2)d
D. λ/2
E. 2 λ
Page 4 of 6 v D. λ 1 + s v λ' vs
v E. λ 1 − s v
P L2 x
d θ
L1 1 soap
2 Questions 24 and 25 are based on the following description: Two point charges of +4.00 µC and
–9.00 µC are placed 1.00 m apart, as shown in the accompanying figure. Assume the potential
goes to zero as R goes to infinity.
+4.0 µC 24. The total electric field due to the two charges is zero at a point:
A. 3.00 m to the right of the –9.00 µC charge.
B. 0.40 m to the right of the +4.00 µC charge.
C. 0.31 m to the right of the +4.00 µC charge.
D. 0.80 m to the left of the +4.00 µC charge.
E. 2.00 m to the left of the +4.00 µC charge. –9.0 µC 1.00 m 25. How much work is done moving the –9.00 µC charge from its original position to a new
position 2.00 m from the +4.00 µC charge?
A. – 0.324 J
B. –0.081 J
C. +0.162 J
D. +0.243 J
E. +0.486 J
26. A point charge Q is placed at the center of a spherical conducting shell,
the shaded part of the accompanying figure. A total charge of –q is placed
on the shell. The magnitude of the electric field at point P1 a distance R1
from the center is _____. The magnitude of the electric field at point P2 a
distance R2 from the center is _____.
A. 0, 0
Q
B. k 2 , 0
R1
Q− q
C. k
,0
R12
Q− q
D. 0, k
R 22
Q
Q− q
E. k 2 , k
R1
R 22 P 2 R2
Q PR
1
1 –q R1 27. R4, shown in the figure to the right, is a variable resistor. In order for
there to be no current through the ammeter, R4 must be equal to:
A. R2
B. R3
C. R1R2/R3
D. R1R3/R2
E. R2R3/R1 R2 R3 A
R4 V
28. A resistor R dissipates power P when connected directly to a
voltage source V, as shown in the accompanying figures. What
resistance R′ must be connected in series with R to decrease the
power dissipated in R to P/2?
R
R
A.
B.
C. R
D. R ( 2 − 1) E. R 2
2
2 Page 5 of 6 P
V P/2
V R R
R' 29. The infinitely long straight wire carries a conventional current I as
shown in the accompanying figure. The rectangular loop carries a
conventional current I′ in the counterclockwise direction. The net force
on the rectangular loop is:
µ II' c 1 1 A. o
−
to the right
2π a b µ II' c 1 1 +
to the left
B. o
2π a b I' I b a µ o II' c c 2(b − a )
++
to the left
c
2π a b
µ II' 2(b − a )
D. o
to the right
2π
c
E. 0 c C. 30. A spatially uniform magnetic field of 0.080 T is directed into the
plane of the page and perpendicular to it, as shown in the accompanying
figure. A wire loop in the plane of the page has constant area 0.010 m2.
The magnitude of the magnetic field decreases at a constant rate of
3.0 x 10–4 T/s. What is the magnitude and direction of the induced emf?
A. 3.0 x 10–6 V clockwise
B. 3.0 x 10–6 V counterclockwise
C. 2.4 x 10–5 V counterclockwise
D. 8.0 x 10–4 V counterclockwise
E. 8.0 x 10–4 V clockwise Page 6 of 6 X X X X X X X X X X X X X X X X 1994 MULTIPLE CHOICE SCREENING TEST ANSWER KEY 1. B 11. C 21. C 2. B 12. B 22. D 3. D 13. D 23. E 4. C 14. E 24. E 5. E 15. A 25. C 6. A 16. B 26. B 7. C 17. D 27. E 8. D 18. E 28. D 9. C 19. A 29. A 10. B 20. A 30. A Physics Olympiad
Entia non multiplicanda sunt praeter necessitatem
1997 MULTIPLE CHOICE SCREENING TEST
30 QUESTIONS—40 MINUTES
INSTRUCTIONS
DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN
This test contains 30 multiple choice questions. Your answer to each question must be marked on
the optical mark answer sheet that accompanies the test. Only the boxes preceded by numbers 1
through 30 are to be used on the answer sheet.
Select the single answer that provides the best response to each question. Please be sure to use a
No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer,
the previous mark must be completely erased.
A handheld calculator may be used. However, any memory must be cleared of data and
programs. Calculators may not be shared.
Your grade on this multiple choice test will be your number of correct answers. There is no
penalty for guessing. It is to your advantage to answer every question.
The values of some possibly useful constants are given below:
mass of electron me = 9.1 x 1031 kg mass of proton mp = 1.7 x 1027 kg electronic charge e = 1.6 x 1019 C speed of light c = 3.0 x 108 m/s Coulomb's constant k = 9.0 x 109 N·m2/C2 permittivity constant ε0 = 8.9 x 10 µ = π 7 C/
T·m/A 11 G 2 N·m /kg 24
E = 6.0 x 10 kg mass of Earth
radius of Earth RE = 6.4 x 106 m gravitational field at Earth’s surface
speed of sound (20 oC) g = 9.8 m/s2
vs = 340 m/s DO NOT OPEN THIS TEST UNTIL YOU ARE INSTRUCTED TO BEGIN
Copyright © 1997, AAPT 1. Starting from rest at time t = 0, a car moves in a straight a in
line with an acceleration given by the accompanying graph. m/s2
6
What is the speed of the car at t = 3 s?
5
A. 1.0 m/s
4
B. 2.0 m/s
3
C. 6.0 m/s
2
D. 10.5 m/s
E. 12.5 m/s
1
0
0 1 23 4 5 6 t in s 2. A flare is dropped from a plane flying over level ground at a velocity of 70 m/s in the
horizontal direction. At the instant the flare is released, the plane begins to accelerate horizontally
at 0.75 m/s2. The flare takes 4.0 s to reach the ground. Assume air resistance is negligible.
Relative to a spot directly under the flare at release, the flare lands
A.
B.
C.
D.
E. directly on the spot.
6.0 m in front of the spot.
274 m in front of the spot.
280 m in front of the spot.
286 m in front of the spot. 3. As seen by the pilot of the plane (in question #2) and measured relative to a spot directly under
the plane when the flare lands, the flare lands
A.
B.
C.
D.
E. 286 m behind the plane.
6.0 m behind the plane.
directly under the plane.
12 m in front of the plane.
274 m in front of the plane 4. A force F is used to hold a block of mass m on an incline as shown in
the diagram. The plane makes an angle of θ with the horizontal and F is
perpendicular to the plane. The coefficient of friction between the plane
and the block is µ. What is the minimum force, F, necessary to keep the
block at rest?
A.
B.
C.
D.
E. µ mg
mg cosθ
mg sinθ
(mg/µ) sinθ
(mg/µ) ( sinθ  µ cosθ ) F θ 5. You hold a rubber ball in your hand. The Newton’s third law companion force to the force of
gravity on the ball is the force exerted by the
A. ball on the Earth.
B. ball on the hand.
C. hand on the ball.
D. Earth on the ball.
E. Earth on your hand.
1997 Physics Olympiad page 1 6. A ball of mass m is fastened to a string. The ball swings in a vertical circle of radius R with the
other end of the string held fixed. Neglecting air resistance, the difference between the string’s
tension at the bottom of the circle and at the top of the circle is
A. mg B. 2 mg C. 4 mg D. 6 mg 7. Three air track cars, shown in the accompanying
figure, all have the same mass m. Cars 2 and 3 are
initially at rest. Car 1 is moving to the right with speed v.
Car 1 collides with car 2 and sticks to it. The 12
combination collides elastically with car 3. Which of the
following is most nearly the final speed of car 3?
A. 0.17 v B. 0.50 v C. 0.67 v v m D. v
E. 2 v 2 3 m 1 m D. 0.80 v 8. A point object of mass 2m is attached to one end of
a rigid rod of negligible mass and length L. The rod is
initially at rest but free to rotate about a fixed axis
perpendicular to the rod and passing through its other
end (see diagram to the right). A second point object
with mass m and initial speed v collides and sticks to
the 2m object. What is the tangential speed vt of the
object immediately after the collision?
A. v 3
B. v 2
C. v E. 8 mg E. 1.0 v axis
L L
vt m v 2m 3
2
3 9. Two artificial satellites I and II have circular orbits of radii R and 2R, respectively, about the
same planet. The orbital velocity of satellite I is v. What is the orbital velocity of satellite II?
v
v
A.
B.
C. v
D. v 2
E. 2v
2
2
10. The gravitational acceleration on the surface of the moon is 1.6 m/s2. The radius of the
moon is 1.7 x 106 m. The period of a satellite placed in a low circular orbit about the moon is
most nearly
A. 1.0 x 103 s B. 6.5 x 103 s 1997 Physics Olympiad C. 1.1 x 106 s D. 5.0 x 106 s E. 7.1 x 1012 s page 2 11. A uniform ladder of length L rests against a smooth frictionless wall. The
floor is rough and the coefficient of static friction between the floor and ladder
is µ. When the ladder is positioned at angle θ, as shown in the accompanying
diagram, it is just about to slip. What is θ?
A. θ = µ
L B. tan θ = 2µ C. tan θ = 1
2µ D. sin θ = 1
µ L E. cos θ = µ
θ 12. Three objects, all of mass M, are released simultaneously from the top of an inclined plane of
height H. The objects are described as follows
I. a cube of side R.
II. a solid cylinder of radius R
III. a hollow cylinder of radius R
Assume the cylinders roll down the plane without slipping and the cube slides down the plane
without friction. Which object(s) reach(es) the bottom of the plane first?
A. I B. II C. III D. I & II 13. A massless rod of length 2R can rotate about a vertical
axis through its center as shown in the diagram. The system
rotates at an angular velocity ω when the two masses m are a
distance R from the axis. The masses are simultaneously
pulled to a distance of R/2 from the axis by a force directed
along the rod. What is the new angular velocity of the
system?
A. ω/4 B. ω/2 C. ω E. II & III
R m D. 2ω m m m
R/2
E. 4ω 14. A meter stick moves with a velocity of 0.60 c relative to an observer. The observer measures
the length of the meter stick to be L. Which of the following statements is always true?
A. L = 0.60 m B. L = 0.80 m C. 0.80 m ² L ² 1.00 m D. L = 1.00 m E. L ³ 1.00m 15. A glowing ember (hot piece of charcoal) radiates power P in watts at an absolute
temperature T. When the temperature of the ember has decreased to T/2, the power it radiates is
most nearly
A. P B. P/2 1997 Physics Olympiad C. P/4 D. P/8 E. P/16 page 3 16. Three processes compose a thermodynamic cycle shown in the
accompanying pV diagram of an ideal gas.
Process 1→2 takes place at constant temperature (300 K).
During this process 60 J of heat enters the system.
Process 2→3 takes place at constant volume. During this
process 40 J of heat leaves the system.
Process 3→1 is adiabatic. T3 is 275 K.
What is the change in internal energy of the system during process
3→1?
A. 40 J B. 20 J C. 0 p 1
2
3
V D. +20 J E. +40 J 17. What is the change in entropy of the system described in Question # 16 during the process
3→1?
A. +5.0 K/J B. +0.20 J/K C. 0 D. 1.6 J/K E. 6.9 K/J 18. A wave is described by the equation: y(x,t) = 0.030 ⋅ sin(5π x + 4π t) where x and y are
in meters and t is in seconds. The +x direction is to the right. What is the velocity of the wave?
A.
B.
C.
D.
E. 0.80 m/s to the left
1.25 m/s to the left
0.12 π m/s to the right
0.80 m/s to the right
1.25 m/s to the right 19. Two sources, in phase and a distance d apart,
each emit a wave of wavelength λ. See
accompanying figure. Which of the choices for the
path difference ∆L = L1 – L2 will always produce
destructive interference at point P?
A.
B.
C.
D.
E. d sinθ
x/L1
(x/L2)d
λ/2
2λ P L2 x
d θ
L1 20. You are given two lenses, a converging lens with focal length + 10 cm and a diverging lens
with focal length  20 cm. Which of the following would produce a virtual image that is larger
than the object?
A.
B.
C.
D.
E. Placing the object 5 cm from the converging lens.
Placing the object 15 cm from the converging lens.
Placing the object 25 cm from the converging lens.
Placing the object 15 cm from the diverging lens.
Placing the object 25 cm from the diverging lens. 1997 Physics Olympiad page 4 21. You are given two identical planoconvex lenses, one of which is
shown to the right. When you place an object 20 cm to the left of a
single planoconvex lens, the image appears 40 cm to the right of the
lens. You then arrange the two planoconvex lenses back to back to
form a double convex lens. If the object is 20 cm to the left of this new
lens, what is the approximate location of the image?
A.
B.
C.
D.
E. 6.7 cm to the right of the lens.
10 cm to the right of the lens.
20 cm to the right of the lens.
80 cm to the right of the lens.
80 cm to the left of the lens. Planoconvex Double
Convex y Q
22. Positive charge Q is uniformly distributed over a ring of radius
a that lies in the yz plane as shown in the diagram. The ring is
centered at the origin. Which of the following graphs best
represents the value of the electric field E as a function of x, the
distance along the positive x axis?
A B
E C
E x E x kq/a2 at an angle 45o above the +x axis.
kq/a2 at an angle 45o below the x axis.
3 kq/a2 at an angle 45o above the x axis.
3 kq/a2 at an angle 45o below the +x axis.
9 kq/a2 at an angle 45o above the +x axis. 1997 Physics Olympiad z
x D E
E x 23. Four point charges are placed at the corners of a square with
diagonal 2a as shown in the diagram. What is the total electric field
at the center of the square?
A.
B.
C.
D.
E. a E x x y
+3q q 2a
x
2q +3q page 5 Both questions 24 and 25 refer to the system shown in the diagram. A
spherical shell with an inner surface of radius a and an outer surface of
radius b is made of conducting material. A point charge +Q is placed
at the center of the spherical shell and a total charge q is placed on the
shell. a
q Q b 24. How is the charge q distributed after it has reached equilibrium?
A.
B.
C.
D.
E. Zero charge on the inner surface, q on the outer surface.
Q on the inner surface, q on the outer surface.
Q on the inner surface, q+Q on the outer surface.
+Q on the inner surface, qQ on the outer surface.
The charge q is spread uniformly between the inner and outer surface. 25. Assume that the electrostatic potential is zero at an infinite distance from the spherical shell.
What is the electrostatic potential at a distance R from the center of the shell, where b ³ R ³ a?
A. 0 B. k Q
a C. k Q
R D. k Qq
R E. k Qq
b Use the circuit below to answer questions 26 and 27. B1, B2, B3, and B4 are identical light
bulbs. There are six voltmeters connected to the circuit as shown. All voltmeters are connected
so that they display positive voltages. Assume that the voltmeters do not effect the circuit.
B1
V1 B3
V2 B2 V3 B4 V0 V4 V5 26. If B2 were to burn out, opening the circuit, which voltmeter(s) would read zero volts?
A.
B.
C.
D.
E. none would read zero.
only V2
only V3 and V4
only V3, V4, and V5
they would all read zero 27. If B2 were to burn out, opening the circuit, what would happen to the reading of V1? Let V
be its original reading when all bulbs are functioning and let V’ be its reading when B2 is burnt
out.
A. V’ ³ 2V B. 2V > V’ > V 1997 Physics Olympiad C. V’ = V D. V > V’ > V/2 E. V/2 ³ V’ page 6 28. A particle with positive charge q and mass m travels along a path perpendicular to a magnetic
field. The particle moves in a circle of radius R with frequency f. What is the magnitude of the
magnetic field?
A. mf
q B. 2πfm
q C. m
2πfq D. mc
qR 29. Two wires, each carrying a current i, are shown in the
diagram to the right. Both wires extend in a straight line for a
very long distance on both the right and the left. One wire
contains a semicircular loop of radius a centered on point X.
What is the correct expression for the magnetic field at point
X? HINT: The magnitude of the magnetic field at the center
of a circular current loop of radius R is µ 0i/(2R).
A.
B.
C.
D.
E. µ 0i/(4a) + µ 0i/(2πa)
µ 0i/(2a)  µ 0i/(2πa) + µ 0i/(2πa)
µ 0i/(4a) + µ 0i/(2πa)
µ 0i/(4a) + µ 0i/(2πa) + µ 0i/(2πa)
µ 0i/(2a)  µ 0i/(2πa) E. µqf
2πR i
a X
a i out of the page
out of the page
into the page
into the page
into the page 30. You are given a bar magnet and a looped coil of wire. Which of
the following would induce an emf in the coil?
I. Moving the magnet toward the coil.
II. Moving the coil away from the magnet.
III. Turning the coil about a vertical axis.
A. I only B. II only 1997 Physics Olympiad C. I & II D. I & III E. I, II, III page 7 1997 MULTIPLE CHOICE SCREENING TEST
ANSWER KEY 1. D 11. C 21. B 2. D 12. A 22. D 3. B 13. E 23. B 4. E 14. C 24. C 5. A 15. E 25. E 6. D 16. E 26. A 7. C 17. C 27. D 8. A 18. A 28. B 9. B 19. D 29. C 10. B 20. A 30. E 1999 AAPT PHYSICS OLYMPIAD
Entia non multiplicanda sunt praeter necessitatem
1999 MULTIPLE CHOICE SCREENING TEST
30 QUESTIONS  40 MINUTES
INSTRUCTIONS
DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN
This test contains 30 multiple choice questions. Your answer to each question must be marked on
the optical mark answer sheet that accompanies the test. Only the boxes preceded by numbers 1
through 30 are to be used on the answer sheet.
Select the single answer that provides the best response to each question. Please be sure to use a
No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer,
the previous mark must be completely erased.
A handheld calculator may be used. However, any memory must be cleared of data and
programs. Calculators are not to be shared.
Your score on this multiple choice test will be your number of correct answers. There is no
penalty for guessing. It is to your advantage to answer every question.
The values of some possibly useful constants are given below:
mass of electron
mass of proton
electronic charge
speed of light
Coulomb’s constant
permittivity constant
permeability constant
gravitational constant
mass of Earth
radius of Earth
gravitational field
speed of sound (200 C) me
mp
e
c
k
e0
µ0
G
ME
RE
g
vs =
=
=
=
=
=
=
=
=
=
=
= 9.1
1.7
1.6
3.0
9.0
8.9
4π
6.7
6.0
6.4
9.8
340 x 1031
x 1027
x 1019
x 108
x 109
x 1012
x 107
x 1011
x 1024
x 106
N/kg
m/s kg
kg
C
m/s
N.m2/C2
C2/N.m2
T.m/A
N.m2/kg2
kg
m DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN
Copyright © 1999, American Association of Physics Teachers.
1999 Physics Olympiad 1 1. A truck driver travels threefourths the distance of his run at one velocity (v) and then
completes his run at one half his original velocity (½v). What was the trucker’s average speed for
the trip?
[A] 0.85v [B] 0.80v [C] 0.75v [D] 0.70v [E] 0.65v 2. Hercules and Ajax horizontally push in the same direction on a 1200 kg crate. Hercules
pushes with a force of 500 newtons and Ajax pushes with a force of 300 newtons. If a frictional
force provides 200 newtons of resistance, what is the acceleration of the crate?
[A] 1.3 m/s2 [B] 1.0 m/s2 [C] 0.87 m/s2 [D] 0.75 m/s2 [E] 0.5 m/s2 3. A uniform 2 kg cylinder rests on a laboratory cart as
shown. The coefficient of static friction between the
cylinder and the cart is 0.5. If the cylinder is 4 cm in
diameter and 10 cm in height, which of the following is
closest to the minimum acceleration of the cart needed to
cause the cylinder to tip over?
[A] 2 m/s2 [B] 4 m/s2 [C] 5 m/s2 [D] 6 m/s2 [E] The cylinder would
slide at all of these accelerations. 4. At right is a graph of the distance vs.
time for car moving along a road.
According the graph, at which of the
following times would the automobile
have been accelerating positively?
[A]
[B]
[C]
[D]
[E] 0, 20, 38, & 60 min.
5, 12, 29, & 35 min.
5, 29, & 57 min.
12, 35, & 41 min.
at all times from 0 to 60 min. 5. A ball which is thrown upward near the surface of the earth with a velocity of 50 m/s will
come to rest about 5 seconds later. If the ball were thrown up with the same velocity on Planet
X, after 5 seconds it would still be moving upwards at nearly 31 m/s. The magnitude of the
gravitational field near the surface of Planet X is what fraction of the gravitational field near the
surface of the earth?
[A] 0.16 [B] 0.39 1999 Physics Olympiad [C] 0.53 [D] 0.63 [E] 1.59 2 6. Two light plastic shopping bags of negligible mass are placed 2 meters apart. Each bag
contains 15 oranges. If 10 oranges were moved from one bag to the other, the gravitational
force between the two bags would
[A]
[B]
[C]
[D]
[E] increase to 3/2 the original value.
decrease to 2/5 the original value.
increase to 5/3 the original value.
decrease to 5/9 the original value.
not change. 7. If a net force F applied to an object of mass m will produce an acceleration of a, what is the
mass of a second object which accelerates at 5a when acted upon by a net force of 2F?
[A] (2/5)m [B] 2m [C] (5/2)m [D] 5m [E] 10m 8. A large beach ball is dropped from the ceiling of a school gymnasium to the floor about 10
meters below. Which of the following graphs would best represent its velocity as a function of
time? [A] [B] [C] [D] [E] 9. A driver in a 1500 kg sports car wishes to pass a slow moving truck on a two lane road. What
is the average power in watts required to accelerate the sports car from 20 m/s to 40 m/s in 3
seconds?
[A] 10,000 [B] 20,000 [C] 100,000 [D] 300,000 [E] 400,000 10. A 40 kg mass is attached to a horizontal spring with a constant of 500 N/m. If the mass rests
on a frictionlless horizontal surface, what is the total energy of this system when set into simple
harmonic motion by an original displacement of 0.2 meters?
[A] 10 J [B] 20 J 1999 Physics Olympiad [C] 50 J [D] 4000 J [E] 100,000 J 3 11. Two skaters on a frictionless pond push apart from one another. One skater has a mass M
much greater than the mass m of the second skater. After some time the two skaters are a
distance d apart. How far has the lighter skater moved from her original position?
[A] d [B] d FG M IJ
H mK [C] d FG m IJ
H MK [D] d FG m IJ
H M + mK [E] d FG M IJ
H m+ MK 12. Given the three masses as shown in the
diagram, if the coefficient of kinetic friction
between the large mass (m2) and the table is µ,
what would be the upward acceleration of the
small mass (m3)? The mass and friction of the
cords and pulleys are small enough to produce a
negligible effect on the system.
[A] b m1 g
m1 + m2 + m3 g
[D] [B] b gµ m1 − m2 − m3
m1 + m2 + m3 b g g b b g g m1 + m2 µ
m1 + m2 + m3 g [C]
[E] b
b g m1 − m2 µ − m3
m1 + m2 + m3 g g b gµ m1 + m2 + m3
m1 − m2 − m3 b g g 13. An object with a mass of 2
kilograms is accelerated from rest.
The graph at right shows the
magnitude of the net force in
newtons as a function of time in
seconds. At t = 4 seconds the
object’s velocity would have been
closest to which of the following:
[A]
[B]
[C]
[D]
[E] 2.2 m/s
3.5 m/s
5.8 m/s
7.0 m/s
11.5 m/s 14. A thin ring of mass m and radius r rolls across the floor with a velocity v. Which of the
following would be the best estimate of the ring’s total kinetic energy as it rolls across the floor?
1
1
1
mv 2
1
r2
[B] mv 2
[C] mv 2
[D] mv 2 +
[E] mv 2 + m 2
[A] mv 2
2
4
2
r
2
t 1999 Physics Olympiad 4 15. A 4.0 kg mass is attached to one end of a rope 2 m long. If the mass is swung in a vertical
circle from the free end of the rope, what is the tension in the rope when the mass is at its highest
point if it is moving with a speed of 5 m/s?
[A] 5.4 N [B] 10.8 N [C] 21.6 N [D] 50 N [E] 65.4 N 16. A mole of ideal gas at STP is heated in an insulated constant volume container until the
average velocity of its molecules doubled. Its pressure would therefore increase by what factor?
[A] 0.5 [B] 1 [C] 2 [D] 4 [E] 8 17. An ideal heat engine takes in heat energy at a high temperature and exhausts energy at a
lower temperature. If the amount of energy exhausted at the low temperature is 3 times the
amount of work done by the heat engine, what is its efficiency?
[A] 0.25 [B] 0.33 [C] 0.67 [D] 0.9 [E] 1.33 18 . If 100 g of ice at 0o C is mixed with 100 g of boiling water at 100o C, which of the following
graphs would best represent the temperature vs time of the two components of the mixture? [A] [B] [C] [D] [E] 19. Two wave sources, G and H, produce waves
of different wavelengths as shown in the diagram.
Lines shown in the diagram represent wave crests.
Which of the following would be true of the
resulting interference pattern?
[A] The perpendicular bisector of GH would be
an antinodal line.
[B] The interference pattern will be a stable pattern but not symetrical left to right.
[C] The pattern of nodal and antinodal lines will sweep from left to right.
[D] If G and H were moved farther apart, the fewer the number of nodal and antinodal lines.
[E] There will not be an organized interference pattern because the sources are not producing the
same frequency. 1999 Physics Olympiad 5 20. Given a wave produced by a simple harmonic oscillator whose displacement in meters is
given by the equation: y = .3 sin (3πx + 24πt), what is the frequency of the wave in hertz?
[A] 3 Hz [B] 7.2 Hz [C] 8 Hz [D] 12 Hz [E] 24 Hz 21. Which of the following wave properties cannot be demonstrated by all kinds of waves?
[A] Polarization [B] Diffraction [C] Superposition [D] Refraction [E] Frequency 22. A source when at rest in a medium produces waves with a velocity v and a wavelength of λ.
If the source is set in motion to the left with a velocity vs, what would be the length of the
wavelengths produced directly in front of the source? FG
H [A] λ 1 − vs
v IJ
K FG
H [B] λ 1 + vs
v IJ
K FG
H [C] λ 1 + v
vs IJ
K [D] λ −v
v − vs [E] λv
v + vs 23. A beam of light strikes one mirror of a pair of right
angle mirrors at an angle of incidence of 45o as shown
in the diagram. If the right angle mirror assembly is
rotated such that the angle of incidence is now 60o,
what will happen to the angle of the beam that emerges
from the right angle mirror assembly?
[A] It will move through an angle of 15o with respect
to
the original emerging beam.
[B] It will move through an angle of 30o with respect
to the original emerging beam.
[C] It will move through an angle of 45o with respect
to the original emerging beam.
[D] It will move through an angle of 60o with respect
to the original emerging beam.
[E] It will emerge parallel to the original emerging beam. 24. What would be the total current being supplied by the
battery in the circuit shown?
[A]
[B]
[C]
[D]
[E] 3.0 amperes
2.25 amperes
2.0 amperes
1.5 amperes
1.0 amperes 1999 Physics Olympiad 6 25. Consider a simple circuit containing a battery and three light bulbs. Bulb A is wired in
parallel with bulb B and this combination is wired in series with bulb C. What would happen to
the brightness of the other two bulbs if bulb A were to burn out?
[A]
[B]
[C]
[D]
[E] only bulb B would get brighter.
both would get brighter.
bulb B would get brighter and bulb C would get dimmer.
bulb B would get dimmer and bulb C would get brighter.
There would be no change in the brightness of either bulb B or bulb C. 26. A coil of wire is moved vertically at a constant velocity through a horizontal magnetic field.
Assume the plane of the coil is perpendicular to the magnetic field. Which of the following
graphs would best represent the electric current induced in the coil if it started somewhat above
the magnetic field and ended equally as far below the magnetic field ? [A] [B] [C] [D] [E] 27. Three metal spheres A, B, and C are mounted on insulating
stands. The spheres are touching one another, as shown in the
diagram. A strong positively charged object is brought near sphere
A and a strong negative charge is brought near sphere C. While the
charged objects remain near spheres A and C, sphere B is removed
by means of its insulating stand. After the charged objects are
removed, sphere B is first touched to sphere A and then to sphere C.
The resulting charge on B would be:
[A]
[B]
[C]
[D]
[E] the same sign but ½ the magnitude as originally on sphere A.
the opposite sign but ½ the magnitude as originally on sphere A.
the opposite sign but ¼ the magnitude as originally on sphere A.
the same sign but ½ the magnitude as originally on sphere C.
neutrally charged. 1999 Physics Olympiad 7 28. Which of the following graphs would best represent the electric field of a hollow Van de
Graff sphere as a function of distance from its center when it is charged to a potential of 400,000
volts? [A] [B] [C] [D] [E] 29. A mass spectrograph separates ions by weight using simple concepts from physics. Charged
ions are given a specific kinetic energy by accelerating them through a potential difference. The
ions then move through a perpendicular magnetic field where they are deflected into circular
paths with differing radii. How would the radius of a singularly ionized common helium atom
4
1+
compare to the radius of a doubly ionized common oxygen atom 16 O 2+ if they were
8
2 He c h c h accelerated through the same potential difference and were deflected by the same magnetic field?
[A]
[B]
[C]
[D]
[E] The radius of the He ion path is 4 times the radius of the O ion path.
The radius of the O ion path is 2 times the radius of the He ion path.
The radius of the O ion path is 4 times the radius of the He ion path.
The radius of the O ion path is 8 times the radius of the O ion path.
The radius of the He ion path is equal to the radius of the O ion path. 30. A bar magnet is thrust into a coil of wire as indicated in the diagram. Which of the following
statements concerning this experiment is correct?
[A] The polarity of the coil at H is North, at G is South,
and the induced current is indicated by arrow X.
[B] The polarity of the coil at H is South, at G is North,
and the induced current is indicated by arrow Y.
[C] The polarity of the coil at H is North, at G is South,
and the induced current is indicated by arrow Y.
[D] The polarity of the coil at H is South, at G is North,
and the induced current is indicated by arrow X.
[E] There is no polarity to the coil and no current in the
coil. 1999 Physics Olympiad 8 1999 AAPT PHYSICS OLYMPIAD
MULTIPLE CHOICE SCREENING TEST
ANSWER KEY 1. B 11. E 21. A 2. E 12. E 22. A 3. B 13. B 23. E 4. C 14. A 24. D 5. B 15. B 25. C 6. D 16. D 26. E 7. A 17. A 27. C 8. D 18. C 28. C 9. D 19. C 29. E 10. A 20. D 30. B 1999 Physics Olympiad 9 Physics Olympiad
Entia non multiplicanda sunt praeter necessitatem
1998 MULTIPLE CHOICE SCREENING TEST
30 QUESTIONS—40 MINUTES
INSTRUCTIONS
DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN
This test contains 30 multiple choice questions. Your answer to each question must be marked on
the optical mark answer sheet that accompanies the test. Only the boxes preceded by numbers 1
through 30 are to be used on the answer sheet.
Select the single answer that provides the best response to each question. Please be sure to use a
No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer,
the previous mark must be completely erased.
A handheld calculator may be used. However, any memory must be cleared of data and
programs. Calculators may not be shared.
Your grade on this multiple choice test will be your number of correct answers. There is no
penalty for guessing. It is to your advantage to answer every question.
The values of some possibly useful constants are given below:
mass of electron me = 9.1 x 1031 kg mass of proton mp = 1.7 x 1027 kg electronic charge e = 1.6 x 1019 C speed of light c = 3.0 x 108 m/s Coulomb's constant k = 9.0 x 109 N·m2/C2 permittivity constant ε0 = 8.9 x 1012 C2/N·m2 permeability constant µ0 = gravitational constant G = 6.7 x 1011 N·m2/kg2 mass of Earth ME = 6.0 x 1024 kg radius of Earth RE = 6.4 x 106 m gravitational field at Earth’s surface
speed of sound (20oC) g = 9.8 N/kg
vs = 340 m/s 4¹ x 107 T·m/A DO NOT OPEN THIS TEST UNTIL YOU ARE INSTRUCTED TO BEGIN
Copyright © 1998, AAPT
1998 Physics Olympiad page 1 1. The graph to the right is a plot of position versus time. For
which labeled region is the velocity positive and the acceleration
negative?
A. A B. B C. C D. D x
AB E C E. E D t
2. A child left her home and started walking at a constant
velocity. After a time she stopped for a while and then
continued on with a velocity greater than she originally had. All of a sudden she turned around
and walked very quickly back home. Which of the following graphs best represents the distance
versus time graph for her walk?
A. B. C. D. E. s s s s s t t t t t 3. In a rescue attempt, a hovering helicopter drops a life preserver to a swimmer being swept
downstream by a river current of constant velocity v. The helicopter is at a height of 9.8 m. The
swimmer is 6.0 m upstream from a point directly under the helicopter when the life preserver is
released. It lands 2.0 m in front of the swimmer. How fast is the current flowing? Neglect air
resistance.
A. 13.7 m/s B. 9.8 m/s C. 6.3 m/s D. 2.8 m/s E. 2.4 m/s 4. A child tosses a ball directly upward. Its total time in the air is T. Its maximum height is H.
What is its height after it has been in the air a time T/4? Neglect air resistance.
A. (1/4) H B. (1/3) H C. (1/2) H D. (2/3) H E. (3/4) H 5. A whiffle ball is tossed straight up, reaches a highest point, and falls back down. Air resistance
is not negligible. Which of the following statements are true?
I.
II.
III. A. I only B. II only 1998 Physics Olympiad The ball’s speed is zero at the highest point.
The ball’s acceleration is zero at the highest point.
The ball takes a longer time to travel up to the highest point than
to fall back down.
C. I & II only D. I & III only E. I, II, & III page 2 6. A pendulum is attached to the ceiling of an elevator car. When the car is parked, the
pendulum exhibits a period of 1.00 s. The car now begins to travel upward with an upward
acceleration of 2.3 m/s2. During this part of the motion, what will be the approximate period of
the pendulum?
A. 0.80 s B. 0.90 s C. 1.00 s D. 1.10 s E. 1.20 s 7. Two identical blocks of weight W are placed one on top of the
other as shown to the right. The upper block is tied to the wall.
The lower block is pulled to the right with a force F. The
coefficient of static friction between all surfaces in contact is µ.
What is the largest force F that can be exerted before the lower
block starts to slip?
A. µW B. (3/2) µW C. 2 µW W F W D. (5/2) µW E. 3 µW 8. An object placed on an equal arm balance requires 12 kg to balance it. When placed on a
spring scale, the scale reads 120 N. Everything (balance, scale, set of masses, and the object) is
now transported to the moon where the gravitational force is onesixth that on Earth. The new
readings of the balance and the spring scale (respectively) are:
A. 12 kg, 20 N B. 12 kg, 120 N C. 12kg, 720 N D. 2 kg, 20 N E. 2 kg, 120 N X 9. A large spool of rope lies on the ground. See the diagram
to the right. The end, labeled X, is pulled a distance S in the
horizontal direction. The spool rolls without slipping. The
distance the spool’s center of mass moves is
A. 2S B. S C. S/2 D. S/3 10. Air track car Z of mass 1.5 kg approaches and collides
with air track car R of mass 2.0 kg. See the accompanying
diagram. Car R has a spring attached to it and is initially at
rest. When the separation between the cars has reached a
minimum, then:
A.
B.
C.
D.
E. E. S/4 Z R car R is still at rest.
car Z has come to rest.
both cars have the same kinetic energy.
both cars have the same momentum.
the kinetic energy of the system has reached a minimum. 1998 Physics Olympiad page 3 11. Two ice skaters, a 200 lb man and a 120 lb woman, are initially hugging on a frictionless level
ice surface. Ten seconds after they push off from each other, they are 8.0 m apart. How far has
the woman moved in that time?
A. 8.0 m B. 6.5 m C. 5.0 m D. 4.0 m 12. The diagram to the right shows the velocitytime graph for
two masses R and S that collided elastically. Which of the
following statements is true?
I.
II.
III. 1.2
0.8 B. II only C. I & II only v(m/s)
R S 0.4 R and S moved in the same direction after the collision.
The velocities of R and S were equal at the midtime of
the collision.
The mass of S was greater than the mass of R. A. I only E. 3.0 m 1 D. II & III only 2 3 t(s) 4 E. I, II, & III Questions 13 and 14 refer to the motion of two blocks along a frictionless level track. Block #1
(mass m1) is initially moving with speed vo. It collides with and sticks to an initially stationary
block (#2) of mass m2 = 9 m1.
13. What is the speed of the two blocks after the collision?
A. vo B. (9/10) vo C. (8/9) vo D. (1/9) vo E. (1/10) vo 14. What fraction of the initial kinetic energy of the system is converted to other forms (heat,
sound, ...) as a result of the collision?
A. 1 % B. 10 % C. 50 % D. 90 % 15. Three identical objects of mass M are fastened to a massless rod of
length L as shown. The array rotates about the center of the rod. Its
rotational inertia is
A. (1/2) ML2 B. ML2 C. (5/4) ML2 D. (3/2) ML2 16 A length of rope is wrapped around
a spool of weight W with inner radius r
and outer radius R as shown in the
accompanying diagram. The rope is
pulled with a tension T at an angle θ.
Which of the following conditions must
be satisfied for the spool to slide
uniformly without rolling?
A. cos θ = r/R B. sin θ = r/R 1998 Physics Olympiad E. 99 %
M M
L/2 L/2 E. 3 ML2 T
r M θ R
r R C. T = W D. T = W sin θ E. T = W cos θ page 4 17. Two identical bricks of length L are piled one on top of the other on a
table. See the diagram to the right. What is the maximum distance S the
top brick can overlap the table with the system still balanced?
A. (1/2) L B. (2/3) L C. (3/4) L D. (7/8) L L E. L
S 18. A gas contains a mixture of He4 and Ne20 atoms. If the average
speed of the He4 atoms is vo, what is the average speed of the Ne20
atoms?
B. (1/ 5 ) vo A. (1/5) vo C. vo D. 5 vo E. 5 vo 19. One end of a metal rod of length L and crosssectional area A is held at a constant
temperature T1. The other end is held at a constant T2. Which of the following statements about
the amount of heat transferred through the rod per unit time are true?
I.
II.
III.
A. II only The rate of heat transfer is proportional to 1/(T1  T2).
The rate of heat transfer is proportional to A.
The rate of heat transfer is proportional to L.
B. III only C. I and II only D. I and III only E. II and III only 20. On all the pV diagrams shown below the lighter curve represents an isothermal process, a
process for which the temperature remains constant. Which dark curve best represents an
adiabatic process, a process for which no heat enters or leaves the system?
A.
p B.
p C.
p V D.
p V E.
p V V V 21. The longest wavelength photon in the visible Balmer series for the hydrogen atom is
A. 0.66 nm B. 6.56 nm C. 65.6 nm D. 656 nm E. 6560 nm
X 22. Waves are produced by two point sources S and S’ vibrating in phase.
See the accompanying diagram. X is a point on the second nodal line. The
path difference SX – S’X is 4.5 cm. The wavelength of the waves is
approximately
A. 1.5 cm B. 1.8 cm 1998 Physics Olympiad C. 2.3 cm D. 3.0 cm E. 4.5 cm S S’ page 5 23. A thin film of thickness t and index of refraction 1.33 coats a glass
with index of refraction 1.50 as shown to the right. Which of the
following thicknesses t will not reflect light with wavelength 640 nm in
air? Hint: Light undergoes a 180o phase shift when it is reflected off a
material with a higher index of refraction.
A. 160 nm B. 240 nm C. 360 nm D. 480 nm n=1.00
n=1.33 t n=1.50 E. 640 nm 24. An object is placed 12 cm in front of a spherical mirror. The image is right side up and is two
times bigger than the object. The image is:
A. 6 cm in front of the mirror and real.
B. 6 cm behind the mirror and virtual.
C. 12 cm in front of the mirror and virtual.
D. 24 cm in front of the mirror and virtual.
E. 24 cm behind the mirror and virtual. 25. A charge is uniformly distributed through a volume of radius a. Which of the graphs below
best represents the magnitude of the electric field as a function of distance from the center of the
sphere?
A. B.
E C.
E a r D.
E a r E.
E a r E a 26. A free electron and a free proton are placed between two oppositely
charged parallel plates. Both are closer to the positive plate than the
negative plate. See diagram to the right. Which of the following
statements is true?
I.
II.
III.
A. I only r a r +++++++++
+
–
––––––––– The force on the proton is greater than the force on the electron.
The potential energy of the proton is greater than that of the electron.
The potential energy of the proton and the electron is the same.
B. II only 1998 Physics Olympiad C. III only D. I & II only E. I & III only page 6 6.0Ω 27. In the electrical circuit shown to the right, the current through the
2.0 Ω resistor is 3.0 A. The emf of the battery is about 6.0Ω A. 51 V 2.0Ω B. 42 V C. 36 V D. 24 V E. 21 V 3.0Ω 28. An ion with a charge q, mass m, and speed v enters a magnetic field B and is deflected into a
path with a radius of curvature R. If an ion with charge q, mass 2m, and speed 2v enters the same
magnetic field, it will be deflected into a path with a radius of curvature
A. 4 R B. 2 R C. R D. (1/2) R E. (1/4) R 29. Which of the following wiring diagrams could be used to experimentally determine R using
Ohm’s Law? Assume an ideal voltmeter and an ideal ammeter.
A.
R
V B. C.
A R
A V D. R R
V
A E.
A V R
V A 30. A vertical wire carries a current upward through a magnetic field directed to the north. The
magnetic force on the wire points
A. south B. north 1998 Physics Olympiad C. east D. west E. downward page 7 1998 MULTIPLE CHOICE SCREENING TEST
ANSWER KEY 1. E 11. C 21. D 2. B 12. C 22. D 3. D 13. E 23. C 4. E 14. D 24. E 5. A 15. A 25. D 6. B 16. A 26. B 7. E 17. C 27. B 8. A 18. B 28. A 9. C 19. A 29. B 10. E 20. C 30. D 1998 Physics Olympiad page 8 19941995 Physics Olympiad Preparation Program
University of Toronto
Problem Set 1: General Physics
Due October 24, 1994 1. Your physics professor you're in university now has babbled all year long in lectures and you couldn't
understand a word he said. It's now exam time and, since you didn't bring a cheat sheet good for
you, you decide to derive the formulas for various physical quantities using dimensional analysis in
the hopes of partial marks. Your babbling professor may give you partial marks but we'll give you full
ones if you get all these as long as you show your assumptions and your work!.
a You remember a 2 but can't remember the rest of the expression for the period of a simple
pendulum. What is the full expression?
b A particle of mass m rotates in a circle of radius r with speed v. The particle has an acceleration
ac called centripetal centre seeking acceleration. What is the form of ac ?
c A gas bubble from a deep explosion under water oscillates with a period T . Known variables are
p, the static pressure, , the water density, and e, the total energy of the explosion. Find the
dependence of the period on the other variables.
2. Being the hip physicsdude dudette you are, you often go to parties with your nonphysics friends.
You like to be the centre of attention, telling them about the physics of quarks and gluons, but that
usually ends up killing the party. You decide, then to pull a di erent beast out of your physics hat:
rapid estimation. You know that sometimes we don't need exact answers but only orderofmagnitude
estimations. At a recent party, your friends asked you the following:
a Jake asked: if I beat everyone on the planet into a pulpy liquid, roughly how deep would the liquid
be on the Earth's surface? Would I need Doc's, rubber boots, hip waders, or a boat to a avoid
getting my new socks dirty? What if I wanted to ll up containers the size of the SkydomeTM in
downtown Toronto with this vile mess; how many containers would I need?
b Jane asked: if the entire debt of the Government of Canada were paid o in loonies, how much
would it weigh? If I stacked them, how high would it reach? If each loonie was turned into a
pound of esh, how would that a ect the answer to Jakes question?
Remember to justify any nonobvious estimations you make; your physics teacher's daughter is at the
party and she'll call your blu on bad estimations, making you look like a fool in front of your friends.
3. Your GreatAunt Edna is in town visiting. She's really rich and is about to kick the bucket, so she will
be drawing up her will soon. You want to be in it, you materialistic !@?$!, you. Now, Auntie Edna,
as you like to call her, made her fortune as a highschool physics and english teacher, so she really
gets PO'd when people use poor grammar and physically incorrect phrases. You decide to impress
her with the following:
a Metro Toronto Police as well as local newscasters on TV often describe the latest tra c fatality
using the phrase the car was traveling at a high rate of speed." What is wrong with this?
b People describe their watch as being ve minutes fast" or two minutes slow". What is wrong
with this? What do they really mean or what should they really say? Is there a correct way of
describing a watch's inaccuracy using the words fast" or slow"?
c Weather announcers and even meteorologists use the phrase normal high" and normal low".
What do they mean by normal"? Is that normal? What might be a better word? Discuss. X Y Figure 1: Rectangular wire mesh of in nite extent.
4. Your Uncle Sal from Italy was in North America for the World Cup. He was terribly disappointed with
their loss, and is in a blue funk the blue that is found on the Brazilian ag, as a matter of fact. You
want to make him a pesto pizza to cheer him up. While pouring the olive oil, you accidentally spill
some into a glass of water, and it spreads out on the surface. Sal, who is watching, is a physics teacher
in Italy and he is reminded of an experiment done by Lord Rayleigh. He tells you that he would be
happy if only you would solve the following.
Rayleigh1 found that 0.81 mg of olive oil on a water surface produced a monomolecular layer 84 cm
in diameter. What value of Avogadro's Number results?
Note: The approximate composition of olive oil is H CH218COOH , in a linear chain with one end
which? hydrophilic and the other hydrophobic. Its density is 0:8 g=cm3
5. Ever the consummate student, you are at Woodbine doing some  studying. Of course, you
betting
brought your $500 camera with a telephoto lens. Watching the thoroughbred racers Eric's Idle and
Paul's Bunyon thunder down the home straight, you decide to take a picture. You are looking head on
through the telephoto lens and you notice that the horses seem strangely foreshortened not as long
from front to back as they gallop towards the camera. Explain this observation.
6. Your little brother has been playing with the wire screens and batteries again. Will he ever learn? He
asks you the following.
a A rectangular wire mesh of in nite extent in a plane has 1 A of current fed into it at point X, as
in the diagram Figure 1, and 1 A of current taken from it at point Y. Find the current in the
wire XY.
b Suppose the wire mesh was made of equilateral triangles joined in the obvious way. What is the
current where X and Y are separated by one wire?
1 Rayleigh, Proc. Roy. Soc., 47, 364 1890; Holy smokes, that's 104 years ago! And you're only learning about it now?
Bonus points to anyone who can get me a photocopy of the article. ...
Figure 2: In nite resistor ladder.
c The screen has been cut up and now resembles the circuit in the diagram. What is the resistance
of the circuit shown, an in nite ladder. 19941995 Physics Olympiad Preparation Program
University of Toronto Solution Set 1: General Physics 1. Dimensional analysis can be used to check to see if an expression is dimensionally correct or to get the
form of an expression if we don't know it. We are going to do the latter.
a Hmm... What could the period of a simple pendulum possibly depend on? There's the mass of
the ideal point bob or is that Bob, founder of the Church of the SubGenius?, m, the length of
the essentially massless ideal string, l, the angle of the swing, , and the acceleration of gravity,
g. We might think to include things like frictional forces, but they're small compared to gravity
and besides, that would be too complicated. Let's assume the period T is a function of these four
variables, each raised to some power:
T = Cmw lx y gz :
C is a dimensionless constant, and w, x, y, and z are exponents for which we wish to solve no
dangling prepositions here. The dimensional equation for this relationship is
T = M w L x L=T 2 z ;
the angle measured in radians has no dimensions. Simplifying,
T = M w L x+z T ,2z :
To have dimensional consistency, we must have
0=w
0 = x+z
1 = ,2z
which gives us w = 0, z = ,1=2, and x = 1=2. Thus the desired equation is
p T = C l=gf
It turns out that C = 2 and f ' 1 for small angles, but we can't get that from this analysis.
b The general equation for this problem is ac = Cmx ry vz , which gives us the dimensional equation
L=T 2 = M x L y L=T z . Simplifying, L T ,2 = M x L y+z T ,z , which gives us the three
equations 0 = x, 1 = y + z , and ,2 = ,z , which are solved by x = 0, z = 2, and y = ,1. Thus
ac = Cv2 =r. It turns out that C = 1.
c The general equation for this problem is T = Cpx y ez , which gives us the dimensional equation
T = M =LT 2 x M =L3 y M L2=T 2 z . Simplifying, T = M x+y+z L ,x,3y+2z T ,2x,2z, which
gives us the three equations 0 = x + y + z , 0 = ,x
, 3y 2z , and 1 = ,2x , 2z , which are solved
+
3 e2 1=6
by x = ,5=6, y = 1=2, and z = 1=3. Thus T = C p5
.
2. For these ones, we quote numerical values of things from memory or make good guesses, multiply
numbers in our head, rounding them o , and in general not worry about being too precise. It may also
help to do the problem algebraically rst, then put in the numbers, since this will allow us to plug in
more precise numbers if we wish.
1 a In the spirit of rapid estimation, we will give the result as a series of approximate calculations.
This way we only have to remember a few numbers at a time.
Presumably the depth of liquid, d, will not be large compared to the radius of the Earth, RE , so
we can use the formula for the volume of a thin shell1; V = 4R2 d. Thus d = V=4R2 . What
E
E
is V? Well, there are nearly 6 billion people on the planet. Their average mass is, say, about 50
kilograms thus the total mass of people is about 300 billion kilograms. The density of a human
body is close to that of water2 , about a tonne per cubic metre, so the total volume is about 300
million cubic metres.
The radius of the Earth is about 4 thousand miles a good number to remember or over six
thousand kilometres. Squared, we have about 40 trillion square metres. Multiplying by 4 ' 12:5
we have the surface of the Earth being about 500 trillion square metres. Dividing the volume by
this we get a depth of about a half a micron m. I don't even know if this would get your socks
dirty, but you could easily get by without the rubber boots and just wearing shoes.
The politicallycorrect restatement of the problem is: if every person on the planet decided to go
skinnydipping in the ocean, then by how much would the ocean level rise? The answer is about
4 3 the answer to the nonpolitically correct question since the oceans cover about 3 4 of the
planet.
What about the number of SkydomeTM's? We just divide the volume found above by the volume
of a building. The rough area of the football eld is 60 m by 130 m or about 8 thousand square
metres. The total area is maybe four times that, and the total height is, perhaps, 30 m. Thus the
total volume is on the order of one million cubic metres. Dividing the volume of liquid by this we
get something on the order of several hundred SkydomeTM's.
b You may have heard that the federal debt per capita is about $20,000. If you hadn't heard that,
then you should read more newspapers! With 27 million people in Canada, this works out to
about $550 billion. Wow! A loonie has a mass of about 10 grams, I believe. On Earth, 454 grams
weighs a pound, so lets say that 50 loonies weigh one pound. Thus the federal debt weighs about
11 billion pounds in loonies.
Stacking them, they are about 2 mm think so that makes about 1.1 trillion metres or a million
kilometres. That's to the moon and back easily!
Converting them to a pound of esh, we have that a pound is about half a kilogram and the
density is a tonne per cubic metre. This works out to 270 billion kilograms or 270 million cubic
metres. This nearly doubles the volume of liquid for the previous problem, thus it nearly doubles
the depth.
3. a What is wrong with the phrase the car was traveling at a high rate of speed"? Well, the word
rate usually refers to a change in a quantity measured with respect to an interval of time. The
de nition of speed is the distance traveled per unit of time, i.e. the rate of change of position
with respect to time. Thus the word rate in the above phrase is super uous. One could say at
a high speed" or one could say at a high rate" where it is implicit that the rate is position with
respect to time, i.e. speed. Note also that we can't really interpret at a high rate of speed" as
acceleration since in that case it would be at a high rate of change of speed".
b As in the previous question, the fast and slow refer to rates as well. What most people mean is
ve minutes ahead" or two minutes behind". One can use rateimplying words when talking
about clocks, but in the following context. Suppose we set a watch to coincide with an atomic
1 You can derive this in the following way. Let r be the radius of a sphere, and r + r be the radius of a slightly larger
sphere. The volume of the shell between the two spheres is the di erence in volumes which is 4r + r3 , r3 =3 =
4r3 + 3r2 r + 3rr2 + r3 , r3 =3. This is then equal to 43r2r + 3rr2 + r3 =3 = 4r2 r1 + r=r + r2 =3r2.
We then use the fact that r=r 1 to get the volume V ' 4r2 r.
2 People oat with the aid of air in their lungs; without that they would sink. Thus people are denser than water, but not
by much. 2 clock. Thirty days later we check and the watch is ahead one minute. We have gained 1 minute
over a period of 30 days. Thus the watch is about one minute per month fast. Strictly speaking
the units can be canceled to get a unitless number, but it is more meaningful to people to use the
above form. Also note that I have seen people with degrees in physics make the mistake of saying
fast when they mean ahead!
c The normal high and normal low are, in more precise language, the average high and low. That
is, they are the averages of the daily highs and lows for a particular day of the year, averaged
over the past 100 years or so for which there are recorded temperatures. What is left out is the
variance of the highs and lows or rather the distribution of the temperatures.
It is possible that the average temperature on August 28 is 25 C but there may be just as many
August 28's which were 23, 24, 26, and 27. Does it then make sense to use the word normal to
mean average? In such a case is 25 normal? Probably not.
It is really a case of experts not wanting to use precise language because it may end up confusing
people, or so they think.
4. We know that N=M = NA =MA where N is the number of molecules in the sample and M is the
mass of the sample and the subscript refers to a mole of the substance. Thus Avogadro's Number
is NA = NMA =M . To get MA , we take the formula and the molecular masses and nd the total;
38 1 + 19 12 + 2 16 = 298 g. To get N , we take the volume of the sample, V , and divide it by
the volume of a single molecule, v. Getting v is the tricky part.
The COOH end of the molecule is hydrophilic while the rest of the molecule is hydrophobic. Thus
the molecules form a layer standing sidebyside vertically on the surface of the water. The diagram
shows approximately what the molecule looks like if we assume that the carbons are at the centres of
tetrahedrons with the neighbouring atoms on the corners. There is an endon view and a side view.
The distance between two atoms joined to a common carbon is xthis is the length of an edge of
the tetrahedron. It turns out that the endon area is nearly squarethe zigzag in the carbon chain
compensates for the di erence between the edge length in one direction and the distance from edgecentre to opposite edgecentre in the other direction. In all three directions a distance of x is added
to account for the fact that neighbouring molecules will not be butted up against oneanother. The x
puts the hydrogen atoms on one molecule a distance x from those on the neighbouring molecules. Thus
the volume of each molecule is about v ' 4t3 =121 where t is the layer thickness. There is a measure of
uncertainty in this last equation since we don't know how much space there is between molecules.
The total volume is V = At, thus 1=v ' 121A3=4V 3 and N ' 121A3=4V 2 . Now A = d=22 and
V = M= ; with a little algebra we get
3 2 d6
NA ' 121 M 3MA :
256
Putting in the numbers,
3
32 84 6
NA ' 121 0:8 g=cm 10,cm3 298 g
3 g
2560:81
: 2 1024:
=
So Avogadro's number is about 1024 molecules per mole. We are o by a factor of 3 from the accepted
value, which is ok considering how many uncertain numbers there were.
5. A telephoto lens, like a telescope, makes far objects appear near by giving rise to an angular magni cation. Let us approximate a horse call him Boxy by a rectangular body with a vertical leg at each
corner. Suppose the separation of the forelegs and of the back legs is 0.5 m, and the distance between
the front and back legs is 2.0 m. If we observe the horse headon when it is 10 m away, the gap between
3 its forelegs will subtend an angle of 1 20 rad at the eye, while the gap between the hind legs subtends
1 24 rad. This di erence gives most of the impression of the length of the horse.
If the same horse is 100 m away the same two angles become 1 200 rad and 1 204 rad, which are not
very di erent. However, if we use a telescope with an angular magni cation of 10, the angles become
1 20 rad and 1 20.4 rad. The apparent size of the front is the same as that of a horse standing 10 m
away, but now its hind legs appear to be only 0.2 m further back. A very short horse!
6. a Break the problem into two parts. Consider the mesh with 1 A going in at X and 1 A coming
out at in nity. By symmetry, the wire XY must carry 1=4 A of current. Alternately, consider
the mesh with 1 A going in at in nity and 1 A coming out at Y. Again, by symmetry, 1=4 A
runs through the wire XY. Add the two cases together and you get the stated problem, with the
solution that 1=2 A ows along wire XY.
b Similarly, the two cases each have 1=6 A owing along XY, giving a solution of 1=3 A.
c Referring to the diagram, you can see that because the circuit is in nite, it can be written as a
circuit with itself being one of the components. If R0 is the resistance of the entire circuit, then,
using the rules for combining resistors in series and parallel, we have
11
1
R0 = R + 2R + R0 :
Multiplying by all three denominators and rearranging we end up with a quadratic equation for
R0 , R0 2 + 2RR0 , 2R2 = p We use the quadratic formula to solve, with the negative root thrown
0.
out. The result is R0 = 3 , 1R. 4 19941995 Physics Olympiad Preparation Program
University of Toronto
Problem Set 2: Mechanics
Due November 28, 1994 1. A crude estimate of the depth of a well can be obtained by dropping a stone down it and listening for
the splash, then using s = gt2 =2. You are asked to make rstorder estimates of the corrections due to
a the nite speed of sound vs = 340 m=s, and
b air drag.
Which of these is the bigger correction for s = 30 m and a stone of radius r = 2 cm and density
3 103 kg=m3? Would your conclusion be signi cantly di erent if a table tennis ball were dropped in
the same way r = 2 cm, m = 1:7 g? You may assume that the splash would still be audible!
Note: the drag for a sphere of radius a depends on the dimensionless Reynolds number R = va=,
where is the uid density and is its viscosity. For R 10, Stokes' law F = 6rv is valid, but for
R 100 the drag force is given by F = CD r2 v2 ; the variation of CD with log10 R is given in the
diagram. For air at 20 C, = 1:3 kg=m3 and = 1:8 10,5 kg=ms.
2. Your physed teacher is a student of the physics of human motion, and asks you the following questions:
a When told that the world record for the pole vault was about 5.5 metres, the fast rising athlete
Rod Fibreglass told the press, Give me a pole long enough, and I will raise the record to 9
metres." Could he manage it? How high might he raise it if he tried hard?
b I obtained the following data one Sunday:
i. There are 238 steps from the basement to the 10th oor of our building.
ii. After measuring four steps, I believe the step height is 18:2 0:3 cm.
iii. I jogged up the steps in 1 minute and 40 seconds, give or take 2 seconds. Well, actually
started o jogging and ended up walking up with a notsosteady pace.
iv. My mass is 82:07 0:2 kg. The uncertainty is due to the fact that I drank some water, ate a
banana, sweated and urinated in the time between doing the climb and measuring my mass.
v. The value of g measured by some geophysicists at the base of the north stairwell is 9:804253 m=s2.
They don't give an uncertainty estimate. Note that they say that g varies with height as
g=g = ,2R=R, where1 R = 6371 km.
How much useful work did I do? What was the average power output? Calculate the uncertainties in these values as well.
c It is proposed to make a humanpowered helicopter with a rotor 10 m in diameter. Assuming
that the rotor blows a cylindrical column of air uniformly downwards, the cylinder diameter being
the same as the rotor diameter, and that the mass of the pilot plus machine is 200 kg, calculate
the minimum mechanical power in Watts that is necessary for the pilot to generate to remain
airborne. Is the system practicable? Compare with my power output above. The density of air
is 1:23 kg=m3.
3. NASA has just contracted you to make the following calculations, you lucky dog. As with many NASA
contracts, you are being paid a handsome sum with bonuses for correctness. No $10,000 toilet seats,
please.
1 I don't know why this number, posted in our undergraduate labs, disagrees with others quoted here 6 38 103 km, except
to say that the Earth isn't a sphere.
: a A small moon of mass m and radius a orbits a planet of mass M while keeping the same face
towards the planet. Show that if the moon approaches the planet closer than rc = a3M=m1=3,
then loose rocks lying on the surface of the moon will be lifted o .
b It is well known that in an orbiting space vehicle the occupants can drift around freely in zerogravity" conditions. Assume that you are in a strong spaceship, 100 m long and fairly narrow,
which is in a circular orbit of 1000 km radius around a neutron star, with its long axis always
pointing towards the centre of the star. There is an inspection tunnel running down the axis of
the ship.
i. What would happen if an astronaut attempted to oat down it?
ii. Calculate likely values of the acceleration observed, assuming that the mass of the star is
3 105 Earth masses and that the radius of the Earth is 6 106 m.
iii. Is an orbit such that the axis of the craft always points towards the star a stable condition?
Explain brie y.
c Two stars in a binary system have a separation 2r and equal masses m, and move in circular
orbits about their centre of mass. One star explodes by expelling a small fraction of its mass very
rapidly, and immediately after its recoil speed is vf . What is the largest value of vf for which the
two stars will remain gravitationally bound?
4. So, Minnesota, you think you know your billiards? Try to make this trick shot. A billiard ball of radius
a rests on a table. It is hit with a cue in such a way that it starts out with speed u0 and backspin
!0 about a horizontal axis perpendicular to the direction of motion. How does the subsequent motion
depend on the ratio u0 =a!0? Discuss.
5. It is sometimes stated that it would be possible to construct a spaceship using a photon drive, which
would be able to travel away from Earth with a speed close to that of light. Assuming that the fuel
consists of equal masses of protons and antiprotons, does one get a greater nal velocity
a by allowing half the fuel to annihilate and using the energy released to eject the remaining half,
or
b by allowing all the fuel to annihilate and ejecting photons?
You may use the relativistic relationship between energy and momentum for a particle of rest mass
m0 , E 2 = pc2 + m0 c22 and for a massless photon, E = pc, where E is energy, p is momentum, and
c is the speed of light.
6. You are on board the USS Enterprise when an arti cial structure known as a ringworld" is discovered
circling a star similar to Sol. It is a large circular ribbonshaped object with the following characteristics. The at side is illuminated by the star; its radius is that of Earth's orbit, RSE = 1:50 1011 m.
The width is 1:00 109 m and the thickness is 100 km. The edges of the interior oor" are lined with
walls which have a triangular crosssection, are 1000 km high and have a base width of 100 km. This
keeps in an earthlike atmosphere. Data is o duty now so Captain Picard asks you the following.
a If the oor is to have the same gravity" as the earth, what rotational period should it have?
Before calculating, give some limits on the period based on simple facts about the Earth.
b Calculate the tensile strength needed for the material in order for it to remain intact. You may
assume the density of the material is that of aluminum, 2:70 103 kg=m3. For comparison,
aluminum's tensile strength is about 2 108 N=m2.
c Remember, there is a star in the centre of the ring. Is the spinning ring stable with respect to
the star? That is, if we were to push the ring a bit, would it return to its original position like a
marble in a bowl or would it move further out of position like a marble on a beach ball? You
may answer qualitatively. 19941995 Physics Olympiad Preparation Program
University of Toronto Solution Set 2: Mechanics 1. a If the real depth is s and the velocity of sound is vs, then the measured time would be tm =
2s=g1=2 + s=vs and the measured depth would be sm = gt2 =2. Neglecting squares of small
m
terms, we obtain
s
r
sm , s ' 2gs = 29:830 = 7:2:
:
2
s
vs
3402
That is, the depth calculated without taking into account the nite speed of sound would be
about 7 percent too large.
b An approximate value for the nal velocity is1 2gs1=2 ' 24 m=s, showing that the Reynolds
number is of the order of 104 most of the time; the drag force is therefore CD r2 v2 , with
CD ' 0:4. The equation of motion is thus ma = mg , v2 , with = CD r2 . For the zeroth
approximation, we neglect the last term so that a = g and thus v = gt. For the rst order
approximation, we insert this value of v into the small last term to give ma = mg , g2 t2 , or
a = g , g2 t2=m. Now, if you knew calculus, you would immediately integrate this to obtain
s = gt2 =2 , g2 t4=12m. You might also just know that whenever you have a term Atn in the
acceleration, it gives you an Atn+1=n + 1 term in the velocity and an Atn+2 =n + 1n + 2
term in the displacement. You should really learn calculus.
Anyway, neglecting the air drag thus introduces an error equal to the last term, g2 t4 =12m. The
relative error is thus gt2 =6m ' s=3m. From the data given, we nd that this is about 6, thus
the air drag and the nite speed of sound give approximately equal errors. Both errors make the
calculated depth too large.
c For the table tennis ball, the drag force is much more important relative to the weight. The
velocities are smaller, but still large enough to make R 103, i.e. C:D is still about 0:4. In
this case, the terminal velocity is given approximately by2 mg= 1=2 = 5:3 m=s. The ball will
reach this speed after a few meters, so that the zeroth approximation, sm = gt2 =2, is a gross
overestimate of the depthby a factor of about 2.6 for s = 30 m. A better rst approximation is
to assume that it travels at its terminal velocity for the whole distance. In this approximation, the
error due to the nite speed of sound is clearly given by the ratio of the terminal velocity to the
speed of sound. The error due to the initial period of acceleration cannot be evaluated without
using the exact solution of the equation of motion. This is too complicated to be evaluated in the
present context.
2. a The dominant consideration is the conversion of the kinetic energy of the running man to gravitational potential energy with something less than 100 e ciency. If, in his approach, he attains
a speed of 10 m=s, the corresponding rise is 5 m. Smaller terms arise from
i. the fact that his centre of mass is already 1 m above the ground when he starts,
ii. the work done by his legs on takeo and by his arms in climbing up the pole ? = 0:5 m;
consider the vertical" of a standing person,
iii. the fact that his centre of mass actually passes below the bar ? = 10 cm.
Adding these terms together gives approximately 6.6 m. The di erence between this and the
observed 5.5 m is due to an e ciency of less than 100or to errors in the estimated quantities.
In any case, there is clearly no hope of Mr. Fibreglass making good on his boast of 9 m since 10
m s is an excellent sprint speed and the other terms can not be improved by much.
Equate initial potential energy with nal kinetic energy and solve for speed.
Write force equation for terminal body, including gravity and drag force but neglecting the small buoyant force, set
acceleration to zero, then solve for speed.
1
2 1 b Work done is the change in potential energy of my body, W = mgh. Uncertainty is
p W=W = m=m2 + g=g2 + h=h2:
The g=g2 is negligible. The result is 34:85 0:4 kJ. Note that the uncertainty in the stair
height dominates. A measurement of the actual distance would improve things immensely.
The average power output is P = W=t, while the uncertainty is
p P=P = W=W 2 + t=t2
p
= m=m2 + g=g2 + h=h2 + t=t2:
The result is 348:5 8 W. This is nearly half a horsepower. Don't be impressed. I only did this
for a couple of minutes. A horsepower is the average power a horse can put out over a work day.
I've been told that for longer times 15 minutes to 2 or 3 hours for people of average mass 65
to 75 kg, useful power output is roughly 275300 W. By that criteria, given how winded I was, I
am out of shape!
c If the rotor, of radius R, gives a downward velocity v to a column of air initially at rest, the
momentum transferred per second will be v where , the mass of the air accelerated per unit
time, is R2v . If the helicopter is to be airborne, this must equal Mg, where M is the mass of
:
the helicopter. Inserting the values given, we obtain v = 4:5 m=s. Neglecting energy losses, the
energy transferred per second to the bulk motion of the air is v2 =2 which equals 4.5 kW in this
case. For the system to be practical, the pilot must thus generate about 5 kW continuously. From
the answer to the previous question, it appears that the maximum rate of energy production by
a human is of the order of 1=3 kW. Thus we are a far cry from being able to run a helicopter by
human power.
3. a For the moon in an orbit of radius r, we have3 GMm=r2 = mr!2 , assuming M m, where G
is the universal gravitational constant and ! = 2=T is the angular frequency with T being the
period. For the rock of mass , in orbit of radius r , a, Newton's Law gives us
GM Gm
2
r , a2 , a2 + F = r , a!
where F is the force of contact between the rock and the moon. If the rock is to be lifted o ,
F = 0. Eliminating ! between the two equations, we obtain
M = r3 r , a2 ' 1 r 3
m a3 3r2 , 3ra + a2 3 a
as required.
b For any spacecraft in a: circular orbit, GMm=r2 = mr!2 . For a body on the surface of the
2
Earth, GME =rE = g = 10 m=s2. With the numbers given, we nd ! = 10:4 rad=s, i.e. the
period is approximately =5 secondssomewhat uncomfortable for the intrepid astronaut! We
can also deduce that the linear velocity in orbit is approximately 107 m=s ' 0:03chigh, but not
relativistic.
i. For an astronaut of mass ma at the mass centre of the craft, GMma =r2 = ma r!2 is also true,
i.e. she will, indeed, oat.
ii. If r changes to r , x, the gravitational force becomes GMma =r , x2. If sideways motion
across the vehicle is prevented, and if the vehicle is large compared with the mass of the
person, then ! will remain constant, that is the ship will force the astronaut to orbit with
3 The gravitational force provides the centripetal force which is set equal to the mass times centripetal acceleration. 2 the same period. The force needed to maintain the astronaut in the new, smaller orbit
will be ma r , x!2. This is no longer equal to the gravitational force. The di erence,
approximately 3ma !2 x, will p
accelerate the person towards the centre at a rate 3!2x. This
4
will give rise to a velocity 3!x. Starting from the end of the tunnel x = 50 m, this
gives rise to a maximum speed of 900 m=s 3240 km=h in more homely units. On the way to
this unfortunate conclusion there will be a Coriolis acceleration sideways across the tunnel
p
equal to 2 3!2 x. This will reach the value g before the victim has moved more than a few
centimetres.
iii. The orbit as described is indeed stable. The di erential radial force calculated abovethe
socalled tideraising forcewill provide a restoring couple if the attitude is disturbed, as
can be seen by drawing a diagram. Also, one can appeal to tides themselves; they are at a
maximum when the graviational source is overhead.
c Initially Newton's Law gives
mv2 =r = Gm2 =4r2:
1
The nal system is most likely to break up if the explosion increases the velocity of one of the
one star in the direction in which it is moving at the time of the explosion consider the diagram
where before and after denote the situation an instant before and after the explosion. Call the
increased velocity vf . The centre of mass now has a velocity v , vf =2 and the velocity of either
star with respect to the centre of mass is, say,
v + vf =2 = v :
2
The system will break up if the new kinetic energy with respect to the centre of mass coordinates
is greater than the work needed to separate the components to in nity, i.e. if mv 2 GM 2=2r.
Using relations 1 and 2, the required condition becomes
p
p
vf = 2 2 , 1v = 2 2 , 1Gm=4r1=2:
The above argument assumes that the mass lost as a result of the explosion is so small that it
has no e ect on the result. If this quantity is denoted by m, then an analysis along exactly the
same lines gives the condition vf =v = 22 , m=m1=2 , 1, thus showing that the assumption
was valid.
4. We choose the positive sense of ! so that the velocity of the point of contact of the ball with the
table, vc , is given by u + a!. The frictional force F = mg is assumed constant. The motion of the
centre is thus given by u = u0 , gt and the rotation about the centre by ! = !0 , amgt=I . Since
I = 2ma2 =5 for a sphere, the latter expression becomes ! = !0 , 5gt=2a. These results combine
to give vc = u0 + a!0 , 7gt=2. This is equal to zero i.e. slipping stops and rolling begins when
t = 2u0 + a!0 =7g = , say. The value of u at t = is 5u0 , 2a!0=7; this gives the speed of rolling
after slipping has stopped. If u0=a!0 2=5, the value is negative and the ball will roll backwards
towards the cue. This corresponds to the imposition of backspin" as speci ed. If u0 =a!0 2=5, then
the ball will roll forwards.
To enquire whether both motions are possible, we consider an impulse Q given to the ball at a point
a , x above the table in the direction making an angle with the horizontal see the diagram. Then
mu0 = Q cos and I!0 = Qp, with p = a sin + and sin = x=a. If we use the relation 5u0 = 2a!0
to determine the boundary condition between the two modes of behaviour, then, after a little algebra,
we get cos = sin + , i.e. 1 , x=a 1=2 :
tan = 1 + x=a
Real values of can be found for any x=aexcept that x ! a would be a little di cult, in practice!
0 0 4 Think of simple harmonic motion with a spring constant over mass ! 2 = k=ma = 3! 2 ; the maximum speed, obtained while
0
passing through the centre, will be x!0 . 3 5. a The annihilation energy of a particle of rest mass m1 will make available an energy E = m1 c2.
If this energy is given to another particle of rest mass m2 , which thereby acquires momentum p,
we have the relation5
E + m2 c22 = p2 c2 + m2 c2 2 :
In the particular case envisaged in the problem, m1 = m2 = m is the mass of a proton or
p
antiproton. This leads to the result p = 3mc. If there are initially N protons and N antiprotons,
then this process can be repeated N times6. If all N ejected particles can be persuaded to leave
p
the spaceship in the same direction, the e ect will be to give a total momentum 3Nmc to the
ship.
b If we consider the mutual annihilation of two particles, each of rest mass m, a similar argument
shows that the scalar total of the momentum thereby produced is 2mc. There will, in fact, be
two photons proceeding in opposite directions since the vector total must be zero. With the
aid of a deep paraboloidal mirror made of a material that can re ect highenergy gamma rays !
all such photons can be collimated into a beam carrying a momentum 2Nmc. The spaceship will
have an equal momentum in the opposite direction. This is larger than the other result.
The above arguments are strictly valid only if the increment in velocity of the spaceship is much less
than c. If this is not the case, the total number N can be conceptually subdivided into small groups for
which it is true, and the total e ect of the momentum increments between successive inertial frames
can then be calculated as an integral sum. The advantage of 5b over 5a applies at every stage
and therefore also overall.
6. a As a rst guess, one thing we can do is to get a maximum on the period. We don't feel the e ect
of the centripetal acceleration of our orbiting the sun very easily on Earth, so it must be less than,
say, a percent of g. Since the acceleration goes as the inverse square of the period, the period
must be less than 10 percent the Earth's year, that is less than a month. It is di cult to say
anything more since we don't have an indication of how restrictive our maximum is.
More quantitatively, the centripetal acceleration of a particle moving in a circle of radius r at a
constant speed v is 2
2
r 1
ac = vr = 2T
r
p
:
:
where T is the period. Thus T = 2 r=ac . Using ac = g we get T = 7:77 105 s = 9:0 days. That
seems like an awfully short time to go around the sun! Note that we've ignored the gravitation
attraction of the star, which is small compared to g.
b Consider a narrow section of the ring as shown in the diagram. The net force on the section is
2FT sin towards the centre. This is the centripetal force and must equal mac = mg where m
is the mass of the section. This is equal to 2rM=2r where r is the radius of the ring and M
is the total mass. The total mass is the volume times the density, M = 2rA , where A is the
crosssectional area. Letting sin ! for small angles, we nd that the tension is given by
FT = rA g. This is just the weight" of the ringworld on Earth if we could weigh it divided
by 2! To get the tensile strength we just divide by the crosssectional area; S = r g. Notice
how we don't need to know A. If we take the density of aluminum as a possible density for the
material, then the required tensile strength is 4 1015 N=m2. This is seven orders of magnitude
greater than most metals.
The lefthand side is the square of the total energy before, while the righthand side is the square of the total energy after,
using the relationship between energy and relativistic momentum.
6 Actually, the more complicated process of annihilating a protonantiproton pair giving the energy to another proton and
antiproton pair which doesn't annihilate but rather is ejected in the same direction is performed N=2 times, but the e ect is
the same.
5 4 c The rotation of the ring will give it orientational stability, but not positional. The positional
stability is depends only on the net gravitational attraction of the star.
If the ring is pushed so that it moves perpendicularly to its plane, then one can see a net gravitational force attracting the ring back to equilibrium. If the ring is pushed in its plane, the situation
is not as easy to see. Consider the force due to four pieces as shown in the diagram. The size of
the force goes as 1=r2 which depends on the position of the piece on the circle. The vertical"
component see gure also depends on the position but in a opposing way. The result is that one
can't nd out whether there is a net gravitational force in the plane by simple arguments. One
must use math and it turns out there is a net restoring force.
Remarkably enough, there are cases where one can have an instability. A Dyson Sphere a
spherical shell of similar dimensions surrounding a star would have no net force on it due to a
star inside, thus it would be unstable. The argument for this is mathematical as well.
I got the idea for the ringworld question from a book by Larry Niven called, curiously enough, Ringworld. It is followed by Ringworld Engineers, and may be found in the science ction section of libraries
and bookstores. The book describes the material as Very dense, with a tensile strength on the order
of the force that holds nuclei together." Anyone care to check on whether that would be enough? 5 19941995 Physics Olympiad Preparation Program
University of Toronto
Problem Set 3: Thermodynamics
Due January 9th , 1995 1. Helga Helmholtz has a cool" summer job working for the Acme refrigeration company. Her neurotic
boss has asked her to test out the latest CarnotDeluxe model, one with a 50 W motor in it. She was
told to ll the fridge with ice water, the liquid part of which weighed 2 kg, and wheel the unit outside
where the air temperature was 27 C. Her task was to nd the smallest possible time necessary to
freeze the rest of the water. Now, Helga also happens to be a bright physics student, and she doesn't
get paid by the hour either. She instead decides to stay in her air conditioned o ce, calculate the
answer, and go home early. Wouldn't you like to work for a fridge company next summer? Well, even
if you don't, you can still try answering this problem. Note: take the latent heat of fusion for water
to be 6.0 kJ mol,1. Be sure to mention any assumptions you make.
2. Imagine that there are two kinds of E. coli bacteria, which are in every way identical except that one
kind is beige" in colour and the other is fuchsia" in colour. Each type reproduces they have no sex
by splitting into halves, beige beige+beige or fuchsia fuchsia+fuchsia, with a reproduction time of
1 hour. A colony of 5000 beige and 5000 fuchsia E. coli is permitted to eat and reproduce. In order to
keep the colony size down, a Pacman c predator is introduced that keeps the colony size at 10,000 by
devouring the bacteria at random.
a Derive an expression for the probability distribution of the number of beige" bacteria existing
after a very long time.
b Estimate how long you would need to wait for the answer in a to be true.
c Suppose the Pacman c predator has a 1 preference for eating fuchsia" bacteria. How would
this a ect your answers to a and b above?
d What do E. coli like to eat?
3. It is safe to say that the most popular drug among physicists is, you guessed it, ca eine. If you make
a cup of hot co ee in which you would like to have cold milk added, and you are called away and
therefore prevented from drinking it for 10 minutes, when should the cold milk be added in order that
the cup's contents are as hot as possible when you return after the 10 minutes are up? Discuss.
4. Now, how about some chemical thermodynamics? Doc" played by Christopher Lloyd in the movie
Back to the Future is building his latest intergalactic spacetime instantaneous travel machine. The
temporal blaster circuit, in order for it to function, needs to be inserted into a vat of hydrogen iodide.
Doc makes his hydrogen iodide by reacting 2.94 gmol of molecular iodine with 8.10 gmol of molecular
hydrogen at a constant temperature. How much hydrogen iodide can he expect to produce at the given
temperature if the equilibrium constant for this reaction is Kc = 0:02?
5. One day while contemplating your morning toast, imagine a long, thin grain of dust shaped like
a needle oating in a box of laughing gas at a xed temperature T . On average, is the angular
momentum vector nearly parallel or perpendicular to the long axis of the dust grain? Explain your
answer.
6. You are a NASA scientist charged with the design of the mission abort circuitry for a remote terrestrial
volcano exploration module. Once conditions get too hot for the module, a bimetallic strip thermostat
trips the abort circuitry and the device is jettisoned from the volcano. The bimetallic strip has a
thickness x and is straight at temperature T . What is the radius of curvature of the strip, R, when
it is heated to temperature T + T ? Take the coe cients of linear expansion of the two metals to be
1 and 2, respectively, with 1
2. You can assume that each metal has thickness x=2, and that
x R.
! ! 19941995 Physics Olympiad Preparation Program
University of Toronto Solution Set 3: Thermodynamics
1. First, calculate the heat released in freezing 2 kg of water at 0 C, Qc :
3
g 6:0 kJ
Qc = 2 10 18 g=mol =mol = 667 kJ
The maximum coe cient of performance, max , is given by: max WQc = T Tc T ;
min
h, c
where Wmin is the minimum work needed to e ect the cooling, and Tc and Th are the temperatures of
the cold inside fridge and hot outer air reservoirs, respectively.
Thus,
,
Wmin = Th T Tc Qc :
c The minimum time, min , is given by P is the power, and note that absolute temperatures must be
used:
Wmin 300 K , 273 K
:
min = P = 273 K50 W 667 103 J = 1:3 103 s:
Assumptions can include:
a The refrigerator, as suggested by the model name, is taken to be a Carnot heat engine, where
Qc =Qh = Tc =Th .
b The ice and water are in equilibrium; therefore, the liquid is at 0 C.
c Only the water is being cooled i.e. no container material or air.
d If the cold reservoir interior of the fridge were completely lled with ice water, then it would
need to be expandable in some way, since water expands as it freezes.
2. a After a long enough time, and if there were no predator, the bacteria would amount to a huge
number, N 10; 000. Since the predator eats either colour at random to keep the population
at 10; 000, then it's the same thing as selecting n = 10; 000 surviving bacteria out of N total
bacteria. Since N n, in every selection the probabilities of surviving beige or fuchsia bacteria
are the same, namely 1 . In other words, there are 2n ways of selecting the n survivors. There
2
n
are Cb combinations of b beige bacteria to survive out of the n total surviving. The probability
distribution of the number of beige E. coli is therefore:
Cbn = 1 n! ; b = 0; 1; . . . ; n:
2n 2n b!n , b!
b To answer this, you basically need to determine how long it takes for N n. Let us require
N=n 102. Since N = 2t n, where t is the number of reproductive cycles, t = 6 or 7 hours would
be su cient. c Let p be the degree of preference of eating fuchsia bacteria over that of eating beige bacteria. The
,
,
probability of eating beige bacteria is then 1 , p , and that of eating fuchsia bacteria is 1 + p .
2
2
The result in a then becomes: 1 , p b 1 + p n,b C n = 1 , p b 1 + p n,b
n!
b
2
2
2
2
b!n , b! The result in b is una ected by the predator preference p.
d Sewage.
3. First of all, we assume that the room temperature falls between that of the milk and the black co ee.
There are 2 principal competing e ects:
a The addition of the milk cools the co ee directly by a simple law of mixtures.
b Newton's Law of Cooling", which basically says that the rate of heat loss is greater when the
temperature di erence between the two heat reservoirs i.e. the co ee or milk co ee mixture and
the surrounding air is greater.
For the simple case where the milk is colder than room temperature, it is in fact better to add the milk
at the beginning of the 10 minutes in order that the milk co ee mixture be as hot as possible at the
end of the 10 minutes. By cooling with milk the cup's contents right at the beginning, the temperature
di erence between the two reservoirs is lessened, and hence the rate of cooling slows. The blackness"
of co ee without milk also enables the co ee to be a better radiator of heat, but this e ect is rather
smaller than those above. Note: For a good article on this subject, see Rees, W.G., and C. Viney,
Am. J. Phys. 56 5 434, May 1988.
4. We are dealing with the chemical equation:
H2 + I2 * 2HI:
P We can write this in the form i iAi = 0, where Ai are the reacting substances and reaction products,
and i are the coe cients of the chemical equation. Hence,
2HI , 1H2 , 1I2 = 0:
Reaction equilibria can be determined by the Law of Mass Action,
Y i Ci = Kc ;
i where Ci are the equilibrium concentrations of the various constituents of the reaction.
Let x be the number of gmol of H2 or I2 used to produce HI . Then:
:1
CH2 = 811:, x
04
94 ,
CI2 = 2:11:04x
2
CHI = 11:x
04
11.04 is the total number of gmol present. Now, using the Law of Mass Action gives: 2:94 , x ,1 8:10 , x ,1 2x 2 = 0:02:
11:04
11:04
11:04 Solving the quadratic equation, and choosing the root that doesn't violate the conservation laws, we
get x = 0:319. Therefore, there will be 0.638 gmol of HI .
5. Let the long axis of the dust grain be parallel with the z axis. From the shape of the grain, we know
that the principal moments of inertia satisfy: Iz Ix ; Iy :
At thermal equilibrium:
1 I !2 = 1 I !2 = 1 I !2 ;
2zz 2xx 2yy
where w the average angular velocity. This is due to the energy equipartition theorem of thermodynamics, which states that in thermal equilibrium each degree of freedom contributes an equal amount
of kinetic energy to the total energy of a molecule. Here the dust grain is being treated like a very
big molecule.
Therefore,
r r j!z j = Ix j!x j = Iy j!y j ;
Iz
Iz
and
Similarly, r I
jIz !z j = Iz jIx !x j jIx !x j :
x jIz !z j jIy !y j : So, the angular momentum vector is nearly perpendicular to the long axis of the dust grain.
6. Assume the initial length is lo . After heating, the average lengths of the 2 metallic strips are: l1 = lo 1 + 1 T
and l2 = lo 1 + 2T :
Assuming that the radius of curvature is R, the subtending angle of the strip is , and the change of
thickness is negligible, then
l2 = R + x=4 and
l1 = R , x=4:
Therefore,
l2 , l1 = x = x l12+ l2 = 4x lo 2 + 1 + 2 T :
2 2R
R But, we know that l2 , l1 = lo T 2 , 1 : Making the substitution, we have
giving lo T 2 , 1 = 4x lo 2 + 1 + 2 T
R
+
R = x 24T 1 +, 2 T :
2 1 19941995 Physics Olympiad Preparation Program
University of Toronto
Problem Set 4: Waves and Optics
Due February 6th , 1995 1. One dark and stormy night you nd yourself lost in the forest when you come upon a small hut.
Entering it you see a crooked old woman in the corner hunched over a crystal ball. You are about to
make a hasty exit when you hear the howl of wolves outside. Taking another look at the gypsy you
decide to take your chances with the wolves, but the door is jammed shut. Resigned to a bad situation
you approach her slowly, wondering just what is the focal length of that nifty crystal ball...
a If the crystal ball is 20 cm in diameter with n = 1:5, the gypsy lady is 1.2 m from the centre of
the ball, where is the image of the gypsy in focus as you walk towards her?
b Describe the image of the crooked old lady, and be more quantitative than just telling me she is
hideous.
c The old lady moves the crystal ball closer to her wrinkled old face. At some point you can no
longer get an image of her, which at this point is no small blessing. At what object distance will
there be no image of the gypsy formed?
2. If any of you people end up going into fysix, you will study about Young's two slit experiment until
you are just about ready to yak. Just to magnify the extent of any future yakking, here we go... The
idea of the experiment is to take a wave and pass it through two slits, and then observe an interference
pattern on a screen see gure 1. The gure uses a laser but the experiment can be done with any
wavelike entity such as electrons or water waves.
a What pattern is observed on the screen if the laser has a wavelength ? The answer should be
expressed as the intensity of the interference pattern at the screen in terms of the variables x,
h, D and , where x is the distance from the middle of the screen. You will need to make two
assumptions to solve this problem, namely D
x and x
h. Start with an electric eld at
the two slits that looks like
E = E0ei
p
where i = ,1 and = pathlength= is the phase of the wave at the slits.
I will give some quick hints on how to handle complex numbers. We introduce complex numbers
in problems like this to make the math easier to handle. At the end of the problem we get rid of
the complex parts of the equations to give us a real answer. Some rules when handling complex
numbers:
i. ei = Cos + iSin
ii. ei ei = ei +
iii. Intensity = E E
iv. E = E0e,i is the complex conjugate
b In the diagram the laser source is centred with respect to the two slits. What changes in the
pattern would occur if the laser was shifted such that the waves going into the two holes were
radians out of phase?
c This theoretical pattern goes on forever in both directions along the screen, whereas the experimental pattern fades. What causes this fading of the experimental interference pattern? 3. You are conducting a feasibility study into the SDI Strategic Defense Initiative missile defense system.
You are to consider the viability of a ground based laser defense against ICBM's, which will use powerful
CO2 lasers to bore holes in the incoming missiles, thus destroying them. The lasers have an output
power of 50 W in a beam diameter of 1 mm. The laser beam is red at the missile when it is 10 km
away and the beam loses 3 of its intensity for every kilometer travelled. The outer skin of the missile
is aluminum that is 3 cm thick. When the laser res the skin temperature of the missile is ,50 C
and must be heated to its boiling point at 2500 C . The density of aluminum is 2:34 g=cm3 and heat
capacity is 0:9 J=g C.
a How long will it take the laser to burn through the outer skin of the missile, thus destroying it?
Assume that all of the laser power that reaches the missile goes into heating the 1 mm diameter
spot.
b What angular accuracy must the aiming system of the laser have in order for the laser not to drift
away from the target spot?
c Suggest three ways that one could use to protect a missile from this laser defense system.
4. We have all seen those really cool holographic images of our favourite cartoon characters on the
cereal boxes, not that I still watch cartoons anymore, well occasionally I see the odd one, actually on
Saturday... anyway, back to the problem. Being a physics nerd, and proud of it by the way, I could
never just enjoy those pictures without trying to gure out how they work. I will now in ict my geeky
inquisitiveness on all of you, heh heh heh.
a Holograms are 3D images that change depending on the viewer's perspective. How is the dimensionality of the image captured on the holographic plate? How is this di erent than normal 2D
lm?
b Sketch an experimental apparatus that could be used to produce a hologram. Be sure to explain
what is going on.
c When normal lm is exposed to light and developed, a negative is produced. It is called a negative
because the dark and light areas are interchanged by the developing process. When light is shone
through the negative onto photographic paper a positive image is produced, which then ends up
in your photo album. In holography it makes no di erence whether the holographic plate ends
up being a negative or positive. Why is this?
5. Now for a really dull question, for which I humbly apologize; it won't happen again. There are two
thin lenses of focal lengths f1 and f2 separated by a distance d. In general, the focal length of a system
of lenses will be di erent depending on which side the image is on. For systems with more than one
lens we de ne a back focal length and a front focal length, which are just the two focal lengths for
these two possibilities.
a Find the back focal length and the front focal length of this system of lenses.
b Something nifty happens when d = f1 + f2 . What is it?
c Find the focal length of the system for d = 0. Does it matter which side of the system the image
is on in this limit?
6. Now for a few small questions that will really get you wondering why you are doing these Olympics in
the rst place. Just give qualitative answers.
a I'm sure most of you have been on trips up into northern Ontario and seen the Northern Lights
rst hand. What causes the Aurora? Why are they strongest in Northern Canada?
b We were all kids at one time or another except maybe Pekka and probably sung Twinkle Twinkle
Little Star. What makes stars twinkle? Do the planets and moon twinkle also? c If you are walking on the beach one day and nd a seashell, and just happen to hold it up to your
ear, you will be able to hear the roar of the ocean. What causes this?
d If lightning strikes a tree, it can either be left unharmed or be completely shattered. Why will
some trees be destroyed and others be ne? It turns out that Oak trees are preferentially destroyed
over other species. Why? 19941995 Physics Olympiad Preparation Program
University of Toronto Solution Set 4: Waves and Optics 1. a To solve this problem we need the equation which describes how a curved surface bends light
n1 n2 n2 , n1
So + Si = R
where n1 and n2 are the indices of refraction of the two media, R is the radius of curvature of the
surface, Si is the image distance and So is the object distance. Solving for the image distance
Si = n , n2 RSo, n R :
n1 So 1
2
At the second surface we have
n2
n1 ,n2 , n1
2R , Si + x = ,R
where x is the image distance we are trying to nd and all of the other variables have the same
meaning as above. Here, the second object distance So = 2R , Si because a virtual image is
formed by the rst surface. Solving for x gives
,i
x = n , n1 R2R, SS, n R :
n1 2R i
2
2
Putting n1 = 1:0, n2 = 1:5, So = 1:2m and R = 10cm gives Si = 36cm and x = 6:9cm. Therefore
the image is 6:9cm from the crystal ball.
b The image is real, inverted and mini ed.
c For the gypsy lady to produce no image means that x ! 1. This corresponds to
n2 , n1 2R , Si = n2 R
or Si = ,10cm and So = 5cm. In other words, that gypsy's mug is awfully close to the crystal
ball.
2. a This problem is solved by looking at the interference that the two waves undergo at a particular
point on the screen. The rst thing to do is work out the path length that each wave undergoes
on its way to the screen see gure 1. Consider rst r1 r Using h
and using D
Similarily x gives
x gives r1 = D2 + x , h 2 :
2 r 2,
r1 = D 1 + x D2hx
2
hx
r1 = D1 + x 2, 2 :
D
2
hx
r2 = D1 + x 2+ 2 :
D b c 3. a b c 4. a The path di erence between the two waves will be r2 , r1 = hx=D and phase di erence =
2xh=D. The electric eld at the screen will be the superposition of the waves that went through
the slits
Etot = E ei + E exp i +
2
2
where the factors of two appear because the wave splits in wave in half when it hits the slits.
Notice that one wave is phase shifted radians with respect to the other. To nd the intensity
we multiply Etot by its complex conjugate, which gives
2
xh
Iscreen = E 1 + cos 2 D :
2
If the laser is shifted so that the wave coming from one slit is out phase by radians with respect
to the other then the electric eld becomes
Etot = E exp i + + E exp i +
2
2
Multiplying this by its complex conjugate gives
2
xh
Iscreen = E 1 , cos 2 D :
2
This new formula has its maximum o centre with respect to the two slits, which is the major
di erence between the interference patterns.
The reason the experimental pattern fades is because the same amount of light doesn't hit all
parts of the screen. Most of the light passes straight through the slits, with far less scattered at
large angles. This means that the interference pattern is brightest opposite the two slits ie at
x ' 0.
To solve this question we must nd out how much energy is needed to evaporate the aluminum,
and then how fast the laser can supply this energy. The volume of aluminum to evaporate
is V = =4d2h where d is the spot diameter and h is the thickness, which means that the
mass of aluminun is M = V with . The heat needed to evaporate this amount of aluminum
neglecting the heat of vapourization is H = CM T where C is the heat capacity and T is the
temperature di erence. Plugging in the numbers gives H = 126:5 J. The laser beam is attenuated
to P = 500:9710 = 36:9 W by the time it arrives at the missle so it will take 3:4 sec for the laser
beam to burn through the missle's skin.
The angular accuracy needs to be enough to keep the laser aimed at the same spot on the missle
for the entire 3:4sec. Assuming the beam can't move by more than its radius gives an accuracy of
,4
tan 5 10
4
10
or the drift in must be less than 8:4e , 7 degrees=sec.
To defend against this sort of attack one could: rotate the missle, thicken the skin, make the skin
re ecting, increase heat transfer over the skin, have dummy warheads, send up cha e, or as some
of you suggested, not re the missles, use a cloaking device, blast laser system and nally, don't
piss the guys at ICBM control o .
The thing that makes holograms di erent from normal pictures is that holograms record phase
information, in addition to amplitude information on the lm. A picture only records how intense
the light is, which gives no information about depth. A hologram on the other hand is just a
recorded interference pattern of the image. The depth information about an image is preserved
on the plate in the form of an interference pattern. b To create a hologram one needs to produce an interference pattern of the desired object. This is
done by splitting the laser beam into a reference beam and an object beam see gure 2. The
probe beam is re ected o the object, and then interferes with the reference beam. Because the
object has depth to it, the re ected beam will not have the same phase across it when it reaches
the holographic plate. It will then interfere with the uniform reference beam, thus producing an
interferomic record of the object.
c It makes no di erence wether the holographic plate is a negative or a positive. Since the information is just an interference pattern, the di erence between positives and negatives will only e ect
the overall phase of the pattern and will not e ect the nal image.
5. a I will rst work out the back focal length using the Lensmakers equation. At the rst lens we can
write
1= 1+ 1
f1 Si1 So1
or
Si1 = SSo1 f1f :
o1 , 1
This will be the position of the image after the rst lens. The same can be done for the second
lens, which gives
Si2 = SSo2 f2f
,
o2 2 Since the lenses are located a distance d away from each other we can write So2 = d , Si1. Using
these three equations to solve for Si2 while eliminating So2 and Si1 gives
f
f
back focal length = Si2 = d 2 d ,, 1f
, f1 2
Doing the same analysis but in the other direction gives
f
f
front focal length = d 1 d ,, 2f :
, f1 2 b If we set d = f1 + f2 then the system acts like it isn't there. Putting plane waves in one end
produces plane waves at the other. Note that the magni cation of the system will not likely be
unity.
c If d = 0 in the above equations then
back focal length = front focal length = f f1f2f
1+ 2
or in other words it doesn't matter which side the image is placed on. We can write
1
11
feff = f1 + f2 :
This is just a modi ed version of the thin lens equation.
6. a The Aurora are caused by low energy solar electrons that are caught by the Earth's magnetic eld.
They are then sped up and enter the atmosphere at the magnetic North Pole. These electrons
excite nitrogen and oxygen which produce the characteristic blue of the Aurora. The reason it is
strongest over Canada is that the magnetic North Pole is centred over ??? Island in the Canadian
Arctic. b Stars twinkle because the atmosphere that the light passes through to get to our eyes is turbulent.
This is due to uneven heating of the atmosphere by the sun. This turbulence causes random
refracting of the light which we see as a twinkle. This twinkle also appears for larger bodies eg.
the moon but is much less noticable.
c The sound from the seashells is actually produced by air currents that pass by the shell and excite
the shell's air volume at its natural resonances. Since these air currents are not constant, the
resonances will be excited sporadically giving the impression of ocean waves.
d When lightning strikes a tree, the current in the lighning stroke needs a path to ground. If the bark
of the tree is wet, then the current will pass over the outside of the tree leaving it undamaged. If,
on the other hand, the tree isn't wet then the current will pass into the tree and descend through
the sap. When this happens the sap is superheated and expands, which ends up turning the tree
into toothpicks. The Oak tree has especially rough bark which makes it di cult for the current
to nd a path to ground along the bark. It is therefore not that Oak trees get struck more by
lightning, just that they are more likely to be destroyed when hit. 19941995 Physics Olympiad Preparation Program
University of Toronto
Problem Set 5: Electricity and Magnetism
Due March 6th , 1995 1. Okay, all you mental gymnasts it's time for the parallel conducting bar problem, an Olympian event
designed to exercise your upper bodies. Consider two long parallel conductors, each carrying currents
in the same direction, as shown in Figure 1. Conductor A carries a current of 100 A and is held rmly
in position. Conductor B carries 147 A and is allowed to slide freely up and down parallel to A
between a set of nonconducting guides. If the equilibrium distance between the two conductors is 2
cm, what is the linear mass density of conductor B?
2. Pretend for a moment that you're an engineer working on the design of a resistor that is to have a
temperature coe cient of expansion of zero at 20 C. The design is a composite of two bars having
octagonal cross sections each side of the octagon having a length d = 0:25 mm as shown in Figure
2. The ratio of the resistivities of the two materials is 1 = 2 = 3:2, and the ratio of the lengths
is l1 =l2 = 2:6. The cross sectional geometry and dimensions are uniform throughout the resistor.
Assuming that the temperature of the two sections remains equal, calculate 1= 2, the required ratio
of the temperature coe cients of resistivity of the two materials.
3. It is the year 2236, and the Federation has just banned all telecommunication optical bre in your
sector for some unknown reason. The replacement con guration, consisting of two coaxial dielectrics,
is depicted in Figure 3. The permittivities are 1 and 2 , respectively, the conductor has potential V0,
and the outer shield is grounded.
a Derive an expression for the capacitance per unit length for the cable.
b Given that V0 = 1:2 kV, r1 = 4:5=36 10,9 F m, r2 = 3:0=36 10,9 F m, and r3 =
2r2 = 4r1 = 40 mm, determine the maximum electric eld intensity in each dielectric.
Note that, for an in nite single dielectric coaxial cable, the capacitance per unit length is c =
2 = lnb=a where is the permittivity of the dielectric with outer radius b, and a is the radius
of the conductor a b. By Gauss' Law, the electric eld intensity for such a coaxial cable is given
by E = q=2r , where a r b and q is the charge per unit length in the region enclosed by r.
4. Consider two conducting spheres, one of radius 6.0 cm and the other of radius 12 cm, where each has
3 10,8 C of charge and is positioned very far apart from the other. The spheres are then connected
by a platinum wire maintained at a temperature of 312.5 K. Find:
a the direction of motion and magnitude of the charge transferred, and
b the nal charge on and electric potential of each sphere.
5. As an accelerator physicist, you have been asked to design a xed target machine to produce antiprotons, p, the antimatter partner of protons, p. The reaction is p + p ,! p + p + p + p, where one
high energy proton hits another proton that is at rest since it's part of the target. In addition to
the original two reactant particles, a protonantiproton pair is created. What is the minimum energy
required by the incident proton for this reaction to occur? Express your answer in terms of the proton
mass m and the speed of light c. Hint: consider the centre of mass reference frame.
p
6. Imagine a long copper conductor with rhombic cross section having sides of length 5 mm and diagonal
dimensions of 2 mm and 4 mm. The conductor is carrying a current of 10 A. Each second, what
percentage of the conduction electrons must leave to be replaced by others a 100 mm length? 19941995 Physics Olympiad Preparation Program
University of Toronto
Solution Set 5: Electricity and Magnetism 1. All forces mentioned in the following are with respect to conductor B refer to Figure 1. The magnetic
force on a conductor in a uniform external magnetic eld is given by
Fmag = IB~ BA ;
l~
~
where ~ is a displacement vector along the conductor B, and BA is the magnetic eld from conductor
l
A. The above expression can be simpli ed to
Fmag = IB lBA sin = IB lBA ;
~
since BA is perpendicular to ~ = 90 .
l
~
From the BiotSavart Law, the magnetic eld BA at a distance r from a wire carrying a current IA is
given by
IA
BA = or ;
2
where o is the permeability of free space.
The magnetic force per unit length on conductor B is therefore
Fmag = o IA IB :
l
2r
The gravitational force per unit length on conductor B, Fg , neglecting the gravitational elds of conductor A and the nonconducting guides, is
Fg = g;
B
l
where B the linear density of conductor B and g is the acceleration due to gravity.
At equilibrium, we set the forces to be equal, which gives:
Fmag = Fg
l
l
o IA IB = g
B
2r
o
B = 2IA IB
rg
,7
2
= 12:566 10 :02N=A :100 sA147 A
2 0 m9 8 m= 2
= 1:5 10,2 kg=m = 0:15 g=cm: 2. Calculate the change in total resistance with respect to the change in temperature, Rt:
RT = Ro 1 + tT
RT , Ro = R
=
ot
T
= Rt = Ro t :
T
Given that t = 0 at 20 C,
Rt = 0:
T
For each of the two individual resistance regions:
R1 = R
o1
T1 1 1
Ro 1 = Al1 ;
where A is the cross sectional area of the octagonal bars.
Therefore,
R1 = 1 l1 1 :
T
A
1 Similarly, R2 = 2 l2 2 :
T2
A
Since the two materials are in a series con guration:
Rt = R1 + R2 = 0:
T T1 T2
Since T1 = T2, R1 = ,R2 and
1 l1 1 = , 2 l2 2 = 1
2 =, 2
1 ,1
ll2 = 3:22:6 = ,0:12:
1 3. a The capacitances per unit length due to each dielectric are given by:
1
c1 = ln2r =r
21 and 2
c2 = ln2r =r :
32 The total capacitance per unit length is calculated by treating the above two capacitances in
series. The total capacitance per unit length is therefore given by:
c = c c1c2c ;
1+ 2
c = lnr =r2+1 2 lnr =r :
2
21
1
32
b
4:5 10,9 F=m
:
c1 = 2 36 ln 2
= 0:36 nF=m; c2 = 3
2 36 10,9 F=m :
= 0:24 nF=m;
ln 2 V2 = c1 = 0:36 = 1:5;
V1 c2 0:24
V1 + V2 = 1200 V:
Solving for V1 gives V1 = 480 V. Therefore
q = c1V1 = 172:8 nC=m:
The maximum electric eld intensity occurs at the inner surface of either dielectric. Hence, at
r = r1 :
172:8 10,9 =m
:
Emax r1 = 20:010 m4:510,C=36 F=m = 69:1 kV=m:
9
Similarly for r = r2 :
172:8 10,9 m
:
Emax r2 = 20:020 m3:010,C==36 F=m = 51:8 kV=m:
9
4. First calculate the initial potentials on the sphere surfaces. For the smaller sphere:
9
,8
1
m2=C2 3
V1 = 4 q = 9 10 N 0:06 m 10 C = 4:5 103 V:
r For the larger sphere: 9
,8
m2=C23
V2 = 9 10 N 0:12 m 10 C = 2:25 103 V: Since V1 V2 , the direction of charge motion is towards the 12 cm radius ball.
When the conducting wire is connected, the spheres seek to assume equal potentials. Therefore,
1 q1 = 1 q2 ;
4 r1 4 r2
q1 = q2 ;
r1 r2
q1 = q2r1 :
r
2 Since q1 + q2 = 6 10,8 C,
q2 r1 + 1 = 6 10,8C;
r2 and the nal charge on the larger sphere becomes
q2 = 4 10,8 C;
whereas the nal charge on the smaller sphere becomes
q1 = 6 10,8 C , 4 10,8 C = 2 10,8 C:
The magnitude of transferred charge is therefore 1 10,8 C.
The nal potential of each sphere is given by:
9
,8
q
m2=C22
Vsmaller = 41 r = 9 10 N 0:06 m 10 C = 3 103 V
and
9
,8
m2=C24
Vlarger = 9 10 N 0:12 m 10 C = 3 103 V: 5. In the frame of the lab, the total energy is E + mc2 , where E is the energy of the incident proton and
mc2 is that for the proton at rest. Let the incident proton have momentum p.
The invariant mass, M , of this lab system is then given by
2
2
2
Mlab = Etotal , p2 :
c4
c In the centre of mass CM frame, the 4 reaction products are at rest for this problem, since we are
only calculating the minimum, or threshold energy, for this reaction. The total energy in this system
is therefore 4mc2 and the invariant mass is
, 4mc2 2
= c4 :
2
2
Since Mlab = MCM and E 2 = p2c2 + m2 c4, we get
2
MCM E + mc2 2 , p2 = 4mc2 2 ;
c2
c2
E + mc2 2 , E 2 + m2 c4 = 4mc2 2;
E 2 + m2 c4 + 2Emc2 , E 2 + m2 c4 = 16m2 c4 ;
2Emc2 = 14m2c4 ;
and
E = 7mc2 :
6. First calculate the number of electrons per unit volume. Take Avogadro's number to be N = 6:02 1026
atoms kmol, the density of copper to be 8:96 103 kg m3, and the atomic weight of copper to be 63.54
g mol. Assuming one conduction electron per atom, the number of electrons per unit volume is
Ne = 6:02 1026 atoms 1 kmol 8:96 103 kg 1 electron
kmol
63:54 kg
m3
atom Ne = 8:49 1028 electrons=m3:
1
The rhombic cross sectional area is 2 2 mm4 mm = 4 10,6 m2.
The number of electrons in a 100 mm length is N = 4 10,6 m20:100 m8:49 1028 electrons=m3 = 3:40 1022 electrons:
A 10 A current requires that
10 C=s1:6 10,19 C=electron,1 = 6:25 1019 electrons=s
pass a xed point.
The percentage leaving the 100 mm length per second is
6:25 1019 100 = 0:184:
3:40 1022 19941995 Physics Olympiad Preparation Program
University of Toronto
Problem Set 6: Electronics
Due April 3rd , 1995 1. You just gotta love electronics. All of those little resistors and capacitors to play with. And just
when you think there can be nothing as exciting as that, out come those digital circuits with all of the
chips that can be used. It's even better than having... sorry about that, just got rambling...on to the
business at hand. This problem concerns itself with a circuit called a tank circuit see gure 1. I
have no idea why it is called that, but whatta name.
a Find the absolute value of the impedance of the tank circuit as a function of frequency. The
impedance of this circuit is called the transfer function because it describes how signals are
transferred across the circuit. At what frequency is the impedance of the circuit in nite? To do
this p will need the resistance of both a capacitor Zc = 1=i!C and an inductor Zl = i!L, where
you
i = ,1. The absolute value of a complex number is de ned as
p
Z = Z Z
where Z is the complex conjugate of Z .
b Sketch this impedance function as a function of frequency.
c How will the transfer function change if the inductor has some resistance? Just determine the
complex transfer function.
2. Often in electronics one uses lter circuits to remove unwanted frequencies in signals. Most of them work
by using the frequency dependent resistance of capacitors and inductors. If a certain frequency signal
is put into a circuit it will either be allowed to pass through or get blocked by the circuit depending on
the frequency dependent resistance of the components. Filter circuits fall into the following categories:
a High pass lters only allow high frequencies through them.
b Low pass lters do the opposite.
c Band rejection lters block an intermediate range of frequencies.
d Not a lter at all.
Classify the ve circuits given in gure 2.
3. You're out with your friends one day and you stumble across an ornately carved lamp lying on the
ground. One of your friends says that it must be a magical lamp with a genie inside. You all tease
him mercilessly until, sobbing uncontrollably, he rubs the lamp and out comes a genie. The ethereal
entity speaks, and I quote:"For teasing my master so ruthlessly, I will ask you one question which you
must get right to live. What is the resistance between opposite vertices of an octahedron, if each of
the twelve edges has the same resistance." You begin to sweat and really wish you hadn't skipped your
electronics class all last term......
4. "Could the Romulin Warbird be hiding in that interstellar gas?" Captain Picard said from the bridge
of the Enterprise.
"Our sensors cannot penetrate the gaseous cloud, but I think it quite likely Captain," says Data in his
usual monotone.
Then Wesley piped up, "The homogeneous material of the cloud has known conductivity c , whereas
the Warbird has a conductivity of w . By measuring the resistance of the cloud we could determine if
the Warbird is in there." He sits in his chair with a stupid smile on his face. Picard's face turns red as he shouts, "Will you shut your insolent little face. Worf, stun Wesley and
throw him in the brig. After Wesley is unconscious, the captain remarks to Data, "Make it so."
a Data assumes that the cloud is cylindrical with radius b and length L, and that the input and
output currents are uniformly distributed across the ends of the cloud. Show that Data computes
the resistance to be
L
:
R=
b2
c b Data decides to model the Romulin Warbird as a metallic cylinder of radius a, length a and
conductivity w in the centre of the cloud. Estimate the relative change in resistance of the cloud
to rst order in c , w if c ' w . Assume that b
a. Model the various parts of the cloud
as resistors in series or parallel with each other.
5. This problem concerns itself with the recti cation of AC voltage. The circuit shown in gure 3 is
called a full wave recti er and is used to turn an AC signal into a DC signal. The funny looking circuit
elements are called diodes and only allow current to pass in the direction of the arrow.
a If the voltage source produces a sinusoidal signal, what signal is expected across the resistor?
What will be seen at the resistor for a square wave signal as input? Give reasons for your answer
and sketch the output and input waveforms. Ignore the capacitor for this part.
b If I add a large capacitor in parallel with the resistor, how will it a ect the voltage signal at the
resistor? To be more speci c, why must the capacitor be large to produce an almost DC signal
at the resistor?
6. This just wouldn't be an electronics problem set without at least one question on digital circuits. See
gure 4 for a type of circuit that shows up when designing computers and such.
a Find the truth table of the circuit. What is this circuit actually doing and how could it be useful?
b Using the circuit from a, design a full adder that can add three one digit binary numbers. Give
the truth table for your circuit. 19951996 Physics Olympiad Preparation Program
University of Toronto Problem Set 1: General Physics
Due October 16, 1995 1. Joe buildem tall" Girders is designing a construction crane did you know he learned his stu working
with Tinker Toys?. He uses two beams with mass m1 = 9000 kg and m2 = 3000 kg with lengths
l1 = 45 m and l2 = 15 m, attached to the vertical trunk" of the crane and supported by cables, as
shown in Fig 1.
a How much mass, mc , should there be in the counterweight located at the end of the second beam
to ensure that the crane is perfectly balanced when the crane is not carrying a load.
b The cable support has a height h = 15 m. What are the tensions on the two cables supporting
the beams?
2. In the original movie Batman, our masked avenger has a cool getaway planned when he and Vicky
Vale, the hotshot reporter, are trying to escape the Joker after the debacle in the museum check out
the movie if you folks forget this scene. He res up one of his awesome cables and attaches it to a
girder above the street. With his supercool winch located on his belt, he begins to lift the two of them
out of harm's way. But wait! He only gets part way up. His winch appears to run out of juice. As he
and Vicky twist in the wind, he ponders his potentially fatal mistake.
a What is the minimum energy required to lift the two of them up 15 m, if we assume Batman tips
the scale at a t 75 kg and Vicky really does weigh 47 kg as she claims?
b They have to be rising at least 1 m s in order to avoid getting plastered by Joker's henchmen.
What is the power required to execute this daring move?
c In the movie, after Batman hooks the winch to Vicky and lets go, she is winched up the rest of
the way to the girder. Can you give a plausible explanation as to what really went wrong with
the winch? Did it really run out of energy?
3. Sven is tooling down Highway 11 17 with a tractortrailer load of pulp wood to be fed into the maw of
the MacBlo mill in Fort Frances. He is doing 120 km h, easy. He comes around a corner and nds
what else? the icon of the northern Ontario bush, a moose, standing in the middle of the road. The
old pro that he is, Sven hits the brakes. He is about 150 m from the moose when his size 13 work boot
hammers the brakes.
a How heavily does he have to brake to avoid bagging a moose out of season and making a mess
of his front end?
b He is still in the curve when he starts braking, so his trailer lled with cord wood makes a small
angle with the tractor unit. He realises as he hits the brakes that he forgot to connect his trailer
brakes, so that all braking is performed by the tractor unit. It has been snowing what else can
go wrong? so the coe cient of static friction between his wheels and the asphalt is s = 0:2.
What is the maximum angle his trailer unit can make with the tractor unit before it starts to
jacknife the truly awesome maneuver when the trailer starts sliding and swings around?
4. Rob Hanks was a wild and crazy man. By day he worked as an electrician for General Motors, but
by night he assumed a secret identity, one that no soul knew a thing about he became a physics
problem solver. Now Rob often works with AWG 12 copper wire with an 80.8 mil diameter. He
already knows that a 50 ft length can carry a current of 20 A, and that copper has a conductivity of
= 5:8 107 ,1 m,1 and an electron mobility of = 0:0032 m2V,1s,1 at 20 C. Can you help Rob
calculate the electric eld intensity E , the voltage drop and resistance across the 50 ft length, and the
time it would take an electron to move a distance of 1 cm in the conductor?
1 h mc
l2 l1 Construction
Crane Figure 1: The construction crane in Problem 1.
5. Wanda Steerbutton was a worldclass magazine photographer. She earned her living taking quality
shots of supermodels, rock bands, movie stars, and politicians for the covers of those glossy magazines
that people only seem to buy in airports and train stations. It was a good life, but she wanted more.
Understanding the secrets of the universe sounded like a cool idea; besides, it would be a nice change
from looking at people's pores through a view nder. So Wanda decided to take up astrophysics. Since
she already owned a fancy camera, and since one of her lab courses involved photographing stars in the
night sky, Wanda wondered whether she could use one of her precision engineered xed focus lenses for
the job. The lens in question had a focal length of 60 mm and was focussed for objects at a distance of
15 m. The problem was that stars are at a distance that is a little more than 15 m, like in nity maybe.
Her lens did, however, have an adjustable aperture. For what aperture diameter would the di raction
blur of visible light = 550 nm be more or less the same as the defocus blur for a distant star?
6. The starship Compromise, the lesser known sister ship to the Federation's Enterprise, is to undergo
an upgrade" to its transporter control system under the supervision of your aging systems engineering professor at Star eet Academy. She has asked you to design logic circuitry that will enable the
transporter to function for the following three Away Team scenarios:
An android A, the captain C , and another senior o cer S are all part of the Away Team;
An android A, the captain C , and another senior o cer S are NOT part of the Away Team;
The captain C is a member of the Away Team.
a Without thinking too deeply about the reasonableness of the above scenarios remember,
your prof is starting to lose it and ought to have retired to a quiet moon at the edge of the
Alpha quadrant years ago, express the process state as a digital logic function in terms of
A, C , and S . Sketch the circuit.
b Algebraically simplify the circuit above so that you use a minimum number of logic devices.
Besides, if you ever were to mass produce this super intelligent control system, you'd surely
want to minimize the manufacturing costs. Sketch the simpli ed circuit. 2 19951996 Physics Olympiad Preparation Program
University of Toronto Solution Set 1: General Physics 1. a We require that there is no net torque about the axis through the point where the two beams
connect to the vertical support recall that torque is ~ = ~ F where ~ is the vector separating
r~
r
the axis about which you are calculating the torque and the point where the force is applied, and
~
F is the force. The torque due to gravity on beam 1 is centred through the midpoint of beam 1.
Ditto for beam 2. The torque due to the counterweight is the cross product of the gravitational
force and the vector between the axis and the end of beam 2. The force diagram for the crane is
shown in Fig. 1a. Thus the net torque is
X
1
2
= ,m1 g l2 + m2 g l2 + mc gl2 = 0
1 2
mc = 1 m1 ll1 , m2
2
2
= 12; 000 kg;
where g is the acceleration of gravity g = 9:8 m s2. Note that we can verify that the equation
for mc makes sense. For example, if m1 = m2 and l1 = l2 , then we would expect mc = 0, which is
the case. Can you think of other sanity checks? What about units? Note that we do not have to
take into account the torque due to the support cable tension on each beam. Can you understand
why not?
b We require that the net torque on each arm to be equal. In this case, the tensions on the cable
supporting the two beams are forces, T1 and T2 that act on the ends of the beams and on the
vertical support. Let 1 be the opening angle between beam 1 and the cable as shown in Fig. 1b.
Then the torque acting on beam 1 is
X
1
= 0 = l1 T1 sin 1 , l2 m1 g
3
4
T1 = 2 m1 g
sin 1
p 2
2
1
= m1 g 2lh + h
5
= 139; 000 N;
where h is the height of the vertical support above the beams. Similarily, for beam 2, the requirement that the net torque on the beam is zero gives
X
2
= 0 = ,T2 sin 2 + m2 g l2 + mc g
6
p2 2 +2
T2 = m2 g + mc g l2h h
7
= 187; 000 N:
Its interesting that the tension is highest on the cable supporting the counterweight. Them cables
are pretty strong. They have to be able to support almost ten tons, and that's without any load
being carried.
2. a To lift Batman and Ms. Vale up h = 15 m, one has to at least expend the work equivalent to
raising them that distance in a gravitational eld it could be more think about the work required 1 a) b)
Cable
Tension mcg θ1 m1g
m2g m1g Beam 1 Figure 1: The forces acting on the two beams around the point at shich they are connected to the vertical
support are shown in a. Note that we do not consider the the force of tension on the two beams from the
support cables. Why not? The forces acting on the end of beam 1 due to the cable and the weight of
gravity are shown in b.
to accelerate them into motion. This is equal to
W = h mB + mV g = 17; 931 Joules;
8
where g is our faithful acceleration due to gravity at Earth's surface.
b Let's assume they accelerate to vh = 1 m s and then rise steadily the rest of the way. The power
expended ignorning the power required for acceleration is just the force velocity, which is
P = mB + mV g
9
= 1; 196 Watts = 1:6 hp:
10
Recall that 1 horsepower is equal to 746 Watts. That's a pretty hefty little motor he has there.
c Let's consider why the winch stopped in the rst place. There are only a few likely scenarios
given that it wasn't really broken:
i. motor overheated or otherwise failed,
ii. the energy source couldn't continue to supply the needed power, or
iii. the winch is based on some funky technology that I know nothing about that gives out at
exactly the right moment in the script.
The fact that it continued to operate and yank Vicky up when Batman let go and the tension
dropped indicates that i is unlikely; the drop in tension itself would not have been correlated
with the moment when the motor became operative. Scenario ii is plausible, as the power level
required to haul Vicky up would be less than that required to haul both of them up. Of course,
I would go for iii. Hollywood rarely cares about conforming to reality.
3. a We can safely assume that Sven and the pulpwood are going to decelerate uniformly, with deceleration a. This is an approximation, as brakes usually fade as they heat up, but we have to start
from somewhere, and it isn't such a bad assumption. We can then employ our usual equations for
an object of mass m decelerating uniformly over a distance d. If v0 = 120 km h is Sven's initial
speed, then the time required to decelerate is t = vo =a. Thus,
1 2 v2
d = 2 a vo = 2o
11
a
a
v2
a = 20 = 5:8 m=s2;
12
d
2 Fb
θ
Tractor
Unit
Trailer Figure 2: The braking force acting on the trailer.
which is over half a g. That's real heavy braking. I'm guessing Sven ended up with the meat.
b The force diagram on the trailer is shown in Fig. 2. The braking force Fb is shown, along with
the component of that force perpendicular to the axis of the trailer, Fp = Fb sin , where is
the angle between the axis of the trailer and the braking force. There is one other force in the
problem; the reaction force of the pavement acting on the trailer through its tires to keep the
angle constant note that it is not shown in Fig. 2. This reaction force would normally be equal
to Fp , so that there is no increase in angular momentum of the trailer note that the trailer is
swinging in uniform motion as it turns the corner. The maximum frictional force is
Ff = s mw g; 13 where mw is the weight of the trailer resting on the back axle, which we approximate as half the
total trailer weight mt . Since the braking force is Fb = mt a, the maximum angle between the
trailer and tractor unit is given by the requirement
Ff = Fp
s mt g = m a sin 2
t = sin,1 sag = 9:7 :
2 Not a big angle in this case. Note that it matters not a whit what the trailer weighs.
1
4. Since a mil is 1000 inch, the crosssectional area of the conductor is:
A= 0:0808 in 2:54 10,2 m 2 = 3:31 10,6 m2:
2
1 in The current density therefore is:
I
20
J = A = 3:31 A ,6 = 6:04 106 A m,2:
10
The electric eld intensity is:
6A
,2
E = J = 56::804 10 ,1m ,1 = 1:04 10,1 V m,1:
107
m
The voltage drop is given by:
V = E` = 1:04 10,1 V m,150 ft12 in ft,10:0254 m in,1 = 1:59 V:
3 14
15
16 The resistance is given by: :59 V
R = V = 120 A = 7:95 10,2 :
I
Since = , where is the charge density:
8 7 ,1 ,1
= = 05::003210 2 V,1 ms,1 = 1:81 1010 C m,3:
m
Since J = U , where U is the drift velocity:
,2
6 04 106
U = J = 1::81 1010 A m ,3 = 3:34 10,4 m s,1 :
Cm
With the above drift velocity, an electron will require approximately 30 seconds to move a distance of
1 cm in the 12 copper conductor.
5. The image distance, v, can be calculated from the lens makers' equation:
1 + 1 = 1;
vuf
which yields a distance of:
v = uuf f ;
,
where f is the focal length and u is the object distance.
22
The Rayleigh criterion, = 1:D , where is the limiting angle of resolution, is the wavelength, and
D is the diameter of the aperture, can be used to calculate the blur diameter:
d = v = 1:22v :
D
The defocus blur is calculated by similar triangles as:
d = v ,ff D :
Equating the di raction blur and the defocussing blur gives:
s
p
D = 1:v22vf = 1:22u:
, f
For = 550 nm and u = 15 m, we end up with D = 3:2 mm.
6. a The digital logic function is an OR expression encompassing the combinations that lead to a
correct process, namely:
X = A C S + A C S + C:
The circuit is sketched in Fig. 3a.
b Operating on the second term of the above expression with one of DeMorgan's laws of Boolean
algebra A B = A + B gives:
X = A C S + A + C + S + C:
Applying the redundancy law of Boolean algebra A + A B = A yields:
X = C + A C S + A + C + S = C + A + C + S:
The simpli ed circuit is sketched in Fig. 3b.
4 a)
A
C
S b)
A•C•S A
C
S x A+C+S
x A•C•S Figure 3: The circuit to handle the logic of the starship Compromise transporter control logic is shown in
a. A simpli ed circuit is shown in b. 5 19951996 Physics Olympiad Preparation Program
University of Toronto
Problem Set 2: Mechanics
Due November 13th , 1995 1. A re broke out in your building and you are trapped on the roof. Upon looking around, you nd a
bungee cord and a harness for it. Being a person who likes to show o , you decided to bungeejump
o agpole at the top of the building of this and release the harness at the bottom.
The initial length of the bungee cord is 30 m and k = 77 kg sec2 . The height of the building+ agpole
is 60 m See Figure 1.
a How much bungee cord do you need if you want to have a perfect landing i.e. v = 0? Assume
that you are a point particle with mass of 55 kg.
b So you jumped, but it hit you that the tensile strength times the cross section of the bolt that
holds the agpole that the bungee cord is tied to is T . What's the minimum T you need in order
not to be squashed on the ground?
2. Suppose you are a high school student well, you are who enjoys working on physics problems. One
day your physics teacher asks you to try physics olympiad preparation problems. And you see the
following dull questions prepared by a desperately dull guy.
a A yoyo is on a level surface Figure 2 a. A gentle horizontal pull is exerted on the cord so that
the yoyo rolls without slipping. The yoyo will roll toward the pull. Why?
b Now a yoyo is sitting on a wedge Figure 2 b. Calculate the force needed on a cord to have the
yoyo stationary when it is released. What is the minimum friction coe cient needed to make the
yoyo stationary in this condition? Assume that the yoyo is a cylindrical object and the width of
its groove is negligible.
The moment of inertia of a cylindrical object that has a rotational axis parallel to the axis passing
through the centre of mass, which is perpendicular to the at surface is IR = IM + aM where a is
the distance of the rotational axis from the principal axis, M is the mass of the cylinder, and IR
and IM are the moments of inertia about the axis of rotation and about the axis passing through
the centre of mass, respectively.
3. For all your life you wanted to be a geek. Now your life long dream is about to come true. You've
been noticed and invited to a prestigious geek party. But you have to answer questions correctly to be
accepted as one of the geeks.
a A marble is resting on the top of an overturned popcorn bowl shaped like a hemisphere See
Figure 3 a. The radii of the popcorn bowl and the marble are R and a, respectively. At what
height, in terms of R, will the marble be airborne if the marble rolls down without slipping?
b Now, the bowl is put upright Figure 3 b. If you let go of the marble near the bottom of
the bowl it will oscillate. Compared to a simple pendulum of length R , a, will the period of
oscillation be shorter, the same, or longer? Why? A big brownie point will be awarded if you can
give a quantitative answer.
4. You are an agent of Her Majesty's Secret Service and are being followed by goons from an organization
whose plan is to conquer the world by taking control of the world's gummybear production. You decide
to shake the pursuit. Your car sensor indicates that a turn with a radius 30 m lies ahead. Your car
1 weighs 1500 kg and its centre of mass is 40 cm above the ground. You can approximate your car to
have the dimensions 500 200 80 cm3 length width height.
a Your car is equipped with a set of SuperSticky tires tm which means the friction coe cient is
very, very big. But you realise that you have a cup of co ee without a lid in your cup holder
and you do not want to spill the co ee on the rather expensive leather seat but you don't want
to throw it away either. The car pool department is going to raise hell if they nd co ee stains
on the leather seat. How fast can you go without spilling your co ee? The co ee cup is 8 cm
in diameter and 15 cm tall and 4 5 full. Assume that all four wheels of the car stay on the road
without slipping.
b Oww, bloody hell, they start shooting at you and, to make it worse, the sensor indicates that
the pavement is wet and the friction coe cient of the tire pavement has dropped to = :23
Good thing your car is equipped with an aerodynamic device that produces a downward force of
150v N v in km h. Forget about ruining leather seats. Your life is on the line. How fast can
you go without slipping and falling from a cli or raising inside wheels probably resulting in
loss of control?
5. Eeck!!! Computers at the satellite control centre are down and you are asked to transfer a $1:0
109 communication satellite from a parking orbit h = 200 km to a circular geosynchronous orbit
manually. Due to engine controller malfunction you can use only two short thruster bursts. You are
to use an elliptical transfer orbit whose perigee meets the lower parking orbit and apogee meets the
geosynchronous orbit. Assume that the thruster expended is negligible. If you succeed, you will get
your boss' eternal gratitude for saving his her its butt; but if you fail, you will lose your job and more.
a What is the v required to put the satellite to a transfer orbit from the parking orbit? Also,
what is the v needed to put the satellite from the transfer orbit to the geosynchronous orbit?
b How long does the transition take?
6. Pat and Chris bought two identical pendulum clocks at X, which is on the equator. Their next
destination is Y, which lies exactly on opposite side of the earth. But their travel agency screwed up
and Pat has to take a westbound plane and Chris has to take an eastbound plane, where both planes'
routes follow the equator. They arrive at Y exactly 12 hours later at X local time.
a Do the two clocks agree with each other? Well, obviously not. If not, what are the e ects that
would make them not agree?
b Find the time di erence between Pat's and Chris' clocks at Y. Consider only the most important
physical e ect. 2 Figure 1: Figure for Problem 1. 3 Figure 2: Figure for Problem 2. 4 Figure 3: Figure for Problem 3. 5 19951996 Physics Olympiad Preparation Program
University of Toronto Solution Set 2: Mechanics 1. a Assume that the bungee cord obeys Hooke's law,
F = Y l
A
l
F = Y lA l = kl;
where A is the cross section of the elastic cord, Y is Young's Modulus, l is the length of the cord,
l is the change in the length of the cord, and k is the elastic constant of the cord. Since Y and
A are constants, kl is constant, i.e. kl = k0l0 .
At h = 0 the velocity of the jumper has to be zero, i.e. v = 0, where h is the vertical distance of
the jumper from the ground. Thus, using energy conservation
mgh = 1 k0 h , l02
2
= 1 kl=l0 h , l0 2
2 where h is the initial vertical distance. This can be rearranged as a quadratic equation in l0 . l0 2 , 2h 1 + mg l0 + h2 = 0:
kl
0 then,
Solving for l
l0 = h !
r
mg 1 + mg 2 , 1
1 + kl
kl Using g = 9:8 m s2 , m = 55 kg, k = 77 kg s2, h = 60 m, and l = 30 m, one gets the only
physical solution, l0 = 21:4 m. Note: In order for l0 to be physical, it must be 0 l0 30 m.
b The force exerted on the agpole at h = 0 is:
F = mg + k0 h , l0
= mg + kl=l0 h , l0 :
The torque exerted on the agpole around the pivot point P is = FL0 cos : But from Figure
1, it is obvious that L0 cos = L cos . Thus
= FL cos :
The maximum torque the bolt has to withstand is
= bT;
1 L F θ
L’ T b φ
P Figure 1: Figure for problem 1
where b is the distance between P and the bolt and T is the tensile strength times the cross
section of the bolt.
Thus equating the above two equations,
T = FL cos =b
= 55 9:8 + 77 30=21:4 60 , 21:4 L=0:5 m kg s2
= 1:16 104 L m kg s:
2. a When rolling without slipping, the bottom point P of the yoyo has zero instantaneous velocity.
Rotation around a xed axis passing throughP is determined by the torque about that axis, but
only F has non vanishing torque. Thus the rotation is clockwise Figure 2a and the centre of
mass moves to the direction of the pull.
b In order for the yoyo to stand still, the net torque of the system should be zero about the contact
point P Figure 2 b. Thus mgb sin = F b , a. Then
b
F = b , a mg sin :
When the yoyo is about to slip, the sum of the torques around the pivot point P 0, tot is
=0
= 1+ 2
= mga sin , mgb , a cos ;
where 1 is the torque due to the centre of mass of the yoyo being pulled down by the gravitational
force and 2 is the torque due to the friction between the yoyo and the inclined surface.
tot 2 F
P
(a) b a F
P’
P θ
(b) Figure 2: Figure for problem 2 3 Thus a sin
= b , a cos a tan :
= b , a Note: The moment of inertia of a cylinder whose rotational axis is perpendicular to the at
surface of the cylinder and lies a distance a away from the centre is IR = IM + Ma2 . There was
a typo in the problem set. Not that it would have made any di erence in solving this problem...
3. a From energy conservation the energy at h = R + a cos is
2
mgR + a1 , cos = 1 mv2 + 1 I!sph
2
2
= 1 mv2 + 1 mv2
2
5
7 mv2
= 10 where we used Isph = 2=5ma2 and v = a!sph .
Note: The trajectory of the centre of mass of the marble rolled to an angle from the top
of the bowl is a + R Figure 3 a. And R = a . Combining these equations, one gets
a + R = a + Ra=R
= a +
= asph ;
which means v = a + R! = a! + ! = a!sph .
For a marble about to take o from the surface of the bowl the force perpendicular to the surface
of the bowl has the same magnitude as the centripetal force on the marble. Thus,
mg cos = mv2 =a + R:
From the above two equations
1 , cos = 7=10 cos Solving for cos gives cos = 10=17. Thus the marble leaves the surface of the bowl at the height
of hc = R cos = 10=17R or hc = 10=17R + a, depending on what your de nition of hc is.
b For a rolling pendulum" due to the lack of a better name the potential energy has to be
distributed between the rotational motion of the marble and the translational motion, whereas
the simple pendulum" potential energy is converted only to the translational motion. Therefore,
the simple pendulum has a higher translational speed than the rolling pendulum at the same
angle, i.e. it takes more time for the rolling pendulum to complete an oscillation than the the
simple pendulum.
Brownie points:
4 a θ φ mg
mg
to the centre
of the bowl R
(a) h
(b) Figure 3: Figure for problem 3
5 For a simple pendulum, the energy conservation equation at h looks like
1
mgh , h = 2 mvs 2
. Then
p
vs = 2gh , h
, where h is the release height of the pendulum.
For a rolling pendulum,
12
2
mgh , h = 1 mvr + 2 I!r
2
72
= 10 mvr
p vr = 10=7gh , h .
p
p
Thus vr =vs = 5=7 for any given height. Which means Ts =Tr = vr =vs = 5=7
4. a The surface of the liquid is perpendicular to the direction of the sum of the forces exerted,
centripetal force and gravitational force Figure 4. Thus i.e. 2
tan = v g=r .
The liquid touches the rim of the the cup when tan sati es the following equation:
2
tan = d H , h
.
Using the above two equations, one can obtain
r 2gr H , h
rd
= 2 9::8 30 0:15 , 0:12
0 08
= 14:8 m s:
b When the inside wheels of the car are about to leave the ground provided that the car does not
slip:
W F = hF
l
2d
where
W = the track width of the car or the width of the car for our purpose;
Fd = downward force;
h = the height of the centre of mass of the car;
and
Fl = lateral force:
v= 6 θ
Fl
θ H
ho
mg d Figure 4: Figure for problem 4
The downward force is the sum of the gravitational force and the force generated by the aerodynamic device, i.e. Fd = mg + cv. Here c has units of N km hr. Converting it into MKS units, it
becomes N km hr = N 1000 m 3600 s = 3:6 N m s. The lateral force is centripetal force,
2
i.e. Fl = mv =r .
Rearrange it as a quadratic equation in v:
Solving for v c
v2 , rW m v , rW g = 0:
2h
2h
p 2
v = ac=m + ac=m + 4ag
2 ,
where a = rW=2h.
Using r = 30 m, W = 2 m, h = 0:4 m, m = 1500 kg, and c = 150 3:6 N m s = 540 N m s,
one gets
vr = 43:8 m s:
However, if mg + cv = mv2 =r, the car is about to slide. Rearrange it as a quadratic equation
in v:
v2 , rc v , rg = 0:
m
Solving for v
p Using = 0:23 one gets 2
v = bc=m + bc=m + 4bg :
2 7 vs = 9:6 m s:
Since vr vs , the car will slide before it lifts its inside wheels. Thus the maximum speed you can
go through the turn is at 9:6 m s.
5. a For a satellite on a circular orbit of radius r,
mv2 = GMm
r
r2
and
p
v = GM=r:
The velocity of the satellite at the parking orbit h = 200 km is then vp = 7:772 103 m s,
where G = 6:673 10,11m3kg,1 s,2 and M = 5:974 1024kg are used.
The angular velocity of the satellite on the geosynchronous orbit is ! = 2=24 3600 s =
7:272 10,5 s,1: The radius of the geosynchronous orbit, rg , can be obtained from the rst
equation using v = r!,
rg GM 1=3
=
!2
= 4:224 107 m: Thus the velocity of the satellite on the the geosynchronous orbit is vg = rg ! = 3:072 103 m s.
From energy conservation
1 mv , GMm = 1 mv , GMm ;
2 pt
rp
2 gt
rg
where
vpt = The velocity at the perigee of the transfer orbit;
vgt = The velocity at the apogee of the transfer orbit;
rp = The distance of the perigee from the centre of the earth
= The radius of the parking orbit;
and
ra = The distance of the apogee from the centre of the earth
= rg :
From angular momentum conservation
~p ~pt = ~a ~at
rvrv ,
and at the perigee and the apogee, ~ and ~ are perpendicular to each other, so
r
v
rp vpt = ra vat
8 or
vat = rp =ra vpt :
Thus the energy conservation equation now looks like
2! 1 mv2 1 , rp
2 pt
ra
giving
vpt = GMm 1 , rp
ra
ra 2GM 1=2
= r 1 + r =r
a
pa
= 1:022 105 m s: So vpt , vp = 2:248 103 m s.
The velocity at the apogee of the transfer orbit is then rp v
ra pt
= 1:597 103 m s:
Thus the velocity change the satellite has to achieve in order to get into the geosynchronous orbit
from the transfer orbit at the perigee of the transfer orbit is vg , vat = 1:475 103 m s.
b The transition takes half the time for the satellite to complete the elliptical orbit. So using the
Kepler's law
vat = T1=2 = T
2 1 42 rp + ra =23 1=2
=2
GM = 1:900 104 sec
' 5 hrs 17 min
It takes 5 hours and 17 minutes to complete the transition.
6. a Of course not, silly. The possible physical e ects are
i. Coriolis force. 2m~ ~
!v
ii. Centripetal force. mv2 =r.
iii. Decrease of gravitational force due to the increase in r.
p
iv. Relativistic e ect time dilation. t = = 1 , v2 =c2 , etc.
International ight normally ies at an altitude of about 10 km. Thus the change in the graviational acceleration at an altitude h is g = g1p R=R + h2 = 3:1 10,3 g. The change in
,
p
p
T is T ' l=g , g , l=g ' 1:6 10,4 l=g, where l is the length of the pendulum. But
it a ects the same way regardless of the direction of the ight.
The speed of the east bound plane that Chris is on from the xed reference frame is v =
p
=12 36006:4106 ' 460 m s. Relativistic e ects are then on the order of T T 1= 1 , v=c2 ,
1 10,11T 10,7 sec :
9 b If we use the inertial system with the origin of the coordinates sitting at the centre of the earth
as the reference system, the Coriolis force vanishes in this system. The e ective gravitational
acceleration at the earth's surface for an object with angular velocity ~ against an inertial system
!
is
^!!~
~eff = , GM R , ~ ~ R:
g
R2
Since Pat and Chris are moving along the equator, their angular momentum vectors ~ are per!
~
pendicular to R. Thus j~ ~ Rj = j!2 Rj. Then the e ective gravitational acceleration looks
!!~
like ,
and if we replace geff = GM , !2 R
R2
GM
R2 with g geff = g , !2 R:
The period of a simple pendulum with the length of the pendulum L is
s T = 2 L :
g
Since Pat is moving westward, the angular velocity is zero. Thus
s L
TP at = 2 g
s = 2 g +L 2 R
!
ss
1
= 2 L 1 + !2 R=g
g
s
1
= T 1 + !2 R=g :
Whereas, Chris is moving eastward, so the angular velocity of Chris is 2!. Thus
s TChris = 2 g ,L !2 R
4
s
= 2 g , L!2R
3
ss
1
= 2 L 1 , 3!2 R=g
g
s
1
= T 1 , 3!2R=g :
10 TChris , TP at = T s s
! 2 1
1
, 1 + !2 R=g = T 2!g R
2 R=g
1 , 3! Using g = g , !2 = 9:8 m s2, R = 6:4 103 km, T = 12 hrs, and ! = 2=T one gets
TChris , TP at ' 298 seconds : Since TChris , TP at 0, Pat's clock goes faster than Chris' clock.
Pat's clock would be 75 seconds faster and Chris' clock would be 224 seconds slower than the
local clock at Y. 11 19951996 Physics Olympiad Preparation Program
University of Toronto Problem Set 3: Thermodynamics
Due December 18th , 1996 1. Let's say you begin a chain email that says send a copy of this message to someone or you will step
on a pile of crap within 24 hours" to a completely random user on a computer network.
Here are the rules facts about the chain email.
i. There are users on the net, including yourself.
ii. These users are a unusually superstitious bunch. Thus when one receives this apocalyptic
message, s he will send it to a random user excluding the immediate sender without failure.
iii. If any user receives the message twice, s he ignores the second message.
iv. Only the directly immediate sender is known to the user that receives the message.
v. The network failure rate is and when it happens, the chain email dies.
Now you know the rules facts of the chain email. Here are the questions.
a What is the probability of the email reaching all the users?
b What is the probability of you, the originator of the email, getting the email back before users
receives the message?
c What is the probability of you stepping on a pile of crap within 48 hours?
2. Jacques Cousteau, a fearless marine explorer, prepares for a dive into the cold arctic sea. The sea
water temperature is 3 C.
a Jacques dons his rubber suit and scuba gear and dives into the sea. He remains underwater for 20
minutes. How much heat does he lose during the diving? Assume that his body surface is 2 m2
and his rubber suit is 0.5 cm thick. Also, assume that Jacques skin maintains a temperature of
33 C throughout the dive.
b Jacques then boards his le Submersible with his le Cat. She, the submersible, not the cat, is
essentially spherical in shape 3 m in diameter and is constructed with 1 cm thick iron lined
with 1 cm of rubber and then 0.5 cm thick aluminium. le Cat gets irritated when the ambient
temperature goes down below 15 C. Jacques is an avid cat lover and he would not want to subject
his cat in such harsh conditions. le Submersible carries 15 kWh of batteries reserved for heating.
How long can Jacques and le Cat remain underwater comfortably? Note: State any assumption
and or approximation you make.
3. After four years of physics education in university, you ditched the idea of having a career in physics
and decided to do the next best thing. Well, now you are working as a road crew for a famous rock'n
roll band Killer Pumpkins and Dead Pumpkins". The subject of your service asks you to make him
a jugful of iced co ee made with 4 scoops of ground co ee brewed at 95 C. No sugar or cream, please.
Oh, and I want it to be served at exactly 3 C. Make it fast, physics dude." So you decide to use ice
cubes to cool the co ee down to 3 C. The fridge at the kitchen has ice cubes at 5 C. The volume of
the jug is 600 ml. How much water for co ee brewing and ice cubes do you need?
4. An inventor claimed that he invented a perpetual motion engine. Your uncle Moe, an investor, came
to you asking about the plausibility of the claim. The following is what the inventor says about the
engine.
N p n 1 You know when water turns into ice, the volume it occupies increases. Now here is a heat
engine to exploit that. Figure ?? shows the sequence of how the engine works.
a Initial state: water is con ned in a cylinder with a piston and under 1 atm.
b A weight is placed on the piston.
c The cylinder is cooled and the water freezes. The expansion on freezing raises the weight,
doing work.
d The weight is moved laterally. No work is done in moving the weight laterally.
e The temperature is returned to the initial value and the ice melts. The state of the
system is now identical to the initial state.
Since ice melts and water freezes at 0 C, which implies that this heat engine can operate at
a single temperature, which in turn implies that this engine can run forever.
Now you know that there is no such thing as perpetual motion machine". What is wrong with this picture? Uncle Moe won't take answers like yeah, you cannot violate the second law of thermodynamics.
that's why."
5. You are a member of a team developing an autonomous volcano exploring vehicle. You are in charge
of the cooling system design. The ambient temperature of the volcano is 530 C. The electronics bay
of the vehicle is insulated and should be maintained at 30 C.
The electronics in the bay produces heat at a rate of 720 Wh and the the heat ow rate into the bay
through insulation is 0 002 , W.
: Tout Tin a What's the maximum coe ciency of performance of the cooling system?
b What's the minimum power requirement for the cooling system?
c The coe cient of performance of the cooling system onboard the vehicle is rated at 0.2 at the
conditions above. The typical mission lasts about 100 hours. What is the minimum energy
requirement for the dedicated power source of the cooling system?
6. You work for CSA Canadian Space Agency and your team is involved with the Mercury Observer, a
long waited Mercury probe spacecraft, project. Your supervisor asks you to do the following preliminary
calculations. By the way, the funding for the project was approved on the condition that the spacecraft
be painted black bureaucracy, you see.
Note: The average radius of the orbit of mercury is 5 79 106 km. The radius of the sun is 6 96 105 km
and the temperature on the surface is 6000 K.
: : a What would be the maximum radiation pressure that the spacecraft experiences from the sun's
radiation when the spacecraft is on the mercury orbit.
b What would be the surface temperature of the craft a long time after it arrives at the mercury
orbit? The spacecraft is spherical in shape, has a 3 m diameter, and its cooling system expels
4 W from the interior to the exterior of the spacecraft. 2 Figure 1: Figure for Problem 4. 3 19951996 Physics Olympiad Preparation Program
University of Toronto
Problem Set 4: Wave and Optics
Due January 22th , 1996 1. As a physicsilliterate rock concert organizer on Lilliput island remember Gulliver's Travels, Donald
Armen is still puzzled by the complaints from the rock fans. In fact, one of the two speakers was
destroyed. He just cannot believe it. He bought a new speaker to increase the loudness of this mono
sound system to the one he had. He placed them 8.00 m apart. The closest audience member, Christie,
was 10.0 m away from each speaker, respectively or 9.17 m from the midst of the both. Can you
help him understand what was going on? Assume the average frequency is 690 Hz being tiny, these
Lilliputs have high pitch. The speed of sound is 344 m s.
a Could Christie hear the concert?
b Albion, standing on Christie's left, was among the ones who got angry and destroyed the speaker.
He just barely listen to the concert. Where did he stand?
c This ction is far from reality. Explain!
2. A group of police cars moving at 80 km hr are chasing O'Samson's white Bronco, which is travelling at
the same speed. They are afraid that this former famous hockey player will commit suicide. It seems
that 3000Hz sirens from the police car do not bother him much.
a What is the wavelengh of the sirens in the air between O'Samson and his chasers?
b What frequency does he actually hear?
The speed of sound is 344 m s.
3. Learning from Donald Armen's story Problem 1, you decide to study more about loudspeakers. You
place two small loudspeakers facing each other at the two ends of a 5.0mbench. The rst loudspeaker
emits a sound of pressure amplitude twice as much as the second one. Both sounds are at 1334 Hz and
are in phase at the speakers.
a At what point on the line joining the two speakers is the intensity a minimum?
b What is the intensity compared to one from the second speaker alone?
c At what other points on the line joining the speakers will destructive interference occur?
d At what point on the line joining the speakers is the intensity a maximum?
e Express the maximum intensity terms of the intensity from the second speaker!
4. Your professor told you that di raction gratings are better than prisms for atomic and molecular
spectroscopy. You don't believe it; you are sceptical .... So you perform experiments using both and
you nally agree with him. Why is your professor right? To support your arguments compare your
calculations for both prism and di raction grating for a source light which emits 434 nm blue, 486
green blue, 589 nm yellow and 656 nm red. Take the indices of refraction of heavy int prisms
n = 1.675, 1.664, 1.650, and 1.644, which correspond to the wavelengths above, respectively. The angle
of the prism is A = 60o and the di raction grating has 8000 lines per centimeter.
5. Kids love to play with soap bubbles and are always fascinated by them. Ok, ... you are not kids
anymore but you need to guess what colour a bubble of 400 nm thickness appears to be. Need some
hints? Alright.
a What colours do not appear in the re ected light? b For what colours does constructive interference occur?
c So the colour of the soap bubble is predominantly ....
The index of refraction of the soap lm is 1.36.
6. Ok, let us have a break now with a few small questions.
a Most large astronomical telescopes in use today are re ectors rather than refractors. Why is this
so?
b You are able to nd a very sharp focal length of a crystal ball. Right or wrong? Explain. 19951996 Physics Olympiad Preparation Program
University of Toronto
Problem Set 5: Electricity and Magnetism
Due February 19th , 1996 1. Pretend that you are an electrical engineer. You supervise the construction of an electric generating
plant system at a rate of 1 GW and a voltage of 12,000 V.
a It is desired to transmit the power at 756 kV. What must be the turns ratio of the stepup
transformer?
b What current would be sent out over the power lines if the transmission were at 12,000 V.
c What is the actual current you expect?
2. Many environmentalists are disturbed by the radiation from power lines. They may not be concerned
with the electric eld because they can buy the argument that human body is almost a perfect conductor. Will you buy it ... ? Anyway, but they worry so much with the e ect of magnetic eld in
the high voltage transmission lines like your design in the previous problem. As an engineer you try
to convince them that the powerful plant like yours 1 GW and 756 kV transmission lines is still safe
because you already put the transmission line high enough. Otherwise, you will lose your job because
they are very persuasive these days. What is the minimum height you have to put your lines so the
e ect of magnetic eld on the ground is only 10 of the natural one? The magnetic eld at Earth's
surface is 0.5 gauss in contrast to the one in the lab which can go up to 1.0 T.
3. You are trapped in a dungeon to assist an old physicist of 1950's who still has his own ancient accelerator. It is a boring job but he pays you very well, better than your previous job as an electrical
engineer. He asks you to accelerate a beam of electrons at a voltage of 10,000 V and this beam enters
a uniform magnetic eld of 0.50 T perpendicular to the motion of the electrons.
a What is the force on each electron due to the magnetic eld?
b What will the speed of each electron be after it has been in the eld for 10 seconds?
4. After working for years as a research assistant, sometimes you have nightmares. One day, you dreamt
that you were an electron travelling parallel to a very long,thin wire at a distance of 20 cm. Suddenly,
you felt you were pushed away from the wire by a strong force well, you are very tiny now because
your boss turned on the electric current of 2.0 A on the wire. How much force did the magnetic eld of
the current exert on you and what was the direction of the force with respect to the current? After you
woke up, you told your boss about your dream and he laughed, Why didn't you tell me to place the
wire between two large plates of a capacitor so I would not have bothered you?" You seemed puzzled.
Now, after many years of enlightment, you know the reason why he said this. What was the voltage of
the capacitor to keep you travelling happily along the wire and what was the direction of the electric
eld?
5. Nuclear fusion is like a long term commitment for a blind mouse couple. Sometimes you need to
interfere by pushing one or both toward the other. But you can attract them by placing a piece of
cheese between them so they will smell it. This kind of cheese is called a catalyst in the fusion reaction.
To make the story simple, you would like to have fusion of two protons.
a If the catalyst is an electron, what is the total kinetic energy of the system at its minimum
potential energy where the electron sits collinearly between these two protons? Assume that all
three are far apart from each other and the minimal distance between the protons is about 0.1
nm. b Instead of an electron, you can also use a muon which is heavier by a factor of 207. This heaviness
makes the distance between the protons smaller by a factor of 207, too. What is the total kinetic
energy of the system now?
Of course, the story does not end here because the two protons will transform into a deuteron, and
emit a positron and a neutrino.
NOTE: This minimal distance between the protons is determined by the Heisenberg uncertainty principle of quantum mechanics.
6. Here are a few questions that do not require any calculations.
a You have been told that wave requires a medium to propagate like sound requires air. However,
light, which is electromagnetic wave, can travel in vacuum. How do you explain this phenomenon?
b Electromagnetic wave, like other waves, can transmit energy from one place to another, say E .
How much momentum does it transmit if it can?
c Why do we all have an AC, instead of DC, power in the houses and buildings? 19951996 Physics Olympiad Preparation Program
University of Toronto
Problem Set 6: Electronics
Due March 18th, 1996 1. The circuit in Figure 1 is one that you can't easily escape from. Whether you're taking in tunage from
O spring, Metallica, or a Bach fugue for harpsichord, this little gem will most likely come into play
oops, no pun intended. It's a wait for it noninverting operational ampli er circuit.
0
a Find an exact expression for the voltage gain ratio, Av = v2 , in terms of R1, R2, Rd the e ective
v
0
resistance between the +" and " terminals, and AOL vd the open loop voltage gain.
v
b Assume that iin is zero, so that vd 0 and v1 v2 . Derive an expression for the voltage gain
ratio for this simpli ed case.
2. Gill Bates is the billionaire founder of Sicromoft Corp., the creator of ScreenDoors 96 a soontobe
popular operating system. It is a little known fact that Gill spent her teenage summer months
designing power supplies for a small Ontario electronics rm. One year, she dreamed up the 5=2 kHz
circuit shown in Figure 2. Her boss wanted it put on the market immediately, and asked her to write
a speci cation sheet for it, including the magnitude and phase of a the open circuit voltage across
the terminals AB and b the driving point impedance at AB with the internal voltage source set to
zero. Solve for these quantities and you too may be a billionaire someday.
3. Let me tell you about a nightmare I once had. It was a dark and stormy night and I had somehow
become trapped in a dripping dungeon guarded by a band of simian subterranean gypsies. Their leader,
Murd, handed me a circuit breadboard and a baggy containing half a dozen NOR gates. He told me
that if I could realize the logic function f = A B + A B using the devices provided, then I would be
free to leave his dungeon. How did I do it in my dream, using all the given gates? Of course, I then
awoke in a cold sweat and realized how stupid I and Murd's band for that matter had been. Why?
4. Aren't CDs just too cool, with their glittering colours and cute size? In the early 80s they were heralded
as the perfect sound" medium, and thanks to the power of advertising many folks still think this to
be true. The shortcomings of the CD format, which are rooted in the fact that the standard was
established way back in the late 70s, could form the basis of many POPTOR problems. CDs are here
to stay for a while, anyway, so let's be positive and examine the medium a little. First of all, the data
words on a CD are 16 bits long. The digital sampling rate is 44.1 kHz, and the nominal frequency
limits of human hearing are 20 Hz to 20 kHz.
a How many di erent levels of sound pressure can be represented with this medium?
b Discuss why the digital sampling rate was chosen to be 44.1 kHz.
c What is the maximum possible dynamic range in decibels?
d What is the data density in bytes m2 on a CD if the angular speed is 500 rpm when the laser
pickup is at the innermost point and 200 rpm when it's at the outermost point. Assume the outer
diameter to be 120 mm, that a CD plays for 74 minutes, and that there's no data redundancy or
errorchecking information.
5. The con guration in Figure 3 contains two AND gates and two NOR gates. The input marked CLK
is connected to a square wave generator that oscillates between the 1" and 0" logic states.
a Give a truth table for this logic circuit.
b How do you think this con guration could be useful?
1 + V
+d V
2 + + R2 iin
+ Vo
i V1
 R1   Figure 1: The noninverting opamp circuit in Problem 1. 2 21 50 5 kHz
2 A 12 B 30
4.8mH Figure 2: Gill's circuit in Problem 2. 3 12mH R Q CLK
Q S Figure 3: The digital circuit in Problem 5. 4 c Can Figure 3 be replaced by a system of NAND gates? If so, how? If not, why not?
6. Your fairy godmother gives you a black box with four terminals on it two inputs, A and B, and
two outputs, C and D. She tells you that the resistance between terminals B and C is equal to the
resistance between terminals C and D, and that there is negligible resistance between terminals B and
D. She also informs you that, when a sinusoidal voltage with a frequency fo is applied across A and
B, the output voltage across C and D is also sinusoidal, but has an amplitude that is 20 less than
the input amplitude. If the input frequency, f , is varied either higher or lower, the output voltage
amplitude is attenuated symmetrically about fo , roughly as a function of 1= jf , fo j. What does the
circuit inside the black box look like and what can you say about the values of the devices inside?
Note: You may assume that there's no active electronics inside only passive devices are used. 5 19951996 Physics Olympiad Preparation Program
University of Toronto Solution Set 6: Electronics 1. a The feedback voltage, i.e. the voltage at the " terminal of the op amp, can be written as: v1 = v0 , i + iin R2:
Since i = v1 =R1 and iin = vd =Rd = v0 =AOL Rd:
R =AOL
v1 = 1 , +2R =R Rd v0 :
1 21
But vd = v1 , v2 = Av0 ;
OL or v2 = v1 , Av0 :
OL Combining and rearranging the above expressions gives:
R1 + R2
Av = v0 =
2
1
v2 R1 , AR1 RRd , RA+R2 :
OL
OL
b If iin = 0, then the currents through R2 and R1 are equivalent and:
v0 , v1 = v1 ;
R2
R1
and
R
Av v0 v0 = 1 + R2 :
v2 v1
1
2. Gill's circuit is a Bridge circuit and the problem is solved by calculating the Thevenin equivalent voltage
and impedance. The impedance of a resistor and inductor in series is given by the complex expression
Z = R + i!L, where R is the resistance, ! is the angular frequency, and L is the impedance. The
quantity XL = !L is called the inductive reactance.
The equivalent impedance at terminals AB with the source set equal to zero is: Z0 =
1
1 ,1 +
1
1 ,1 ;
12 + i5 kHz 4:8 mH + 21
30 + i5 kHz 12 mH + 50 which simpli es to: Z 0 = 47:3 at 26:8 :
With the circuit open at AB , the current on the left side of the bridge is:
V
I1 = 21 + 12 +205 kHz 4:8 mH ;
i
1 and the current on the right side of the bridge is:
20 V
I2 = 50 + 30 + i5 kHz 12 mH :
If we assume that point A is at a higher potential than point B , then we have: V 0 = VAB = I1 12 + i5 kHz 4:8 mH , I2 30 + i5 kHz 12 mH:
Substituting and simplifying gives: V 0 = 329 mV at 170:5 :
3. Just like in many other nightmares, things here didn't proceed optimally. The function, f = AB +A B ,
can be simpli ed to B , so no NOR gates are needed and the output can simply be wired to the input
B . The question, however, insists that we design a logic circuit using NOR gates, so here goes.
Since f = A B + A B , then:
f = A B + A B:
Applying deMorgan's Laws A + B = A B , and A B = A + B gives:
f = A B A B ;
and:
f = A + B A + B ;
and:
f = A + B + A + B :
Therefore:
f = A + B + A + B :
The circuit of NOR gates is given in Figure 1.
4. a Since levels of sound pressure, or intensity, are related to signal voltage levels in the electronics,
and the data word length for a single sample is 16 bits, the number of levels of sound pressure is:
216 = 65; 536 levels:
b The Danish mathematician, Sven Nyquist, proved what is now called Nyquist's Theorem, which
states that a sampling rate of at least double the highest recorded frequency is required for
the reproduction of sine waves of the highest recorded frequency, and of any frequencies below
that point. Since any waveform can be decomposed into a superposition of sinusoidal waves,
the sampling rate for frequencies detectable by the human ear needs to be at least 40 kHz 2
20 kHz.
2 A B
f Figure 1: The NOR gate con guration in Problem 3. 3 S
0
0
0
0
1
1
1
1 R CLK Q Q
0 0 QQ
0 1 QQ
1 0 QQ
1 1 01
0 0 QQ
0 1 10
1 0 QQ
1 1 ?? Table 1: The truth table for the gated RS ip op in Problem 5.
c The dynamic range of CDs is nominally said to be at least 90 dB. In this case, the decibel is a
logarithmic measure of the relative highest and lowest power levels possible with the CD format.
The decibel, when used to compare two power levels, is de ned as 10 log10 P2 , where P represents
P1
power. We know that CDs o er 65,536 voltage levels, and thus, since P = V 2 =R V is voltage
and R is resistance, the maximum dynamic range is:
:
20 log 65; 536 = 96:3 dB:
10 d For vinyl long playing records, the stylus spirals along a groove from the outer radius to the inner
one while the turntable rotates at a xed angular velocity. For CDs, the laser pickup instead
tracks a spiral from the inside radius to the outer one, and the disc is rotated with a variable
angular velocity to provide a constant linear speed di erence between the pickup and the track
for any given radius on the disc. The linear relative speed at the outer radius is: 120 mm 200 rpm = 12 m=s:
2 Equating this with the linear relative speed where the magnitude of the angular velocity is 500 rpm
gives an inner radius of 24 mm. Now that we have the inner and outer radii, we can calculate the
active area of a CD as:
, 60 mm2 , 24 mm2 = 9:5 10,3 m2:
The storage capacity is approximated by using the maximum playtime and the digital sampling
rate:
74 min60 sec=min44:1 103 words=sec2 bytes=word = 3:92 108 bytes:
The data density is therefore:
3:92 108 bytes = 4:1 1010 bytes=m2 :
9:5 10,3 m2
5. a The truth table for this sequential logic circuit, which is a gated RS ip op, is given in Table 1.
b The principle of this RS ip op can be used to store information in a clocked computer memory
register.
c Refer to Figure 2 for an equivalent con guration made up of NAND gates.
4 R Q CLK
S Q Figure 2: The NAND gate equivalent to Problem 5. 5 6. Combining the the given attenuation and interterminal resistance information suggests that there's a
passive bandpass lter in the black box. Refer to Figure 3 to see the circuit. Since we're only concerned
with amplitudes here we can ignore the phase, we can deal with the equation V = I jZ j, where Z is
the impedance. The ratio of the output over input voltage amplitudes is therefore given by: Vout = r
Vin R
: 1 2 + R2
2fL2 + r2 , 2fC The resonant frequency from the above equation, fo , is calculated to be:
r 1
fo = 21 LC :
We know that the resistance r exists because of the nonzero attenuation 20 at the resonance
frequency fo . At resonance: Vout = 80 = p R :
Vin
R2 + r2
The relationship between r and R is therefore: r = 0:75R: 6 L r C A C
R B D Figure 3: The passive LRC circuit in the black box of Problem 6. 7 teX?.aux teX?.aux 19961997 Physics Olympiad Preparation Program
– University of Toronto – Problem Set 1: General
Due November 1st , 1996
1. While playing a game of LaserTagTM , Daniel is trying to target his pal Yiorgos with his laser beam.
Daniel (from his sonar display) knows that Yiorgos is hiding in the next room separated from him by
an opaque wall and a thin window. Daniel wants to ﬁre at Yiorgos, but he can’t aim without exposing
himself. Suddenly he gets a great idea! He pulls a GasBomb from his belt and pulls the trigger. The
invisible gas ﬁlls the room, slowing the speed of light to one half of its normal value in air. Figure 1: For problem 1
(a) What is the smallest lateral distance (i.e. parallel to the window) Daniel must move to target Yiorgos? (Remember that for physicists, ’thin’, ’small’, ’minuscule’, and similar adjectives
generally mean ’negligible’.)
(b) Daniel, delighted with his ingenuity, moves this distance and ﬁres his laser gun through the
window. He is surprised to see the beam exit the window and continue straight. What went
wrong?
2. An Ontario apple of mass M = 0.2 kg rests on fence post of height h = 3 m. A bullet with mass
m = 10 g, traveling at v0 = 500 m/s, passes horizontally through the centre of the fruit. The apple
reaches the ground at a distance of s = 10 m. Where does the bullet land? What part of the kinetic
energy of the bullet was lost as heat when the arrow passes through the apple?
3. The prototype spaceship Stealth (mass: 250 000 kg) is in a solar orbit(r=1.5×108 km), testing out its
new cloaking mechanism. The cloaking device makes the ship perfectly absorbing of all electromagnetic
radiation.
(a) Assuming the ship has a crosssectional area facing the sun of 20 000 m2 and knowing that the
sun produces 3.8×1026 W of power , what is the energy the Stealth absorbs after being cloaked for
ten minutes?
(b) Assume that the energy the ship radiates is negligible. Before the test period, the average temperature in the Stealth is 18 ◦ C. What will be the temperature after the ten minute test period?
Assume that all the absorbed energy is going into heat, and the average speciﬁc heat of the Stealth
and its crew is 2 J/g◦ C.
1 (c) In reality the ship does radiate energy. Taking this into account, what will be its ﬁnal temperature
if the ship stays cloaked? (you might want to use Stefan’s constant: σ =5.67×10−8 Wm−2 K−4 ).
What is the peak wavelength of the spectrum that the ship emits?
4. Wesley (of Star Trek: Next Generations fame) is given a box with three electric terminals protruding
from it labelled A, B , and C . He is told that the box contains three resistors and, using an ohmmeter,
ﬁnds out that the resistances across AB , AC , and BC are exactly 8 Ω each. He is asked to describe the
conﬁguration and values of resistances in the box, knowing that if he gives the wrong answer, phasers
on “kill” will automatically wipe him out. He’s under a little stress. Find the resistances. Is the answer
unique?
5. In her laboratory, MarieEve ﬁnds an unknown component with two electrical leads. To determine
what it is, she applies diﬀerent voltages across the leads and measures the following current going
through the device: Figure 2: Device for problem 5
(a) What is the eﬀective resistance of the device at 10V? at 4V?
(b) MarieEve then uses the unknown device in the following simple circuit: Figure 3: Circuit for problem 5
Using the IV graph, circle the point (or points) that represents the voltage and current that could
be passing through the unknown component in the above circuit.
(c) Marie Eve replaces the 10V supply with a variable power supply and an ammeter (current measuring device with negligible resistance). Starting a 0 V, she increases the power supply voltage
to 15 V and then decreases it slowly back to 0 V. Roughly sketch a graph of the current she would
observe as a function of this voltage.
2 6. For some roll dispensers of toilet tissue, it is possible to drag the end of the roll over and ﬂush the entire
roll down the loo like spaghetti, with the tissue spinning freely oﬀ the roll. Presumably to interrupt
such physics experiments, manufacturers often construct the holder to have a springloaded pressure
plate at either end of the holder, which press ﬂat against the ends of the cardboard core. The friction
is enough to keep the paper from spinning oﬀ the roll, but as the roll becomes smaller, it soon becomes
impossible to pull the paper directly oﬀ the roll without breaking. Figure 4: Roll dispenser for problem 6
(a) Consider a cardboard core 4 cm in diameter and with walls 1mm thick, and take that the pressure
plates push only on circular ends of this core with a force of 13 N on either end, and that the core
has a static coeﬃcient of friction µs = 0.27. Empirical studies (!) show that most toilet paper
segments will burst at the perforations for a typical tension of T = 3.5N . If the toilet paper can
be pulled oﬀ the roll without breaking at the perforations, what is the minimum diameter of the
toilet paper for this to be possible?
(b) The kinetic coeﬃcient of friction µk is measured to be 0.18. If the paper is pulled oﬀ starting
with a full roll (diameter 11.5 cm) and the motion never stops, what now will be the diameter of
the roll of paper at the point where the paper tears?
(c) Take that the paper on the roll wraps at 10 layers per millimetre. What length of paper remains
on the roll for the critical diameter you found in (a) and (b) above?
(d) Suppose that the pressure plates pushed on the whole side of the roll, cardboard core and rolled
paper together, so that the contact area changed as the roll got smaller. How might the whole
problem be changed? Assume you can use the same static coeﬃcient of friction µs . 3 PROBLEM SET #1 Ð GENERAL
1. LaserÐTag This is a standard refractiontype question. Since Daniel is in a medium with a
different (i.e. higher) refractive index than air, the laser beam will ÔbendÕ around the
corner.
Let x be the minimum distance Daniel must move to target Yiorgos. The only other
variable to worry about is at what angle Daniel will aim (i.e. where will the beam hit
the window). For minimum x, Daniel will aim to hit the very edge of the window.
From SnellÕs law
n i sinθ i = nr sin θ r
We know that after the GasBomb is used, ni = 2nr
4 sinθr = √ 42Ê+Ê12 = 4 1
√7 Also
sinθi = ∴ 2n r x x2Ê+Ê2
√2 x x2Ê+Ê4
√ = nrÊ = x x2Ê+Ê4
√
4 1
√7 17x2 = 4 (x2 + 4)
13x2 = 16
x = 1.1 m
(Negative root extraneous).
Daniel must move at least 1.1 m to the right to target Yiorgos.
(Common error: In Snell's law, the incident and refracted angle are measured
between the ray and the normal to the interface, not the ray and the interface.)
b)
Nothing went wrong, if Daniel was looking along the beam with the lasergun held high to aim. Daniel sees his laser beam due to light scattering back to him
from dust or other particles in the air. This scattered light, along nearly the same
path, but backwards, will also obey SnellÕs law, and as such the light that reaches his
eye will follow a similar path as the laser beam. Therefore, to Daniel it looks
straight.
(Many of you came up with ingenious ways that would result in the laser beam not
bending, and, if it was physically possible, got partmarks for your efforts. As you
can see from the solution, none of these were necessary.) 2. Apple shoot The total momentum of the system remains constant in a collision. Thus,
mv 0 = mv + MV (2.1) Here, v is the velocity of the bullet and V that of the apple after the collison. Both
objects fall a distance h to the ground; their time of flight is t = √ = 0.78 s. 2h/g
During time t, the apple covers a horizontal distance of s = 10 m, so its horizontal
velocity is V = s/t = 12.8 m/s. Thus, the conservation of momentum equation yields
v = 104 m/s for the horizontal velocity of the bullet immediately after the collision.
Since the bulletÕs time of flight is also t, it travels a distance vt = 105 m horizontally
from the column.
Initially, the kinetic energy of the system is Ei = mv02/2 = 1250 J. After the collision,
the kinetic energy is Ef = mv2/2 + MV2/2 = 93.2 J. Thus,
1250 Ð 93.2 = 1156.8 J (2.2) is lost as heat (92.5% of the original energy). Note that the collision is n o t
completely inelastic. In a completely elastic collision, the kinetic energy is
conserved. In a completely inelastic collision, the bullet would remain inside the
ball.
3. Stealth In the world of sciencefiction, a ÔcloakedÕ ship has to be perfectly transparent; or,
nextbest, absorbing all radiation that strikes it but emitting only the same radiation
as the molecules it displaced would have. Our current knowledge of physics doesnÕt
allow this, in steady state, but letÕs assume the second case.
a) We can assume that the sun radiates uniformly over a sphere. What is the energy the Stealth absorbs per second?
The sun radiates P = 3.8 × 1026 W over itÕs entire sphere. At r = 1.5 × 108 km, this
corresponds to a flux of P/(4¹ r2) = 1.3 × 103 W mÐ2.
∴ Stealth absorbs a total power of 1.3 × 103 W mÐ2 ¥ 20 000 m2 = 2.6 × 107 W.
Over 10 minutes, Stealth absorbs 2.6 × 107 J sÐ1 ¥ 10 min ¥ 60 s minÐ1 = 1.6 × 1010 J.
b)
With no radiation losses: ∆t = 1.6 × 1010 J / { (2.5 × 108 g)(2 J gÐ1 ¡CÐ1)} = 32¡ C,
∴ tfinal = (18 + 32) = 50¡C) c)
Unfortunately, the question does not give enough information to really
answer this problem. We need the t otal surface area Asurf , not just the crosssectional area. We can make some guesses at this, though. Since the crosssectional
area =Ê20Ê000Êm2, the surface area must be at least 40 000 m2. There is no upper limit
to this number.
(Common error: many people just plugged in the only area value they were given,
i.e. 20 000. A ship that has a crosssectional area of 20 000 cannot have a surface
area of 20 000, this answer is impossible.)
At steadystate, power absorbed = power radiated. Using StefanÕs law: H = eσ A surf T4
(T in Kelvin), and since ship is perfectly absorbing, emissivity e = 1, ship absorbs 1.3
× 103 W/m2, ∴ at steadystate 1.3 × 103 W mÐ2 ¥ 20 000 m2 = σAsurf T4 , 2.6×107 Ð1/4
Τ= σAsurf Since we do not know Asurf, we can determine only a rough range of T.
Maximum for Asurf = 40 000 m2 (the minimum): 2.6×107 Ð1/4
Τ=
= 389 K = 116¡ C
σAsurf Since Asurf can be any larger surface area, Tmin can approach zero K.
A possible ship shape would be a sphere. The radius of such a sphere is
πr2 = 20 000 m2
⇒ r = 80 m
Asurf = 4πr2 = 80 000 m2
giving T = 275 K → 2¡C
No matter what your answer for T, you can determine the corresponding peak
wavelength of the radiation distribution. WienÕs Displacement law is the quickest
way to the answer. It states:
λmaxT = 0.002898 m¥K
Substitute for T (in Kelvin) and solve for wavelength in meters. 275K gives λ max =
20.5 µm.
4. WesleyÕs test A typical first guess at the contents of the black box is a triangular configuration of
resistors of resistances Rab, Rbc and Rca shown in Fig. 4.1(a). Since Wesley is such a
stereotypical thinker, he immediately grasps at this solution. When the resistance across terminals a b is measured, the resistance of the parallel circuit formed by
resistors Rab and Rbc + Rca is
1
1 Ð1
[4.1]
Rabtotal = R abÊ+ÊR caÊ+ÊRbc Similarly, the resistances measured across terminals bc and ca are
1
1 Ð1
Rbctotal = RbcÊ+ÊR caÊ+ÊR ab [4.2] Ð1
1
1
=
Ê+ÊR Ê+ÊR R ca ab
bc [4.3] and
Rcatotal
respectively.
Since all three measurements give the same result, Rabtotal = Rbctotal = Rcatotal .
From the similar equations (4.1), (4.2) and (4.3) it can easily be shown that
Rab = Rbc = Rca ≡ Rp [4.4] The total resistance for each pair of terminals, measured in terms of Rp, is
2
1 Ð1
1
= 3 Rp
R total = Ê+Ê2R R p p [4.5] and so, if Rtotal is to be 8Ω, then Rp = 12 Ω . Figure 4.1: Two black box circuits. a a (a) (b)
Ra Rab b Rca Rb
c Rbc b Rc
c Is the answer unique? This is the tough part of the problem (assuming you got past
the ÔguessingÕ part). Well, if the configuration above were the only solution the
answer would certainly be yes. However, there is another possibility, which is
shown in Fig. 4.1(b). The resistors are connected together and to each terminal is
associated a single resistor named R a , Rb , and Rc . In this circuit, a measurement
across terminals ab will give the resistance of the series circuit formed by Ra and Rb.
As above, it can be shown that equal measurements across ab, bc, and ca must
require Ra = Rb = Rc = Rs and for Rtotal = 8 Ω, Rs = 4 Ω.
Wesley, that impetuous fellow, immediately thought that the answer was unique.
Too bad. Shows that one should put oneÕs mind into gear before thinking. Would
you have been able to confidently answer this question?
5. Nonlinear electronics a)
For a linear device, R = V / I. But this is not a linear device, so this
relationship is not useful. A more general definition is R effective = dV/dI, the local
rate of change of voltage with current Ñ the slope of the tangent. When the device
is linear, V vs. I is a straight line through the origin, so this definition comes back to
R=V/I then.
Even if you have no calculus experience, you can find an intuitive solution for
VÊ=Ê10 V and V = 4 V. At both these points, an increase or decrease in voltage
corresponds to no change in current, i.e., Reff is huge, approaches ° .
b)
We do not know the voltage drop across either component or the current.
Start by listing the unknowns.
VR :
V? :
iR :
i?:
Right away, we know Voltage accross resistor
Voltage across unknown device
Current through resistor
Current through unknown device the current flows through one then the other
i R = I?
V R + V? = 10 Volts
V R = iR ¥ R the resistor is a linear device
I? = F (V?) where F() is the function shown in fig. 2, PS#1. So we have 4 equations with 4 unknowns. Thus we can find a solution(s). Start with V R = iR R
but VR = 10 Ð V?
∴ 10 Ð V ? = iR R
but iR = I?
∴ 10 Ð V ? = I? R ⇒ i? = Ð V?/R + 10/R Now we have 2 equations with 2 unknowns, but we must solve it graphically.
The intersections between the two lines are possible operating points. Current (mA)
25 possible operating points Voltage (V)
10
The actual voltage/current across/through the unknown device depends on how
MarieEve started the circuit (as shall be discussed in the section below).
c) At V = 0 V Current (mA)
25 Voltage (V
10
i?=v?/R+0/R
This line moves as MarieEve increases the voltage. Since we assume there is no
noise in the system, the operating point moves up the curve: Current (mA)
25 Voltage (V
10
At this point (Vvar. P.s. = 13 V), the operating point jumps to Current (mA)
25 new operating
point 10
V var p.s. continues up to 15 V with I? = 0A
As MarieEve decreases from 15 V: Current (mA)
25 Voltage (V
10 At this point (Vvar p.s. = 9V), the operating point jumps to: Current (mA)
25 10 Voltage (V LetÕs put this all together. i? (mA) 25 15 Vvar power supply (V) This is an excellent case of hysteresis!
6. TP Pull Comment: see the POPTOR web pages, under Problem Set Extras for a photograph
of such a rolldispenser. Also a frictionlock, using stacks of many plates to increase
the drag torque, in the quickrelease adjustments of an adjustable officechair.
a)
Knowing the normal force on each of two faces of the roll, we can find the
total frictional (drag) force:
FnÊ=Ê13N Ê
ÊÊ⇒ÊÊFDÊ=ÊµsÊFNÊ=Ê0.27Ê×Ê13NÊ
µsÊ=Ê0.27 = 3.51N @ each end The question is properly a matter of torques, since it is rotational forces
Torque from drag: = 2 ⋅ FD ⋅ ro = 2.351 N ⋅ 2 cm
= 14.0 N⋅cm
(0.14 N⋅m ) TD Torque from pulling: Tpull = Fpull ⋅ b where Fpull ² 3.5 N. Tpull = Tdrag while
pulling, and Tdrag is constant in the static case (i.e. testing from rest at different ÔbÕ)
So, 14.0 N ⋅ cm = Fp u l l ⋅ b unless it breaks b;
maximized; the last b for not breaking is
b= b is minimized where Fp u l l is 14.0N ⋅cm
= 4 cm
3.5N 60
Tdragroll Torque (N•cm) 50 Tpull 40
30
20
10
0 0 2 4 6 8 10 12 14 16 b Figure 6.1: Comparing the constant dragtorque from the friction plates on the
cardboard roll (red dashed line) to the maximumdeliverable pullingtorque, as the
roll size increases (black solid line). The paper is pulled only with as much force as
is necessary to overcome the dragtorque, but as the roll becomes smaller, even the
maximum tension on the paper (burststrength) is not enough, and the paper breaks
at b = 4 cm, as the lines cross.
b) T D = 2 µ sF N ⋅ ro = 2 (0.18) 13 (2) = 9.4 N ⋅ cm (0.094 N ⋅ m) T pull = Fpull ⋅ b
b= 9.4N•cm
= 2.7 cm
3.5N TD = Tpull ⇒ 2.7 cm, i.e. 0.7 cm left on roll.
c)
Approximate, rather than finding the spiral length from an integral: consider
separate layers, starting from the core (not a perfect spiral anyway ...)
length layer = 2πr 2 cm ²r ² 5.75 cm (d/2)
increment in r is ∆r = 1mm/(10 layers) = 0.1 mm layer1
= 0.01 layer1
Thus we have (4 cm Ð 2 cm) * 100 layers cm1 = 200 layers
200 ∑
n=0 2π(ro + n ∆r) j
1
where ∆r = 0.01 cm. Using ∑ n = 2 j(j + 1) to simplify, we get
n=1
200 200 = ∑ 2π o + 2 π ∆ r ∑ n
n=0 n=0 = 200(2πro) + 2π ∆r (200(201)/2)
= 200 2π 2 cm + 2π 0.01 (100) (201) cm = 3776 cm
= 37.76 m
Likewise for b = 2.7 cm (2.7  2) * 100 = 70 layers
length = 70 (2πro) + 2π ∆r (70(71)/2)
= 70 (2π) 2 cm + 2π 0.01 ¥ 35¥71 cm = 1036 cm
= 10.36 m Compare this with the answer you can get by dividing the area of the side of the roll
by the thickness of a single layer Ñ giving a length. A = (¹b2 Ð ¹ro 2 ), ∆ r = 0.01 cm;
for b = 4 cm, A = 37.70 cm2, so A/∆r = 37.70 m; for b = 2.7 cm, A = 10.34 cm2, so A/∆r
= 10.34 m. This is probably just as good as the approximation above (thanks, Gordon
Cook!)
d)
In this case the frictional force is the same (area increases but normal pressure
decreases) BUT the torque changes. The torque changes because the frictional force
is applied at a changing effective radius. To best do the problem requires adding up all the torques contributed by all the
similar regions (r = const).
Plan ÔAÕ: If you can integrate, it is not hard:
Consider a roll of diameter b, and core size ro . With a normal force FN evenly
distributed over this area, the force per unit area is PN = FN/(πb2  πro2). Then for an
annulus (ring) very thin of width dr set at radius r, the force on the annulus is
dFdrag = µsPNdA = µsPN2πrdr
then the torquecontribution due to this annulus is
dTdrag = r⋅dFdrag = µs PN 2πr2 dr
The whole dragtorque on each side is then the sum of torques over all annuli
1
2 Tdrag = T(b) ∫ b
d Tdrag = T(ro) ∫ µs PN ro 2π2 d r
b = µs PN 2π ∫ r2 dr = µs PN
ro b
1
2π 3Êr3 r o
2π
3
= µs PN 3 (b3  ro)
So, since Fn = Pn π(b2 Ð r o2 ) = Pn π(b Ð ro )(b + ro ), then we can write πPn(b Ð ro)
Fn
= (bÊ+Êr ) and substitute for PN
o
3
(b3ÊÊro)
4
Tdrag = 3 µs FN 2 3
b ÊÊr 4
= 3 µs FN (b2Ê+Êbr o
2
oÊ+Êro) (bÊ+Êro) now as you can see, the drag torque is not constant but scales with the outer
diameter of the roll, b. To find out if the paper breaks Ñ ever, never, or always Ñ
you have to compare this drag torque to the pulling torque you can produce from the maximum (breaking) tension of the paper applied as a force at the radius b of the
roll. 80
Tpull
Tdragarea
Tdragareakinetic 70 Torque (N•cm) 60
50
40
30
20
10
0 0 2 4 6 8 10 12 14 16 b Figure 6.2: Comparing the c hanging dragtorque from the friction plates on the
decreasing roll (dashed lines: redÐµstatic , greenÐµ kinetic ) to the maximumdeliverable pullingtorque (solid black line), as the roll size increases. In the static
case, the maximum pulling torque is never enough Ñ the paper always breaks;
with a rolling start (the kinetic case), there can be barely a solution for a little while
until the paper roll becomes too small.
Plan ÔBÕ: If you do not yet integrate, you can also do it by series summation:
∆Fdrag (r) = µs PN ∆(Area) = µsPN2πr∆r
the torque about the cylinder axis from just this ring is then
∆Tdrag (r) = r ⋅ ∆ Fdrag = µsPN 2πr2∆r
we can take each ring of width ∆ r to be an additional layer, then ∆ r = 0.01 cm as in
part ( c ) above. Then for a diameter b there are (b  ro)/∆r layers, each contributing a
torque from the drag at its particular radius; adding all the torques up on o ne side
gives us half the total drag torque, so 1
2ÊTdragÊ
2∆r) + ... = ∆Tdrag (ro) + ∆Tdrag (ro + ∆r) + ∆Tdrag (ro +
(bro)/∆r = ∑ ∆Tdrag (ro + n∆r) , n=1 (bro)/∆r = ∑ µs PN 2π(ro + n∆r)2 ∆r, n=1 = µs PN 2π∆r (bro)/∆r ∑ (ro + n∆r)2 , n=1 (br )/∆r o
= µs PN 2π∆r∑ n=1 2
(ro + 2ro ∆r •n + (∆r)2 n2) , (bro)/∆r = µs PN 2π∆r ∑ 2
ro + ∑ 2ro ∆ r •n + ∑ (∆ r)2 n2 n=1 = µs PN 2π∆r { 2
ro ∑ 1 + 2ro ∆r ∑ n + (∆r)2 ∑ n2Ê } Now
N ∑ Ê1 = N
n=1
N ∑
n=1
N ∑
n=1 n= N(NÊ+Ê1)
2 n2 = N(NÊ+Ê1)(2NÊ+Ê1)
6 where N is number of layers = 100 (b  ro), n is layer indexnumber. So
N(NÊ+Ê1)
N(NÊ+Ê1)Ê(2NÊ+Ê1)
Tdrag = 4πµs PN∆r {ro2 N + 2 r o ∆ r
+ (∆r)2
}
2
6
where N = (b  ro)/∆r
∆r = 0.01 cm
b = diameter of roll (cm)
ro diameter of core (cm)
If now you let the layers become very thin, so that ∆r → 0 and N → °, this becomes
TdragÊ = 4πµ P r2 (N∆r) + r (N∆r) 2 + 1 •2(N∆r)3
sNo
o
6
with N∆r = (b Ð ro ), 1
= 4πµs P n r2 (b Ð ro) + ro(b Ð ro)2 + 3 (b Ð ro)3
o
4
= 3 πµs PN(b Ð ro)3 + 3ro(b Ð ro) 2 + 3r2 (b Ð ro)
o
4
2
= 3 πµs Pn (b Ð ro) (b Ð ro)2 + 3ro(b Ð ro) + 3r o
Fn
Then with πPn(b Ð ro) = (bÊ+Êr )
o as above, (bÊÐÊro)2Ê+Ê3ro(bÊÐÊro)Ê+Ê3r2
4
o
TdragÊ=Ê3ÊµsÊFnÊ
(bÊ+Êro)
(b2Ê+ÊbroÊ+Êr2 )
4
o
=Ê3ÊµsÊFnÊ (bÊ+Êr )
o
The same answer as the integral, because this is in fact doing the integral from first
principles. 19961997 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 2: Mechanics
Due December 13, 1996
1) Pinball pastimes
A pinball machine is one of the ancient ÔarcadeÕ games (see figure). As a good pinballer,
you do not need to pull the spring knob too hard. In this machine, the table is tilted at 15¡,
and the ball has a
0.60 m
diameter of 3.0 cm and a
mass of 100 g. The
pinballplunger spring
constant is 200 N¥mÐ1 .
C
The ball is assumed not to
F
B
D
roll and the contact
A
E
plunger
between the ball with the
1.40 m
20 cm
table is frictionless.
ball
Assume
that
all
thicknesses
not
15 deg
specifically indicated are
knob
side view
negligibly small.
3.0 cm
a) What is the minimum
d is ta n c e fro m the
top view
equilibrium point of the
spring that one has to pull the knob so that the ball could reach the top of the semicircle?
b) After it reaches the top, where will the ball go: bumper A, B, C, D, or E? Explain.
c) Pinball semiwizard Elton does not pull the spring hard enough, and the ball does not
even get into the semicircle (it barely touches F) after he releases it. How far did he pull
the knob? [Chairul]
2) Net result
Martina is 6 feet tall and is a very good tennis player. Amanda is also a good tennis player,
at 5Õ 3Ó tall. In tennis, a good serve can make the difference between a mediocre player and
a very good one. For a serve to be a legal one, it must bounce in the service area after
clearing the net (see figure). We model the service stroke of both players as a circular arc
with the arm and racquet forming a straight line from the shoulder to the tip of the racquet.
1 Assume that they each can generate the same racquethead speed, and further that the
angular velocity remains constant throughout the stroke. Take it that the ball speed after
leaving the racquet is 1.8× the racquethead speed.
Here are Martina and AmandaÕs vital statistics:
Statistic Martina Amanda Height 1.82m 1.60m Height from ground
to shoulder 1.57m 1.35m Arm length 0.71m 0.56m length of racquet from handle to middle of racquet head: 0.46m. Height of the net 0.91 m.
a) Serves must land in the service area, which
includes the lines. The fastest way to get the
ball past your opponent is to maximize the itÕs
velocity component parallel to the ground, vp.
At what point in the stroke should each player
hit the ball to maximize vp? Now assume both
players hit the ball at this point. If they want
to hit a power serve touching the centre
service line, what is the maximum v p w ith
which each player can hit a legal serve? Point of service 23.7 m Service Area 6.40 m 8.23 m b) Contacting the ball at the top of their
stroke, what is the minimum v p with which
each player can hit the ball and still clear the
net? (Ignore air resistance.)
c) What is the maximum speed with which
each player can hit the ball at the top of their
stroke and still have it land in the service area?
(Ignore air resistance.) d) In a match, Amanda tried a trick serve in
which she struck the ball early Ñ she hit the
ball with her arm at an angle of 50¡ to the
ground. What would the angular velocity of
her arm need to be in order that the ball would
make it over the net into the service area? How high in the air would the ball go? (Ignore
air resistance.) 2 e) If Amanda wants to increase the speed v p with which she can serve the ball into the
service area, explain qualitatively how she could apply spin to the ball and use it to her
advantage. [Nipun]
3) Mind for rocks, or rocks for brains?
Some of us students here at the University of Toronto have taken up gym climbing (indoor
rock climbing). For safety reasons, a rope is attached to the harness of the climber. It runs
through a metal loop bolted to the top of the wall, and down to a device
attached to the harness of a partner (or ÔbelayerÕ) standing on the floor. A good
belayer never allows any slack on the rope. For the sake of this problem
assume that climbing ropes donÕt stretch.
a) Assume no friction in the system, what is the maximum mass of climber
with whom you would want to be partnered? Why?
b) When we started climbing, we were told that the general rule for relative
masses between partners was ÒMore than double, youÕre in troubleÓ. If this is
the case, what would be the coefficient of friction between the rope and the
metal loop if this is the only friction in the system?
c) There is also friction between the floor and the belayer. For simplicity
ignore the friction between the metal loop and the rope for this part of the
question. Does standing farther away from the wall allow you to support more mass than
in part (a)? Is there an optimum angle? Assume the floor friction coefficient is 1.0. [James]
4) Dense and denser
Diane walked into her physics lab to find her demonstrator Erica standing in front of a
cylinder floating upright in a tank of a clear fluid. Diane asked her what she was doing.
ÒIÕm measuring the density of the fluid. I know that the 10 cm long
cylinder has a density of 0.63 g¥cmÐ3 Ó.
5 cm
a) Diane, trying to impress her demonstrator, remarked that since the
cylinder was half submerged, she knew what the fluid density was.
What was her answer?
b) ÒGoodÓ, responded Erica,Ó now try something more challenging.
Cause the cylinder to sink farther into the fluid without touching it.Ó
Diane looked around the lab and spotted a compression chamber. ÒI need a force, and
pressure is force per unit area, so IÕll let air pressure do my work for me.Ó She put the tank
with the cylinder in the chamber and increased the air pressure to 100 atm while keeping
the temperature constant. What happened to the floating cylinder? Assume that the fluid is
incompressible, and that air is an ideal gas (density= 1.29 kg¥mÐ3 at 1 atm). 3 c) What would Diane observe if she could increase the air pressure in the chamber to 1000
atm? (Keep the totally silly assumption that the gas remains ideal.) [James]
5) ÔI crush your moon Ñ crush crush!Õ
a) The moon takes about 29 1/2 days to orbit the
earth, as related to the earth (the synodic period) or
about 27 1/3 days as related to the stars (the sidereal
period). Knowing the constant of acceleration on the
surface of the earth, and the earthÕs radius of 6.38 ×
10 6 m, show how to estimate how far the moon is
from the earth.
b) Knowing the result of part (a), the neighbourhood
tyke, O. Bray, takes a ruler out at night and uses it to
figure out the approximate diameter of the moon.
Explain how he might best have made his
measurement. [Robin]
¥ For Hubble Space Telescope info & pictures, try http://www.stsci.edu/public.html
¥ For NASA info, pictures and teaching materials, try:
http://spacelink.msfc.nasa.gov:80/Instructional.Materials/Curriculum.Materials/Sciences/Astronomy/Our.Solar.System/ 6) WhatÕs the buzz?
P ART I: Sepsis is a housefly. Interestingly, when houseflies land on a ceiling, it goes
something like this: they fly up close to the surface while still upright, put up their forelegs
up above their head, and stick Ôem to the ceiling Ñ then they swing around that way, doing
a sort of flip, until they smack all their legs onto the ceiling and they are upside down. a little
like...
Take it that Sepsis weighs 0.3 gm, and that he is about 15 mm long from ÔtailÕ to compound
eyes. We need a simplified model of a fly, so take Sepsis to have legs that reach 1 cm from
his centreofmass, and take that his main legs let him bounce about 1.5 mm on landing.
Say that Sepsis is cruising along at a good 1 m sÐ1, when he decides to alight on the spot of
the last ceiling spaghettitest. He sticks his forelegs to the ceiling when his centreofmass
is 7 mm below the ceiling, and pivots rigidly around that stickingspot to a landing.
a) If you ignore any slowingdown by drag in the air, how quickly does he land Ñ how
much time between sticking his forelegs and touching down all his legs?
b) How fast is Sepsis going toward the ceiling when he finally ÔsticksÕ his landing?
c) If he has only 1.5 mm to bounce, then he has only 1.5 mm to stop Ñ what is the
minimum acceleration Sepsis suffers at the end?
4 d) Given his small weight, what is the minimum force to stop Sepsis at the ceiling? What
would be the force to similarly decelerate a 70 kg human in the same distance? Is that a
practical force Ñ would the ceiling hold up? Would the human?
PART II: Automobiles are made nowadays to be energyabsorbing, in the event of a crash.
It means that components of the body are designed to be anything but elastic, and to be
sacrificed while dissipating kinetic energy. Sometimes this is described loosely by the size
of the crumplezone ahead or behind the passengers of the car.
A Datsun I once met had a crumple zone of roughly 1.5 m, and about 1 m of it had been
used up when it hit a concrete wall at about 30 km/hour.
e) What is the minimum acceleration needed to bring a car to rest in this distance? If the
car weighs 1,000 kg, what is the minimum force required?
f) At about 50 km/hour, your lap and shoulder belts stretch considerably in a hard
collision. Take a uniform deceleration, thanks to a 1.5m crumple zone. If the belts stretch
like a spring so that a passengerÕs centre of mass moves an additional 25 cm, then what is
the overall peak force on the passenger? With a belt 5 cm wide, and contact over about 35
cm of length, what is the average pressure of the belt on the skin? [Robin] 5 19961997 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 2: Mechanics 1. Pinball pastimes We can consider the ball as a point concentrated in its centre of mass. This assumption is valid
since the ball is not rolling and the contact between the ball and the table is frictionless.
a) When the ball reaches the top of the semi–circle, it already has speed v; thus, it also
experiences a centripetal force:
mv2/(0.285 m) = mg sin 15° [1.1] v = √ ° = 0.850 ms–1 (0.285)(9.79) sin 15 [1.2] or Let x be the distance between the ball with the equilibrium point of the spring. Using the
conservation of energy (with the equilibrium of the spring as the reference point, x = 0):
1
2 1
kx2 – mgx sin 15° = 2 mv2 + mg(1.170) sin 15° [1.3] or
100x2 – 0.253x – 0.333 = 0 [1.4] Solving this equation, one obtains x = 0.0590 m or 5.90 cm. (The other root is not used for
obvious reason).
b) After it reaches the top, the ball continues to pass A since it has momentum. Due to the inertia,
the ball tends to go straight but the curve bends the path. One can also get the same answer using
the symmetry argument.
c) Since the ball barely touches F, its speed v = 0 and its position from the equilibrium is s =
0.870 m.
1
2 kx2 – mgx sin 15° = mg(0.870) sin 15° [1.5] or
100x2 – 0.253x – 0.220 = 0
Solving this equation, one obtains x = 0.0482 m or 4.82 cm. [1.6] 2. Net Result To maximize vp, want to maximize the velocity component of the ball parallel to the ground.
Therefore, one should hit the ball at the top of the stroke, i.e., when the racquet head is
perpendicular to the ground.
a) The "power serve" hitting the centre service line with the maximum vp (and still being a legal
serve) would land at the back of the service area.
Horizontal distance travelled by ball:
ball’s trajectory ⇒ x = 18.7 m x2 = a2 + b2 = (4.12)2 + (18.25)2 40 vertical distance travelled by ball:
Martina: 1.57 m + 0.71 m + 0.46 m = 2.74 m
Amanda: 1.35 m + 0.56 m + 0.46 m = 2.37 m
50 12
gt ⇒ t =
2
= 0.75 s
= 0.70 s time ball is in the air: d =
Martina: tM
Amanda: tA 2d
g ∴ maximum vp for:
Martina: v = x/ tM = 18.7m / 0.75s = 25m/s or 90 Km/h
Amanda: v = x/tA = 18.7m / 0.70s = 26.9m/s or 97 Km/h
We, of course, have to make sure that balls with these velocities will clear
the net! (see part b)
b) (Note that a more precise solution to this problem can be obtained by the methods of
calculus. We sketch here an approximate solution.)
Using the same trajectory for the ball as in part (a), we need to find the distance from the server to
the net.
By similar triangles:
4.12
x
18.25 = 6.40
2 net ⇒ x = 1.44m ∴ y2 = (6.40)2 + (1.44)2 y1 ⇒ y2 = 6.56m ∴ y1 = 18.7 m – 6.56 m = 12.14 m
To find the minimum vp for each player, time for
ball to fall from point of stroke to height of the
net: 4.12 m
y2
x
centre line
18.25 m tm′ = ta′ = √ √ 2(d m – h net )
=
g 2 (274 – 0.91)
= 0.61 s
g √ √ 2(d a – h net )
=
g 2 • (2.73 – 0.91)
= 0.55 s
g vmin:
Martina:
Amanda: 12.14m / 0.61 s = 19.9 m/s or 71.5 km/h
12.14m / 0.55 s = 22.1 m/s or 79.5 km/h So, the serves in part (A) are fast enough to clear the net.
c) maximum speed is for longestpossible horizontal trajectory, as
horizontal distance
vp = time for ball to fall to ground
so we want the ball to land in for rear corner of service area:
Horizontal distance travelled by ball = ((8.23)2+ (18.25)2)1/2 = 20.0m
Maximum speed:
Martina:
Amanda: 20.0m / 0.75s = 26.7 m/s, or 96.0 km/h
20.0m / 0.70s = 28.6 m/s, or 103 km/h d) We need to know:
1) how high above the ground the ball is when Amanda hits it
2) how much vertical distance between the release point of the ball and the top of the net — gives
time available to clear net
3) the horizontal velocity component of the ball needed to reach the net at the closest distance to the
point of service such that the ball will land in the service area. (See part b)
1) ball height at release point: 1.35 + (0.56 + 0.46) sin 50° = 2.13 m
2) d = (2.13 m – hnet) = 1.22 m
∴ Time available to clear net = (2d/g)1/2 = 0.50 s
3) From part (b): distance = 12.14 m
vhorizontal = 12.14m / 0.50s = 24.3 m/s
vtotal (tangential to the arc of the racquet): vhorizontal = vtotal cos 40°
vtotal = 31.7 m
Racquet head speed × 1.8 = ball speed
vracquet = 17.6 m/s angular velocity of arm = ω = v / r = 17.6 / (0.56+0.46) = 17.3 rad/s
vertical velocity of ball: vy = vtotal sin 40° = 20.37 m/s
So, the maximum height above the ball’s release point is
2gy = vy2 ⇒ y = 21.1 m
Max. height of ball = 1.22 m + 21.1 m = 22.4 m 3. Mind for rocks, or rocks for brains? The system has been simplified since ropes don't stretch and so no slack is allowed on the line.
This of course is not physically reasonable (climber would almost always be supported by belayer
and so could never actually ‘climb’!) but is a good first approximation.
a) Since belayer does not want to accelerate off the ground:
Σ Fb = 0 Tc Since there is no friction Tc = Tb mcg Tb Let 'c' indicate climber
Let 'b' indicate belayer
Let R be force by floor on belayer ∴ Τb + R = mbg
Tc + R = mbg R Climber is also not accelerating.
∴ Τc = mbg mbg Maximum mc for a given mb occurs for R = 0, i.e., belayer barely touching floor.
∴ as expected, max. mass of climber is your mass, as belayer
(Of course in real life, you would not want to be right on the edge of stability, but within the
assumptions given the correct answer is ‘maximum mass = your mass’, not ‘< your mass’ which
some people stated. This is a minor point.)
b) Same pix as before except for friction on metal loop. Friction force opposes motion.
Again we are trying to find a max. mc for a given mb. Start by setting R = 0.
Note that forces on rope must balance
∴ mcg = f + mbg
(direction of f choosen since we are opposing motion of climber dropping, not belayer dropping). f Frictional force is standard static friction, i.e.
f ≤ fmax
≤ µs (mcg + mbg)
For maximum mc, set f = µs (mcg + mbg)
∴ mcg = µs (mcg + mbg) + mbg
mcg (1 – µs) = mbg (1 + µs)
∴ mc = mb (1+ µs) / (1− µs) Tc
mcg but question states that mc = 2mb for maximum mc
∴ 2 = (1+ µs) / (1− µs)
⇒ µs = 1/3 <— cofficient of friction between loop and rope. Tb c) Since there is no friction in the rope, Tc = Tb. Since climber does not fall Tc =
mcg, ∴ the forces on the belayer look like:
f ≤
≤ µs R
1.0 R For maximum mc, f = 1.0 R
Since there is no acceleration, Σ F = 0. Standard way would be to
break this up into horizontal and vertical forces but more elegant way is
to consider vector sum of forces: (since Σ F = 0, can draw triangle) but we know: mcg
mbgR θ
f=R 4. mbg θ
mcg (mbg – R)2 + (1.0 R)2 = (mcg)2 And write: R R
f
m bg (a + b)2 ≥ a2 + b2
∴ (mbg – R)2 + R2 ≤ (mbg – R + R)2 ≤ (mbg)2
∴ (mbg)2 ≥ (mcg)2
∴ max. mc is mb
∴ friction of floor does not help! Dense and denser a) Effective buoyancy force = g (mdisplaced air + mdisplaced fluid), but the masss of the displaced air
is very small, so can be ignored.
Net force = 0, ∴ g Vc /2ρfluid = g Vc ρc
∴ ρfluid = 2 ρc
= 1.26 g/cm3 b)
This problem explores what happens when you take a model and go to a silly extreme.
You are required to make one simplification. The question states only the pressure for one point in
the chamber. Of course, as the gas becomes more dense the pressure increase due to gravitational
forces (ρgh) becomes more important. It is this very pressure gradient that causes the cylinder to
rise. For this part of the question, since the pressure is increased only a small amount one can
assume that the gas it displaces would have a minor pressure gradient across the length of the
cylinder. Thus one can assume a constant density and that Diane’s measurement was taken at a
height similar to the height of the cylinder.
Let x be the distance that the cylinder is submerged.
Net force = 0
∴ g Vc ρc = g (x Vc ρfluid +(1 – x)Vc ρair)
You cannot ignore the displaced air.
ρc = (l – x) ρair + x ρfluid
ρair is calculated from PV=nRT .
∴ x = 4.4 cm
Cylinder moves up 6 mm.
Note: cylinder floats higher in fluid. Diane should have reduced air pressure, not increased it.
c) Cylinder + fluid 'float' to top of chamber. 5. ‘I crush your moon – crush crush!’ a) Knowing the period of the moon’s orbit, we can figure out its distance from the earth, because
the period depends on the orbit radius r and orbital speed v:
Torbit = distance 2 π r M , m
=
speed
v rM, m = earthmoon separation, v = speed of moon [5.1] The orbital speed and radius are also related by the centripetalforce formula:
FMm m v2
=
rM ,m m = mass of moon [5.2] Where the centripetal force is gravitational attraction:
FMm = GMm
2
rM ,m M = mass of earth, m = mass of moon, G = gravitational const. [5.3] Equations [5.2] and [5.3] both give the force, so with righthandsides equal, we can eliminate the
moon mass m, to get:
2 2π rM ,m GM
= v2 = , where the last equality uses [5.1].
rM ,m Torbit Then we have
rM ,m 2 GM Torbit = 4π2 1/ 3 For G = 6.67 × 10–11 N m2 kg–2, M = 5.98 × 1024 kg, Torbital = 27.333 days * ( 8.64 × 104
seconds/day) = 2.36 × 106 s, this gives rM,m = 3.83 × 108 m (cf. 3.844 × 108 m).
b) Ah, now that we have the distance rM,m we can figure out the diameter of the moon by
constructing similar triangles:
r earthmoon
rmeasure disk, or ruler
moon
By holding a small paper disk, and moving it toward and away from the eye until it exactly covers
the moon, then recording both the disk size ddisk and the distance held away from the eye rmeasure,
the earthmoon distance from (a) can be used to figure out the moon’s size:
dmoon
d
= disk
r M , m rmeasure
dmoon = ddisk or finally rM ,m
rmeasure When I did this, I found roughly ddisk = 5 mm at rmeasure = 50 cm, and so dmoon = (0.5/50)3.83 ×
108 m = 3.83 × 106 m. Compare this with a referencetable value of 3.476 × 106 m.
Small Shortcut: Instead of using the values of G and of M, you might note that at the earth’s
surface (rearth,m = rearth) we must have FMm = m g, where g = 9.8 kg m–1 s–2, so: FMm 2
GMm
rearth
2
=2
= m g 2 , i .e. , GM = g rearth , constants easier to remember
rM ,m
rM ,m [5.4] Synodic vs. sideral: How should you measure a complete rotation of something in space?
Suppose the earth did not rotate at all. If it made one orbit of the sun, every spot on earth would
have daytime and nighttime, both, so the earth would see one revolution or one day in terms of the
sun, though in terms of the ‘fixed’ stars it had not rotated (no sidereal days at all). So the synodic
and sidereal periods for the earth are not the same. The same holds for the moon and its orbit of
the earth.
As the moon orbits the earth, the earth orbits the sun. If the moon ‘startfinish line’ for an orbit is
an imaginary line between the earth and some very distant star, then it will orbit and come back to
this line again in a sideral period. If the startfinish line is a line between the earth and the sun,
then while the moon is orbiting, this line will change along with the earth’s orbiting, so the
finishingline also will move through some angle in space. With a retreating finishline, a complete
revolution in these terms will take longer.
(Of course, during the time it takes to catch up, the line will have moved a little bit again, making a
bit more to catch up, and so on, and so on. The problem ends up like figuring out from 8:00 pm
exactly when the minute hand ends up opposite the hour hand: in 10 minutes it is opposite where it
used to be, but the hour hand has moved an angle 1/12th as much as the minute hand just moved.
So the minute hand reaches the corrected place 1/12 of 10 minutes later, but then the hour hand in
that time moved 1/12th of 1/12th of 10 minutes... The answer can be found using the sum of the
series (1/12)n, which has a wellknown formula.)
6. What’s the buzz?
a) Need some approximation
0.7cm
angle θ : 1.0cm = s in θ = 0.7
⇒θ = 0.775 radians = 44.4 ° Momentum: 0.3 g = 3 × 10–4 kg; mv = 2 × 10–4 kg · 1 m s–1 = 3 × 10–4 kg · m s–1 . The
momentum resolves along the rigid rod and along the ⊥ to the rigid rod.
p⊥ 0.7cm
p = 1cm = p⊥ = 0.7 p
Angular momentum is conserved; with the freepivot action, it sees no torque and so is conserved
without contribution from the ceiling, earth, etc..
L = P⊥ • r = 0.7 p •10–2 m ,
= 2.1 × 10–6 kg m2 s–1 It’s undefined what the final angle is when he reaches the other side — he could swing around
until he is parallel to the ceiling, or maybe until his 1 cmlong legs, halfway down his body, hit the
roof. Either is good for the sort of rough estimates we are making on behalf of flies! You figure.
Say, take 1st case: the final angle is π radians. Then he swings through (π – 0.775) radians =
2.37 radians or 135.6°
Arc length: ∆θ · r = 2.37 rads · 1 cm = 2.37 cm = 2.37 × 10–2 m
Speed: v⊥ = 0.7 · v = 0.7 × 1 m s–1 = 0.7 m s–1
d 2 .37 × 1 0 –2 m
t=v=
= 3.39 × 10–2 s = 33.9 ms
–1
0.7 ms
we’ll accept 15 – 35 ms, about one frame of a video camera.
b)
Again depends on final angle, leg length, etc., but approx v⊥ = 0.7 m s–1; we’ll
accept answers 0.35 – 0.7 depending on final angle
u2
v = 0 ⇒ –2d = a
c) v2 = u2 + 2ad
a=– d) 0.49
(0.7)2
=
= 0.16 × 103 = 1.63 × 102 = 163 m s–2
– 3 3 × 10– 3
21.5 × 1 0 F = Mfly • a
= 0.3 × 10–3 kg (–163 m s–2)
= –49 × 10–3 kg m s–2
= –4.9 × 10–2 kg m s–2
= –4.9 × 10–2 N F = Mhuman • a
= 70 kg (–163 m s–2)
= –1.14 × 104 kg m s–2
= –1.14 × 104 N PART II
30 km/hr = 30 × 103 m / 3600 s = 8.3 ms–1
e) v2 = u2 + 2ad
v=0⇒
(8.3ms–1)2
–2
a=–
21.5m = –23 m s , much less than for the fly
F = m • a = (1,000 kg) (–23 m s–2)
= –2.3 × 104 kg m s–2 f) not very significant that belts stretch a little, if assume constant acceleration force (unless you let
the passenger bounce) total distance is 1.75 m
∆x = 1.75 m
50 km/h = 50 × 103 m / 3600 s = 13.9 m s–1
u2
( 13.9 m s –1 )2
a = – 2d = – 2(1.75 m) = – 55.2 m s–2
F = m • a = 80 kg • (–55.2 m s–2)
= – 4.416 kN
Then 0.35 m • 0.05 m = 0.0175 m2 area for belt. Pressure is
(4.416 kN / 0.0175 m2) = 2.52 × 105 Pa
[1 ATM = 1.01325 × 105 Pa]
– about 2.5× atmospheric pressure; can bruise heavily where the belt cuts in
– or, say my fist is 7 cm × 9 cm = 63 cm2 = 0.0063 m2, so to produce the same
pressure I would need a force of about 1.6 kN, the weight of about 162 kg —
much more than I weigh, and probably a really decent heavyweight prizefighter
punch if the followthrough is strong.
Or, to compare another way, the whole force on the person is F = 4.416 kN corresponds to the
weight of ~450 kg – almost half the weight of a small car, so this is a serious force. How long
does it last?
13.9 m s – 1 t = – = 0.126 s = 126 ms 2(–55.2) m s –2
Perhaps this is something like being directly hit by a motorcycle? You can hit the belt, or you can
hit the windshield, but much better the belt! Buckle up — it saves lives. 19961997 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 3: Thermodynamics
Due January 13, 1997
1) Great balls oÕ fire!
The sun may be modelled as a ball of hydrogen gas. Assume
its main energy source is the conversion of hydrogen (atomic
weight = 1.0078 amu) to helium (atomic weight = 4.0026
amu) roughly by the process 4H → He (the real process is
much more intricate than this). The energy emitted is given
by the famous Einstein equation, E = ∆m c 2 , where c is the
speed of light.
a) How much energy is given off during the life of the sun if
only 10% of its hydrogen is converted to helium?
b) The power radiated from the sun and arriving at the
earth has a density today equal to 1400 W¥mÐ2 . If we
assume that this rate is constant throughout the sunÕs life,
estimate the life expectancy of the sun. Is this value
plausible? [Nipun] EIT image in the He II emission line
at 304•, upper chromosphere/lower
transition region, ~60,000 K.
http://sohowww.nascom.nasa.gov
/gallery/EIT/index.html 2) Running Amok
The Amok IV space probeÕs initial design included no insulation or shielding. The
electronics require a minimum operating temperature of 200K. Since it is a deepspace
probe the engineers had to provide it with a heat source.
a) What would be the minimum power requirements for the heat source for a black
spherical satellite of radius 2 m? Assume that outer space is really,really cold.
b) To reduce heating bills, the engineers decided to put a nice flat black outer layer around
the satellite and stuff it with Fibreglass Fuchsia insulation. This insulation has an R value in
vacuum of 18.8 ft2¥F¡¥h¥BTUÐ1. (Believe it or not, this is the unit often used for insulation,
even in our SIenlightened society.) What is the new power requirement to keep the
electronics working properly? The radius of the shield is only slightly larger than the
radius of the probe. (1 BTU=1.054 kJ, 1ft = 30.48 cm, 1F¡ = 5/9 K)
c) Unfortunately, Alain the coop student forgot to pick up the insulation before the
launch, and somehow also the outer layer never got painted. Assuming that the emissivity
1 of the silver shield was 0.2, what was the steadystate temperature of the interior of the
satellite? [James]
3) Too blue to be true?
Light from a star has the broad spectrum pictured below, and it fits a blackbody
distribution pretty well over much of the frequency range recorded.
6
fitted blackbody spectrum
star's emission spectrum Relative Brightness 5
4
3
2
1
0 0 300 600 900 Wavelength (nm) 1200 a) From the distribution, what is the
temperature of this star?
b) Travelling in a spacecraft toward
this star at 0.1 c, all the frequencies of
the light from the star will be Doppler
shifted, and so the peak of the
spectrum shifts.
By WienÕs
displacement law, λmax [nm] = 2.8978 ×
106 / T [K], what is the new recorded
temperature of the star? Present your
arguments about whether or not this is
really a new temperature for the star,
as measured in the frame of reference
of the traveller. [Robin] 4) Huddle up!
Edna Hillary and her team have set up a research station in the antarctic. Their base is a
structure of dimensions, 20×20× 3 metres. While they have brought along with them many
of the conveniences of home, they have overlooked one very important detail Ñ they
forgot to insulate the walls of their cabin from the cold! As soon as they set up, they
powered up their generator to heat the room to a cozy 20¡ C and then shut the generator
off, only to find that the temperature then dropped at a rate of 1¡ C every 5 minutes!
a) Assume that the research team is wellinsulated (they're all wearing parkas), so that the
heat capacity inside the cabin is due only to the air. Furthermore, treat the air as a diatomic
ideal gas. If the air pressure is 1 atmosphere, how much power would the generator have
to emit to stabilize the temperature?
b) Bicycle pumps get hot by adiabatic compression of air: if they wanted to try and
stabilize the temperature by compressing the volume of the cabin, at what rate would they
have to ÔshrinkÕ the cabin. Express your answer in m3¥sÐ1. [Nipun]
5) They chilled out and got it together
The first realization of BoseEinstein condensation was achieved last year by researchers in
Colorado. By cooling rubidium atoms to extremely low temperatures, they underwent a
2 phase change resulting in the group of atoms condensing into one single state Ñ a kind of
ÔsuperatomÕ. Much work has been done in the quest to create this new form of matter since
it was originally proposed by Albert Einstein and Satyendra Nath Bose. A rough
understanding of this new state can be understood in the following way: de Broglie
proposed that all matter can be viewed as waves, with wavelength: λ = h / m v . BoseEinstein condensation occurs when the de Broglie wavelength exceeds the interatomic
spacing. Effectively, the atoms overlap in space.
a) To begin, estimate the wavelength for a car speeding down a highway. What about a
nitrogen molecule at room temperature and pressure (vavg = 400 m¥sÐ1)? While you are
sitting in your chair doing this problem set, is your wavelength infinitely long?
b) By recalling the equipartition theory (each degree of freedom contributes on the average
(1/2)kT of energy), estimate the temperature one must cool a group of rubidium atoms
(ideal gas) down to achieve BoseEinstein condensation. Assume that you start at room
temperature with a pressure of 1 atm and that you maintain constant volume. You should
realize that the equipartition theory does not apply well to these low energy atoms, but it is
adequate to get a rough idea of the required temperature.
c) If you could reduce the atomic spacing, you wouldnÕt have to go to as low
temperatures. Why do you think the first experiments didnÕt use very high pressures?
d) Obviously this new state of matter is much more ordered than your original state. Have
you broken the second law of thermodynamics, which says that entropy (and randomness)
must always increase? [James]
6) The curious case of the cold calculating counterfeiter
Svend just arrived in Toronto. When he got a ÔtooneyÕ in his change from the newstand at
the Pearson airport, he got an idea, ÒIt is a real challenge for me, I will duplicate it.Ó Svend
was very good at counterfeiting coins. So he bought two plates of metal: brass and nickel
with the same thickness, 2.00 mm. Cooling them to liquid nitrogen temperature (77 K), he
drilled a hole in the nickel plate in and fitted it with a brass centreplug. The diameter of
the hole was 12.00 mm and the outer diameter of the nickel is 28.00 mm. He made sure
that the brass could barely fit into the nickel while cold.
a) What is the approximate pressure between the two metals at room temperature (293 K)?
Assume the hole expands in the same proportion as the rest of the plate.
b) Would police catch Svend if he used this ÔtooneyÕ based on its weight? Compare with
an actual one. (Measure one yourself!)
(Note: the thermal coefficients of linear expansion brass and nickel are 19 × 10Ð6 KÐ1 and 11
× 10Ð6 KÐ1; their densities at room temperature 8.87 g¥cmÐ3 and 8.85 g¥cmÐ3 ; their YoungÕs
moduli 9.1 × 1010 N ¥ mÐ2 and 19.0 × 1010 N ¥ mÐ2). [Chairul] 3 19961997 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 3: Thermodynamics
1. Great balls o’ fire! a) Total mass of the sun = 1.99 x 1030 kg
If sun is made of hydrogen, then amount of hydrogen converted to helium = 10% of (1.99 x 1030
kg) = 1.99 x 1029 kg
For reaction 4H → He, mass loss is:
∆m = 4 x mass of H – mass He = 4(1.0078 amu) – 4.0026 amu
= 0.0286 amu = 4.75 x 10–29 kg
Energy given off by 4H → He:
E = ∆mc2 = (4.75 x 10–29) x (3.00 x 108 m/s)2
= 4.27 x 10–12 J
Total # of hydrogen atoms in reaction:
199 × 10 29 kg
.
1.0078 amu / atom × 1.6606 × 10 − 27 kg / amu .
= 119 × 10 56 atoms 4 hydrogen per reaction = 2.97 x 1055 groups of 4 H
∴ total energy: (2.97 x 1055) x (4.27 x 1012 J per reaction) = 1.27 x 1044 J
b) mean radius of earth’s orbit = 1.49597 x 1011 m = re
∴ "surface area" of orbital sphere = 4Bre2= 2.8 x 1023 m2
∴ energy flux from sun: 1400 W/m2 (at earth’s distance)
∴ total power = 1400 x 2.81 x 1023 = 3.94 x 1026 W
1.27 × 10 44 J
= 3.23 × 1017 s = 1.02 × 10 10 years
∴ life of sun =
26
3.94 × 10 W
The life expectancy of the sun is ~10 billion years, so the calculated value is pretty good!
2. Running Amok
a) The probe will lose heat due to radiation. The net rate of heat loss depends on its temperature,
emissivity (e.g., black, white or grey), surface area, and the background temperature. 1 Power radiated from a surface follows Stefan’s Law:
P = σ A e T4 where σ = StefanBoltzmann constant
A = area
e = emissivity (between 0 (white) and 1 (black)
T = temperature (Kelvin) (we are neglecting background temperature since question states “outer space is really, really cold,”
i.e., To.s.= 0).
A = 4π r 2
= 50.3 m2
Since it is black, e = 1.
Minimum power radiated corresponds to minimum temperature, i.e. Tmin = 200K
∴ Pmin = 4.6 kW
b) We must find the temperature of the shield. It is radiating power as in part (a).
Energy is being transferred from probe to shield at a rate determined by R value, Tprobe, Tshield, area P1
TPROBE P2 At steady–state P1 = P2
P1 = σ Ashield εshield Τ4
(T
 Tshield )
∆T
= A probe
P2 = A
R
R
Thus σAshield e shield Ts4 =
Question states A(Tprobe − Tshield )
R Ashield = Aprobe
eshield = 1 ft 2 o F h
Thus,
R = 18.8
BTU
m2 ⋅ oC ⋅ s
= 3.31
J
We want minimum power requirement, thus Tprobe = 200K
Need to solve: Tshield = Tprobe – R σ Tshield4 — cannot isolate Tshield easily...
Many ways to solve this: 1) computer or fancy calculator (HP28s or better)
2) graphically
3) by iteration, substituting each successive approximation back in
I will demonstrate method (3): numbers [x] indicate the iteration being made
1) Start by assuming Tshield << Tprobe, thus take Tprobe = R σ Tshield4 and solve for Tshield:
1 /4 Tprobe Tshield = = 181 K Rσ 2 Plug this in and find new Tshield, then repeat as it converges and a selfconsistent value emerges:
2) 200 − 181 1/4
= 100K
Tshield new = Rσ
200 − 100 Tshield new = Rσ 1 /4 3) 200 − 152 Tshield new = Rσ 1 /4 4) Tshield 200 − 126 = Rσ 1 /4 new Tshield 200 − 141 = Rσ 1 /4 new 7) Tshield new 200 − 133
= Rσ 8) 200 − 137 Tshield new = Rσ 9) 200 − 135 Tshield new = Rσ 5)
6) = 152K
= 126K
= 141K
= 133K
1/ 4 = 137K 1/ 4 = 135K
1/ 4 = 136K 200 − 136 =
Tshield = 136K Rσ ∴ Tshield = 136 K
∴ P = σ A e T4
= σ (4π (2)2) (1) (136)4
= 0.97 kW
1 /4 new 10) This is larger than part (a) by more than a factor of 4.
c) All heat transfer here is radiative. Consider all possible radiated power processes: P3 P1
P4
P5
P2 P1:
P2:
P3:
P4:
P5: from outer shield to outer environment
from inner shield to inner environment
from inner environment to shield
from probe to inner environment
from inner environment to probe Right away we can write (‘p’ = ‘probe’, ‘s’ = ‘shield’):
P1 = σ Aes Ts4
P2 = σ Aes Ts4 = P1
P3 = e s P4
P4 = σ Ae p Tp 4 P5 = P2 + P4 − P3
(since we are at steady state and eprobe = 1) 3 For shell, at steady–state:
P 1 + P2 = P3
2σAeshieldTshield4 = eshieldσAeprobeTprobe4
∴ 2Tshield4 = Tprobe4
Power radiated out by cabin
= P4 – P5
= P 4 – P2 – P4 + P 3
= P 3 – P2
= eshield σATprobe4 – σAeshieldTshield4
= eshield σA(Tprobe4 – Tshield4)
= eshield σA(Tprobe4 – 1/2Tprobe4)
e σA 4
∴ power radiated by cabin: s
Tp
2
Power source is still the same
e shield
σ A Tprobe4
∴ 0.97 kW =
2
∴ Tprobe = 242 K
3. Too blue to be true? a) For a black–body distribution, Wien’s Law gives the peak of the distribution, according to:
λmax T = 2.9 x 10–3 m · K
In the figure the peak is around λmax = 460 nm, so
T= 2.9 × 10 −3 m ⋅ K
= 6,300 K
460 × 10 −9 m b) The relativistic Doppler–shift means a change in frequencies:
1
v
(1 − cos φ)
2
2
c
1− v / c
where N is the angle between the direction of the photon and the spacecraft motion in the star’s
frame. ν′ = ν Here N = 0 so ν ′
=ν
=ν 1− v / c 1−v / c
2 2 =v 1− v / c
(1 − v / c)(1 + v / c ) 1− v / c
1+ v / c with v = –0.1 c this gives ν ′ = 1.11 ν 4 c
ν
c
c
c
so λ’ = =
.
= 0.90 = 090λ
ν ′ 111
.
λ νλ=c ⇒ λ= This means a shift in the peak to shorter wavelengths, and so, nonimally, from Wien’s law, a higher
temperature T’ = 1.11 T = 6965 K
Is this a ‘real’ new temperature? Well, does the blackbody distribution still fit?
8π hν 3
uν = 3
c [exp( hν / kT − 1)]
hν ′ hν
=
if ν ′ = 1.11 ν and we take T’ = 1.11 then
, and (ν’)3 = 1.37 ν’, so a distribution of this
kT ′ kT
type still fits, with an overall factor of 1.37.
Since the overall amplitude of the distribution uν changes, it is no longer normalized properly.
However, this is not really an issue in a typical real measurement — it may simply seem like a larger
star radiating more power. If it were not a perfect blackbody but included some spectral lines, the
patterns of these lines would be shifted from their usual positions, helping you guess what had
happened.
In thermal terms, fitting a MaxwellBoltzmann distribution, adding a constant velocity to all particles
does not change its kinetic temperature. 4. Huddle Up! a) Temperature drop: 1°C / 5 mins Cv = 5/2 R for a diatomic ideal gas (J • K–1 • mole–1)
In finding total # of moles, let’s assume we deal with the gas only, and that the volume of the
researchers is negligible (probably not true)
1200m 3
= 12 × 10 6 l
.
∴ volume = 20 × 20 × 3 =
1 × 10 −3
(1l = 1000 cm3 = 1 x 10–3 m3)
at S.T.P., 1 mole of ideal gas occupies 22.4 l 293K
∴ at 20 °C: 22.4l = 24.04 l • mole–1 273K
Or use PV = NRT which is right for an ideal gas and often OK for a diatomic gas too. (Diatomic
nature of gas doesn’t change the equation of state, as diatomic molecules add to the molecules’
internal energy but don’t affect the entropy so the pressure isn’t affected. Therefore we have the
same equation of state.
1.2 × 10 6 l
= 4.99 × 104 moles
∴ Total # of moles:
24.04l / mole 5 5
5
R = k B = 20.7 J ⋅ mole −1 ⋅ K −1
2
2
1°C / 300 s = 3.33 x 10–3 °C • s–1
∴ Power = (3.33 x 10–3 °C • s–1 ) × (4.99 x 104 mole) × 20.7 J • mole • °C
= 3.46 x 103 J • s–1
= 3.46 x 103 W
For a diatomic ideal gas, CV = b) Adiabatic compression:
For an ideal gas pV = nRT,
∴ at constant temperature and particle number,
pV = constant
Instead, assume the gas is thermally insulated from its surroundings (adiabatic conditions)
Changing the volume of the gas will do work on the gas and the internal energy of gas will change,
so the temperature will change. We need the relationship between volume and temperature.
From 1st law of thermodynamics, heatenergy transfer is
∆ Q = ∆ E + p∆ V
(1)
In terms of heat capacity at constant volume, ∆E = n CV ∆T, where n = # of moles. For an adiabatic
process, there is no heat transfer, so ∆Q = 0, and thus from (1)
(2)
0 = n CV ∆T + p∆V
We can write the differential of pV:
∆(pV) = p(∆V) + (∆p)V
p∆V + V∆ p
so, ∆T =
nR
into (2): (CV + R) p∆V + CVV∆p = 0 (3) Dividing by CVpV:
( CV + R ) ∆ V ∆p
+
=0
CV
V
p
∆V ∆p
⇒γ
+
=0
V
p
where (4)
(5)
C
C +R
γ= V
=p
CV
CV 7
Cp 2 R 7
For diatomic ideal gas,
=
=
CV 5 R 5
2
Integrating equation (5), with ∆V as a differential dV:
γ ln(V) + ln(p) = constant
γ
⇒
pV = constant
γ–1
⇒
V T = constant
⇒
V2/5 T = constant (γ = 7/5 for diatomic ideal gas) 6 For a change in temperature of 3.33 x 10–3 °C from 20 °C
V12/5 T1 = V22/5 T2
∴ T1 V2 2/5
20 o C
V2 2 / 5
= ⇒
=
T2 V1 20 − 3.33 × 10 −3 (1200m 3 ) 2 / 5 Take the difference: V2 – 1200 m3
∴ Need to compress at a rate of 0.5 m3 • s–1 [0.034?]
5. They chilled out and got it together
a) DeBroglie wavelength, λ = h/p = h/mv
assume car v = 100 km • h–1 = 28 m • s–1
assume car mass = 1000 kg
∴ λ = 2.4 × 10–38 m
N2 mass is 14 g/mol, ∴ for one molecule,
m = 2.3 × 10–26 kg
v = 400 m • s–1
∴ λ = 7.1 × 10–11 m
My net velocity over time might be zero, but my instantaneous velocity is not
∴ my λ does not approach ∞. b) Since these are atoms, they have 3 degrees of freedom (corresponding to 3 directions of motion),
ignoring internal degrees of freedom for now.
3
∴ Ekin = kT
2
But this also equals:
1
E kin = mv 2
2
3
= kT
2
3kT
∴v =
m
Mass of a rubidium atom: 85.5 g • mole–1 ⇒ 1.42 × 10–25 kg atom–1
λ=
= h
h
m
=
mv m 3kT
h
3 kT m 7 What is the interatomic spacing as a function of T? Since question states that we start at room
temperature, and maintain a constant volume as we cool, spacing does not change.
From ideal gas law: PV = nRT
n
P = RT
V
n
P
N
P
=
⇒
=
V RT
V kT where now N = number of atoms and k = Boltzmann’s constant. Assuming atoms are evenly
distributed in space, x distance apart from each other:
density is 1 atom/x3
1
N
P
∴ 3=
=
x
V kT atoms x with P = 1 atm = 1.01 × 105 N • m–2
k = 1.38 × 10–23 J • K–1
T = 300 K
∴ x = 3.4 × 10–9 m, independent of T for this question For condensation, x = λ (spacing between particles is about one de Broglie wavelength)
h
∴ x=
3kTm
h2
T= 2
x 3k m
T = 6.5 mK
This is actually much warmer than what is really required. (Our problem is that the equipartition
theory breaks down for low temperatures.)
c) Relatively vague question, possible answers:
1)
increasing P often increases T1 which is undesirable
2)
methods to cool atoms lose the warmest atoms, thus decreasing pressure
d) Overall entropy of system does go up.
6. The curious case of the cold calculating counterfeiter
The assumption that the hole expands in the same proportion as the metal is valid. From the
microscopic point of view, when one considers that the plate is composed of atoms, the inter–atomic
distance increases in all direction for each atom. (It is similar to scaling up the grid points). One can
also look at it from macroscopic view point. 8 Mathematically, it is a scale transformation, like a photocopier set to enlarge to, say, 110%.
Suppose we have a situation that the brass is NOT inside the nickel. The brass will expand
according to thermal expansion:
r o → ro + ∆ B
where ∆B = ro αB (T – To)
Similarly the hole (of the nickel) will expand
r o → ro + ∆ N
where ∆N = ro ∆N (T – To)
a) Now, the brass is inside the nickel. Since ∆B > ∆N, there is a contact force (consequently a pressure)
between the two, as action = reaction forces, so that both the brass and the hole expand with the
same expansion:
r o → ro + ∆
Of course ∆N < ∆ < ∆B, and thus using different materials creates pressure. Since the pressure
forces act perpendicular to the contact surface (we are dealing with the geometry of annulus), the
pressure (Jpress) is primarily determined by Young moduli YB, YN. The pressure which inhibits the
thermal expansion of the brass:
∆ −∆
pB = YB B Ι
ro + ∆
where ro + ∆I is the radius of the brass after an equilibrium temperature T (room temperature).
Similarly for the nickel,
p N = YN ∆ − ∆N
r + ∆ II where ro + ∆II is the radius of the hole. Both ∆I and ∆II are negligible compared to ro and
pB = pN = p
due to action = reaction. So we have two unknowns and two equations:
(∆B – ∆)/ro = p/YB
(∆ – ∆N)/ro = p/YN
1
1
Y (∆B – ∆N)/ro = p + YB YN which we solve for p:
p= Thus p= (∆ B − ∆ N ) / ro (α B − α N )(T − To )
=
1
1
1
1
+ + YB YN YB YN (19 − 11) × 10 −6 (293 − 77)
= 1.06 × 10 8 Pa
1+ 1 −10 × 10 9.1 19.0 = 1050 atm 9 b) The expansion of the brass at T = 293 K
p
∆ = ∆B −
ro
YB
= [α B (T − To ) − p / YB ]ro 1.06 × 10 8 = 19 × 10 −6 ( 293 − 77) −
× 6.00 mm
91 × 1010 . = 17.65 µm
Thus the radius of the brass r = 6.0177 mm, and so is the inner radius of the nickel (the hole).
The volume of the brass, where r is radius, t is thickness:
VB = π r2 t = π (6.0177)2 2
= 227.53 mm3 = 0.228 cm3
The outer radius of the nickel also expands
R = Ro [1 + αNi (T – To)]
= 14.00 mm [1 + 11 × 10–6 (293 – 77)]
= 14.0333 mm
The volume of the nickel annulus
VN = π (R2 – r2)t = π [(14.0333)2 – (6.0177)2] 2
= 1009.84 mm3 = 1.010 cm3
Hence the mass of tooney:
m = (VB • 8.87) + (VN • 8.85)
= 10.95 g
compared to the one I measured: 7.31 g. This is about a 30% difference in weight — easy enough
to tell the coin.
In fact, Svend would NOT get away with his coin even by its rough appearance. The brass centre
piece is actually 16 mm wide in a real tooney, not 12 mm! 10 19961997 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 4: Optics and Waves
Due February 14♥, 1997
1) JesseÕs ÔantimusicÕ
Jesse has a problem with his roommate Bongo. BongoÕs stereo is way TOO LOUD! In fact,
Bongo has already blown his right speaker, and is pumping his TooMuch Music over the
left one. So Jesse decides to fight sound with sound. He places an identical speaker of his
own, on the left channel of the s ame station, facetoface with BongoÕs and 6.0 m away.
Jesse then sits with his one (remaining) good ear 1.5 m from his own speaker, on a direct
line between the two. For the following, you can assume that Jesse and Bongo are
pumping out exactly the same signal, in amplitude and in phase, and that JesseÕs head
doesnÕt interfere with the sound. The speed of sound in his apartment is 331.45 m s Ð1.
a) At what frequency(ies) in the music would Jesse hear minimum sound at his position?
b) Are there points along the line joining the speakers at which Jesse cannot hear sound of
any frequency?
c) What ways might Jesse set up his speaker, better to cancel BongoÕs racket? [Chairul] θ 2) A new twist on light
Like any vector, the electric field of linearly polarized light can be
considered as the sum of two components. A linear polarizer allows
through only the component that is aligned to its ÔaxisÕ, and absorbs the
other component. This results in MalusÕ law, which states that the light
transmitted by a polarizer is: I = Io cos2θ, where I is the transmitted light
intensity, Io is the incident light intensity, and θ is the angle between the
Êv
axis of the polarizer and the input E field direction.
a) Why does MalusÕ law involve cos2θ, not cosθ? b) Kimberly wanted to verify MalusÕ Law. She shone a flashlight on a photodetector and
measured 1.0 mA of current generated by the detector due to the light intensity. She
inserted a linear polarizer between the flashlight and the detector and measured the
current. What was this value? 1 c) She inserted a second polarizer and rotated it to minimize the current generated by the
photodetector. What was this current, and what was the angle between the polarizersÕ
axes?
d) On a whim, she placed a third polarizer after the first two and aligned it with its axis 45¡
to the first one. Without changing the orientation of the first two polarizers, she measured
the current. What current did she measure? She exchanged positions between the second
and third polarizers. What was the new current generated?
e) As a result of her findings in section (d), she decided to extend her studies. With the
first and last polarizers aligned perpendicularly, she inserted two polarizers aligned at 30¡
and 60¡ to the first one. What current did she measure? Write down the general expression
for the amount of current generated for a total of N polarizers, all aligned at 90¡/(N1) to
each other. For really big N, what do you expect this value to be?
f) For large N , Kimberly effectively rotated the polarization of the beam by 90¡. What
problems would you face if you attempted part e) in real life? What is a much easier way to
produce vertically polarized light from a horizontally polarized beam, using only a few
standard mirrors? [James]
3) People see through Claire
Claire (not her real name) plans to smuggle a tiny glowing radioactive pellet out of Ukraine.
She safely implants the pellet in the centre of a solid transparent cube of index of refraction
n = 1.75, but realizes the glow of the pellet will give her away. She decides to paint the cube
black, but knows that Customs officers will be suspicious of such an obvious concealment. .
Therefore she wants to paint only some parts of the surface.
a) What is the least surface she must paint Ñ and in what shape?
b) Unfortunately, at Customs she accidentally drops the cube in a glass of water and the
inspectors catch her. What is the smallest surface she should have painted? (The index of
refraction of water is 1.33.) [Chairul]
4) NothinÕ but blue sky...
A plane wave of light has a wavelength λ. It scatters off a small sphere of radius a << λ.
The light scattering off of the sphere causes the charges in it to oscillate and thus induces an
electric dipole moment p in the sphere. p increases as a function of the volume of the
sphere, p ∝ a3. Oscillating dipoles radiate electromagnetic waves, so that the intensity of
scattered radiation Iscatt from such a sphere scales as the square of p: Iscatt ∝ p2.
If we know the intensity of the incident wave, Io , we should be able to calculate the
intensity of the scattered wave Iscatt as a function of the distance R from the sphere. In
particular, we want to know how the scattered intensity varies with the wavelength of the
2 incident radiation. Note that the three lengthdependent variables in the problem are λ , R ,
and a.
a) By conservation of energy, the amount of radiation flowing through any spherical shell
at any distance R from the sphere must be a constant. How does Iscatt scale with the
distance R from the scattering sphere?
b) The ratio Iscatt /Io is dimensionless. As a result, how must Iscatt depend on λ? Use the
dependence of Iscatt on the other lengthdependent variables (which you know from above)
to obtain your answer.
c) The situation above can be used to model the sky, where visible sunlight of all
wavelengths scatters off of water molecules in the air. The human eye has evolved so that
it is most sensitive to that part of the spectrum for which the intensity of light given off by
the sun is maximal. As a result, we are most sensitive to green and this sensitivity falls off
as we move to either the red or violet ends of the spectrum. Use this knowledge and the
result of part (b) to explain why the sky appears blue, rather than, say, green. [Nipun]
5) Just tweezing...
Focussed laser light can be used to push tiny spheres around. It is
possible to coat these tiny spheres with biological antigens and make
them glue themselves to particular polymer molecules, even the end
of DNA strands. Then in moving the spheres with a laser, the
polymer molecules can be manipulated, stretched, moved.
Sometimes this is referred to as optical tweezers
The force doesnÕt come from light pressure, exactly, but through
refraction. Consider a plastic sphere 25 µm in diameter and with an
6 µm
index of refraction n = 1.4. Visible laser light can be focussed to a
spot a few micrometers across, through a microscope, so letÕs
approximate the light by a single ray as illustrated, incident on the sphere at 6 µm off the
normalincidence axis.
a) Find the approximate path of refraction through the sphere, and analyze the change in
momentum for a photon of wavelength λ = 530Ênm.
b) For a laser with power 1 W, what is the net force exerted on the sphere as it changes the
momentum of all the photons in the beam? You can assume the sphere doesnÕt move
[Robin]
Web reference: http://wwwleland.stanford.edu/dept/news/relaged/940509Arc4280.html
(a sort of pressrelease on related work at Stanford University); perhaps also see ÔOptical Levitation by
Radiation Pressure,Õ A.ÊAshkin and J.M. Dziedzic, Appl. Phys. Lett. 19, p.182 (1971) 3 6) Photons get the bends
In one kind of description, lenses work by changing the curvature of the wavefronts of an
incoming beam. When a plane wave passes through a positive lens, the flat wavefronts
come out curved concaveforward. Then by HuygenÕs
construction, it is not hard to see that the beam must collapse
to a focus.
The speed of the wavefronts is the phase velocity vφ = c/n,
where n is the index of refraction.
a) Consider a simple lens, flat on one side and with a
spherical surface on the other, 50 mm in diameter, and 5 mm
thick in the middle (see figure). If the glass has an index of refraction n = 1.66, find out how
much the wavefronts in the middle of the beam are delayed in time relative to the part of
the same wavefront that passes through the very edge of the lens.
b) What is the radius of curvature of the new wavefronts, and where do they come to a
focus?
Something similar can happen in an ordinary flat block of glass when very intense light is
incident on it. It turns out that the index of refraction depends also on intensity I, as:
n = no + n2 I
where n is the actual index of refraction, n o is the ordinary index of refraction for lowintensity light, I is the intensity of light in W cmÐ2, and n2 is a constant equal to 5 × 10Ð15 W
cmÐ2 for glass. Where light is very intense, it increases the index of refraction it encounters.
c) Consider light which has flat wavefronts as before, but is more intense in the middle
than at the edges, with the distribution: ( 2 × 1013 −b + 10 6 – x 2 I [ W cm ] = 0 –2 where b ≡ ) for x ≤ 2.5
else 10 6 – (2.5)2 = 999.996875 Because the light is more intense in the middle, the
wavefronts are delayed in the middle of the beam
relative to the edge, and again the beam will come to a
focus. For this intensity and for a block of glass 10 cm
thick, find the approximate focal point of the beam.
[Robin] 4 19961997 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 4: Optics and Waves
1. Jesse’s antimusic This problem here is a simple standingwave problem. Each speaker is at the antinode position.
To have minimum sound, Jesse has to be at one of the nodes of the standing wave pattern.
Wave patterns Mode n fundamental 1 L = 1/2 λ 2 L= λ 3 L =3/2 λ 4 L=2λ 5 L = 5/2 λ 6 6 L=3λ 7 7 L = 7/2 λ 8 8 L=4λ 9 9 L = 9/2 λ 1 2 3 4 5 0.0 1.5 3.0 4.5 6.0 m where L=6.0 m is the distance between the two speakers and λ the wavelength. 1 a) Analytically, the appropriate wavelengths can be obtained by imposing the antinode boundary
conditions:
cos ([2π/λ] L ) = ± 1, thus [2π/λ] L = n π for n = 1, 2, 3, … At position x =1.5 m, the intensity must be a minimum, meaning
cos([2π/λ] x ) = 0 ⇒ cos([2π/λ] Lx / L) = 0 ⇒ cos([2π/λ] L [x / L ]) = 0
or
0 = cos ([n π]x /L ) = cos (0.25 n π) for the acceptable n.
Thus n = 2, 6, …, and
f = v / λ = v (2 m – 1) / L for m =1, 2, 3, …. So the frequencies would be minimum for f = 55.24 Hz, 165.73 Hz, 276.21 Hz, ….
b) No. He, in fact, would hear some frequencies louder (at maximum with intensity four times
larger than its original without his own speaker).
c) If Jesse has only one speaker, he could not do any better no matter what direction his speaker
faces. If he has more speakers he could arrange them as such that more frequencies can be
cancelled out. Nevertheless, there is some complication due to the interference between his
own speakers.
2. A new twist on light This problem can be answered with different levels of rigour. To try and be as ‘correct’ as
possible, while avoiding integration. I will define intensity as the time average of the square of the
electric field:
r
I = <  E (t)2>
This is necessary since for visible light fields, the electric field is oscillating very fast (≈ 6 × 10
Hz), much faster than a detector can follow. This method can also be used for a slowlyvarying
–12
field by taking the time average over a length of time corresponding to detector response (≈ 10 –
–6
10 s). We do not require ‘I’ as a function of time for this question.
14 PLEASE NOTE: Many of you may have taken a simpler approach (more likely to be found in an
OAClevel course) by writing:
r
r
I =  E 2
(ignore oscillating nature of E (t) )
This is o.k., but not as rigourous as my solution.
a) Consider a polarizer with a set axis and a general (instantaneous) electric field vector.
r
r
r
where  E   =  Eo . Transmitted electric field is E . Thus transmitted intensity is:
r
IT = <  E  2 > 2 r2
= < Eo cos2 θ >
r2
= < Eο > cos2 θ
r2
but Ι ο = < Eo >
∴ IT = Io cos2 θ
b) Required assumption: current on photodetector ∝ intensity of light.
A flashlight is an unpolarized source. This means the polarization of the EM (light) field does not
have a preferred direction. A really rough answer to this would be that since there is no preferred
direction, 1/2 of the intensity is polarized parallel to the polarizer, and 1/2 is orthogonal.
∴ she measured 0.5 mA. E
polarizer
axis E θ E Let’s do this a little more rigourously. We can model the flashlight as a very large number of
linearly polarized sources, all with the same amplitude, and with their polarizations pointing in all
directions.
What does Kimberley measure without the polarizer?
Nr
2
ITOT = < ∑ En  >
n=0 where N is very large.
This can be written as:
Nr
Nv
ITOT = < ∑ E n + ∑ En⊥ 2 > r
where E n :
N: n=0 n=0 (1) is a parallel component of the nth source.
a very very big number (I’m working hard to avoid calculus here. If using
calculus, N → ∞ and all the sums become integrals). 3 With the polarizer she oberves
Nr
2
Ipol = <  ∑ E n  > (2) n=0 We need I pol
ITOT Take (1) ⇒
Nr
Nr
Nr
Nr
2
2
ITOT = < ∑ E n  +  ∑ En⊥  + 2( ∑ E n  ) ⋅ ( ∑ En⊥ )>
n=0 n=0 n=0 n=0 Due to the absolute value brackets, we can write.
Nr
Nr
Nr
Nr
2
2
= <  ∑ E n  > + < ∑ En⊥  > + 2 < ( ∑ E n  ) ⋅ ( ∑ En⊥ )>
n=0 n=0 n=0 n=0 We know that the flashlight polarization had no preferred direction. Therefore no matter how
Kimberly placed the polarizer, she would get the same result. Thus ‘parallel’ and ‘orthogonal’ are
arbitrary and the first two terms must be equal.
rr
To simplify the third term, consider a case with only two sources. A, B
The third term looks like
r
r
r
r
<( A  + B  ) ⋅ ( A⊥ + B⊥ )>
r
r
=0
since A  ⋅ A⊥ = 0
r
A  ⋅ B⊥ = 0
r
r
B  ⋅ A⊥ = 0
r
B  ⋅ B⊥ = 0
Therefore (1) ⇒ Nr
ITOT = 2 <  ∑ E n 2 > (since first two terms are equal).
n=0 I
∴ Ipol = TOT
2
She observes 0.5 mA of current.
c) Only linearly polarized light is transmitted by the first polarizer. Therefore the second polarizer
transmits
2
I2 = I1 cos θ
For minimum I 2 , chose θ = π
, ∴ I 2 = 0 and current = 0.0 mA. The angle between the axes is
2 90°. 4 d) Let I1 be intensity of light passing through first polarizer. Let I2 be intensity of light passing
through second polarizer.
I2 = I1 cos2 90° = 0
And the final polarizer is aligned at 90° – 45° = 45° to the second one:
I3 = I2 cos2 45° = 0
∴ she measured 0 mA.
After exchanging:
I1
2
I
2
I3 = I2 cos (90° – 45°) = 1
4
∴ she measured 0.13 mA.
I2 = I1 cos2 45° = e) Output intensity is 2
2
2
Ifinal = I1 cos (30°) cos (30°) cos (30°) 3
= I1 2 6 ≈ 0.42 I1
Therefore she measures 0.21 mA. By adding more polarizers, she is outputting more light!
For N polarizers:
N 2 90o I FINAL = I1 cos ( N − 1) By ‘subbing’ in increasing values for N, we see this coefficient is approaching 1.
Slightly more rigourously
90
=0
lim
N →∞ N − 1
2N
lim cos 0 = 1 N →∞ Therefore she measures 0.5 mA. (Recall I1 is intensity of light passed by one polarizer).
f) Real polarizers are not perfect. They still absorb some of the EM wave polarized parallel to their
axes. Thus very little light would make it through N polarizers, where N is a large number.
To change the polarization direction using two standard mirrors, recall what happens to a polarized
beam in reflection: 5 polarization parallel to
surface of paper polarization orthogonal
to surface of paper So arrange two mirrors as follows: 3. People see through Claire A
ray with incident angle
less than θc θi
θc θr
ray with incident angle θc
ray with incident angle
θi > θc B The lightray from the pellet
will be refracted out if the angle
of incidence (θ I), measured
from the normal, is less than
the critical angle (θ c). Due the
geometry of the cube (shown
as a square in the figure at left),
θc has to be less than 45 °.
Otherwise the whole surface
has to be covered. Any direct
rays fall with θ i > 45° in one
side will fall on the adjacent
one. If θc < θi < 45°, the ray will be
totally reflected and it will be
reflected forever! To see this effect refer to the figure. Coming from the source, the light will be
reflected by side A with θ i . The second reflection will be θr = 90° – θ i > 45° which is certainly
6 greater than θ c , and the third will be θ i , and so on alternately. In this way a cube is a good
container for hiding an object.
The light rays can also run in a direction not parallel to the edges of the faces of the cube: the
extreme case is one in which the rays are directed along the planar diagonal of the cube. In this
case θi can be as large as 54.73° (i.e., tan–1 ( 2) ), with θr = 35.26°. To contain this ray also, we
need θc < θr, i.e., θc < 35.26°.
a) A point source will emit light spherically. So Claire must paint in a circular shape on each side.
The area she covers must be such that θ i < θ c cannot penetrate. The radius of each circle will be
determined from θc.
θc = sin–1 (nfluid / ncube)
For air nfluid = 1.00, thus θc = 34.85° and the radius of the circle r =
d is the vertex of the cube. So the area she must cover is 1 2 d tan θc = 0.349 d , where R = π r2 / d2 = 1 4 π tan2 θc = 0.3808
i.e., 38.08% of the total surface.
b) First we have to check whether in water, θc < 35.26°. Using the same formula, θ c = 49.46°
so Claire has to paint it all!
4. Nothin’ but blue sky ... a) Assume that the radiation is emitted symmetrically for 2 radii, R and R . We must have same
1
2
radiation flux through surfaces 4π R12 and 4π R22, so
1
I∝ 2
R
Iscatt
∝ p 2 ∝ ( a 3 )2
Io b)
and 0.014 Iscatt
1
∝
Io
R
Iscatt
Io to be dimensionless 1
dependence. As
(length)4
∞ is the only remaining length–
dependent variable, we have
requires a Iscatt
1
∝ 4.
Io
λ Power (arb. units) So for scattered power
(arbitrary units) 0.012
0.01 much more blue is scattered
out of the beam than red is 0.008
0.006
0.004
0.002
0
200 300 400 500 600 700 800 900 Wavelength (nm) 7 This 1 dependence is known as ‘Rayleigh
λ4
scattering’. 101
Relative Sensitivity c)
From part (b), longer wavelengths
experience smaller scattering. Thus especially
at sunset (when the sunlight passes through the
longest stretch of atmosphere), looking at the
sun, the blue end of spectrum is preferentially
scattered away by water molecules and the sun
appears reddish or redorange. The shorterwavelength, bluer light — sent sideways from
the transmitted beam of sunset light — has
showered sideways to become someone else’s
blue sky. 100 102 rods
(scotopic) cones
(photopic) 103 104 Our eyes’ sensitivity peaks near green and
sensitivity falls off towards red and violet.
So, superposing the eye’s sensitivity and the
Rayleigh scattering law, we see a peak near the
blues. 5. 105 Wavelength (nm) Just tweezing ... As the beam enters the sphere, its direction deviates, due to
refraction at the interface
sinθ1 = n sinθ2 (Snell’s Law) The amount the beam deviates is then ∆θ
⇒ 400 450 500 550 600 650 700 α = θ2 + ∆θ = θ1
∆θ = θ 1 – θ 2 θdev
θ1 (opp. ∠ th’m) when the beam leaves the sphere the angle of incidence at
the exit is again θ 2 and then θ 1 at the exit, and again the
angular deviation is ∆ θ .
Thus the overall deviation in angle is
θdev = 2∆θ = 2(θ1 – θ2)
b
b
sin θ1 = sin α = → θ1 = sin −1 r
r
1
b
b
→ θ 2 = sin −1 sin θ 2 = sin θ1 = nr n
nr
b
b
Thus θ dev = 2 sin −1 − sin −1 r nr θ2
r r θ2 ∆θ
α θ1
b
8 b = 6 µm n = 1.4 ⇒ 6 µm 6 µm θ dev = 2 sin −1 − sin 25 µm 1.4 25 µm = 2 (13.88° – 9.87°)
= 8.03° This new direction means photons have a new xcomponent of momentum.
momentum is pγ originally, then
px If the photon = pγ sin(θ dev ) b
b = pγ sin 2 sin −1 − sin −1 r nr here
px = pγ sin(8.03°)
= pγ (0.140)
For a photon: p = so hν h
–34
–7
= , where h = 6.62 × 10 J • s. Here λ = 530 nm = 5.30 × 10 m:
c
λ
6.62 × 10 −34 J s
[but J: kg•m2•s–2 ]
p=
−7
5.30 × 10 m
= 1.25 × 10–27 kg• m•s –1
∆px = 1.75 × 10–28 kg•m•s–1 b) Laser 1W = 1 J/s
1 photon has energy hν = hc (6.62 × 10 −34 J s) 3 × 108 ms −1
=
λ
5.3 × 10 −7 m
= 3.75 × 10 −19 J so 1 J = 2.67 × 1018 photons per second
dp
18
–28
–1
FTOT = x = Nx • ∆px = 2.67 × 10 ⋅1.75 × 10 kg• m•s
dt
= 4.67 × 10–10 kg•m•s –2
= 4.67 × 10–10 N 6.
a) Photons get the bends
air only
glass only vφ = c / n phase fronts move slower in glass where n = 1.66
9 nd 1.66 5 × 10 −3 m
=
• time through 5 mm glass t = d / vφ =
c
3 × 108 ms −1
= 2.77 × 1011 s
= 27.7 ps (picoseconds)
• time through 5 mm air t = d
nd 1.00 5 × 10 −3 mm
=
=
vφ
c
3 × 108 ms −1
= 1.67 × 10
= 16.7 ps –11 s ⇒ time difference is (27.7 – 16.7) ps = 11 ps
b) In one way of describing it, lenses focus by putting a spherical curvature on phase fronts, then
Huyghen’s Principle shows the wavefronts converge to a point — the focus
8
–1
–12
Right after lens, the 11 ps relative delay means distance ct = 3 × 10 ms • 11 × 10 s = 3.3 mm
behind edges: b
3.3 mm 50 mm R
a
R This is a chord of a circle. What is the radius of the circle?
b = 25 mm
2
2
2
a + b = R
a = R − 3.3 mm 2
2
2
so R = (R – 3.3 mm) + (25 mm)
2
2
R = R – 6.6 mm⋅R + 625 mm2
625 mm 2
R=
= 94.7 mm
6.6 mm
= 9.47 cm
So the radius of curvature is 9.47 cm, and this is
where the beam comes to a focus 9.47 cm after the
lens! 91 3/8"
.47 cm
10 c) The form of I looks complicated. That is because it is precooked to give an easier answer!
n = no + n 2 I
∆n = n2I across the beam, making wavefronts curved following intensity changes across beam.
Let’s find wavefront delay according to x
∆t = n( x )d
c n(x) = no + n2 I(x)
d = 10 cm = 0.1 m
c = 3 × 108 ms–1 no d n2 I ( x )d
+
c
c
constant for all ↑
↑ causes curvatures in wavefronts
∆t = and the physical displacement of the wavefronts is:
c∆t = no d + n2 I(x)d
constant (0.17 m) ↑
↑ depends on x
So all the curvature in the beam comes from the second term:
∆z = n2 I(x) d = 5 × 10–15 • 2 × 1013 W • cm–2 • (–b + 10 6 − x 2 )
= 0.01 (–b + 10 6 − x 2 )
= 0.01 ( 10 6 − x 2 – b)
2
–1
Note the units of n2 : cm W — there was an error in the question! (eek! ) Then (100 ∆z + b) = 10 6 − x 2
(100 ∆z + b)2 = 106 – x2 Take y = 100 ∆z, then
x2 + (y + b)2 = 106 cm2
the equation of a circle, offset by –b along y, with R = 103
What is the ydisplacement?
y = yo where x = 0
2
6
(y + b) = 10
3
y + b = 10
y = 103 – b
= 1000 – 999.996875
–3
y = 3.13 × 10
so ∆z = y/100 = 3.13 × 10 –5 m 11 Use the same method as in (a) to get the curvature
a 2 + b2 = R2
2
6
2
R = 10 c m
3
R = 10 cm = 10 m a = 2.5 cm
b = 999.996875 cm The beam focuses in 10 m. 12 19961997 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 5: Electricity and Magnetism
Due March 7, 1997
1) Gauss visits Planesville...
The (very obscure) town of Planesville, Manitoba is much like any small Canadian town,
except that the town is twodimensional Ñ it is lacking the third dimension that we
experience everywhere else on earth. We want to see how GaussÕs law might be different
in Planesville. For an arbitrary charge distribution, GaussÕs law takes the form of an
integral, but recall that for a point charge Q, we can draw a gaussian sphere at radius R
from the charge and use the radial symmetry to write GaussÕs law as:
Esphere ¥ (area of sphere) = Q/εo
a) If CoulombÕs law applies in Planesville in the same form it does everywhere else, would
GaussÕs law still apply in its usual form? (Hint: Derive the form of GaussÕ law in a 2D
world for a point charge.)
b) How could we ÔfixÕ CoulombÕs law so that GaussÕs law would apply in Planesville?
Write the expression for the new force law.
c) For the force law of part (b), consider a solid disc of charge of radius 10 cm and charge
density 0.5 mC mÐ2. What is the electric field of such a disc in Planesville as a function of
the radial distance from the centre of the disc? Sketch a graph of this field as a function of
the radial distance from the centre of the disc. [Nipun]
2) ItÕs not just a good idea... itÕs OhmÕs law! Êv
Êv
Êv
OhmÕs law for conductors can be written in the form V = IR or as J = σ E , where J is the
current per unit crosssectional area flowing in a wire and σ is the electrical conductivity of
the material. Ê For our purposes, the important thing is that the current is proportional to the
v
electric field E , so a constant electric field produces a constant current. Note also that the
current in a wire is proportional to the average velocity with which the charges are moving.
a) A single point charge Q is moving freely under the influence of a constant electric field
Êv
E . Is such a charge moving with a constant velocity, or is it accelerating? Does this charge
obey OhmÕs law? 1 b) Now, instead of free point charges, we consider electrons flowing through a wire. Such
electrons undergo occasional collisions with ions making up the metal. On average, an
electron travels a distance d before colliding with an ion and stopping altogether. For a
Êv
constant applied field E , what is the average time between collisions? What is the average
velocity of an electron? Does such a result agree with OhmÕs law?
c) In part (b), we neglected the fact that the electrons (at room temperature) have a lot of
thermal energy. Therefore, they are constantly jiggling about in random directions with a
thermal velocity v t. If this thermal velocity determines the time between electronion
collisions (because vt is much greater than the electron velocity due to any applied field),
Êv
what is the new time between collisions? If we then apply an electric field E to the wire,
what is the average electron velocity? Does this result agree with OhmÕs law?
d) If the model of conduction in part (c) was accurate, what would happen to the current at
Êv
very high and very low temperatures for a constant applied field E ? [Nipun]
3) Total Recoil
After seeing a recent Schwarzenegger film ERASER, Noah was impressed by Arnie's nasty
railgun type weapon. The handheld gun fired metal projectiles at nearlight speed with
rather destructive results. Noah decided to try to build one himself. He recalled a physics
demonstration, in which a metal ring is placed on a metal post, attached to the top of a
solenoid. The solenoid consists of a wire wrapped many times around a cylindrical metal
core. When a sudden voltage was
v
applied across the ends of the
B wire, the metal ring was launched
away from the solenoid. Noah
decided to make use of this effect
ring post
to build his own rail gun.
switch
Noah's first design contained the
following components: solenoid (core wrapped with 2000 turns of wire, length = 50 cm, µ
= 500 µo), metal post (length = 5 cm, µ = 500 µo), metal ring (inner diameter = 1 cm, outer
diameter = 1.1 cm, thickness = 0.5 cm, resistance = 10Ð3 Ω , mass = 0.5 g ).
a) With the metal ring anchored to the post so it could not slide, Noah turned on the
voltage source and found that the current did not reach its maximum value immediately.
He measured the current (in amperes) as a function of time and determined that it satisfied
the following equation:
I = 1.0 (1 Ð exp(Ðt / T))
with T = 0.04 s. What caused this time lag? 2 b) From a physics text, Noah determined that the magnitude of the magnetic field at the
end of a solenoid is:
B = µ N I /2l
where N is the number of turns of the solenoid, l is the length of the solenoid, and I is the
current in the wire. What value of magnetic field does Noah generate, as a function of
time?
c) Recalling Faraday's law of induction, what is the magnitude and direction of the current
induced in the anchored metal ring as a function of time? You can assume that the current
response of the ring is instantaneous.
d) If the solenoid creates a uniform Êmagnetic field, pointing exactly parallel to the metal
v Ê v Êv
Êv
ringÕs axis, what is the Lorentz force ( F = q v × B , where v = velocity of particle of charge
q) the ring feels as a function of time and in what direction? If the ring weren't anchored to
the post, what would be its resulting motion?
e) The magnetic field is of course not uniformly pointing parallel to the solenoid axis but
starts to diverge as it exits the solenoid (as shown in the diagram). For simplicity, assume
that at the fixed position of the ring, the field lines leave the post at 15¡ to its surface. What
is the new net force on the anchored metal ring, as a function of time?
f) Noah turned off the voltage supply and removed the anchors so that the metal ring
could slide freely on the post. He closed the switch again and the ring was projected
forward. To get an approximate value for the resultant speed, approximate the force
applied to the metal ring for arbitrary time t to be the same as calculated in part e) at time
t = T. Use this constant force to get a rough idea of the velocity of the ring as it left the
metal post.
g) Noah decided to improve his design. In the movie, Arnie's gun was supposed to fire at
nearlight speed, but Noah didn't want to be greedy. He redesigned his device to launch
the projectile with a velocity of 0.01 c (c = speed of light, 3 × 108 m¥sÐ1). Somehow Noah
succeeded in building it and though it worked perfectly, Noah was seriously injured. What
had he forgotten to take into consideration? [James]
Êv
4) The Lambda particle: outstanding in its ( B ) field.
In the dawn of the field of particle physics, the discovery of the Λ Ð particle played a
significant role. It is relatively longlived. One of the most important decay modes of this
particle is the decay to proton and pion:
Λ → p+ + ¹ Ð
The minus sign for ¹ denotes that the charge of this pion (¹) is Ðe. Suppose the free
kinematics of this decay is such that all particles before and after the decay move in the 3 same direction (see figure). Now a magnetic field of 1 T,
perpendicular to the motion of the particles, is applied to
separate the moving proton and the pion. If the radius of the
proton path is 3Êcm, what is that of the pion? (restmasses: MΛ
= 1115 MeV ¥ cÐ2, Mp = 938 MeV ¥ cÐ2, M¹ = 140 MeV ¥ cÐ2,
where c is the speed of light) This problem can be solved in a
nonrelativistic way (even though it will not be accurate),
however, one should take into consideration the mass missing
after decay, which will contribute to the kinetic energies of the
proton and the pion. [Chairul]
5) Bohring after the truth p Λ π
decay
(no magnetic field) p
Λ
π
decay
(under magnetic field) In the Bohr theory of the hydrogen atom an electron circles the
proton in an orbit of radius 0.053 nm. The electrostatic
attraction of the proton for the electron furnishes the centripetal force needed to hold the
lectron in its circular orbit. Find, in this classical model:
(a) The force of electrical attraction between the proton and the electron.
(b) The speed of electron. What is this speed as a fraction of the speed of light c? [Chairul]
6) Plasma, plasma, on the wall
Practically all of the mass of the universe is not solid, not liquid, and not quite gaseous, but
in the form of plasma. Plasma is any ionized matter, whether ionized by heating (as in the
sun), by electric discharge (as inside a fluorescent light tube), or by dissociation in solution
(as in blood).
When spacecraft reenter the earthÕs atmosphere, the heat of reentry produces an envelope
or sheath of ionized air and ablating heatshield surrounding the craft. The density of this
plasma increases as the craft enters thicker atmosphere, and later dissipates as the vehicle
slows down and cools. This plasma sheath can actually act like a mirror for radiowave
communications between vehicle and groundcontrol, reflecting the signals and blocking
communication for a period during reentry. A similar kind of reflection takes place for
shortwave radio waves reflecting off the right part of the ionisphere. This happens
because plasma is a good conductor and because plasma of a certain density has a
characteristic resonant frequency, the plasma frequency.
Consider a hydrogen plasma, made of a mix of electrons and protons only, with equal
density of electrons and protons. If in a region of plasma (as illustrated) we pull the
electron component to the left, away from the proton component, it leaves two slabs of
excess charge, like a parallelplate capacitor. 4 b) Find the force on the electrons and ions within
this parallelplate capacitor. +
+ +– – –
–
++
+ –––– ––
–
––– – –
–
+
–– ––
–
+
–
++ + – – –– + –– –––– –
+–
+ – –+ +– – – –
–
–
+ ++ + –
++ + +– – +– ––––
+ + + +–– +–+–– – – –– – – –
–+ – +
–
++++ ––+–+–– – – – ––
+–+
+ + ++++ +–– – ––– –
–
– –– + a) Find the amount of charge in each charged slab,
assuming a displacement x of the electrons relative to
the protons, and a particle density of N cmÐ3 for the
protons and the the electrons, each. – – –++ + +
– – – + +++ ++
++ ++
++
+++
–
++ +
– – + + ++ + ++ ++ +
–– – + + – +– + + ++ +
––– – ++––– + +++ + +
– + –++– + + + +
–– + + +
– – – –– – –+– ++ + +++ +
+
+
+–+– + +
–
–– – +– +–+++ +++ + ++
–
– –– –– + +
+–
+
+
– c) Show that this is a restoring force F = – kx, like a
mass on a spring. Under the assumption that the
protons have so much more mass that they hardly
move, show that the force constant k leads to an oscillation frequency for the electrons
sloshing back and forth given by: νp = 1
1
ωp =
2π
2π N e2
ε o me where N is the particle density (electrons or ions), e is the electron charge, and m the
electron mass. This is the frequency of best reflection of electromagnetic waves by the
plasma Ñ any frequency lower than this is reflected.
d) Find the minimum density N of plasma needed to reflect shortwave radio at λ = 30 m.
[Robin] 5 19961997 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 5: Electricity and Magnetism
1. Gauss visits Planesville ... a) for a pt. charge in 2D:
GAUSS’ LAW: E • (circumference of circle) = Q
4 πε o R 2 2 πR = Q
1
Q
⋅
≠
2ε o R
εo which is what one would expect from Gauss’s law
b) to ‘fix’ this, make Coulomb's law ∝ 1
R , i.e.,
E=
then Q
1
⋅
2 πε o R R (point charge) E ⋅ (circumfere nce of circle ) = Q
1
Q
⋅ ⋅ 2π R =
2πε o R
εo (satisfies Gauss’s law) c) inside disc:
2
E • 2π • R = σ • πR
–2
where σ = 0.5 mC•m
σ
R , R < Ro,
2 (Ro = 10 cm) outside disc:
2
E • 2π R = σ • π R o
–2
where σ = 0.5 mC•m
⇒ E= E(R) ⇒E = Field needed to satisfy
Gauss's Law in 2D 2
σ Ro
2R
disc radius Ro 2. It's not just a good idea ... it's Ohm's law! a) F = ma = QE, R J ∝ Vavg = ∴a= qE
m ∴ charge experiences constant acceleration qEt
∴ current should increase linearly with time! (‘nonOhmic’)
m 1 b) d= 12
at ⇒ t =
2
∴ vavg = 2d
2d
∝
a
E
1
1
at = a
2
2 2d
∝ a∝ E
a ∴ it is nonohmic
c) if v thermal >> v due to field , t= d
vthermal 1
1
d
then, v avg = at = a
∝E
2
2 vthermal
∴ it is ohmic
d) σ∝ 1
v thermal from part (c) so, as T increases, vthermal increases so the resistance is greater. If T decreases, vthermal decreases,
so the resistance decreases.
3. Total Recoil a) The voltage increase causes an increasing current in the wire of the solenoid. This increasing
current creates an increasing magnetic field through the solenoid. This increasing magnetic field
acts back on the solenoid wires to create an ‘induced’ voltage which opposes the original voltage
change, and will continue until the system reaches steadystate. In brief, the effect is the
‘induction’ of a ‘reverse’ electric field (and current) in the solenoid.
µNI µN
(1 − e − t / T )
=
2l
2l
500µ o
(2000 )(1 − e − t / 0.04 )
=
2(0.5)
= 1.3(1 − e − t / 0.04 ) Tesla B=
b) c) Faraday’s law of induction
dΦ
E =−
dt
where ε is the induced electric field
rr
M is the flux = B ⋅ A
r
B is the magnetic field
r
A is orthogonal to the loop area
r
A = area of loop
r
Since the B direction is effectively orthogonal to the loop
2 rr
Φ = B⋅ A = A B
so
dB
dt
µN d
0.01 2
= –A
(1 − e − t / T ), a = π m2 2
2l dt
–3 –t/T
V
= 2.5 × 10 e
ε
..i=
R
–t/–0.4
= 2.5 e
•A ε = –A (in opposite direction to current in solenoid.)
(We can ignore backinduction from the ring since it is very small.)
rr
d)
F = qv × B
Let n be the charged particle density per unit length. Thus q = –neR
rr
F = − nelv × B
r
F
rr
= − nev × B
l
r
r
But within this model B is orthogonal to v and –ne v = i (current).
r
F
r
=i B
v
l
µN
= 2.5 e − t / 0.04
(1 − e − t / 0.04 )
l
2
= 3.3 e − t / 0.04 (1 − e − t / 0.04 ) F r
rrr
Direction of F is parallel to v × B ; v
r
is parallel to the ring; B is orthogonal
r
to the ring. Thus F is on the
surface enclosed by the ring,
pointing towards the centre of the
ring. out of page B ring, looking down solenoid This results in NO motion (unless the ring implodes). 3 r e) Even in this case v is still
r
r
orthogonal to B , but now F
is not on the surface enclosed
by the ring.
r
The vertical components of F
cancel, and we are left with a
horizontal component facing
to the right.
r
F
FNET =
l sin
l
–t/.04 v
F post solenoid
v –t/.04 F(t = 0.04 s) = 2.7 × 10
–3
= 6.3 × 10 N
but out of page
into page F (1 – e
) (.01) π sin 15
= 3.3
2 –t/.04
–t/.04
= 2.7 × 10 e
(1 – e
)N
f) B –2 –1 e B ° –1 (1 – e ) F = ma
F
∴a=
m
2
= 13 m/s For a constant acceleration:
2 ⇒ 2 v2 – v1 = 2ad
v2 = 2ad
= 1.1 m/s but v1 = 0 The ring would leave the post going 1.1 m/s.
g) Noah forgot about the conservation of linear momentum.
Consider m1 to be the mass of the ring
m2 to be the mass of the gun and Noah
v1 to be the final velocity of the ring
v2 to be the final velocity of the gun and Noah Since the system starts at rest:
m 1 v 1 + m2 v 2 = 0
v 1 = –v 2 m 2
m1
Guess that Noah and gun mass is on the order of = 100 kg. Their final velocity is 4 v1 = − 3 × 10 6 0.0005
100 = 15 m / s
= 54 km / hr
Being accelerated to > 50 km/hr in such a short time could hurt quite a bit. Even if Noah wasn’t
hurt in the acceleration, the deceleration would probably be quite disastrous.
4. The Lambda particle: outstanding in its (B) field We can determine the speed of proton from the radius of its circular path
Rp = pp/e B = Mp vp / e B (for nonrelativistic case). So, its speed
–2 6 –2 vp = Rp e B / Mp = (3 × 10 m) e (1 T) / (938 • 10 eV • c )
6
–1
= 2.88 × 10 m • s = 0.00959 c. (4.1) The assumption of nonrelativistic proton is applicable here.
We apply the conservation law of energy and momentum:
EΛ = Ep + Eπ
p Λ = pp + pπ
which can be written nonrelativistically as
2 2 2 2 2 2 MΛ c + 1/2 MΛ vΛ = Mp c + 1/2 Mp vp + Mπ c + 1/2 Mπ vπ , (4.2) MΛ vΛ = Mp vp + Mπ vπ. (4.3) vΛ can be eliminated by taking the square of (4.3) and substitute the term in (4.2)
2 2 2 2 2 2 MΛ c + 1/2 (MΛ) (Mp vp + Mπ vπ) = Mp c + 1/2 Mp vp + Mπ c + 1/2 Mπ vπ
2 (1115 − 938 − 140) MeV + (938 × 0.00959 + 140 vπ/c) /(2 × 1115) MeV −
2
2
938/2 MeV (0.00959) − 140/2 MeV (vπ/c) = 0
2 61.21 (vπ / c) − 1.13 (vπ / c) − 36.99 = 0
6 –1 So we can obtain vπ = 0.787 c = 2.36 × 10 m • s . This is a relativistic problem and the
assumption of nonrelativistic pion is incorrect. Nevertheless, let us find out the radius of
the pion path.
6 –1 Rπ = pπ / e B = Mπ vπ / e B = (140 × 10 eV • c • 0.787)/ e (1 T)
= 0.367 m = 36.7 cm
The full relativistic treatment (especially for pion) gives the radius of 40.0 cm. 5 Use the following formulae:
2 2 2 2 MΛ c + 1/2 MΛ vΛ = Mp c + 1/2 Mp vp + Eπ,
MΛ vΛ = Mp vp + pπ,
2 4 2 E π = Mπ c + p π c ,
to solve for pπ and Rπ = pπ / e B.]
5. Bohring after the truth a) Applying Coulomb’s law:
22 9 F = k e /r = (9.00 ×10 ) • (1.602 × 10
–8
= 8.22 ×10 N
b) –19 2 –9 2 ) / (0.053 × 10 ) 2 The centripetal force F= m v / r. Hence the speed of the electron is
–8 –9 v = √ (F r / m) = {(8.22 ×10 ) • (0.053 × 10 ) / (9.31 × 10
6
–1
= 2.19 × 10 m • s = 1/137 c
6. –31 )} 1/2 Plasma, plasma, on the wall a)
Consider a slab of plasma, as in the question, which measures x × y × z. If the negative
electrons are pushed off the positive protons to one side, by a tiny amount ∆x, then there will be a
thin slab of excess charge on either side — one positively charged and one negatively charged.
The volume of each little excess slab will be:
(∆x) × y × z [1] If the density of electrons is N [cm–3], then the amount of excess charge will be
N e (∆x) × y × z = N e (∆x) × A [2]
where A is the area of the side of the slab In other words, what we have is something like a parallelplate capacitor, with the two thin excesscharge slabs as the two charged plates of the capacitor.
b)
For this, we can find the field between the two plates of a parallelplate capacitor. In the
middle, where there are both electrons and ions, the charges of each cancel each other out (unless
you give them time to move around and redistribute themselves), so the net field in the middle is
just what is produced by the thin slabs of excess (unbalanced) charge.
E = 4 π kc σ where kc = 9 × 10–9 N m2 C–2
σ = charge per unit area on each plate [3] 6 [The easiest way to see this is to use Gauss’s Law (if you know it), drawing little rectangular
boxes with sides parallel to the plates, and having one side between the plates and the opposing
side outside the plates. The field is perpendicular to the plates, by the symmetry of the situation,
and it is quick to find the contribution each parallel plate makes to the overall field.]
Then from the second part of [2], we can find the charge per unit area of the two slabs of excess:
σ = (N e (∆x) × A) / A = N e (∆x) [4] and the field between the plates is:
E = 4 π kc N e (∆x) [5] Now the charges in the middle see the field produced by the excess charges, and experience a
force:
F=qE which depends on q, +ve or –ve [6] This then gives us the force on each charge within the thin slab:
F = q E = q 4 π kc σ = e 4 π kc σ = e 4 π kc (N e (∆x) ) [9] = 4π kc N e2 (∆x)
c)
The whole block of electrons in the diagram of the question is free to move (the ions, being
more massive, have a certain ‘right of weigh’) in the electric field they see. Most of them see the
electric field between the two plates, and this pulls them back onto the ion background. So the
force on an electron between the chargeexcess thin slabs is a restoring force against the separation
of charge.
Fr = – k (∆x) where k = 4π kc N e2 [10] This is the restoring force of a simple harmonic oscillator (SHO) — it means the electrons mostly
will slosh back and forth past the ions, barring collisions, once they are released from their initial
displacement. Since the ions are relatively massive, they hardly move, but the electrons oscillate
sinusoidally, as does a SHO, with a frequency which depends on the electron mass:
ωp = where ε o = 1
4π kc [11] Then νp = ωp / 2π to get the result shown. This is called an electron plasma wave.
d)
When an electromagnetic wave is incident on a plasma, the Efield can begin to drive
electrons back and forth, causing something like the excess charge in the model above. The
electrons then oscillate at their own frequency, given just above in [11].
If the EM wave is at exactly the same frequency, it is in resonance with the electron plasma wave,
7 and oscillations can grow very large. This means that a large AC current flows back and forth in
the plasma, and that current can radiate EM waves itself, cancelling the light that comes in and
sending it back as a reflection. At what frequency does this happen for shortwave radio at λ =
30m?
c = λ ν ⇒ ν = c / λ = 3 × 108 cm•s–1 / 30 m = 1 × 10 7 s–1
v = 1 × 10 7 =
⇒N= 1
2π Ne 2
εo m e vp = 1
1
ωp
2π
2π [12] Ne 2
εo me (2 πv) 2 ε o m e
= 124 × 1012 m −3 = 1.24 × 106 cm − 3
.
2
e Compare this to gases at sealevel: 1 mole of ideal gas occupies 22.4 l @ STP,
22.4 l = 22.4 × 103 cm3
6.022 × 10 23
cm − 3 = 2.69 × 1019 cm − 3
3
22.4 × 10
Much less dense than this — shortwave radio is usually reflected not from fully ionized gas at
atmospheric pressure, but from a layer of partially ionized highly rarefied gas high up, in the
ionosphere. 8 19961997 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 6: AC Circuits and Electronics
Due March 28, 1997
1) WhatÕs a gigawatt? Thirty of Ôem??
The province of Ontario requires something like 3 × 1010 watts of electrical power
(aÊplausible guess?: a kilowatt or so for every person, and then twice as much again for all
of industry), at 110 volts AC ( r.m.s.). A heavy power cable might be an inch in diameter.
Say the whole province was supplied by a single cable a metre in diameter Ñ what would
happen? Take it that the cable has a resistance of 0.15 µΩ mÐ1. Calculate:
a) the power lost per metre from Ô I2R losses,Õ
b) the length of cable over which the whole 3 × 1010 watts would be lost, and
c) how hot the cable will get, if it loses all this power by blackbody radiation.
Your answer ought to be preposterous Ñ why is our electrical power system not ridiculous
in this way? [Robin]
2) Waffling on the issues... the issue of waffles
A diode is an electronic device that allows current to flow only one way. An ideal diode
has zero resistance for positive voltage and huge
resistance for negative voltage. A slightly more realistic
I(A)
diode has a turnon voltage, as shown in the figure at
right. One typical use of a diode is as a light source.
These lightemitting diodes (LED) produce visible light
0.1
when a voltage greater than the turnon voltage is
applied across the device.
2
2
Voltage (V)
a) After many serious burns, Anton (the Waffle Czar)
Duzzleeater pledged to improve current wafflemaker
technology. He decided to incorporate an LED as an indicator that would light up
whenever the waffle iron is plugged in. To begin, Anton decided to characterize his
voltage source Ñ normal household AC Ñ so he measured the voltage as a function of
time. Graph what Anton measured (2 cycles is sufficient). DonÕt forget to indicate your
axis and scales.
1 b) Without bothering to think about it, Anton wired
the LED in series with the waffle heating element
(schematic at right). If the diode had a similar I –V
curve to the one shown in figure 1, what would be
the voltage drop across it as a function of time? Does
the LED emit light? When? LED heating coils
(R=10 Ω) c) Unfortunately for Anton as soon as he plugged the circuit in, the diode flashed a brief
burst of light and began smoking. The diode couldnÕt take the excessive current. Draw a
more intelligent circuit that would avoid this problem. [James]
3) 2B ∨ (~2B) ← Q! [...to be or not to be, that is the question!]
Wendy, a Physics Olympiad student, is given a bunch of digital gates.
Unfortunately, she only has three types of gates: AND, OR, and NOT. OR Her supervisor said to her that she can make any combination. Lo and
behold, she finds
AND
((NOT A) AND B) OR ((NOT B) AND A)
circuit is a very useful combination. Draw the circuit for that
combination. Make a Boolean table for that circuit. Can you name that NOT
simple circuit? [Chairul]
4) Connect the dots!
Who said that physics wasnÕt all fun and games? See the figure on a separate page: the
challenge is to connect the dots using the components given, to simultaneously achieve the
desired voltage and current. Note that you may not need to use all the dots (or
components) and you may want to attach more than two components to the same dot. The
symbols ÔAÕ and ÔVÕ correspond to ideal ammeters and voltmeters. [James]
5) Impeding oneÕs reactance
In a DC circuit, the total resistance is a sum of the individual resistances in the circuit, the
nature of the sum depending on whether the resistors are in series or in parallel. In an AC
circuit, the impedance, Z, plays the role of resistance and sums in the same way as
resistance, however it has both real and imaginary parts. For ideal resistors, inductors and
capacitors, the contributions to the total impedance are as follows:
resistors: ZR = R inductive reactance: Z L = iω L capacitive reactance: ZC = (iωC)–1 2 Here, ω = 2¹ f , where f is the frequency of the input AC signal.
a) What is the total impedance of the circuit in the figure below?
b) We are interested in only the amplitude of V out, which is given by
V out = ( Vout ¥ V out*)1/2
where Vout* is the complex conjugate of Vout. What is V out for this circuit? Sketch
V out/Vin as a function of log10 ω.
R c) How would such a circuit affect a highfrequency input signal? A lowfrequency
signal? What do you conclude it might be
used for? [Nipun] V in V out C 6) Say ÔcheeseÕ! (...Zap !!)
In highpower lasers pumped by a
flashlamp discharge, large capacitors are charged to several kilovolts and then a fast, highvoltage switch closes to discharge them through the flashlamp. It is very much like an
electronic camera flash, but hugely larger. These are the facts about the flashlamp circuit:
¥ the electric current pulse through the flashlamp must last for a certain amount of time τ
in order best to ÔpumpÕ the laser
¥ if only a flashlamp and capacitor are connected in series, the current will oscillate back
and forth through the lamp before dying away, and this can damage the lamp or cause it
to explode
¥ the best shape results when the capacitor voltage is a simple
decaying exponential L ¥ the exponential should be the fastestdecaying one possible
To make these things possible, an inductor like a coil of heavy
wire is added to the circuit, which can then be represented as
shown at right: C Ð capacitor; L Ð inductor; R Ð resistance of
flashlamp; i(t ) Ð timedependent current through flashlamp. C R i(t) a) What is the equation for the charge q on the capacitor? What is the voltage across the
capacitor in terms of the charge q on the capacitor? Show your approach.
b) Assuming the voltage is the fastestpossible decaying simple exponential, and that C
and R are fixed, what must be the value of L? What is the exponential decay time, given
RÊ= 1 Ω, C = 200 µF?
c) Find the current i (t) through the resistor R . If the capacitor is charged to 2.5 kV, how
much energy is discharged through the lamp? [Robin] 3 Figure for ÔConnect the Dots!Õ e.g.) Set voltage=5V, I=5A A V A
5V V wires 1Ω 1Ω 5V 10Ω a) Set voltage=4V, I=.2A A 6V 10Ω V
wires 5Ω b) Set voltage=6V, I=.2A A
4V
6V 6V 6V 10Ω V c) Set current I=0A
(note: shortcircuiting batteries is baaaaaad!) 10Ω 5V A
10Ω 50V 90Ω
wires 4 19961997 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 6: AC Circuits and Electronics
1. What's a gigawatt? Thirty of 'em?? a) Power lost per meter, with P = IV (i.e., power and voltage are preset)
2
P loss = I2 R = (P/Vrms)2 R = (3 × 1010 W / 110 V ) • (1.5 × 10–7 ohm • m–1)
= 1.1 × 1010 W • m–1 b) length for loss: 3 × 1010 W / 1.1 × 1010 W • m–1 = 1.1 × 1010 W • m = 2.7 m
(a pretty pathetic distance!) c) power loss by StefanBoltzmann law:
P= σ A T4
trivial power gain from 20°C room temperature, i.e., likewise:
P= σ A (293 K)4
so that net:
P= σ A [(T4 – (293 K)4] ≈ σ A T4
Area of surface of 2.7 m of cable is
A = 2π r L = 2 • 3.14 (0.5 m) 2.7 m = 8.48 m2
T = (P/(σA))1/4
= (3 × 1010 W / (8.48 m2 • 5.670 × 10–8 W m–2 K–4)) 1/4
=15,800 K This doesn’t happen, basically because of Nikolai Tesla: with AC current we can easily manage
stepup transformers, so the power is delivered at much higher voltages (like 250,000 V, I recall)
with smaller losses, and then steppeddown several times, the last time quite near the delivery site.
As well, the total diameter of cables is far in excess of 1m.
2. Waffling on the issues ... the issue of waffles a) Normal household voltage in about 120 V (rms) which corresponds to 170 V peaktopeak
170 voltage (V)
1/60 time (s)
2/60 170 1 b) The LED emits light only when it is forwardbiased, as indicated by the asterisks in the graph
below. Thus it flashes at 60 Hz.
170
voltage across 2
L.E.D. (V) * *
time (s)
2/60 1/60
170 c) One possible circuit is shown at right,
where
R b is a large resistor. This reduces the
amount of current that flows through the
LED. 3. 2B ∨ (~2B) → Q! Rb
AC heating coils [... to be or not to be, that is the question!] The circuit of this combination is
A C E G B D F Boolean table
Input
A
T
T
F
F B
T
F
T
F C
F
F
T
T intermediate nodes
D
E
F
F
T
T
F
F
T
F F
F
F
T
F Output
G
F
T
T
F Clearly, this circuit is an exclusive OR gate (XOR). 2 4. Connect the dots!
a) Set voltage=4V, I=.2A A All R=10Ω V
wires 5Ω 6V A b) Set voltage=6V, I=.2A V 6V 6V 6V 10Ω
10Ω 4V c) Set current I=0A
(note: shortcircuiting batteries is baaaaaad!) 50V 5V 10Ω 90Ω A
10Ω wires 5. Impeding one's reactance a) As they are seen by VIN, the impedances are in series. Therefore, the total impedance is
ZT = R + (iω C) –1 Similarily, as seen by VOUT
Z=
b) 1
iω C The ratio of VOUT to VIN is given by the ratio of the impedances. 3 VOUT ZT
1
1
=
= R+ VIN
Z
iω C iω C −1 = 1
1 + iω CR 1 VOUT
VIN 1 1 *
V VOUT 2 2 2
1
1
1
OUT
=
×
× = =
2 2 2 VIN VIN
1 + iω CR 1 − iω CR 1 + ω C R 1 so, VOUT = VIN 2
1 2 2 2
1 + ω C R 1 c) for ω <<
 Vout  / Vi n 0.8 for ω >> 0.6 0.2 –log(RC)
1 ω (arb. units) 1
, all voltage is dropped
RC So this is a filter which allows only low
frequency AC signals to pass. The
cutoff frequency can be changed by
adjusting R and C 0.4 0
0.1 1
, all signal is passed.
RC 10 100 6. Say 'cheese'! (... Zap !!) a) Each element has a chargedependent voltage associated across it:
resistor:
∆VR = I R
capacitor:
∆VC = q/c
dI
inductance:
∆ V R = –L
dt dq
, and since the potential drops around the circuit must sum to zero, we can establish a
dt
relation in terms of change of charge. Be careful about the signs:
Since I = • choose allocation of +q/–q on capacitor
• +q plate is at higher potential, so sketch current direction (flow of +ve charge)
consistent with dropping potential
Then the current carries charge away from +q
dq
so
I=
dt 4 d2q
q/c + L
+
dt 2 dq
=0
R or
L
Then b) d2q
dq 1
+ R + q =0
dt C
dt 2 a ‘secondorder ordinary differential equation’ ∆Vc (rise) + ∆VL (drop)+ ∆VR (drop)= 0
dI
q/c – L
– IR = 0
dt
assume the form V(t) = Voe
so q(t) = CVoe –t/τ –t/τ and substitute into differential equation
2
–t/τ 1 + R CV − 1 + Vo}e = 0
{LCV o
o τ τ
e –t/τ 0 must mean 2 1 – RCV
LCV o o τ 1 + Vo = 0 τ 2 or multiplying by τ both sides
2 τ – RC τ + L C = 0
then
−b ± b 2 − 4ac
τ=
2ac quadratic formula + RC ± R2 C 2 − 4 LC
=
2 4L 1± 1 − 2 RC
= RC 2 The solution should be a simple exponential, meaning that the discrimant (root here) should be
nonnegative (else get imaginary numbers, which lead to cosine/sine terms — oscillations)
1− 4L
>0
R 2C or 2 4L < R C 5 R2 C
L>
4
Then there are two solutions because of ±
q(t) = CVo (a1e –t/τ1 –t/τ2 + a2 e
)
a 1 + a2 = 1 4L 1+ 1− 2 RC
τ1 = RC 2 4L 1− 1− 2 R C
τ 2 = RC 2 One of these τ1, τ2 is always the longer time — the fastest decay is as τ1 → τ2, i.e. as
1− 4L
→ 0+
2C
R RC
then τ1 , τ 2 →
2
R = 1Ω, C = 200µF i.e., from above ⇒ T = 100µs c)
This was rather (too?) tricky: one way to see the solution is to start with τ 1, τ2 and the
condition
0 = I (t ) = − so but 1
dq(t )
1 −t / τ 2 = − CVo − a1 e − t / τ1 − a2
e dt
τ2 τ1 I(t = 0) = –CVo − a1 − a 2 τ1 τ 2 a 1 τ 2 = –a 2 τ 1
a 1 + a2 = 1,
τ1
⇒ a1 =
τ1 − τ 2
− τ2
a2 =
τ1 − τ 2 6 1
1
e − t / τ1 −
e −t / τ 2 τ −τ
τ1 − τ 2 1
2 I (t ) = + CVo =
as L → CVo
− t / τ1
−t / τ 2
(e
−e
)
(τ1 − τ 2 ) R2 C
0
and τ1 → τ2 this gives a
indeterminacy, which we can work with l’Hôpital’s
4
0 rule:
numerator
demominator 1 2
d
(e − t / τ1 − e − t / τ 2 ) = e − t / τ1 −t − dτ1 τ1 d
( τ1 − τ 2 ) = 1
dτ1 So the answer
I (t ) = CVo t −t / τ
e
τ2 CVo
−t / τ
2 te
( RC)
t2
Vo
–
I (t ) = 2 t e RC
RC
= (close enough) 1
2
CV
2
1
3
2
= 200µF (2.5 × 10 V)
2
= 625 J of energy. E= 7 19971998 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 1: General
Due November 3, 1997
1) Connect the dots!
Who said that physics wasnÕt all fun and games? See the figure on a separate page at the
end of this set: your mission, should you decide to accept it, is to connect the dots using
the components given, to simultaneously achieve the desired voltage and current. Note
that you may not need to use a ll the dots (or components) and you may want to attach
more than two components to the same dot. The symbols ÔAÕ and ÔVÕ correspond to ideal
ammeters and voltmeters. [James]
2) Out with the bad air, in with the good...
In homes which meet the R2000 standard of energy conservation, there are virtually no
drafts which let in (cold) fresh air, and there might arise a risk of pollutionbuildup from
building materials, from cooking and cleaning, or from the occupants themselves. So for
such houses, designers may include a heatrecovery ventilation system Ñ basically a kind
of heatexchanger in which stale air exits the house but first passes much of its heat on to
the incoming cooler air. Because the outgoing and incoming air streams are kept separate
as they pass each other, but still allowed to exchange heat, fresh air enters the house fairly
warm and stale air exits the house fairly cool. In fact, the efficiency is surprising.
It is a bit too complicated for POPTOR to calculate efficiency for a heatrecovery ventilation
system, but consider this problem:
Say you have 1L of hot water (say, dyed red), of temperature 100oC, and 1L of water (say,
dyed blue), of temperature 0¡ C , presenting the warm and cool air above. Say also that you
have many speciallyinsulated containers, any sizes you wish. These containers prevent
heat loss to the air, but exchange heat perfectly when two (or more) touch each other. The
question is like warming the incoming air in the house: without mixing the two together,
to what temperature can you heat the initially cold (blue) water, and to what temperature
can you cool the initially hot (red) water, by arrangements of letting them swap heat?
[Peter]
[Canada Mortgage and Housing Corporation (CMHC) is a remarkable Canadian goldmine of housing research and public
information of every imaginable sort, including energyefficient R2000 houses, heat recovery ventilators, indoor pollution,
etc. To order publications, check out: www.cmhcschl.gc.ca/InfoCMHC/contact.html#Publications]
[Wonder what a heatexchanger looks like? There are some photographs of heatexchangers for industrial processes at:
www.souheat.com/gallery.htm] 1 3) Waves ˆ la mode
If you have ever sung in the shower, or maybe a marble washroom stall, you may have
noticed that certain frequencies really pick up: they resonate within the space. For sound,
the air molecules moving cannot move into the wall, so the walls are ÔnodesÕ or zeropoints
for the molecular motion. Only certainfrequency waves naturally meet this condition, so
they can be resonant within the space, and their reinforced amplitudes can be larger than
that for other frequencies. These are Ôstanding waves,Õ and sometimes called Ômodes.Õ
In making a laser, several items are brought together: a laser cavity (consisting of two
mirrors facing one another) and an optical amplifying laser medium are the bare minimum.
The laser cavity provides ÔfeedbackÕ Ñ a way to recirculate the light many times through
the amplifying medium to make it more intense. Metallic mirrors also set a condition on
the electric field in light: the transverse electric field must vanish at the surface of a
conductor (see question below). Therefore, standing electromagnetic waves are formed in
the laser cavity, which is for that reason sometimes called a Ôlaser resonatorÕ.
a) Luciano Pavarotten is an operatic showersinger, with a threeoctave range starting at C
below middle C (middle C is 256 Hz). You may not know that a note an octave above
another has twice the frequency. If his shower stall is 1.5 m × 1 m × 2 m, how many
standing wave modes in the shower can his voice excite? (Alas, only the standing waves
are excited by his voice) How many different frequencies is this? You should ignore the
issue of how his (large) body interferes with the sound in the shower.
b) Krystal has a Nd:glass (Ôneodymium glassÕ) laser, which typically operates at about 1
Ð6
micrometer (1 µm = 10 m) wavelength. It can amplify any wavelength within ±5
Ð9
nanometers (5 nm = 5 × 10 m) of this wavelength. How many standingwave modes of
light are amplified in a laser resonator which is 80 cm long? [Robin]
4) Opposites attract, but a narcissist always loves himself...
Consider a grounded cylindrical conductor, of great length compared to its radius r. In
orbit around this fat wire is a small charged particle of mass m and charge q . The distance
from the point to the surface of the cylinder is d << r. Why should a charged particle orbit a
grounded conductor? Explain the phenomenon, and find the period of the orbit. Neglect
gravity, ignore air particles, and pay no attention to the man behind the curtain. [Peter]
HINT: For this quiteneat question, you may need to know these two things that you may
not already know:
a) A perfect electrical conductor has no electrical field inside.
Not obvious? Say that there were a field inside: then the free charges in the conductor would see that field
(and so a force), and would begin to move Ñ any positive charges in one direction and any negative ones in
the exactly opposite direction. If the conductor is of finite size, the charges canÕt keep on forever, but when 2 would this whole thing stop? It would stop when the conductor ran out of free charges to move (phenomenally unlikely, for realsized fields) or until the separated + and Ð charges made their own field
which exactly cancelled the original field. The net result is zero DC field. b) Therefore the surface of a perfect electrical conductor is a surface of constant electric
potential.
If the field everywhere in the conductor is zero, then a small ÔtestÕ charge could be moved anywhere inside
without seeing a force. Therefore the potential energy of the test charge is the same anywhere in the
conductorÑ no force, no work, no potential energy change. So a charge a tiny distance inside the surface of
the conductor is everywhere at the same potential energy: this means, then, that the surface of a conductor is
an electrostatic equipotential. 5) Fermi Questions
The famous and fabulous Italian physicist Enrico Fermi was known for a curious type of
question he sometimes posed his students. One of them was Ôhow many piano tuners are
there in Chicago?Õ. The students were expected to work out the answer without resorting
to any reference materials. The questions were an exercise in a very important problemsolving talent: developing the judgement that lets one make reasonable assumptions. With
some notion of how many people are in Chicago, one might estimate roughly what
percentage would have pianos, how much a pianotuning costs, or how long it takes, etc.
Here are some ÔFermi questionsÕ to try. The idea is to use only what is already in your
head, and to make plausible guesses for the rest. For each, give in brief point form the
steps you followed in your reasoning. Estimates within a factor of 3 might be considered
good, for some of these.
a) about how many piano tuners are there in Toronto? (DonÕt just check the Yellow Pages)
b) Claudio ChiapucciÕs bicycle has reflectors attached to the spokes of his wheels. He is
fussy, and starts out each ride with the wheels set so each reflector is in the 12 oÕclock
position, at the top of the wheel, but they never stay that way. Roughly how long does he
ride before they end up opposite each other, e.g., 10 oÕclock and 4 oÕclock?
c) Jacques Villeneuve went shopping with his car one recent Saturday, but found his
favorite small lot at EatonÕs was full. He parked at the top of the lot, and from there he
could see, and quickly get to, about fifty parking spaces. On average, about how long
would he have to wait for someone to leave?
d) About how many pounds of cigarettebutts are dropped on sidewalks in Canada each
year?
e) Halftime during an exciting Grey Cup football game: trips to the kitchen and
washroom. How much, roughly, are the sudden increases in electrical power and water
demand? [Robin]
3 6) ÔStacking the Deck Ñ Friction in the House of Cards,Õ by Kitty Kelly
PART I Ð Consider the problem of making a structure out of cards. It's pretty difficult. Step
One: consider the simplest problem of leaning two cards against each other on a flat
surface. Let θ be the angle between the cards (we're assuming symmetrical stacking), µ1 be
the static coefficient of friction between a card and the smooth tabletop, and µ2 the
coefficient between the two cards. I'm using a brand new Bicycle deck, so the cards are
slippery and µ2 < µ1 . θ When two cards are stacked together at some angle θ,
there is some minimum force F min required in order to
collapse them. Find out whether there is some
optimal angle, θbest Ñ an angle at which the cards will
be most stable (i.e., the largest Fmin) Ñ and determine
that angle.
(θ = 180¡, with the cards lying flat on the ground is
stable, alright, but useless Ñ that special case does
NOT count!) PART II Ð Suppose that you want actually to predict the value of the angle θbest, using
results from PART I. How do you get µ1 and µ2 Ñ just look them up in the CRC Handbook
of Playing Card Constants? Doh! You could measure it, you know.
a) Find a deck of playing cards Ñ a couple of baseball or hockey cards (or perhaps even
computer diskettes) will do. For your cards, and your smooth tabletop, measure µ 1 and
µ2 . Describe what you do, and why, and figure out estimates of uncertainties about your
measurement process. Give an errorrange for your final answer; this is essential because
it represents to others how closely they can trust your exact answer.
Note that everyone will (most likely) get a different answer Ñ it's how you do it that
matters for POPTOR.
b) Now knowing µ1 and µ2 , predict θbest. Does it seem reasonable? Explain why. [Peter] 4 For Question #1:
e.g.) Set the voltage: V=5V, current: I=5A using only the given components. A V
5V 5V V wires 1Ω 1Ω A 10Ω a) Set the current measured by A1 to be I=2A, using only the given components.
What is the current measured by A2, A3 and A4? A2 A3 1 A4 A 2 wires
All R=5Ω 15V A b) Set the current: I=3A using only the given components. 2 wires 13V
each R=1Ω 3 V 2 V V 1 c) Set voltages: V1=11V, V2=7V, and V3=5V using only the given components. 17V 6V
3 unknown R
4V 2V 7 wires 5 19971998 Physics Olympiad Preparation Program
Ñ University of Toronto Ñ
Solution Set 2: Mechanics
1) The Pluto Problem
KeplerÕs Second Law results from the conservation of angular momentum of the
planet about the sun. According to this law, the straight line joining the sun and a
given planet sweeps out equal areas in equal intervals of time. Thus, the ratio of the
time interval t, during which the PlutoÕs elliptical orbit is situated inside the
NeptuneÕs circular orbit, to the PlutoÕs orbital period T is equal to
t/T = S/(πab)
The denominator of the right side is the area of the ellipse representing the PlutoÕs
orbit, and the numerator S is the area of the sector of subtended angle 50 degrees.
We can estimate the value of S as the area of a circular sector (50/360) πR2, where for
more accuracy we can substitute R by (rmin + R)/2. Therefore,
t = (50/360) x (((rmin + R)/2)2 x T)/(ab) = (5/36) x ((4.4 + 4.25)/2)2 x
(248)/(5.9 x 5.73) = (5/36) x (18.71 x 248)/(5.9 x 5.73) = approx. 20 years.
It turns out that the question was in error, and Pluto became closer to the sun i n
1979, not 1969. Therefore, Pluto will be again the ninth planet from the sun in 1999.
[http://seds.lpl.arizona.edu/nineplanets/nineplanets/pluto.html]
2) Free falling, at $4.50 a throw
a) Let l be the length of a nonstretched rope (l + h < H), k is its elasticity constant,
m the mass of a person who is jumping down and v is a value we are looking for.
It is obvious that at the equilibrium height h your velocity had the maximum
value. We can write the following three equations:
1) the equilibrium condition at the height h :
1 mg = k (H Ð l Ð h ) [2.1] 2) the energy conservation law at the height h :
mgH = m gh + (k /2)( H Ð l Ð h )2 + (1/2)mv 2 [2.2] 3) the energy conservation law on the ground level :
mgH = (1/2) k (H Ð l)2 [2.3] Divide [3] by [1] and obtain
l = √ H (H Ð 2 h ) [2.4] Then rewrite [2] using [1] in the following form :
mg (H Ð h ) = (mg /2) (H Ð l Ð h ) + (mv 2/2)
and get
v = √ g (H Ð h + l ) [2.5] or using [4],
v = √ g (H Ð h + √ H (H Ð 2h )) [2.6] If H = 25 m, from [6] we have v = 16.2 m/sec or 58 km/hour. If H = 50 m, then v = 28
m/sec or 101 km/hour.
b) Only the third equation will be different compared to case a) :
mg (H Ð h 1) = (1/2) k (H Ð l Ð h 1)2 [2.3Õ] Hence,
l = √ H (H Ð 2 h ) + h 1 ( 2 h Ð h 1 ) [2.4Õ] Use [4Õ] in [5] and get the new value for v . If H = 25 m, then v = 17.1 m/sec or 61.4
km/hour. If H = 50 m, then v = 28.2 m/sec of 101.6 km/hour.
c) Elasticity constant can be found from [1] : k = (mg )/( H Ð l Ð h ).
For H = 25 m, h = 10 m, m = 100 kg and l = 11.18 m (from [4]), we have k = 261.8 H/m.
For H = 50 m, h = 10 m, m = 100 kg and l = 38.73 m (from [4]), we have k = 787.4 H/m
(the rope must be 3 times stronger for the height H = 50 m compared to the one for
the height H = 25 m).
3) ÔBobÕs your uncleÕ, or sometimes heÕs a simple harmonic oscillator (SHO)
a) submerged length of the cylinder
force of gravity on the cylinder:
volume of water displaced by the cylinder Y
F=mcylg
Ya
2 So the force of gravity on the cylinder = Yaρwaterg
At equilibrium, the two forces are equal, so:
a
m cyl g = Yaρ water g Y= mc yl Bob ρ waterπa h
1'1" y
8"
b) For a small displacement from equilibrium, y,
there is a small difference between the buoyant
force and gravity. Assume we push the cylinder
slightly deeper into the water than the equilibrium point. Then the i ncrease in the
buoyant force is an excess force, in the amount:
F = − mg = − ayρ water g and likewise if we pull the cylinder up, the excess is in the opposite direction.
c) (Note: in this section, A is acceleration, while a is the cross sectional area of Bob)
F = m cyl A = mcy l 2 dy
= − ρ water ayg
2
dt [3.1] so,
d 2y
ρ ag
= − water
y.
2
dt
m cyl [3.2] d) Just by substituting, it is easy to show that y(t) = y0 cos(ωt+φ) is a solution of a ny
equation of the form
m(d2y/dt2) = Ðky,
as in part (c). We will use this same form in future POPTOR problems!
The maximum/minimum value that y can take occurs when cos(ωt+φ) = ±1. So, y0
is the maximum amplitude of Bob, measured from the equilibrium point.
If we let go of Bob at an amplitude y0, at time t=0, then this should be his maximum
amplitude. Then
y (t = 0) = y 0 cos φ = y 0 [3.3] holds, whether we start off by lifting Bob or by pushing him down a little. This
requires cos(φ) = 1, which is true as long as we make φ = n 2π, where n is any integer.
3 For convenience, we simply set φ = 0. (If however t=0 is chosen at some other point
in BobÕs oscillation, we will need φ to take on some other value.)
dy
= −ωy 0 sin(ωt + φ )
dt [3.4] To determine ω for Bob, note that,
2 dy
k
= −ω 2 y0 cos(ωt + φ ) = −ω 2 y = − y
2
dt
m [3.5] so, ω = (k/m)1/2. From part (c),
k
ρ ag
= water
mcyl
mcyl [3.6] therefore, ω= ρ water ag
mcyl [3.7] e) F = buoyancy + gravity a = ρwga(yw Ð yB) Ð mBg
at equilibrium F = 0, so Bob 0 = ρwga(ywo Ð yBo) Ð mBg
mBg = ρwga(ywo Ð yBo) [3.8]
yw
2' yb 1'5" where ywo and yBo are equilibrium
(rest) values.
In general, A mB¥ accelÕn = F,
so,
m BØ B = ρwga(yw Ð yB) Ð m Bg,
equilibrium: but we substitute for m Bg the value from mBØB = ρwga(yw Ð yB) Ð ρwga(ywo Ð yBo)
= ρwga((yw Ð ywo) Ð (yB Ð yBo))
= ρwga(∆yw Ð ∆yB); ∆yw ≡ yw Ð ywo
4 since (ØBo) = 0, we have
ØB = (∆ØB), (where by (∆ØB) I mean the second derivative in time of (∆yB) ).
Thus,
mB(∆ØB) = ρwga(∆yw Ð ∆yB) [3.9] The overall volume of water in the tank is constant, and this is the same value at
equilibrium:
V = A⋅yB + (A Ð a) (yw Ð yB) = const
= A yBo + (A Ð a) (ywo Ð yBo) [3.10] thus
A(∆yB) + (A Ð a) (∆yw Ð ∆yB) = 0
or
(A Ð a) (∆yw) + a(∆yB) = 0 [3.11] Substituting this into (A Ð a) × I we get
mB(A Ð a) (∆ØB) = ρwag[(A Ð a) ∆yw Ð (A Ð a) ∆yB]
= ρwag[Ða∆yB Ð (A Ð a) ∆yB]
= ρwag[ÐA∆yB ]
So
m B(∆ y B)+ aA
gρ ∆y = 0
(A − a ) w B [3.12] Thus
ωo2 = aA gρw
(A − a ) m B Note: as A → ° , i.e., as the cylinder becomes huge, ωo2 →
T = 2π [3.13]
a g ρw
as above!
mB ( A − a) m B
aA g ρw [3.14] i.e., the period is smaller if the water level also rises and falls. 5 4) Full of fury, and signifying nothing...
a) The system w ill annihilate. There is a net force of
3kee
[towards centre of square]
2a2 (k = 1
).
4πε o on each particle
To see this, consider any particle Ñ the net force is, by symmetry, directed towards
the centre of the square. It has a magnitude of:
FNET = kee kee kee 3kee
+ 2− 2=
2a
2 a2
a2
a Since this result holds for any a, and the masses of each particle are equal, and they
start from rest, they will accelerate toward the centre of the square at the same rate.
Since they all start from rest, their velocities along the diagonals will be equal, and
hence their positions will be equal. So the formation will be preserved, and they will
keep accelerating inwards.
Their final velocities [same number] could be found from:
potential = 1 kqi q j 4kee 2kee
=
+
∑
a
2 i ≠ j rij
2a Hence, m V2 4kee 2kee 4kee 2kee
+
−
−
= 4 e b
a
2b
2a
2
where b is the final size of the box. Note that this gives infinity for b = 0.
b) This is not as easy as it looks.
The particles will obviously be repelled to some large distance, where their
interactions will be virtually zero. By symmetry, the final velocity of each of the
positrons will be equal and opposite; the same is true for the protons.
Conserving energy
E= kqq kqq
kqq
mV 2
MU 2
+
+4
=2
+2
a
2
2
2a
2a where V is the final speed of the positrons, U is the final speed of the protons and q
is the magnitude of the charge of an electron (k = 1/(4πεo)). The terms on the left are
due to: the energy of the positronpositron pair, the protonproton pair, and the
positronsÕ with their neighbouring protons (respectively).
6 Conserving momentum yields no new information Ñ it is conserved because of the
symmetry of the system.
Note, however, that because M >> m, the accelerations of the protons = 1/2000 of
accelerations of positrons Ñ thus, the positrons will escape before the protons will
have moved. I.e., the protons are virtually stationary. This gives
E= kqq kqq
kqq
mV 2 kqq
+
+4
=2
+
a
2
2a
2a
2a Solving, mV 2 = kqq
kqq
+4
a
2a [1] Now, that the positrons have escaped, we can consider the protons. We get: MU 2 = kqq
2a [2] Dividing [1] by [2] gives
mV 2
+1= 4 2 + 2
MU 2
V
M
=
4 2 +1
U
m ( ) Or, 115 plugging in M/m = 2000.
5) Sikorsky meets Newton
From Newton's 2nd law, the force is dp/dt.
Now, we'll put ourselves in a frame rotating with the blades.
At a distance r from the origin, an oncoming air particle has a horizontal velocity of
r*ω m/s.
We'll assume that the collision is partially elastic with a parameter k. I.e., after the
collision the air particle has a velocity k*r*ω (for perfect elasticity, k = 1).
After a collision, the particles will move downward with some speed U and
backward with some speed V. Note that the magnitude of their velocity is
krϖ = U 2 + V 2
By NewtonÕs second law there must be equal an opposite forces acting on the blade
Ñ thus there is a vertical component (lift) mU, and a horizontal one mV, slowing
7 the rotation of the blade. This would make the problem complicated, but luckily we
assumed that the rotation rate is constant (the engine compensates). (m = mass of
one air particle)
Now, after rotating through an angle dα, the area swept out by a blade is:
area = 1 / 2 * dα * a 2
The volume thus swept out is:
volume = 1 / 2 * dα * a 2 * b sin(θ )
And the mass hit is:
mass = ρ *1 / 2 * dα * a 2 * b sin(θ ) θ The downward component of k*r*ω (U) is: k rϖ cos(θ )sin(θ )
Note that this varies this r Ñ technically speaking, we should integrate here, but
weÕll make life easy and use the average r Ñ that is, R/2.
Thus, for 1 blade: ( ) F = ρ *1 / 2 * dα / dt * a 2 * b sin(θ ) * (kR / 2ϖ * cos(θ ) *sin(θ ))
For N blades: ( ) F = N ρ *1 / 2 * ϖ * a 2 * b sin(θ ) * (kR / 2ϖ *cos(θ )* sin(θ ))
BONUS:
Well, tiliting the rotor axis seems like a good idea (so part of the lift is sideways) but I
suppose too hard to build. What is done, instead, is something much more
interesting Ñ during a part of the cycle (say, when passing over the front) each
bladeÕs pitch (θ) is changed Ñ this adds an extra force, as necessary.
Electronic circuits control that ω is steady and the helicopter does not spin around. 8 6) Pulseheight analysis
a) The data in the table can be plotted as a histogram Ñ a range of values of the
output, and then a count of how many times the values fell into that range. Here is
such a table, and two different ways of plotting it:
14 N 137 Cs Pulse Height Distribution 12
10 Count 0
2
2
3
2
4
2
3
3
2
1
2
1
1
1
1
0
1
0
0
0
4
12
3
0 8
6
4
2
0
0 7 14 21 28 35 42 49 56 63 70 77 84 Range
14
137 Cs Pulse Height Distribution 12
10 Counts Xo
1.80
5.40
9.00
12.6
16.2
19.8
23.4
27.0
30.6
34.2
37.8
41.4
45.0
48.6
52.2
55.8
59.4
63.0
66.6
70.2
73.8
77.4
81.0
84.6
88.2 8
6
4
2
0
0 20 40 60 80 Pulse height range The first plot better shows how there are bins, and that the columns marked indicate
how many count datapoints fell into the range of each box of pulseheights. 9 For comparison, here
is how a Ôpulseheight
analyzerÕ package does
the job:
In the foreground is
the
Cs
gammaradioactive
source,
and the package which
holds
the
NaI
scintillator and the
photomultiplier tube.
In the background is
the
electronic
apparatus which sorts
pulses as they happen,
and adds them to bins
much as you did by hand.
b) The raindrops will have a t erminal velocity , i.e., some maximum constant
velocity. If the velocity is terminal (constant) there must be no net force on the
droplet. So we can find the terminal velocity by figuring out the balance of forces:
gravity and wind resistance.
Working from:
g
acceleration due to gravity:
ρ w density of water:
ρf density of air:
r
droplet radius Ð1 980 cm s
Ð3
1 g cm
Ð3
Ð3
1.2928 × 10 g cm at NTP we need to subtract the buoyant force of the air Ñ a small correction given the
density of water compared to air.
Fg = mdrop g − mair g = (ρw − ρ f )V g = (ρw − ρ f ) 4π 3
rg
3 [6.1] For wind resistance, with density of air ρf, drag coefficient CD, speed v and crosssectional area S:
Fd = 1
2 ρ f CD v 2 S = 1
2 ρ f CD vterm2 πr 2 [6.2] Thus we start with Fg = Fd so that: 10 1
2 ρ f CD vterm2 πr 2 = (ρw − ρ f ) 1
2 ρ f CD vterm 2 4π 3
rg
3 or, cancelling, 4π
= (ρw − ρ f )
gr
3 [6.3] If we assume that we have a sphere, we can consider our two limiting cases for CD:
we assume each case to start, and then see from the results what it will take to justify
the assumption.
Case I: Re < 80;
CD ~ 24/Re.
First we find vterm:
1
2 ρf 24
4
vterm2 = (ρw − ρ f ) g r
3
Re 6 η vterm
4
= (ρw − ρ f ) g r
3
r
2 (ρw − ρ f )g 2
vterm =
r
9η and substituting for Re
thus, solving for vterm , [6.4] Then this goes into the formula for the Reynolds number:
Re = ρ f vterm 2r
η (ρw − ρ f )2 g r 2 2 r
⋅
= ρf 9η η = [6.5] ρ f (ρw − ρ f )4 g r 3
9 η2 Now, using this we require that Re < 80, letting us find the corresponding condition
on r:
80 > Re = ρ f (ρw − ρ f )4 g r 3
9η2 then solving for r , 180 η2
=
r < 80 •
ρ f (ρw − ρ f )4 g r 3 ρ f (ρw − ρ f ) g
9η2 3 [6.6] 180 η2
r<
ρ f (ρw − ρ f ) g
3 ⇒ r < 0.0166 cm ; vterm = 333.8 cm s ±1
11 So this approach works for droplets smaller than 0.17 mm, for which the terminal
Ð1
velocity will be around 3.3 m s .
Case II: Re > 1000; CD ≅ 0.4
First we find vterm:
1
2 4
ρ f 0.4 vterm2 = (ρw − ρ f ) g r
3 vterm = 4
(ρw − ρ f ) g r
3
=
1
ρ f 0.4 thus, solving directly for vterm ,
[6.7] (ρw − ρ f ) 20
gr
ρf
3 2 Then this goes into the formula for the Reynolds number:
Re = ρ f vterm 2r
η
2 ρ f vterm 2r (ρw − ρ f ) 20
=
gr 3
η
ρf = ρ f (ρw − ρ f ) 80
2 η 3 [6.8] g r3 Now from this we require that Re > 1000, letting us find the corresponding r:
2
Re > 106
ρ f (ρw − ρ f ) 80 η 2 3 g r 3 > 106 r3 > 3η2106
80 g ρ f (ρw − ρ f ) r>3 then solving for r , 3η2106
80 g ρ f (ρw − ρ f ) [6.9] ⇒ r > 0.098 cm ; vterm = 703.4 cm s ±1
So this approach works for droplets larger than 0.98 mm, for which the terminal
Ð1
velocity will be around 7 m s . 12 M omentum
The microphone will approximately record the impulse, or momentum deposited
by the droplets, so it is the scaling between terminal m o m e n t u m mdrop¥ vterm and
droplet radius which is needed:
∆p = mdrop vterm = ρw V vterm = ρw ρw =
ρ
w 4π 3
r vterm
3 4 π 3 2 (ρw − ρ f )g 2
8π
ρw (ρw − ρ f )g r 5
r•
r=
3
9η
27η
= 5.07 × 106 r 5
4π 3
r•
3 (ρw − ρ f ) 20
4π
•
gr =
3
3
ρf (r < 0.016 cm ) ρw (ρw − ρ f ) 20
g
ρf
3 = 9.39 × 103 r 3.5 [6.10] 7 r2 (r > 0.98 cm ) We can plot this; for points inbetween, there is a plausible smooth curve to sketch
the transition between the two regimes.
101 9
14 momentum (g cm s– 1) 10 R e <80
R e >1000 9 10 104
101
106
1011
1016
1021
1026
0.001 0.01 0.1 1 radius (cm)
In order to find r adius from e xperimental m o m e n t u m data, we start on the yaxis
(ordinate) and look up the radius on the xaxis (abscissa). 13 For your interest, a fuller set of data for CDs for spheres Ñ found by experiments Ñ
looks like this: 1 02 CD 1 01 1 00 1 01
1 01 1 00 1 01 1 02 1 03 1 04 1 05 1 06 R
adapted from: A PhysicistÕs Desk Reference, H.L. Anderson, ed., American Institute of Physics, New York (1989). 14 19971998 Physics Olympiad Preparation Program
Ð University of Toronto Ð
Solution Set 3: Thermodynamics
NB: in thermodynamics, some of these questions draw on the relatively advanced
international syllabus of physics used in the International Physics Olympiad (IphO)
Ñ not all of which is taught in North America.
1) The heat is on at the Pentagon
Ð8
a)There was an error in one of the given constants. StefanÕs constant is 5.6696E
24
W/(m K ). All I can say is that you didnÕt notice the sign error in the constant
during the question, you would have found that your face would be radiating more
power than Pickering Nuclear Power Station can generate.
Even though the numbers were wrong, the given constants were meant to help you
if you have never seen this type of question before. To find the peak in the blackbody radiation curve, you could actual plot the black body radiation curve as a
function of wavelength for each temperature, but this is too much work. A little
calculus provides WienÕs displacement law, which you can use directly: λmax = 2898/ T
where T is temperature measured in Kelvin, and λmax is the peak wavelength in the
black body emission curve, measured in micrometers.
37°C human emits at 9.3um.
75°C car peak emission at 8.3um.
5°C terrain peak emission at 10.4um.
What is the power/area emitted? We look to StefanÕs law for that:
P/A= σ eT 4
Ð8 2 4 where σ is StefanÕs constant = 5.6696E W/(m K )
e is the emissivity (for a black object, e=1)
and T is measured in Kelvin
2 37°C human emits 520 W/m .
2
75°C car radiates 830 W/m .
2
5°C terrain radiates 340 W/m .
How about a flashlight? Assume the flashlight has a an excellent lens system that
provides a uniform beam of radius 5cm. A guess of a bulb wattage of 0.1W is not
2
bad, thus the power per unit area is 13 W/m . Surprising eh? Remember also that
the vast majority of this power is also emitted in the infrared (i.e. as heat). 1 b) Missile time!
In a very short time the missile surface heats up and starts emitting. How much
power is it radiating at equilibrium? The same as it is absorbing: 5MW (ow!). LetÕs
use the same expression as used above, assuming that the missile is painted a nice
dull black so that emissivity =1:
P/A= σ e T 4 For spot size of diameter 1m: T=3300K or 3000°C.
5
For a spot size of diameter 1mm: T=10 K ! Kappoey!!
2.
Planet X
a) Let us find first the value of the gravitational field constant for the new planet.
From the universal gravitation law we have for any mass m
2 mg = G (mM/r )
2 g = GM/r = ((6.7)(10 (1)
Ð11 25 14 )(10 ))/((9)(10 ))=0.74 N/kg The atmospheric pressure on the surface of the planet is equal to
p = ρ gh (2) where ρ is the atmospheric density. Assume that g is constant since h << r.
From the equation of state for ideal gas
pV = nRT = (M/(µ ))RT (3) ρ = M/V=(p(µ ))/(RT) (4) we find and from (4) and (2)
T = (µgh)/R = (10)(0.74)(5)/(8.3) = 4.5 K = Ð268°C (5) Unfortunately, it is too cold to live on this planet !!!
b) For the molar mass of our atmosphere (78% of nitrogen and 21% of oxygen), we
have µ = (0.78)(28) + (0.21)(32) = 28.56 g/mol. The EarthÕs atmospheric density
decreases at the same rate as the atmospheric pressure (1/2 per each h 0 = 5.6 km).
Ðx/ho
We can express this dependence as the function ρ(x) = ρ0Ê2
=Êρ0Êexp{(Ðln2)(x)/h0}, where x is the distance from the surface of the Earth. The
good approximation for the height of the atmosphere equivalent to ours but of
constant density can be found from the following condition:
∞ ∫ ρ (x) dx = r0 h 0 From this condition we obtain the value of h = (h0)/(ln2)=(5.6)/(0.693)=8.1 km. Now
we can use the formula (5) to calculate the temperature on the EarthÕs surface:
T = (µ)(g)(h)/(R) = (28.56)(8.1)(9.8)/(8.3) = 273 K = 0°C.
2 3.
Blow ye winds and crack your, er, skin ...
a) From the equation of state for ideal gas
pV = nRT = (m/µ )RT ,
where n is the number of moles and µ is the molar mass, we obtain ρ = m/V = (µ p)/(RT) = (µα p s)/(RT), (1) where ps is the saturation vapor pressure for given temperature T and α is the
relative humidity. The molar mass of water µ = 18 g/mol.
In November,
3 ρ1 = ((µ)(α1)(ps1))/((R)(T1)) = ((18)(0.95)(600))/((8.3)(273)) = 4.5 g/m .
In July,
3 ρ2 = ((µ)(α2)(ps2))/(R)(T2)) = ((18)(0.4)(5500))/((8.3)(308)) = 15.5 g/m .
Thus, in dry July (α 2 = 40 %), the air contains 3.4 times more vapor than in humid
November (α 1 = 95 %).
3 b)
The mass of water vapor in 1 m of outside air is equal to (0.25)(3.48) = 0.87 g.
When the outside air is heated from Ð4°C to 20°C, its volume increases. Therefore,
pV 1 /T 1 = pV2 /T 2 ,
3 V 2 = V1 (T2 /T 1 ) = 1.089 m .
3 3 The density of water vapor in the heated air is equal to 0.87 g / 1.089 m = 0.8 g/m ,
3
3
so the relative humidity will be (0.8 g/m ) / (17.3 g/m ) = 0.046 = 4.6 %. The air will
be uncomfortably dry. When the Santa Ana winds blow about this dry into Los
Angeles, some people get nosebleeds.
4.
Jessica cycles heat
a) The gas presses on the pump handle with a force = P A where A is the crosssectional area of the pump and P is normal atmospheric pressure. The distance the
handle moves is δx. The definition of work is the force vector dot product with the
displacement. In this situation, this gives that the work done by the gas is P A (δx).
You should be careful about signs here: I have implicitly defined that compression
yields negative δx and expansion corresponds to positive δx.
The work done on the gas is the opposite of this, thus:
W= ÐP δ V, where δ V=( δ x A) , where A is the crosssectional area of the bike pump.
b) Given : P′ = 1.0 atm (at steadystate with the room)
V′ = 1 l
V″ = 0.5 l 3 Using adiabatic compression expression we find that:
P ′/P ″ = (V″ /V ′) 1.4 so P
2.6 P″ = 2.6 atm
Using the ideal gas law: PV=nRT or for a constant
number of molecules: PV/T = constant. Thus we can say: 1.0 P ′V ′/(P ″ V ″ ) = T′/T ″
so that the final temperature is: T″ = 381K or about 110
¡C . This assumes that the system started at room
temperature (293K). 0.5 1.0 V c)Using part (a), the work corresponds to the area under the graph give in part (b).
Integration of this function will provide an accurate value, but you can approximate
the curve as a straight line connecting the beginning and final states. Thus the area
is a square + a triangle:
Work = ÐArea = Ð(Ð0.5 l*(1 atm)+(1/2)*(Ð0.5 l)*(2.6 atm Ð 1 atm)) = 0.9 atm l
or in the more usual units for work:
Work = 91 J P
d) Using the ideal gas law again:
P ′V ′/(P ′″V ′″) = TÕ/T′″
but here
TÕ = T′″
so 2.6
2.0
1.0 P′V′ = P′″V′″ giving 0.5 P ′″ = 2 atm
There is no work done on the gas, as can be seen in the
PV graph (area under the curve is 0). 1.0 V P e) Given : P″′ = 2.0 atm (at steadystate with the room)
V ′″ = 0.5 l
V ″″ = 1.0 l 2.0 Using adiabatic compression expression again we find
that: 1.0
.76 P″″ = 0.76 atm 0.5
4 1.0 V Using the ideal gas law again:
PV/(P ″″ V ″″ ) = T/T″″
so T″″ = 223K or Ð50°C Wow eh!
Using a linear approximation:
Work= ÐArea= Ð.69 atm l = Ð70 J
Thus the gas does work of 70 J (the minus sign indicates that the gas is doing work
on something, not that work is being done on the gas).
P
f) Easy one:
final pressure:
final T: 1 atm
room temperature (293K) No work done on the gas, since the area under the curve o n
the PV diagram at right is again 0 (as in section d). 1.0
.76 g) Net work on the gas is 91 J + 0 J Ð 70 J + 0 J = 21 J
0.5
h) No heat is transferred in the adiabatic compression and
expansion. It is only transferred when the gas is held at constant volume.
δT for the cooling stage (stage 2): 381KÐ293K=88K δT for the warming stage (stage 4): 1.0 V 223KÐ293K=Ð70K The total number of moles in the gas is: n=PV/(RT)=0.041 mol
Thus the total heat transferred from the gas to the surroundings is:
total heat = 18K * 2.98 cal/(mol K) * 4.19J/cal * 0.041 mol =9 J
We would expect this to equal the work done on the gas (calculated above as 21 J).
The reason for the discrepancy is due to the linear approximation. If the area was
calculated accurately, these two numbers would agree.
5.
Batboy chills
a)
Heat transfer:
adiabatically insulated.)
Work: Q=0 (No heat flows into the system, as it is W = 0 (No net work done by the system) The energy change in the system after opening the valve is
U 2 Ð U1 = ∆U = ÐW + Q (1st law of thermodynamics) = 0
So
U 2 = U1, or U(T1,V 1) = U(T2,V 2). 5 For an ideal gas, U is independent of volume, so U(T1) = U(T2), meaning that T1 = T2.
Unless the gas in V 1 was initially very cold (which wouldÕve required fancy
refrigeration), Batboy would have nothing to fear.
b) As for part (a), U(T1,V1) = U(T2,V2)
2
For our interacting gas, ∆ U = cv ∆ T Ð a/v ∆ V, where here the ∆ now refers to the
difference in T and V in each compartment before and after the valve was opened.
Relative to some arbitrary initial temperature To and volume Vo,
T dV 1
o (V 1 )2 V U1 = ∫T 1 Cv (T 1 )d T 1 + a∫V
o T V U 2 = ∫T 2Cv (T 1 )dT 1 + a∫V 2
o
o dV 1
, where CV (T) = heat capacity
(V 1 )2 1
dV1
T
V dV
= ∫T 2Cv (T 1 )dT 1 + a∫V 2 1 2
o (V 1 ) 2
o
o (V ) T V U1 = U 2 ⇒ ∫T 1 Cv (T 1 ) d T 1 + a∫V
o Assuming Cv doesnÕt change with temperature,
CvT1 Ð a/V1 = CvT2 Ð a/V2
so
(T2 Ð T1) = a/cv(1/V2 Ð 1/V1)
With T2 = Ð5οC
a = 0.15 T 1 = 1 5ο C
Cv = 21 J/(mol K) V 2/V1 = 2800 V2 +1, so V2 would have to be very large!
6.
The Ôthermal physicsÕ (statistical mechanics) of computing
a) For a system of 2 magnets, U = Ðm l, l2. The possible configurations of the
magnets are:
Both up: ↑
l = +1 ↑
l = +1
↑ first up, second down:
l = +1 ↓
↑ l = Ð1
both down: U = +m
U = +m l = Ð1
↓ first down, second up: U = Ðm l = +1 ↓
l = Ð1 ↓
l = Ð1 U = Ðm So we want spins to be parallel to one another, to minimize the internal energy U. 6 For N magnets, all nearest neighbours have the same interaction as for 2 magnets:
Ex.
↑
1 ↓
2 ↓
3 ↑
↓
(iÐ1) i ááá Interaction energy for the i th ↑
ááá
(iÐ1) ↑
↑
↓
(NÐ2) (NÐ1) N magnet: U l = Ðm (liÐ1 lI + lI lI+1)
Interaction energy for the (i+1) th magnet: U l+1 = Ðm(lI lI+1 + lI+1 lI+2)
Note that U l + U l+1 counts the interaction between l and l+1 twice, which is
incorrect. If we follow this through for the whole chain, we find:
N −1 UTOT = − M ∑ li li + l
i....= 1 So UTOT is maximal if all magnets are aligned in the same direction. ∴ (all spins up
or all spins down.
b) Energy difference = (energy before flipping a spin) – (energy after flipping a spin)
If the flipped magnet was at the end of a chain, the change in energy is due to 1
bond.
∴ energy increase = +m
Otherwise, a flipped spin affects 2 bonds.
∴ energy increase = +2m
So, flipping a spin from the parallel state increases the internal energy.
c) For 3 magnets:
i) U = Ðm(1+1) = Ð2m: ↑↑↑ ii) U = Ðm(1Ð1) = 0: ↑ ↓ ↓ ↓↑↑ iii) U = Ðm(Ð1Ð1) = 2m: ↑↓↑ ↓↓↓
↓↑↓ ↑↑↓ 2 configurations
↓↓↑ 4 configurations
2 configurations i) → lowest energy, low entropy
ii) → moderate energy, high entropy
iii) → high energy, high entropy
d) F = U ÐTS
i) If U > TS, a state with lower internal energy, which has a low entropy, is
preferred. This means that internal energy determines the state of the system and so
all the spins will tend to align in parallel. This is often called the Òordered stateÓ. 7 ii) If TS > U, a state of higher entropy is preferred; in fact, the system will want to
be in the energy state with the most number of configurations available to it. This is
typically the most ÒrandomÓ state of the system.
iii) U = TS defines the ÒorderdisorderÓ transition
e) This depends on temperature. If U > TS, then the configurations
↑↑↑↑↑
1 1 1 1 1 = 31 and ↓↓↓↓↓
0 000 0 =0 are the easiest to code. Conversely, if TS > U, then the ÒrandomÓ states, such as
↑↓↑↓↑
1 0 1 0 1= 21 or ↓↑↓↑↓
0 101 0 = 10 are easiest. Other values cost energy and entropy Ð you can see how much of a free
energy cost there is in constructing other states by evaluating U and S for a given T.
BONUS:
For 1 spin, state possible = ↑ + ↓
2 spins, state possible = (↑ + ↓) (↑ + ↓) = ↑↑ + ↓↓ + ↑↓ + ↓↑
3
3 spins, state possible = (↑ + ↓)
N
N spins, state possible = (↑ + ↓) Let x = ↑ and y = ↓
By binomial theorem, we would know right away for x and y:
N
1
N!
N ( N − 1) x N − 2 y 2 + ... + y N = ∑
x N − t yt
2
t = 0 (N − t)!t!
N −s
and sum over s
Let s = Nup Ð Ndown. Replace t by
2 (x + y) N = x N + Nx N − 1 y + (x + y) N N! s= −N N N + s ! N − s ! 2 2 =∑ 1
(N + s)
x2 1
(N − s)
y2 well, the same form also applies to writing out our combinations of ↑ + ↓
(↑ + ↓ ) N N N! s= −N N + s ! N − s ! 2 2 =∑ 1
1
(N + s)
(N − s)
2
↑
↓2 The number of states with given energy is determined by # of up and down spins.
This is given by the coefficient of the binomial term with value s.
∴ g(N, s) = N!
N+s N−s
!
!
2
2 ( )( ) 8 19971998 Physics Olympiad Preparation Program
Ð University of Toronto Ð
Problem Set 5 Solutions: Electricity and Magnetism
1) The Earth as Cosmic Doorknob
The charge accumulated by the Earth during the time interval t is equal to
2
2
qÊÊ=ÊaSqpt, where a = 1 proton/cm .sec is the flow density; S = 4πR is the area of
6
Ð19
the EarthÕs surface; R = 6.4 × 10 m is the radius of the Earth; qp = 1.6 × 10 C is
the charge of proton. The critical charge q of the Earth, which will prevent
9
protons with the kinetic energy E < 4 × 10 eV from reaching the Earth, can be
found from the following equation:
qqp /(4 πε 0 R) = aStqp /(4 πε 0 R) = E ,
where
( 4 πε 0 )−1 = 9 • 10 9 Nm 2 C −2
From this equation we have:
2
2
t = ( 4 πε 0ER)/( aSqp ) = (Eε 0 )/( aqp R) = 108 /((3.14)(0.9)(1.6)(6.4)) = 3.45 • 106 sec = 40 days 9 This value is tiny compared to the age of the Earth, which is equal to 5 × 10
years. Protons from the cosmic rays continue to reach the Earth because their
charge is compensated by the charge of electrons also reaching the Earth. Since
their energy is extremely small, they are not considered as a component of
cosmic rays.
2) C.C.? No! No!
a) At a distance of 5 cm from a large sphere, we must treat only that section of
the sphere near where you are standing (your closest point to the sphere). This is
for the same reason that we can treat the earth as being flat locally at its surface
even though it is really a sphere.
E field of a spherical conductor with charge Q.
NOTE: the charge on a conductor distributes itself uniformly on the conductorÕs
surface.
GaussÕ Law
charge enclosed
ε0
Q
E • 4 πr 2 =
ε0 ∫ E • ds = 1 ∴E = Q
4 πr 2 ε o ∴ at 5 cm:
0.8 × 10 3 V
Q
=
⇒ Q = 2.2 × 10 −10 C
2
m 4 πε o (.05) b) For the large sphere,
1 q1
V1 =
4 πε o r1
Small sphere,
V2 = 1 q2
4 πε o r2 But since the spheres are part of the same conductor, the potentials of the spheres
must be equal:
1 q1
1 q2
V1 = V2 ⇒
=
4 πε o r1 4 πε o r2
∴ q1 q2
=
r1 r2 The electric field at the surface of each sphere is given by
q1
q2
E1 ∝ 2 , E2 ∝ 2
r1
r1
With the same constant of proportionality for each.
∴The ratio of the electric fields is E1
=
E2 q1 / r12
2
q2 / r2 q1 r =1 q2 r2 1
r1
1
r2 but
q1 q2
=
r1 r2
E
r
∴ 1= 2
E2 r1 So the field is higher on the surface of the smaller sphere. A more sharply
curved or ÒpointyÓ surface on a conductor has a higher electric field and
therefore reaches spark discharge for a lower applied charge.
I have heard that there was an ongoing disagreement between Benjamin Franklin
in the U.S. and European investigators as to what the best shape for a lightning
rod should be: sharp spike or small ball. The argument went like this: all this,
above, is true, but the strong field around a sharp spike might cause air to ionize 2 as the rod began to charge up. If so, there would be glob of ionized air (plasma)
near the rod, and plasma is a conductor Ñ so the effective conductor would have
a larger radius than if a lightning rod had a ball on the top to start with!
3) Goats & Sheep Ð the Hall Effect in
plasma
– – –++ + +
The point here is that moving charges in a
magnetic field see a force Ñ the Lorentz
– – – + +++ ++
++ +++
++
force:
++
v
v
–
v
++ +
FLorentz = q v × B
– – + + ++ + ++ ++ +
–– – + + – +– + + ++ +
The Lorentz force, on each particle, will
––– – ++––– + +++ + +
– + –++– + + + +
be:
–– + + +
– – – –– – –+– ++ + +++ +
+
Ð19
Ð1
– –– + +
+
1.6 × 10 C ¥ 0.3 m s ¥ 0.3 T
–– – ++++– + +++ + ++
––+
– –– –– ++ + +
+–
Ð19
+
–
(opposite
= 1.44 × 10 N
directions for electrons, ions)
A block of plasma, with the electrons
pulled aside from the ionsÉ
So when the flowing gas becomes
ionized, the electrons and ions will
separate from each other in a Bfield Ñ the Lorentz force pushes opposite ways
for the opposite charges.
Do they end up going in circles in the Bfield? Probably not, because as the
charges separate, they are still attracted to one another by the Coulomb force:
v
v
FCoulomb = q E
Then the net force is going to be:
v
vvv
F = q (E + v × B)
Basically, stream of electrons will move aside from the gas stream, and the
stream of positive charges will also move a little in the opposite direction. The
separated charges lead to an Efield, and to an attraction between the two
streams. The two will balance, so that F = 0
What will be the electric field to cancel this?
Ð19
1.44 × 10 N = q E, so
Ð1
E = 0.9 V m
How much will the charges separate from each other, if we assume they keep
their original density, and just separate as a stream? We can figure out the
23
details: air at STP is 22.4 l per mole, and 1 mole = 6.022 × 10 particles. From
this we find the number of particles per unit volume: ++ +– – –
–
++
+ –––– ––
–
––– – –
–
+
–– ––
–
+
––
+ + – – –– – –– – –
+ + + –+ ++–– – ––– –
+ – –+
–
–
++ + + +
+ + + +– + – –
+ +–– +–+–– – – – – – –
– + – +– – – –
–
+
++ + ++–
+ –––– +–– – – – ––
+ + ++++ +–– – ––– –
–
–– + N
6.022 × 10 23
=
= 2.69 × 1019 cm −3
3
3
V 22.4 × 10 cm
17
3
With 1% ionization, this means we have 2.69 × 10 electrons per cm , or 2.69
23
3
× 10 electrons per m , and the same density of positively charged ions. If the
electrons are just pushed somewhat off the ions (the ions are much more
massive, and move less from the forces), what we have is something roughly like
n= 3 a parallelplate capacitor, with the two thin excesscharge slabs as the two
charged plates of the capacitor.
For this, we can find the field Ñ the same as between the two plates of a parallelplate capacitor. In the middle, where there are both electrons and ions, the
charges of each cancel each other out (unless you give them time to move around
and redistribute themselves), so the net field in the middle is just what is
produced by the thin slabs of excess (unbalanced) charge.
E = 4 π kc σ where kc = 9 × 109 N m2 CÐ2
σ = charge per unit area on each plate
The charge per unit area of the two slabs of excess comes from the charge density,
and the distance the electrons are pushed off the ions:
σ = (n e (∆x) × A) / A
= n e (∆x)
and the field between the plates is:
[5]
E = 4 π kc n e (∆x)
or
∆x = E/(4 π kc N e )
Ð16
= 1.84 × 10 m
A very tiny displacement (actually nonsensically tiny, given the size of an atom!).
Consider, though, that if the jet is 1mm across, the potential measured across the
jet from one side to the other would be 0.9 mV.
4) Shepherding fields
a)For the four cases suggested in the problem:
i) v=0 There is no force acting on the particle and
therefore no acceleration. B
B ii) vparallel nonzero: The crossproduct is zero,
therefore same result as first case. iii) vperpendicular nonzero: The crossproduct
simplifies to a multiplication and the resulting force
is always acting perpendicular to the motion. The
particle moves in a circle. From the given equation
and knowing that for perfect circular motion (such
2
as a satellite in orbit) the acting force is F = mv /r we find the radius of the circle
is: r = mv/qB
v out of page B v out of page B iv) We combine the results from sections iii) and ii)
and get helical motion. 4 Since in all cases the velocity of the particle is always orthogonal to the acting
force, the work done on the particle is 0.
b) As given in the question, kinetic energy is conserved. Thus the square of the
magnitude of the total velocity at any given point in time is constant:
2
2
2
v(z) =vper(z) +vz(z) =constant
where v is the total velocity, vper is the component perpendicular to B (i.e. in the
XY plane) and vz is the component parallel to B (i.e., parallel to the z axis). We
want to find vz(z):
2
2
2
2
vz(z) = vper(0) +vz(0) Ð vper(z)
We also know that flux is conserved:
B(0) A(0) = B(z)A(z)
where B is the magnitude of the magnetic field and A is area enclosed by the loop
that is orthogonal to the field. Combining this with the result from part a):
2
2
vper(z) /B(z) = vper(0) /B0
This gives:
2
2
2
2
vz(z) = vz(0) Ð vper(0) (b0 z ) / B0
c) Shades of Bob, itÕs simple harmonic oscillator! KE is transferring from parallel
motion to perpendicular circular motion. The particle swings back and forth
along the Z axis, in a similar way to a weight on a spring, with the added circular
motion around the Z axis
2 2 2 d) vz(z)=0 when z =(B0 vz(0) ) / (b0 vper(0) )
Total KE is conserved. Energy from translational KE transfers to rotational KE.
5) Science on a shoestring
First we suspend the magnet from the stand (using the string). We tilt it by a
small angle and let it oscillate.
It undergoes oscillations (much like a
torsional pendulum) such that:
⋅⋅ I θ = −µ o mBz sin(θ) ≈ −µ o mBzθ
here, Bz is the horizontal component of the
EarthÕs magnetic field, m is the magnetic
moment, and I is the moment of inertia of
the magnet. Treating it as a thin stick,
I= M d2
12 Now, the motion is simpleharmonic with a period of 5 T = 2π I
mBz µ o Now, we align the compass to the north. Then, we place the magnet some
distance r >> d away from the compass and measure the deflection angle β.
North Clearly, then,
β tan(β) = 2m
B
=
Bz 4 π r 3 Bz Combining, we get
r >> d Bz = 2π I
µ o T r tan(β)
23 Ideally, one would collect a set of βs
by varying r, and plot the equation (on
loglog paper), and get Bz that way, but that is a lot of work (when done by
hand).
And thus, plugging in one set of numbers will do.
To measure T, it is best to time say 10 periods, and divide the result by 10 (as
long as the time does not decay too much). r is simply measures with a ruler.
Measuring β might be difficult, and this might be the biggest error here. In fact, it
ο
might be best to set β at, say 45 , and then change r to obtain that deflection (to
make measuring it easier) Ñ but one has to be careful that r >> d!
To get the error, we note that δBz/Bz ≈ δX/X, where δX/X is the biggest error
found (most likely for β or r; δX means error in variable X).
Ð4
The accepted value is about 0.15 × 10 Tesla, BUT it varies a lot because of
interference, etc. Doing the experiment outside would thus be ideal.
6) How much do you charge for a free ride?
a) B circles around a straight
wire in a direction dictated by
the righthand rule (which
comes from the BiotSavart law). B I b) For a timevarying magnetic field in the z direction, the induced electric field
circles the magnetic field direction, like the magnetic field around a wire. The
direction of the induced E field is such that it resists the change in the magnetic
field (Lenz' Law). (figure next page) 6 c) ∫(E•dl) around the loop
= Ð∫ ((dB/dt)•ds on a surface
bounded by the loop
= ÐdΦ/dt, where Φ(t) is the magnetic
flux through the loop at time t.
In our case, making our loop of radius a,
centred at the centre of the wheel,
2
∫(E•dl) = Ð πa (dB(t)/dt)
d) The total torque on the wheel about
its axis:
N = r × F over the whole wheel.
For a piece of the wheel of length δl, r
and F are perpendicular and F = qE =
(λδl)E. Therefore
δN = rλδlE
For the total torque, we integrate around
the whole wheel
dB(t) N = r λ ∫ Edl = r λ − π a2 dt B(t) increasing direction of induced B(t) decreasing direction of induced where we have used the result of part (c).
The total angular momentum is
L = ∫Ndt
where the integral is from an initial time to just before the B field is changed to
the final time t just after the B field has stopped changing.
2
L = Ðρλ πa ∫ ((dB(t)/dt)*dt)) (from to to t)
2
= Ð ρλ πa ∫ (dB) (from Bo to 0)
as initially B=Bo, and finally B=0.
2
so L= ρλ πa Bo
e) From part (d), we see that the value of the final angular momentum is
independent of how fast the field is switched off; it depends only on the initial
field value, or, more generally, it depends on the difference between the initial
and final field strengths
BONUS
f) Marks were given for any interesting answer Ñ the question is relatively
sophisticated.
The torque on the MerryGoRound wheel comes as a result of the Efield
produced as the Bfield produced by the solenoid vanishes (dB/dt  0). This
induced Efield accelerates the charges into circular motion, taking the wheel to
which they are bound along too.
7 XX: Some might argue that this new circulating current produces a Bfield of its
own; the charges (already moving) in the solenoid see this changing Bfield from
the MerryGoRound disk (dB/dt  0) as it induces an Efield. However, this is a
bit of a redherring Ñ the solenoid has equal numbers of positive and negative
charges, so it doesnÕt pick up momentum this way. Someone might argue that
the Bfield pushes the electrons to one side of the wire (the Hall effect, see Q. 3)
so that the angular momentum they pick up doesnÕt match the positive chargesÕ,
but thereÕs no hope here!
ÃÃ: Basically, the wheel will accelerate when it sees the Bfield changes, even if
the solenoid is a long, long distance D away. Conservation of angular
momentum canÕt Ôhold offÕ, or wait, until the solenoid ÔlearnsÕ that it is pushing
on the wheel Ñ a time D/c at the earliest, where c is the speed of light. So the
angular momentum must be in the electromagnetic field; this changing
electromagnetic field must have angular momentum which is ÔleftÕ with the
wheel. The changing fields the wheelÕs motion creates will reduce the angular
momentum of the overall electromagnetic field, as the (kinetic) angular
momentum of the wheel increases. 8 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 1: General
Due November 6, 1998
1) Pushy photons
In an episode of Star Trek: Deep Space 9 , Capt. Benjamin Sisco makes a hobby out of
recreating an early spacecraft, from alien archaeological records. This spacecraft uses
huge sails to move the ship along on the solar wind (a stream of charged particles) or by
light pressure from the sun.
Consider such a craft the same distance from the sun as is the earth, and only consider
light pressure.
a) What is the best colour for the sail?
b) For a sail area the size of a soccer field on a small ship of 2,000 kg, how fast will the
ship be moving after 24 hours? After 1 year?
In my research laboratory we have a laser which produces a terawatt of power (1012 W)
in a pulse of light about one picosecond (10Ð12 s) duration. This pulse can be focussed
to a spot about 10 µm across.
c) How much force from light pressure does such a pulse exert? Roughly how much
momentum is transferred? What is the pressure exerted by the light? [Robin]
2) Out of sight, hope you donÕt mind
In electronics, a Ôblack boxÕ is a mystery package with electrical terminals. You try to
figure out what is inside it by making measurements across the terminals.
a) Is the following black box possible, if only passive elements are used (i.e., resistors,
capacitors and inductors, only)?
Two pairs of terminals come from the box. If a battery of voltage V is connected across
the terminalpair ÔAÕ, the voltage measured across terminalpair ÔBÕ is V/2. But Ñ
reversing things Ñ when the battery is connected across terminalpair ÔBÕ, the voltage
measured across ÔAÕ is V. If it is possible, what is inside? [123 Tricky] 1 3) LeonardoÕs camera obscura
Simple ray optics can be demonstrated using a pinhole camera . This consists of an
opaque sheet with a small hole in it, and a viewing screen (see layout below). Rays
from an illuminated object travel through the pinhole and form an image on the
viewing screen. [You can make a pinhole camera yourself: take a shoebox and make a
small hole in the centre of one of the small sides. Cut out the side directly opposite it
and replace it with wax paper. In a darkened room put a candle in front of the pinhole.
You will be able to observe the inverted image of the candle on the wax paper.]
If you use something other than a small hole, the image you see will be distorted. In the
following five cases, you are shown two of the three components (object, opaque sheet
with hole in it, and image) and your task is to draw the missing component that is
consistent with the other two. To make things simpler, I have drawn the objects from
the observer's point of view (see below). Don't worry about the exact proportions or
intensity, just draw the shape that is consistent with the other two shapes.
(For you fancypants out there, ignore diffraction and anything other than simple ray
optics, i.e., take it that light travels in straight lines).
Example of imaging with a pinhole camera:
i) View from side:
Object Opaque sheet Screen Observer ii) From observer's point of view: Illuminated object Opaque sheet Image on screen 2 All the following are from the observer's view: a) Illuminated object Opaque sheet Image on screen b) Illuminated object Opaque sheet Image on screen c) Illuminated object Opaque sheet Image on screen d) Illuminated object Opaque sheet Image on screen 3 e) Illuminated object Opaque sheet Image on screen [James]
[Artists used pinhole cameras in their invention of perspective drawing. But their cameras were entire
rooms, and the hole was in a wall. They could sketch on the opposite wall. This was the camera obscura.] 4) Heat works
Consider two identical spheres made of aluminum. One of the spheres lies on a heatinsulating plate, while the other hangs on a heatinsulating thread. The same amount of
heat is transferred to the two spheres. Which one will have the higher temperature?
For spheres of 100g each, and a heat of 250kJ, can you estimate the difference in
temperature? [123 Tricky]
5) Death of an Atom
Before quantum mechanics saved the day, it was noticed that there was a pretty basic
problem with the atom. If one thinks of the atom as an electron orbiting the positively
charged nucleus Ñ as a planet orbits the sun, but with electric fields instead of gravity
Ñ it means that the electron is accelerating because of centripetal force. But it is wellknown that an accelerating charge radiates electromagnetic waves: light, xrays, etc.
This means the electron will lose energy, and eventually be drawn into the nucleus, so
atoms cannot survive by classical physics alone. How long does it survive?
The power radiated by an accelerating charge is given by:
dE K q 2 2
= 3a
dt
c 9 2 Ð3 Ð1 where K = 6 × 10 N¥m ¥C , c = speed of light [m¥s ], q = electron
Ð2
charge [C], a = instantaneous acceleration [m¥s ]. Assuming the electronÕs orbit is a circle (atom is Hydrogen)
a) Find the energy radiated during 1 cycle (frequency is f)
b) At this rate, how many cycles before the radius is down to practically 0?
4 c) Therefore estimate the classicalphysics lifetime of a hydrogen atom. [Peter & Robin]
[This radiation by accelerating charges is the principle of a broadcasting antenna for radio, TV or
microwaves.] 6) Stupid bet tricks
Consider the following
arrangement of balls on a
pool table: 4 balls are lined
up (touching each other)
next to a pocket. Next, the
cue ball is aimed exactly
between the two middle
balls. My friend claims that
he can sink all the balls this
way. Is he right? Assume an
ideal table (no rolling
friction) and slippery balls.
Calculate the final velocities of all balls, given the initial velocity of the cue ball is V
Ð1
m ¥s , that all collisions are perfectly elastic (energyconserving), and that all balls are
identical.
As a matter of fact, if the balls are hit sufficiently hard they will all sink. If you found
this to be impossible, above, then explain how it actually happens. [HINT: what if the
slippery assumption doesnÕt hold?] [Peter]
[Try www.physics.utoronto.ca/~POPTOR, to discover if we succeeded in putting a movie of this there!] Useful Bits of Information
Al:
Ð5 linear expansion coefficient = 2.31 × 10
heat capacity = 24.4 J K mole
3Ê density = 2.7Ê×Ê10 kgÊm K Ð1 Ð1 Ð3 Light intensity from sun at earthÕs orbit
Ð2 Solar constant = 1.353 kW m 5 19971998 Physics Olympiad Preparation Program
Ð University of Toronto Ð
Problem Set 6 Solutions: AC Circuits and Electronics
1) MishaÕs MomÕs Medallions
Faraday law of electrolysis states that the mass of substance liberated in
electrolysis is proportional to the charge passed:
m = kq = kIt = (m0/F)It,
Where k is the electrochemical equivalent of the substance; m0 is a molar mass;
F = NAqe is the Faraday constant which is equal to the product of the AvogadroÕs
number and the electron charge. The mass of medallion m = Vd = Ahd = m0It/F.
Thus the period of time for silverplating the medallion t = FdhA/m0I =
((9.65)(1.05)(5)(5.7)(10))/((1.1)(1.8)) = 1458 sec = 24.3 min. Therefore, Misha has
to pay 4 dollars for his momÕs birthday present.
2) Tesla Yes! Edison No!
The electrical power delivered to the factory P = VI = 55 kW. The transmission
line conductors have a resistance of 3Ω ⇒ the voltage drop along the line
U = 1,500 V. Therefore, the output voltage must be 1,500 V + 110 V = 1,610 V to
deliver 110 V to the factory. The power loss is P1 = 2 UI = 1,500 kW, which is 27
times greater than the delivered power.
If we use transformers, we can claculate the turns ratio n2/n1 = V2/V1 = 500 and
the current in transmission line I2 = I1 . (n1/n2) = 1A. The voltage drop along the
Ð3
line U = 3 V, which is only 5 • 10 % of V2 = 55 kV. The power loss in the line
Ð2
now is equal to P1 = 6 W, which is only 1 • 10 % of the value of the delivered
power.
3) Analog differentiation and integration
a)we have
q
dq
+ R = Vin
C
dt
dq
R = Vout
dt [1]
[2] Now, for low frequencies, the capacitor is charging/discharging all the time, and
most of the voltage drop occurs across it [Rcapacitor >> R, if you like]. Hence, we
can drop the Rdq/dt in eq. [1]. Differentiating eq. [1] and plugging into [2] gives: 1 q
≈ Vin
C
d q dVin
=
dt C dt
dq
dV
V
= C in = out
dt
dt
R
Or, dVin
dt
[condition: low frequencies; input frequency << 1/RC]
Vout = RC Plots
i) iii) dotted line: output (cosine)
normal line: input (sine) dotted line: output
normal line: input ii) 0line: output
normal line: input NOTE: plots are not to scale!
b)Circuit:
R Vin C Vout we have
q
dq
+ R = Vin
C
dt
q
= Vout
C [1]
[2] Now, for high frequencies, there is not
enough time for the capacitor to be
2 charged, and most of the voltage drop occurs across the resistor [Rcapacitor << R,
if you like]. Hence, we can drop the q/C in eq. [1].
Differentiating eq. [2] and plugging into [1] gives:
dq
R ≈ Vin
dt
d q dVout
=
dt C dt
dq
dV
V
= C out = in
dt
dt
R
Or, 1
∫ Vindt
RC
[condition is: high frequencies; angular frequency of input signal >> 1/RC]
Plots
i)
iii)
Vout = dotted line: output (cosine)
normal line: input (sine) dotted line: output (parabolas)
normal line: input ii) dotted lines: output
normal line: input NOTE: Plots are not to scale!
4) What goes around, comes around V a) If the voltage across the battery is V, then
the voltage across the capacitor is ÐV in
steady state. 3 b) From the rather direct hints in the question,
the voltage across the capacitor looks like:
dVc/dt ∝ I(t), where ∝ indicates proportional,
I(t) is the total current running through the
circuit as a function of time, and Vc is the
voltage across the capacitor
Again, from the ohsosubtle hints, the voltage across the inductor is:
Vi ∝ dI(t) / dt
From KirchhoffÕs voltage law, the sum of voltages in a circuit must be zero. For
this to be true for all time, the change in voltage across one device must equal the
opposite change in voltage across the other device. This allows us to write:
dVc/dt = ÐdVi/dt
2 2 I(t) ∝ Ðd I(t) / dt
This is the same as
2
2
x(t) ∝ Ðd x(t) / dt
which is the equation for a simple harmonic oscillator, as seen in a previous
problem set featuring everyoneÕs favorite oscillator, Bob (no relation to the
author).
The solution to this is I(t) ∝ cos(kt) where is k is some constant dependent on the
inductance and capacitance in the circuit. If you are not sure if this is a proper
solution, substitute it into the equation and verify that the LHS=RHS.
(I am being a little lax here about the boundary conditions, i.e. what is the
current at I(0). By writing the solution as cos(kt), I am assuming that the initial
current is nonzero, so that we are starting the clock just after the battery has
been shorted and removed, not before.)
c) Energy transfers from the electric field in the capacitor to the magnetic field
around the inductor and back again.
d) As soon as Frido tries to do something with his circuit, he will no longer have
a perfect LC circuit. He must put a resistor in series with it:
2 As you know, electrical energy is turned into heat in a resistor according to RI
where R is the resistance and I is the current flowing through the circuit.
Without even solving the circuit equation, you can show that energy is leaving
the circuit. Alas, poor Frido has been boondoggled.
5) Operation: ÔAmplifierÕ
a) i) 0V
ii) Ð15V
iii) 15V b) With any opamp circuit there are two possible results: either the inputs are
different and the opamp is in saturation, or else the inputs are driven so that
4 they are at the same voltage. Feedback allows the the second result in each of the
given cases.
i) The current flowing through the feed back loop must be the same current
flowing out the V line. We try to find a selfconsistent situation were VÐ =0. For
this to happen, the current must be 1V/100 Ohms. Thus Vout = Ð1V. This circuit
inverts the input voltage
ii) Same reasoning as in part (i). Vout = Ð5V. This circuit inverts and amplifies
with a gain of five, the input voltage.
iii) Vout = Ð1.9V. This circuits adds and inverts the input voltages.
iv) Vout = 2V. This circuit doubles the input voltage (no inversion)
6) FlipFlopping on a counter proposal
a) If R and S are both set to 1, there are 2 equally stable output configurations:
Q=0 and Qbar=1, or Q=1 and Qbar=0. If we switch R to 0 and back to 1, Q and
Qbar will switch from the stable configuration they started in to the other stable
configuration. Thus a pulse in R causes the outputs to "flip flop" between the
two stable configurations.
b)
pulse A B C 0
1
2
3
4 0
1
0
1
0 0
0
1
1
0 0
0
0
0
1 We can see that if we let the output of A be the 20 binary digit, B be the 21 digit,
and C the 22, this counter gives the binary value for the number of pulses that
have been fed into the input (e.g., 4 = 100 base2) 5 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 1: General
1) Pushy Photons
a) Black is good Ñ it would absorb all the light, and therefore transfer all the
momentum of the light. But silver would be better Ñ when the light reflects back
to where it came from, it reverses its momentum. So this would mean (pÐ (Ðp)) =
2p for the momentum transferred to the sail.
b) From the useful bits of info at the bottom of the question sheets, the solar
Ð2
constant is 1.353 kW m . For light, E = pc, so by taking derivatives we also have
that: dE dp
=
⋅c
dt
dt
However, power P is defined as the rate of change of energy with time, on the
lefthand side, and force F is defined as the rate of change of momentum with
time, on the righthand side. So we can write:
P= dE dp
=
⋅c = F⋅c
dt dt or
F= P
c or actually F = 2P/c if we let the light be reflected.
A soccer field isnÕt always the same size, but has to be in a certain allowed range
of sizes; roughly 100m × 50m might be reasonable. The power P incident on the
Ð2
2
field is intensity times area: P = IA = 1.353 kW m ¥ 5000 m = 6.765 MW (quite
a bit of power), so the force on the whole large area would be
8 Ð1 F = 2¥ 6.765 MW / (3 × 10 m s ) = 4.51 × 10 Ð2 N Also, F = ma, so the acceleration of the 2,000 kg spacecraft would be:
a = 2.25 × 10 Ð5 Ð2 ms The velocity after constant acceleration for a time t depends on initial velocity
(which is zero, here): v = u + at A day of seconds is (24 hr) ¥ (60 min/hr) ¥ (60 sec/min) = 86,400 s. So the
Ð1
velocity after a day would be 1.94 m s , which is jogging speed. After a year, it
Ð1
would be 365 times this value, about 710 m s , which is roughly 10% of the
orbital speed of a loworbit satellite.
8 Ð1 3 c) Again, if the light is reflected, F = 2P/c = 2 ¥ 1012 W /(3 × 10 m s ) = 7 × 10
N, which is the gravitational force of the smallest automobile. The force is only
applied for a trillionth of a second, so the momentum transfer (the impulse) is:
3 Ð9 p ¥ ∆t = (7 × 10 N) ¥ (10Ð12 s) = 7 × 10 Ð1 kg m s which is small, like a big dustmote moving in a sunbeam, or a raindrop in a very
misty rain.
The pressure is a different story, though, because this force is applied over a tiny
spot:
3 Ð6 P = F / A = (7 × 10 N) / ( (10 × 10
3 2 13 m) ) = 7 × 10 Ð2 Nm Ð2 100 kPa = 100 × 10 N m is roughly one atmosphere of pressure. So this
pressure, briefly exerted by the focussed laser pulse, is something like 700
million atmospheres of pressure. [Robin] 2) Out of sight, hope you don't mind
This simple circuit works, using any two
identical resistors. Worth noting: since
the terminalpairs ÔAÕ and ÔBÕ donÕt
behave the same, the circuit must not be
symmetric for the pairs of terminals.
[Gnädig/Honyek] A R
R 3) Leonardo's camera obscura
a) Illuminated object Opaque sheet Image on screen B b) Illuminated object Opaque sheet Image on screen c) Illuminated object Opaque sheet Image on screen d) Illuminated object Opaque sheet Image on screen e) Illuminated object Opaque sheet Image on screen [James]
4) Heat works
We begin with two identical spheres at some uniform temperature. One rests on
an insulating table and the other is suspended by an insulating thread. To each
sphere the same amount of heat energy, Q Joules say, is transferred.
In the absence of external forces we would expect each sphere to have an equal
temperature change. However, since gravity is present and the volumes of the
sphere will increase therefore their centres of mass will move. Since the ball on the table will have its centre of mass rise, some heat energy will
be used to do work against gravity. For the hanging ball, however, the situation
is reversed. Therefore the resting ball will have a lower final temperature than
that of the hanging ball.
Suppose each sphere is 100 g, initially at some temperature T I K (the density of
aluminum is 2.7 × 103 kg mÐ3) and 250 KJ of heat is added to each. Let ∆rA and
∆TA denote the change in radius of the sitting sphere, sphere A and ∆rB and ∆TB
denote that for the hanging sphere, sphere B.
The initial radius of each sphere can be calculated with 1
kg
[mass of aluminum]
43
10
πri =
=
3
[density of aluminum] 2.7 × 10 3 kg m −3
1 3 × 10 −3 3
ri = ≈ 0.0207 m 27 ( 4 π) then Assuming a linear expansion coefficient to apply over the entire uniform heating
we have (to first approx.),
∆rA = [2.31 × 105 ∆TA]ri = 4.777
and
∆rB = [2.31 × 105 ∆TB]ri = 4.777
Hence
∆TA = 1
2.31 × 10 −5 ∆rA
ri ∆TB = 1
2.31 × 10 −5 ∆rB
ri and The heat capacity, C, of 100 g of aluminum is simply
C = 24.4 J
27
⋅
⋅ (100 g)
K ⋅ mol 27 g/mol = 90.3704 J
K Where we have used the molar mass of aluminum to be 27 g/mol.
Clearly the centre of mass of sphere A will rise by a distance equal to ∆rA and the
centre of mass of sphere B will fall a distance equal to ∆rB. Since the heat energy transferred must balance the change in potential energy
plus the energy due to the temperature change we have
250 000 = mg ∆rB + C∆TA
= 9.81
( 4.777 × 10 −7 ) ∆TA + 90.3704 ∆TB
10 Thus
250 000 = [4.777 × 10Ð7 + 90.3704] ∆TA
Similarly
250 000 = [4.777 × 10Ð7 + 90.3704] ∆TB
Therefore, the final difference in temperatures is
∆T = ∆TB − ∆TA 1
1
= 250 000 − −7
90.3704 + 4.777 × 10 −7 90.3704 − 4.777 × 10 [ ≈ 250 000 1.170 × 10 −10
≈ 2.925 × 10 −5 Therefore the difference in temperature is approx 2.87 × 10
[Gnädig/Honyek & Peter] Ð5 K. Pretty small! 5) Death of an Atom
The energy given off in a time ∆t is roughly kq 2 a 2 ∆t
c3
In 1 orbit (time T),
∆E = kq 2 a 2T .
c3
We will assume the radius of an orbit stays constant for any one cycle (the
electron is actually spiralling in, so this is not really true). The radius is R ≈ 0.5
Angstroms. The electron is kept in orbit by the attraction between it and the
∆E = 2 nucleus. We have m V = kqq = ma , where m is the mass of the electron and V is
R
R2
the orbital speed. The charge of the nucleus is +q for Hydrogen. Plugging all this
in yields the nice expression: Ecycle = 2 πkq 2 V 3
.
R c Here, this equals about 4.4*1024 J / cycle [this should be negative, but we will
remember that this is the energy lost]
The total energy of the orbit is: − kqq
2R compared to that for a gravitationally orbiting object:
− Gmm
.
R The electron will presumably ÒdieÓ when all this energy is exhausted. If we
assume that the energy loss per cycle is as in i) (not really true), we will have: 2 πkq 2 V 3 kqq
= n
,
2R R c where n is the number of cycles. Plugging in the stuff, n = 3.4*105.
The time, then, for this to happen is simply nT (T is the period) which comes out
to about 6*1011 s. It is not really meaningful to carry the first digit because of all
the approximations we made. Surprisingly, doing a more accurate analysis leads
a very similar answer. Either way, this is a very short time, and we have to
conclude that the classical answer cannot be right. [Peter] 6) Stupid bet tricks
i) First, notice that when the balls collide all forces between them (and hence the
momentum transferred) will be normal to the surfaces (that is the only point of
contact between them). This is because there are supposed to be no frictional
forces.
Let the labeling be as follows: If we tried to do the question directly, we couldnÕt Ð we have 4 unknowns (U, B,
A, direction of B) and only 2 equations (energy, momentum). However, since the collisions last a very small amount of time we can split the
problem into the two middle balls interacting with the cue ball and THEN the
two middle balls interacting with the outside balls.
Doing this we get: where the final speeds of the pair of balls are C (symmetric).
Additionally, we see that the momentum of each ball must be at a 30o angle to
the normal: So, Cx
1
= tan(30) =
.
Cy
3 Solving, C = 23
1
V (direction is 30o to yaxis), U = V [negative y direction].
5
5 Now, the two remaining collisions (which are symmetric, so we only solve one Ð
the right hand side one). We could write down all equations and solve this by brute force, or we could
simply notice that Cy will be unchanged because there are no tangential forces
(the balls are slippery Ð they donÕt ÒstickÓ). Hence, B y = C y .
Solving the remaining equations yields
C
3
=
V
2
5
3
B= V
5
A= (the direction of B is [positive y], that of A is [positive xaxis]).
So we have: And it seems the trick shot didnÕt work Ð only the outside balls have been sunk.
ii) When you hit the balls hard they deform slightly and do ÒstickÓ to one
another. That means that there is a tangential force between the outside most and
middle balls (remember in the solution above we assumed no tangential forces to
solve this collision). Numbering the balls 1 to 4 from left to right [see picture
above] yields the following: 2 has a counterclockwise spin and 3 has a clockwise
spin; moreover, the 2 centre balls now also have a horizontal velocity component
[why ?]. Upon hitting a bank, which is fairly soft, the spin of ball #3 will make it
rebound at more or less the same angle as it came in at Ñ so 3 will go into
another corner pocket. Now that you are completely confused look at the picture
to make some sense of this: [Robin is still trying to get a video picture of this poolshot onto the website,at
www.physics.utoronto.ca/~poptor ]
[Peter] 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 2: Mechanics
Due December 11, 1998
1) Football fizix
In the dying minutes of the World Cup finals, your team is awarded a direct free kick
near your opponentsÕ goal (see figure). The defense has set up an impenetrable wall of
defenders 8 m from the ball such that
21m
the only way to score is to kick it over
their heads and into the 3mhigh net.
you
Plante
Also facing you is Pele Plante, the
8m
superb goalie that has kept your team 10m
off the score board all game. You want
to keep the ball as far away from him as
possible. What initial velocity (speed
crazed fans
and direction) do you give the ball?
Remember the highlight tapes you
viewed before the game indicated that
the defense can jump to a head height of 2.5 m and your hardest kick corresponds to an
Ð1
initial ball speed of 30 m s . (Ignore air effects for the purpose of this question.)
Also remember that if you do not score, your team loses the Cup and Plante takes over
your Nikke promotional contract. The pressure is on...
BONUS: Describe how your answer would change when you also consider the effects
of air on the ballÕs trajectory. [James]
2) Chow, baby
Cat food is what Kit buys at the corner store Ñ in 175 g flat cans. She knows from sad
experience that the plastic carrybags that the store provides will hold well enough for
15 cans, but will tear through right away if she puts in 16 cans.
Coming home with 15 cans in one plastic carrybag, Kit realizes she has a problem.
Between her and her apartment lies The Elevator. Kit knows from physics class that the
acceleration in the elevator will add to the acceleration due to gravity, and that the cans
will weigh more as the elevator accelerates (though their mass remains constant). 1 This elevator starts with constant acceleration over two floors between the ground floor
and the second floor, and then rises at a constant speed of three seconds per floor. All
floors are 4 m apart. Will the bag split open Ñ should Kit have doublebagged?
What should be the maximum number of cans that Kit can load up with? [Robin]
3) Sink, sank, sunk
At some point in the movie ÒTitanicÓ it is said that the pressure at its sunken depth is
2
5500 pounds/inch . Given this, find the depth at which the Titanic supposedly came to
3
Ð3
rest. The ÔrealÕ answer is 12600 feet, density of water is 10 kg m (ignore air pressure).
Comment on the results, i.e., truth in filmmaking.
At another point in the movie someone estimates that the Titanic will sink in a time of
roughly half an hour. Approximating the Titanic as a rectangular box of crosssectional
2
area A = 882 × 34 feet , mass M = 40 tons (loaded), total height 80 m, and a horizontal
2
hole at the bottom of a side of the box of area AÕ = 100*0.5 m , estimate the time it takes
sink Ñ that is, when the top of the box is even with the water line. Ignore air.
[HINT: assume the flow stays practically at equilibrium as water goes in; you may use
BernoulliÕs equation, but if you do, you have to prove it.] [Peter]
4) When bowling, always signal your lanechangeÉ
Consider the problem of throwing a bowling ball with an initial spin, so that its path
curves. Take the xaxis as the centreline down the bowling alley, and y as the
horizontal sideways direction. Say that when the ball is released, it has an initial spin ω
r
with the top of the ball moving sideways (i.e., ω points along the xaxis), and a velocity
r
v at an angle θ to the centreline.
Find the path of the bowling ball, by giving the xcošrdinate and ycošrdinate as time
goes by, i.e., x(t) and y(t), where t is time (this is called Ôparametric formÕ). Assume the
point of contact between the ball and the floor is very tiny. The moment of inertia of the
ball will be that of a sphere.
Plot some nicelooking representative paths (ω = 0, ω > 0, ω < 0) [Peter]
5) A matter of some gravity
i) The moon rose in the east today at 7pm. If the lunar cycle (number of days between
full moons) is 29 days, when will the moon rise tomorrow? (One other subtly ÔhintfulÕ
piece of information: if the earth were moving more slowly, in orbit around the sun, the
lunar cycle would be shorter.)
ii) The spaceship GalileoÕs mission is to study Jupiter and its surrounding moons. To
get the vessel into a Jovian orbit, NASA had the vessel flyby both Venus and Earth to
2 increase its speed. This type of trajectory which provides acceleration without the
burning of expensive fuel is called a slingshot maneuver. Consider one of these
interactions in which the EarthÕs large velocity was used to accelerate the vessel. What
is the maximum speed that the vessel could have attained after this interaction, if its
Ð1
speed before the encounter was 5 km s ?
Galileo (mass: 1000 kg) actually did this type of maneuver twice with Earth on its way
to Jupiter. How many such slingshots would have to occur for the length of our year to
decrease by one second?. [James]
[Check out www.jpl.nasa.gov/galileo/index.html for some great pictures and information on this billion
dollar space mission] 6) Balloon tugowar experiment
HereÕs a neat experiment you are welcome to do with a friend (put both your names on,
for POPTOR marking). ItÕs not too hard, but itÕs full of physics. You will need:
2 goodquality round balloons
1 thick drinking straw
some scotch tape
Have your friend blow up one balloon about 1/4 full and the other about 1/2 full.
Pinch the neck of each balloon so that no air escapes. Carefully insert one end of the
straw into the neck of each balloon and tape it airtight, while still pinching. It will look
like a balloon barbell for weightlifting.
Before actually trying it, what do you think will happen when your friend stops pinching
the balloons so that air can flow between the balloons? Argue for your predictions on
the basis of what physics you can figure out or guess. Then, hold the straw gently and
have your friend release the balloons. Were your predictions right? Try squeezing the
balloons, one at a time, and see what happens.
To examine this interesting phenomenon further, do the following measurement: Try to
determine the air pressure inside the balloon, as a function of the size of the balloon.
You will need:
clear flexible tubing about 3 to 4 metres in length (see INFOBITS, below)
1 clear flexible tube about 1/2 metre in length
1 twohole stopper with holes to fit the tubing snugly
some duct tape
2 metre sticks or long rulers
You might want also to use (optional)
1 500 ml (or larger) kitchen measuring cup, or beaker with volume markings
two large soup cans
1 large bucket or plastic laundry tub (a nearby sink or bathtub will do fine)
3 To set up:
1. Seal one end of each piece of tubing into the stopper
2. Make the long tube into a rough
ÔUÕ shape, and tape each side onto
a meter stick or long ruler. Make
the end farthest from the stopper
to be sloping, as pictured. Be
sure your setup is stable.
3. Half fill the long tube with water.
This will be a manometer for you
to measure pressure with. You
may need more adjustments once
you begin.
∆h
4 . N ow fit the balloon onto the
stopper.
Gently fill up the
balloon, blowing air into it
through the short tube. If you
notice air leaking anywhere you
might use some Vaseline or duct
tape to seal it up. balloon With the right amount of water in the long tube, you can measure the difference in height
of the water on each side of the ÔUÕ. With the ruler or metre stick at a sloping angle you
can measure along the tube, and then use trigonometry to figure out the height. What is
the relationship between air pressure in the balloon and difference of water levels in the
manometer?
Now make a series of measurements of air pressure, for the balloon inflated to different
sizes (including not inflated). How can you best measure the size of the balloon: Cast a
shadow onto paper, and measure the shadowsize? Use a tape measure to find the
circumference? Use the measuring cup and tub to trap air released from the balloon,
and read its volume? WeÕre interested in your best ideas Ñ change anything you want.
Finally make a graph of your data with height vs. balloon size, and connect the points
with a smooth line. Be sure to label your axes with dimensions. From your results, can
you explain the outcome of the first 2 balloon experiments?
[Bonusmark ÔsequelÕ experiments: It would be interesting to see if the pressure in the
balloon, as a function of size, varied d ifferently depending on whether you were
inflating or deflating the balloon. Do you get a different answer if you measure volume
of air and if you measure diameter of balloon? Would it make a difference if you filled
the balloon not with air but something incompressible, like water? (How could you
change the manometer in order to use water in the balloon?)] [Simal and Robin]
4 INFOBITSª Ñ Useful Bits of POPTOR Information
Mass of Earth: 6. × 10
Mass of sun: 2 × 10 30 24 kg kg
11 Distance between Earth and Sun: 1.5 × 10 m
G = 6.7 × 10 Ð11 2 N m kg Ð2 Where to get parts for experiments:
there is plastic tubing at Canadian Tire, in the Plumbing section, for 17¢ per foot. DO
NOT attempt to use glass tubing, unless you have teacher supervision. It is pretty easy
to break it and have it go right through your hand, if your technique is wrong Ñ
especially when putting it into rubber stoppers.
¥ polyethylene, 0.25Ó O.D. (outer diameter); translucent, not quite clear
(stock no. 7 78380 022022 1; SKU: 039166056347)
¥ vinyl, 3/16Ó O.D.; clear, but soft, and hard to stuff into a rubber stopper
(stock no. 7 78380 02032 0; SKU: 039166056224)
2hole rubber stoppers: from chemistry class, we thought
C HECK THE POPTOR WEB PAGE for other hints, and any corrections we might post:
www.physics.utoronto.ca/~poptor 5 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Marker’s Comments on Set 2: Mechanics
General Comments from POPTOR:: Our apologies on how late we are in getting
material back to you. We hope that you find the solutions and these comments
are so good that they partially make up for the delays.
We have had a change in our team: Carolyn MacTavish joins us, to help us
mark, make up solution sets, and run the POPTOR invitational weekend. WeÕre
rushing to catch up, and you should have #3 Ð Thermodynamics returned with
solutions pretty quickly this time. If you havenÕt bothered with #3 because weÕre
so late, please jump right back in for #4 Ð Waves and Optics! Lots of people miss
handing in one set.
Peter sez:
With problem set 2 finished, I am now 2 for 2 in getting the
wording of my questions mixed up. The wrong revisions of questions going out,
me not being able to multiply, etc. have contributed to this. Either way, you all
have my apologies. In all cases, however, the mistakes have not changed the
core of the questions (yet). The marking, and my comments, follow questions 3,
4 and 6 as you had them (not necessarily as I planned).
Robin adds: Er, well, actually that was my fault. Sorry, Peter & all POPTOR
participants. 1) Football fizix
Carrie sez:
Good job! This problem was done fairly well by all (avg. 6.7/10).
The most common mistake was in assuming an initial velocity of 30 m/s.
However, this is not the best choice since it does not yield the maximum
horizontal velocity (given the constraints of the defence wall and that you are
aiming for the top corner of the netfurthest from Plante).
2) Chow, baby
Carrie sez:
Excellent job! The majority did very well on this problem (avg.
8.2/10). The most common error was in the assumption that the bag would
break if it contained exactly the weight of 16 cans (at rest). 1 3) Sink, sank, sunk
Peter sez:
Question 3 (ÒTitanicÓ) was done remarkably well Ð we at POPTOR
have spent weeks trying to figure out the best approach for this, and I thought
this question would be a killer (how it ended up a ÔmoderateÕ #3 I have no clue). I
solved the question using a force approach, and thatÕs where I was trying to lead
you with my hint. What happened, of course, was that no one solved the
question the way I intended, but instead found much more elegant (and shorter)
ways. One person even tried to correct for the fact that the flow of water is
initially nonsteady Ð very impressive! All mistakes made on this question were
minor.
4) When bowling, always signal your lanechange
Peter sez:
Question 4 (ÒBowlingÓ) was a bit disappointing, probably because
of the miswording. In the ÒrealÓ version of this question I actually gave
numbers, which I wanted you to use to figure out the path of the ball Ð this way
you would have to do some algebra. What happened instead was that nobody
calculated the paths, they just sketched some graphs. This is fine in principle, but
you must explain how you arrived at your answer! Also, most people missed the
fact that even though a ball thrown with some spin will turn initially, it will
eventually stop slipping and follow a straight line. You can check that the next
time youÕre in a bowling alley (I recommend 5 pin, unless you work out on a
regular basis) Ð or just watch bowling on TSN, which is what I doÉ The included
solution shows the path I wanted you to obtain (which is the case ω > 0. Other
cases are similar. The data I used follows: ω = 5 rev/s, Vx = 1 m/s, Vy = 5 m/s, µk
= coefficient of kinetic friction = 0.1, R = 40 cm. The bowling lane is about 10
meters long and 2 meters wide Ð the bowl was thrown in the middle of the lane).
5) A matter of some gravity
Carrie sez:
Bravo to all those who attempted this question!! (avg. 2.2/10) Most
seemed to avoid this one. For part i) the common error was not taking into
account the earth's axial rotation and/or realizing that the moon rises later every
day. For part ii) most tried to solve using conservation of energy Ñ which was
good Ñ but then forgot about momentum comservation Ñ which was not so
good.
6) Balloon tugowar experiment
Peter sez:
Question 6 (ÒExperimentalÓ) was done extremely well, by those
who attempted it. I was particularly impressed by the people who plotted their
2 stuff in Excel (or something similar) and typed the whole thing up. Most did not
attempt the experimental question Ð which is perfectly understandable, since
experimental questions donÕt have a long tradition at POPTOR (more like none
at all). Also, they do take extra time an effort.
On the other hand, I would argue that experimental questions teach you the
most physics, because to do well you not only have to know your equations, but
also understand what they mean. This is a purely philosophical argument, so if
youÕre not into that kind of stuff you should realize that at both the national and
international levels experiments are a major component of your mark (50%).
To encourage people to do the experimental questions (which we will have more
of) they will be weighted more than the other questions Ð each will be equivalent
to two theoretical questions (plus a bonus for any interesting ideas / setups you
come up with). Final Comments:
Peter sez:
Concluding I once again want to stress how impressed I was by
peopleÕs solutions to these questions considering how hard they really were. I
hope you keep trying, and speaking of trying we finally got the movie of the trick
shot (problem set #1) online (digitizing it was harder than making the shot!).
Check it out, at:
www.physics.utoronto.ca/~poptor/Com98/98GenXtra.html
Proof positive that physics worksÉ (rules!) 3 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 2: Mechanics
1) Football fizix
Your best bet is to go for the corner of the net farthest from Plante (see figure
given in question). We can reduce this to a twodimensional problem by looking
at the crosssection along the ballÕs trajectory. Since there are no air effects, the
ball heads straight from your
foot, over the heads of the
defenders and into the upper
3m
2.5m
corner of the net, as in the
net
you
figure at right. (Yay! Cheers!)
8m
23.3m
If you kick it at too high an
angle, it will easily clear the
heads of the defenders but will take longer than necessary to make it into the net.
This gives Plante more diving time to intercept the ball. To minimize the flight
time, you want the ball to have the maximum horizontal speed (covering the
distance to the net the fastest), but a corresponding vertical speed that will allow
it to barely miss the defenders and still go into the top corner of the net. As
always in these projectile motion problems: separate the problem into horizontal and
vertical components.
Constraints:
1) At x = 8 m, y > 2.5 m
2) At x = 23.3 m, y < 3 m.
3) At t = 0 s, you kick the ball from x = 0 m, y = 0 m.
Ð1
4) Max speed = 30 m s
The equations of motion are horizontal at constant speed, and vertical motion
with gravity:
vx(t) = vx(t=0) = vx
vy(t) = vy(t=0) Ð g * t = vy Ð g * t (note how I am defining vx and vy) So the ballÕs path is:
x(t) = vx * t
y(t) = (vy Ð (g/2) * t) * t 1 Constraint 3 is satisfied by how I set up the equations.
Constraint 1:
Let t = t1 when x = 8 m.
8 = vx * t1.
y(t1) = (vy Ð (g/2) * (8/vx)) * (8/vx) but y(t1) > 2.5
Thus 2.5 / (8/vx) < vy Ð (g/2) * (8/vx)
vy > 2.5 / (8/vx) + (g/2) * (8/vx)
vy > vx * (2.5/8) + (4*g) / vx <=Eqn 1 Constraint 2:
Let t = t2 when x = 23.3.
23.3 = vx * t2.
y(t2) = (vy Ð (g/2) * (23.3/vx)) * (23.3/vx) but y(t2)<3
Thus 3 / (23.3/vx) > vy Ð (g/2) * (23.3/vx)
vy < 3 / (23.3/vx) + (g/2) * (23.3/vx)
vy < vx * (3/23.3) + (23.3/2) * g / vx <=Eqn 2
Constraint 4:
2 2 vx +vy ≤ 30 2 <=Eqn 3 largest vx solution
3 Graph the three inequalities, to get a
clear picture of how they can be
satisfied, in the region above curve
Ô1Õ, below curve Ô2Õ and below curve
Ô3Õ: see that the maximum horizontal
speed v x occurs very close to the
intersection of eqn 1 and eqn 2.
The intersection is at: 1
2 vx * (3/23.3) + (23.3/2) * g / vx
= vx * (2.5/8) + (4*g) / vx
or
Ð1
vx = 20.2 m s
Ð1 This corresponds to vy = 8.25 m s .
To make sure that you are not hitting the crossbar, reduce vx slightly to
Ð1
20.1ÊmÊs , say. This will still meet the constraints imposed by equations 1 and 2. 2 Ð1 Therefore you must kick it with an initial speed of 21.7 m s at an angle of 22
Ð2
degrees above the horizontal. (Soccer is a science!). (I am using g = 9.8 m s
throughout this problem.)
BONUS:
Air effects can help you. By giving the ball the proper spin, you can cause the
ball to deflect downward from its normal parabolic path. This will allow you to
increase the initial horizontal velocity but still have the ball go in the net. If you
are really fancy, you might be able to curve the ball enough so you go around the
defenders instead of over (ÔcurlÕ or ÔswingÕ the ball left or right). [James] 2) Chow, baby
Find acceleration a:
Ð elevator has constant acceleration over two floors,
Ð final velocity is one floor per 3 seconds,
Ð initial velocity is zero,
2 d = 2¥4m = 8 m
Ð1
v = 4m/3s = 1.33 m s
u=0 2 v = u + 2ad
2
2
(1.33) = (0) + 2a(8 m)
Ð2
a = 0.111 m s
Adding this to the acceleration due to gravity, the total acceleration will be:
Ð2 Ð2 gÕ = 9.80 m s + 0.11 m s
Ð2
= 9.91 m s The ratio of the acceleration in the elevator to that standing on the ground is
therefore: 9.91/9.80 = 1.01. The acceleration is only 1% greater.
Find weight of cans with this additional acceleration:
Then 15 cans of cat food accelerated in the elevator would weigh on the bag the
way that 15 ¥ (1.01) = 15.15 cans would at rest. This is less than 16.
Does it mean the bag would not break? No! We only know that 16 will break the
bag, but we donÕt know whether a little bit over 15 would. ThatÕs because cans
come in units of 1 can Ñ there was no way to test whether 15.15 cans would
break the bag. Until now: Kit can be the first person to test out the bags at this
weight, without having to cut up a can of cat food in the bandsaw.
Recommendations? Double bag, or else carry only 14 cans of cat food. The 14
would weigh in the accelerating elevator the same as 14.14 would on the ground. 3 And any weight 15.00 or under in the elevator is definitely safe, because 15 cans
under normal acceleration was! [Robin] 3) Sink, sank, sunk Consider a pill box of water of area A and height h.
For equilibrium, PA + m g = P ′A
PA + A h ρ g = P ′A
P′ = h ρ g + P
Now extend the pill box to be a tall cylinder of height h (depth h to water level).
Ignoring air pressure, P = 0, and we get:
h= P
ρg Here, h Å 12600 feet [roughly] for 5500 punds/inch2, so thereÕs good agreement.
As the system starts without any water inside, the water must be accelerated
(thatÕs why you cannot use BernoulliÕs equation inside the ship, which only
applies to steady flows). Later, it reaches steady state.
Let us use the hint and use a force approach:
First note that outside of the hole the water will hardly be moving Ñ the water
comes in from all directions, so that the velocity at any one point is almost 0 (but
because the ocean is so big this adds up to give a large flow in the hole).
Now, we will consider the system in a small time interval ∆t. Since we are in
equilibrium, the water inside is stationary at first. For a small time ∆t an external
force (due to external pressure) and a negative force (due to pressure of water in
the ship) acts at the hole. The net force will force in a small amount of water ∆m.
Additionally, a part of the force will push some of the water upwards, but as
∆tÊÐ> 0, ∆x Ð> 0 and this is negligible.
∆m is hence pushed an amount 4 1
a(∆t)2
2
1 (ρ g h − ρ g x)A′
∆s =
(∆t)2
2
∆m
∆s = 0 + Since the pressure due to a height z of water is ρgz (from i))
∆s = 1 ρ g A′(h − x)
1 g(h − x)
(∆t)2 =
( ∆t ) 2
2 ρ ∆s A′
2 ∆s From conservation of mass, and amount ∆m of water at the hole must be equal to
the amount of water that causes a rise of the level in the ship, i.e.,
∆m = ρ∆sA' = ρ∆xA
Using this gives:
2 A
1
(∆s)2 = (∆x)2 = g(h − x)(∆t)2 A′ 2 (in the limit) ∆x A′ 1
dx
g ( h − x) =
=
∆t A 2
dt As the water flows in, the system comes to equilibrium. To compensate for an
increase ∆m in mass the ship must have moved (down) an amount
∆h = − ∆x Note that this does not imply that h = x! This simply says that x and h change by
the same amount. In fact, at time = 0,
x=0
And h = M/(ρA)
(M is the mass of the ship Ñ 40, 000 tons; this is just ArchimedesÕs principle)
And so
h(t) = x(t) + M/(ρA)
Plugging this into the expression for dx/dt yields: dx A′ 1 gM
=
dt A 2 ρA
So the water level rises at a constant rate.
From this we find: 5 dh A′ 1 gM
=
dt A 2 ρA A′ 1 gM M
h(t) = t + ρA A 2 ρA Putting h = 80 m, we find t ≈ 40 minutes. We didnÕt catch them this timeÉ
_______
This answer is pretty good Ñ only the Ò1/2Ó factor above may have to change [I
did an experiment sinking a shoe box, and it looked more like it should be a Ò1Ó,
which is fairly close]. Using BernoulliÕs equation gives nonsense for the time it
takes the ship to sink (something on the order of 1 minute).
Note that you can use BernoulliÕs equation, if you set it up the following way:
take a streamline from the top of the water to the hole in the ship Ñ this flow is
steady and you get something like: Mg 1
= ρV 2 (where V is the speed of the water coming in, P the pressure
A
2
at the holeÕs depth).
P= Using conservation of mass we now get: gM
dx A′
=
2
dt A
ρA
which is the same as the equation above, except for the factor of 2 Ñ which is just
as good because we used approximations in either solution. Note also that the
flow of the water is different while the hole is not completely covered with water
Ñ thatÕs why the height of the hole (0.5 m) is so small, to make this effect
negligible.
Speaking of BernoulliÕs equation, the proof may be found in virtually any
textbook (which is too bad) Ñ see e.g. Halliday & Resnick.
[Peter]
4) When bowling, always signal your lanechange
The math here isnÕt too hard, but one has to think about whatÕs going on or itÕs
very easy to be off be a minus sign (the International Olympiad judges kill (well,
almost) for errors like this). First, since we know there is friction, we see that it
will oppose the ballÕs motion. Now we resolve the motion to two directions Ñ x
and y Ñ and two motions Ñ translational and rotational. 6 ω
y
x Y:
Translational: (friction slows motion) M a = − µ k ( Normal) = − µ k Mg
a = − µk g
V = Vy − µ k g t
Rotational: (friction causes ball to spin faster, until
Iα = τ
2
M R 2α = µ k ( Normal)R
5
5 µk g
α=
2R
5 µk g
ϖ=0+
t
2R
The ball will continue to skid until itÕs rotating fast enough so that it can roll.
Then, V = ωR. This will occur at a time given by:
Vy − µ k g t* =
t* = 5 µk g
t*
2 2 Vy
7µ k g The motion thus consists of two phases Ñ with skidding and without:
1
y(t) = Vy t − µ k gt 2 ,
2
y ©t) = y(t*) + V(t*)t =
( t≤t*
12Vy 2
49µ k g + 5Vy
7 t≥t* t, X:
This is very similar: V = Vx − µ k g t
5 µk g
ϖy = ϖ −
t
2R 7 When skidding stops Ð V = ωR. But we have to be careful! In this case, V changes
sign, and so the above should actually read V = ωR!
[How do we know it changes sign? Well, you can do what I did Ñ go through
the whole calculation carrying the wrong sign. Then do the plot I asked for and
get nonsense. Or you can be a lot smarter Ñ assume that V changes sign (or not),
and then go back and see if your assumption holds]
t2 * = 2(ϖR + Vx )
7µ k g 1
x(t) = Vx t − µ k gt 2 ,
2
x©t) = x(t2 *) + V(t2 *)t =
( t ≤ t2 *
(ϖR + Vx )
5V − 2ϖR
t, t ≥ t2 *
(12Vx − 2ϖR) + x
49µ k g
7 Now we can finally
proceed to plot this. We
have to calculate t* and
t2* to know which
motion occurs when. It
turns out that the ball
stops slipping after
about 3 seconds in the xdirection, and after
about 8 in the y, but by
that time the ball has
long left the bowling
alleyÉ
Plotting yields
figure at right. the
[Peter] 5) A matter of some gravity
i) From the information given we know that if the earth was rotating more
slowly, the lunar cycle would be shorter. This means the direction of rotation of
the earth is the same as the direction of the moon around the earth Ñ it has to
ÔchaseÕ it instead of meeting it ÔheadonÕ. Thus, the moon rises later every day.
By how much? 8 In the time the moon has moved by
angle a, the Earth has moved by angle
b. The two triangles are set so that the
moon rays are parallel to the surface of
the earth. They are similar triangles so
a = b Ð 360 ¡ ray from moon
a Also (where t is measured in hours):
earth a=t / (29*24) *360¡ moonrise b b= t /24 *360 ¡.
Thus the time between moon rises is: 24.86 hours. Thus if the moon rose today
at 7PM, tomorrow it will rise at 7:52PM.
ii) This is a standard twoparticle collision problem. The fact that the interaction
is through gravitational forces does not change how one solves the problem.
Both energy and momentum of the entire system must be conserved. We
consider the system before the interaction (far from the earth) and after the
interaction (far from the earth).
Variables: ve, ve+dve: velocity of the earth before and after the interaction
vs,vs+dvs: velocity of the ship before and after interaction
me: mass of earth
ms: mass of the ship Conservation of energy gives:
2 2 2 me * ve + ms * vs = me * (ve + dve) + ms * (vs + dvs) 2 For maximum increase in speed for the ship, the initial velocities should be
parallel. We can drop the vector notation:
me * ve + ms * vs = me * (ve + dve) + ms * (vs + dvs)
==> Ð me * dve = ms * dvs
Substituting this into the above gives and solving for dvs gives:
dvs = 2 * (ve Ð vs) / ( (ms/me) +1)
We need to know the speed of the earth relative to the sun. We can write:
2 2 Fg = me * ve / r = G * msun * me / (r )
4 Ð1 4 Ð1 This gives ve = 3 × 10 m s . Thus the final speed of the ship is: 5.5 × 10 m s .
7 There are 3.1 × 10 s in each year. For this to change by 1 s, then we must change
7
the momentum of the earth (i.e., its speed) by 1/(3.1 × 10 ) of its total. One
slingshot maneuver changes the momentum by ms * dvs . The total initial
9 momentum is me * ve.
be: Thus the total number of slingshots, N, would have to
7 N * ms * dvs = 1/(3.1× 10 ) * me * ve
N ≅ 1× 10 14 Huge. The earth is not in peril from this, yet! [James] 6) Balloon tugowar experiment
This experiment is great Ñ it shows all kinds of different physics relationships,
even though it is pretty simple. It is even a little bit easier to do than to describe!
What did you think would happen between the two balloons fitted to opposite
ends of a straw? The first time I thought of this, I thought perhaps the balloons
would end up being the same size, because the balloons are identical and used
the same way, so I figured they should do the same thing Ñ and that means they
would keep the same amount of air in them. Wrong!
Instead, what happens is that one balloon will be bigger and the other smaller.
And if you squeeze the bigger one, and make it a bit smaller than the other one, it
will almost collapse, and become the smaller one while the other is the bigger
one. So, they each do the same thing, but not at the same time! The balloon you
squeeze to be the smaller one always expells most of its air. So there are two
ÔstableÕ or fixed arrangements: the left balloon small, or the right balloon small.
These will stay that way, but you can switch between them. This is technically
called ÔbistabilityÕ, and it also describes the kind of roomlight switch that you
can snap on or snap off Ñ it can be one or the other, but not much inbetween,
and it wonÕt just change by itself.
Bistability can be described with a potentialenergy diagram that has two local minima:
there is equilibrium in one minimum or in the
other, but it will cost a little energy to switch
between them. ThatÕs where squeezing the
balloon comes in!
The rest of the experiment is about showing
why there are two minima, or maybe even
showing what the potential energy curve
looks like for two balloons on a straw. A ball can come to rest at either of
two places on a curve like this.
Therefore it is ÔbistableÕ (bi = two).
It is the same for any system with a
similar potential energy curve Ñ
like the two balloons on a straw. The picture below shows my own setup for
this experiment. I guessed it would be a bit hard to read the fairly small pressure
changes of a balloon, so I tilted the manometer arm, taped it to a meter stick, and 10 measured the surface of the water as it moved along. Since the height difference
gives me the pressure in the balloon, over atmospheric pressure, tilting the tube
made the water move farther along, ∆L, for a given height change ∆y (i.e., ∆L/∆y
= 1/sinθ ).
I blew up the
balloon, through the
extra smaller tube,
and measured the
difference in height
∆h of the water
surfaces (meniscus).
Actually, I measured
∆L, and converted it
to a height raised,
adding it to the
height ∆ L by which
the water level in the
other arm dropped.
This gave me the
pressure inside the balloon. To measure balloon size, I tried two methods:
deflating the baloon by letting air out into a beaker turned upside down
underwater, then measuring its volume; also I used a tape measure to measure
the balloonÕs circumference. The second method was good Ñ it was sensitive, it
was pretty easy, I could use it as I inflated the balloon or deflated the balloon,
either way, and I didnÕt get as wet. Below is the curve I graphed from the data I
took. Each symbol marked on the graph is a measurement I took.
5000 4000 Pressure (Pa) ItÕs interesting that the
curve for the balloon as
it is being inflated is
not identical to the
curve for the same
balloon as it is being
deflated.
This is
because the balloon is
not perfectly elastic Ñ
it has a sort of
ÔmemoryÕ because it
takes a little while to
recover itÕs shape after 3000
inflating
2000
wrinkled &
deflated
1000
deflating
0
0.02 0.04 0.06 0.08 Radius (m) 11 0.1 0.12 being stretched. So the deflated balloon ends up wrinkly, since it has not gone
back to its original shape. This means that even though it returns to zero Ôgauge
pressureÕ (pressure above atmospheric), the balloon is larger after being inflated.
The balloon may recover its shape after a halfhour or so.
This kind of general effect Ñ having a curve which is different going up than it is
coming down, for the same values of radius Ñ is called hysteresis. You may see it
in the future in things like magnetism (say, a nail used in an electromagnet,
which stays a bit magnetized afterward), and it comes up in many places in
physics.
The balloon tugofwar doesnÕt depend on hysteresis though Ñ it is due mostly
to the curve of the inflating balloon, which shows a maximum pressure as the
balloon is being inflated. Pressure (Pa) 5000
When
connected
together by the straw,
4000
the two balloons must
have
the
same
same pressure
3000
pressure. Draw a line
across the graph for
2000
this pressure, and
youÕll find that t w o
1000
different sizes of balloon
can have that same
0
pressure Ñ these are
*
0.02 *
0.04
0.06
0.08
0.1
0.12
the sizes of the two
Radius (m)
balloons. When you
squeeze the larger balloon, the pressure of both balloons increases (they are still
connected together!), so the line of the shared pressure rises, on the graph above.
The large balloon is squeezed smaller, and the small balloon inflates because
there is a (nearly) constant volume of air for them to share Ñ they approach each
other in size. With a perfect setup, when the line touches the maximum, the
balloons have the same size, and can switch which one will be smaller, then the
balloon collapses under your hand to become the small balloon. Since the
balloon becomes small, you arenÕt squeezing it much anymore. The pressure
drops back to what it was originally Ñ only now the large balloon has become
the small one, and the small one the large one! MORE FOR THE VERY INTERESTED:
You can also describe all of this in terms of the potential energy of the twoballoon system, which is to be minimimized. This potential energy has the same
12 type of curve as shown above for bistability, and the system rests in one of the
local minima of the curve. You can see how the sizechangeover works, in the
following way:
The work done when squeezing, to compress a balloon is force × distance moved
(W = F ¥ ∆ r), and the force is the pressure × area (F = P ¥ A). So the potential
energy change, for the balloon as it changes size, is the work done on it:
∆U = Ð W = Ð F ¥ ∆r = Ð P ¥ A ¥ ∆r = Ð P ∆V,
since A ¥ ∆r is the change in volume of the balloon, for a small change ∆r. This
potential energy change is positive, because ∆V < 0.
You do work, in squeezing the balloon and increasing the pressure, so the
potential energy of the system increases. You need to increase it to get it over the
hump in the potential energy, and then the system of two balloons can go over to
the other minimum. Because each local minimum is ÔstableÕ, and there are two of
them, the system is bistable Ñ it can be at equilibrium in two different states.
[Robin] 13 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 2: Mechanics Erratum
There are minor errors in the version of Question #3 in this problem set, as we rushed to
distribute it. We will mark student solutions based on the question we distributed, but
you may wish to know that the data, especially, should look more like this!
The change in the hint is interesting, but does not very much alter the answer. 3) Sink, sank, sunk
At some point in the movie ÒTitanicÓ it is said that the pressure at its sunken depth is
2
5500 pounds/inch . Given this, find the depth at which the Titanic supposedly came to
3
Ð3
rest. The ÔrealÕ answer is 12600 feet, density of water is 10 kg m (ignore air pressure).
Comment on the results, i.e., truth in filmmaking.
At another point in the movie someone estimates that the Titanic will sink in a time of
roughly half an hour. Approximating the Titanic as a rectangular box of crosssectional
2
area A = 882*80 feet , mass M = 40,000 tons (loaded), total height 80 m, and a horizontal
2
hole at the bottom of a side of the box of area AÕ = 100*0.5 m , estimate the time it takes
sink Ñ that is, when the top of the box is even with the water line. Ignore air.
[HINT: assume the system comes to equilibrium every time a bit of water goes in;
you may assume that the flow is steady (i.e. has a constant velocity) Ð you may then
use conservation theorems (e.g., Bernoulli's equation) Ð but be careful, as this does
NOT hold initially, when the water is accelerating!] [Peter] 1 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 3: Thermodynamics
1) Newton's Law of Cooling
a) dQ
= κi A⋅ ∆T
dt With ∆T kept constant by the heater system ( (90 Ð 70) ¡F = 20 ¡F = 11 K), the heat loss is
constant, too. There are different heat conductivities Ñ κ 1 for the (bottom + sides), and
κ 2 for the for the top Ñ and they each contribute to heatflow out.
2 Top, bottom area (each): 5Õ × 7Õ = 35 ft = 3.25 m
2 2 2 2 2 Sides area (each): 7Õ × 9" = 5.25 ft = 0.49 m
Ends area (each): 5Õ × 9" = 3.75 ft = 0.35 m
dQ
= (κ1 A1 +⋅ κ2 A2 )⋅ ∆T
dt
2 2 2 = { 0.04ÊJÊKÐ1Êm Ð2 sÐ1 ¥ (3.25 m + 2¥ 0.49 m + 2¥ 0.35 m ) +É
2 Ð1
Ð2 Ð1
+ 0.4ÊJÊK Êm s ¥ 3.25 m } ¥ 11K = { ( 0.20 + 1.3 ) JÊKÐ1Ê sÐ1 } ¥ 11K = 16 J sÐ1 = 16 W
The heater must be delivering an average of 16 W, then. DoesnÕt sound like too much,
but this is day and night Ñ like leaving the oven light on. Or like leaving a reading
lamp on for five hours a day all month. In a month, then (which is 24 ¥ 30 = 720 hrs),
this amounts to 11.5 kWhrs. At a cost of about $0.125 /kWhr, for me, this amounts to
roughly $1.44 per month Ñ much less than a refrigerator. (Hmm, I thought it cost me
about $10 per month for this waterbed!)
b) Newton's Law of cooling tells us the rate at which heat will leave the water bed Ñ
proportional to the temperature difference ∆T with the room. As the water bed loses
heat, it will cool down, and ∆ T will decrease, and so the rate of cooling will, too. So
what we need, to start, is the relationship between temperature and heat content, to go
with what we can figure out for heat flow vs. time.
This relationship between heat content and temperature is what we call the specific heat:
how much heat do we have to add or remove in order to make a kilogram of water go
up by one degree? From INFOBITSª at the bottom of the problem set, we see that this
value is:
specific heat of water: C = 4.2 kJ kgÐ1 KÐ1
1 And ∆Q = C¥M¥∆T. The mass of water in the waterbed is:
Ð3 M = 1 g cm ¥ Ð1 3 3 (5 ft × 7 ft × 9/12 ft) ¥ (0.305 m ft ) = 10 kg m Ð3 3 0.74 m = 740 kg and so the heat capacity of the whole waterbed is
C ¥M = 4.2 kJ kgÐ1 KÐ1 740 kg = 3.1 MJ KÐ1 .
Therefore, we can write ∆T = ∆Q/(C¥M) , for ∆Q a change in heat energy of the water in
the waterbed. This lets us rewrite NewtonÕs Law of Cooling as:
dT
dQ
=Ð
/(C¥M) = Ð ∆T ¥ (κ 1 A1 +⋅ κ2 A2 )⋅/(C¥M)⋅
dt
dt
= Ð ∆T ¥ 1.5 JÊKÐ1Ê sÐ1 /(3.1 MJ KÐ1 )
Ð7 Ð1
= Ð ∆T ¥ 4.8 × 10 s = Ð ∆T / (574 hrs) where ∆ T is the temperature difference between bed and room air, ∆ TÊ=Ê{TÊÐÊTroom}.
(Heat flowing out of the bed is a drop in the energy in the bed; it gives a drop in
temperature, so we have a minus sign.)
Now, we have that Troom is constant so:
d ∆T
Ð7
= Ð ( 4.8 × 10 sÐ1⋅) ¥ ∆T = Ð∆T / (574 hrs.)
dt
and you may know that the exponential function A ¥exp{Ðαt} is a function with just such
a derivative. In fact, this is a good solution:
∆T = (∆T)o exp{Ð t / (574 hrs.) }
where (∆T)o indicates the initial value of the temperature difference. In our case, (∆T)o
= (90 Ð 70) ¡F = 20 ¡F. I turned the heat off on my
own waterbed at home, and
measured the temperature
as time went by.
The
exponential curve at right
shows NewtonsÕs Law of
Cooling!
(Actually, I made 90.0
Water temp (°F)
85.0 Water temp (°F) So temperature differences
decay, as the energy of the
hotter object will flow over
to the cooler one, until they
are each the same
temperature ( ∆T = 0). y = m1+m2*(1exp(m0/m3))
Value
89.057 0.37342 m2 25.295 1.1604 m3 102.14 10.104 Chisq 6.8694 NA R 80.0 Error m1 0.99638 NA 75.0 70.0 65.0
0 50 100 150 200 250 Time (hours) the
2 numbers κ up from this data, so that the answer would work out to match the curve.
But I messed up, and confused a calorie with a Joule, so all the numbers are 4.2 × too
large or too smallÉ)
According to the numbers given, after two weeks (336 hrs.) the waterbed will be
warmer than the room by:
∆T = (∆T)o exp{Ð t / (574 hrs.) }
= 20 ¡F ¥ exp{Ð (336) / (574 hrs.) } = 20 ¡F ¥ 0.58
= 11 ¡F
so, it will have dropped by 9 ¡F, or 5 K. To reheat the bed will then take:
5 K ¥ 3.1 MJ KÐ1 = 15.5 MJ
which is 4.3 kWhr.
On the other hand, 16 W average for two weeks to keep the bed warm would have cost
5.4 kWhr, or 19.4 MJ. The difference is about 25 cents worth of electricity Ñ maybe not
worth the trouble of unplugging the bed.
The increased energy, if the bed is left plugged in, is lost because in keeping the bed
warmer, over those two weeks, the average rate of heat lost is higher. Water temperature (°F) 90.0
In reality, the cooling is
faster than these numbers I
misinvented Ñ instead of
85.0
574 hrs., the number is
more like 135 hrs., in the
80.0
formulae above.
As it
works out, the bed would
75.0
be at room temperature
within the two weeks, and
it would cost a little more
70.0
Water temperature (°F)
than twice as much to
reheat the bed Ñ about 50
65.0
cents. The plot at right
10
0
10
20
30
40
50
shows how my waterbed
Time (hours)
reheated, after I plugged it
back in again. I have to remember to plug it in about 2 days before my guests arrive!
[Robin] 3 2) Spears into pruning hooks
a) In order to ignite the ship part we'll need a temperature above a 'dull red heat',
about 500 ¡C or 773 K. Using Stefan's constant σB we acquire the rate of radiation
I = σ BT 4 ∼ 20 kW ⋅ m −2
Without knowledge of the thermal constants of the wood we can only estimate how
much should be added to this to account for heat losses by conduction and convection.
We also shouldn't assume that the shields are perfect reflectors and that the wood is a
perfect absorber. A factor of 2 should give a reasonable margin of safety and
)hopefully) account for all this. So an intensity greater than 40 kWámÐ2 is required in
order to ignite the ship part.
We are given that the shields are curved, so lets consider the effect of focussing one of
the shields on a ship part. Assume the round shields are each 1m2, thus each has a
diameter, D, of about 1.13 m. The focal length, f, is taken to be 100 m (distance to ships)
and assume a wavelength of incident light, λ , on the shield of 500 nm (about where
sun's energy peaks). Since light diffracts any time it goes through a slit, the worse the
narrower the slit, light cannot be focussed to a tight spot either, without diffraction. The
limit of a focal spot is given by:
y = 2.44 fλ /D
(derived from principles of light diffraction) yields a value y = 1.08 × 10Ð4 m. Where y is
the radius of the bright central spot, created by the shield, on the ship part. So, the area
of the spot is approximately 3.66 × 10Ð8 m2. The power on the tight spot is the whole
power the shield intercepts, 1 kW. Thus, the intensity of light produced from focussing
one shield is,
I shield = 1 kW
3.66 × 10 −8 m 2 = 2.7 × 107 kW ⋅ m −2 This is much, much more than the 40 kW ¥mÐ2 loss.
To determine the time consider a surface layer about 1 cm thick. Now to calculate the
rate of heating lets assume that wood has a density and a specific heat equal,
respectively, to 1/2 and 1/10 the corresponding values for water (since the ship is
floating!) Then the thermal capacity, Ctotal, of the wood is,
Ctotal = mC = ( 500 kg ⋅ m −3 )(0.01 m)(0.42 kJ kg 1K 1 )
= 2.1 kJ m −2 K −1
∴ The rate of heating is 2.7 × 107 kW¥mÐ2 ÷ 2.1 kJ mÐ2KÐ1 or 1.3 × 107 K sÐ1 4 Assuming a temperature change of 480 K it will take about 4 × 10Ð5 sec to set fire to a
ship part of area 3.66 × 10Ð8 m2 and thickness of 1 cm. If the shores are lined with
soldiers holding these shields, those Romans are toast!
b) First, lets look at the power required to replace that radiated from the front of the
disk at its melting point.
Prad = σ B AT 4
= ( 5.7 × 10 −8 W ⋅ m −2 K −4 )(7.85 × 10 −5 m 2 )(1811 K )4
= 48 W
(We gave the wrong sign in the
exponent, in the value for Stefan's
constant at the back of the problem
set. Sorry about that, and hope you
found it by the crazy wrong answer
it gives!)
The power produced by just one
mirror is 2 m2 × 1 kW/m2 = 2 kW.
This should be more than enough to
make up for the radiation losses.
To melt the disk, first heat energy,
Q 1 , is required to increase the A solarcollector farm: The mirrors on the ground
temperature of the disk to the are adjustable, and redirect sunlight to the
focussing mirror in the tower.
melting point and then the latent
heat, Q2, must be supplied in order to produce melting, where
Q1 = mC∆T (assume ∆T = (1811 Ð 239) = 1518 K = (6.18 × 10Ð3 kg) (0.444 kJákgÐ1KÐ1)(1518 K)
= 1.53 kJ
and
Q2 = (moles iron) E = 1.53 kJ
∴ Qtot = 5.7 kJ
The power supplied by 100 mirrors is 200 kW and the power radiated at the melting
point is 48 W. So we can estimate,
t= 5.7 kJ
Q tot
=
= 0.03 s
Ptot (200 kW − 4.8 × 10 −2 kW) The above assumes that Prad is constant, when it actually changes as the temperature of
the disk changes. Initially radiation losses are unimportant compared with input
5 power. But as the temperature of the disk increases the radiation losses become more
important. However, in the case of our disk, even at the maximum temperature, the
melting point, radiation losses are negligible compared to input power. We have also
assumed that the mirrors are perfect reflectors and that the iron is a perfect absorber Ñ
which may be close to true for the mirrors but won't be true for the disk. Finally we
have also ignored convection losses, while these are not negligible, they are difficult to
calculate. [Carrie]
3) Sounds coolÉ
i) We tried to write the question so as to lead you through this standard derivation.
You can find it in first year physics and introductory thermodynamic books. For an
ideal gas with no heat transfer, we can write:
ii) ÐPdV = CdT (1) From the ideal gas law: PV = nRT, and taking the differential (like the derivative):
PdV + VdP = nR dT
Using (1)
VdP = (nR + C) dT (2) Divide (1) by (2) to get rid of the temperature variable:
PdV
−C
=
(VdP) (nR + C)
P
C
V
=− dV (nR + C)
dP
γ To confirm that P¥V =constant is a solution, we substitute it into both sides and verify
they are equal, if:
γ= (nR + C)
C This is fine since these are all constants.
dP
we get:
ii) Using the above expression and B = − V
dV
B = γ P and thus
γP
ρ
nM
ρ=
where M is the molar mass of air. So,
V
v= v= γ PV
nM
6 Using the ideal gas law: PV = nRT we get finally
v= Tγ R
M Given that v =
γR
= 300
M 300 m
at S.T.P. (i.e., T ≅ 300 K), then
s so we can say that v = 17 T . Therefore the colder the air, the slower the speed of
sound.
iii) From personal experience, I can not observe a time lag with a friend who is 20 m
away, but I do notice it with sound source that is at least ~50 m away (i.e., the firing of
a gun at a track meet). So I can observe a timelag of around 0.17 s. For me to notice a
time lag with a friend 20 m away, the speed would have to decrease by a factor of 2.5,
corresponding to a temperature of 48 K. That is pretty cold. In fact, all the
nitrogen/oxygen/carbon dioxide/etc would be frozen, so, not only would sound not
travel (something the Star Trek/Wars folks havenÕt figured out yet) but you physically
wouldnÕt last too long. [James]
4) Ice ice baby
i) Temperature of the vapour is the same as the surface of the water. Before you turn
on the pump it is 20¡C . Just before ice is formed the temperature is 0¡C.
The boiling process requires heat thus cooling the remaining liquid until it freezes.
Consider x kg freeze while y kg are changed into vapour. Consider that only the x kg
had to be cooled from 20¡C to just above 0¡ C.
Conservation of energy:
6 5 2.3 × 10 y = 3 × 10 x + 4200 ¥20 ¥ x
But there is 1 l of water, which has a mass of 1 kg so:
y=1Ðx
and thus x = 0.86 kg
Now consider that all the water (1 kg) had to be cooled from 20¡C to 0¡C:
6 5 2.3 × 10 y = 3 × 10 x + 4200 ¥20 ¥ (x+y)
and thus x = 0.85 kg.
Of course neither case is correct, but they are some sort of upper and lower limits for
the correct answer. Since we require the answer to only one significant digit, it is 0.9 kg. 7 iii) Water evaporation is used in many instances to cool things, from wineskins to your
own perspiration. For a glass of water, this process does cause the water to cool but
there are other factors that do not allow it to get very cold. There is heat transfer with
the rest of the room so you cannot cool the water without also cooling the room. Within
the vacuum system, there is very little heat transfer from the vessel to the chamber since
vacuum is a superb insulator [James]
5) Reduce, reuse, recycleÉ
We have: ∆E = Q − W
W = ∫ pdV i) W=0
3
V1( P2 − P1 )
2
3
3
∆E = nR∆T = V1( P2 − P1 )
2
2
Q = ∆E = (NOTES: all results are obtained from PV = nRT = RT here. Slope of the line in TÐP
graph is R/v1)
ii) W=1J
Q = ÐW = Ð1 J (edÕs. note: I think this should be +W, i.e., +1J)
∆E = 0 (cyclic process)
iii) W = P2 (V2 − V1 )
8 Q = ∆E + W = 5
P2 (V2 − V1 )
2 3
∆E = P2 (V2 − V1 )
2
iv) W = − P1(V2 − V1 )
5
Q = − P1(V2 − V1 )
2
3
∆E = − P1(V2 − V1 )
2
v) W=0
3
Q = − V2 ( P2 − P1 )
2
3
∆E = − V2 ( P2 − P1 )
2
vi) The net work is just the area inside the box (only on a PÐV graph!) Ð or you could just
add up the work from the previous parts of the question.
W = ( P2 − P1 )(V2 − V1 )
The heat is positive during the first 2 processes, so
3
5
Qincoming = V1( P2 − P1 ) + P2 (V2 − V1 )
2
2
The efficiency is 9 e= Wnet
Qinco min g 2 =
3 V1
P2
+5
V2 − V1
P2 − P1 2 =
3 1
x
+5
1− x
1− y Here, V1 = xV2 , P1 = yP2, so 0 ² x, y ² 1.
To make e big, we have to make the denominator small. To do this, we have to make the
denominator of the x and y expressions big. Clearly, the maximum occurs for x = y = 0.
Then,
e = 2/5 = or 40 %.
For a Carnot engine we would get
T
e = 1 − c where T c and Th are the coldest / hottest temperatures (respectively)
Th
occurring during the cycle. Here,
PV
e = 1 − 1 1 = 1 − xy
P2V2
The maximum occurs for either x or y = 0, and is equal to 1.
This is much higher than our cycle, and so I wouldnÕt recommend rebuilding your car
engine, or anything like that. [Peter] 6) Keeping your cool
a) We're given: X moles of air inside fridge at temperature TK
y moles sÐ1 escape
molar heat capacity of air is cp
(which we shall assume constant)
the room temperature is T R In a small finite time ∆ t y∆t moles of air escapes. Therefore y ∆ t moles of warm air at
temperature TR enters the fridge.
The warm air and cold air will mix and we model the final temperature by supposing a
reversible flow of heat. Since the heat gained by cold air must equal the heat lost by the
warm air we have
c p (X − y∆t) (T∆t − T) = c p y∆t (TR − T∆t )
Where T ∆T is the final temperature of the mixed air.
Therefore 10 T∆t − T =
⇒ y∆t
(TR − T∆t )
X T∆t − T y
= (TR − T∆t )
X
∆t Then, as ∆t → 0, we have, by definition of the derivative
T −T
dT
= lim ∆t
dt ∆t → 0 ∆t
= lim ∆t → 0 = y
(TR − T)
X y
(TR − T)
X Thus the rate of change of temperature inside the fridge is
dT y
= (TR − T)
dt X y
−t
)e x b) Verifying T(t) = TR + (T(0) Ð TR is a solution of 1. The L.H.S. is
y dT y −Xt
e
= (T(0) − TR ) − X
dt
y −t
y
= (TR − T(0)) e X
X The R.H.S. is
y − t
y TR − TR + (T(0) − TR ) e X X y −t
y
(TR − T(0))e X
X Thus the L.H.S. equals the R.H.S. and so T(t) = TR + (T(0) Ð TR) y
−t
ex is a solution c) Since the second term above decays to zero exponentially as t increases, T(t)
approaches the asymptote TÊ=ÊTR. This is very reasonable for the unplugged fridge Ñ it
goes to room temperature!
d) In a small finite time interval ∆t the temperature change is
T∆t − T = y∆t
(TR − T∆t )
X 11 To make this change zero the refrigerator must do work and remove the heat
responsible for this change:
y∆t
(TR − T∆t )
X
= c p y∆t (TR − T(0))
∆Q = c p X Where T ∆t = T(0), since the temperature inside is constant.
For a Carnot refrigerator working between the cold air inside at T(0) and to warm air at
room temperature TR the Ôcoefficient of performanceÕ is given by
η= ∆Q
T(0)
=
W TR − T(0) Hence T − T(0) W= R ∆Q T(0) = TR − T(0)
c p y∆t (TR − T(0))
T(0) Thus, by definition, the power required is
P = lim ∆t → 0 (T − T(0))2
W
∆t
= lim R
cpy
T(0)
∆t ∆t → 0
∆t (TR − T(0))2
=
cpy
T(0)
Since the Carnot refrigerator is the most efficient, therefore we conclude that this value
is a minimum for the required power. [Simal] 12 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 4: Optics and Waves
Due February 12, 1999
1) The land of the midnight sun (is where, exactly?)
While travelling through the Yukon this past summer, we travelled across the Arctic
Circle (latitude 66.7 degrees). If the earth was a simple sphere (like a billard ball)
circling the sun while rotating on its own axis, a line at this latitude would experience
exactly one day a year of unending daylight. North of this line would experience more
than one day of continuous daylight and south of this line would have a sunrise and
sunset every day of the year.
a) Using this information, find the tilt of our sphere's axis relative to the plane it
sweeps out as it circles the sun.
b) In reality, it is possible to see the midnight sun at points south of the Arctic Circle. In
each of the following cases, calculate how much farther south than the Arctic Circle you
could position yourself and still be able to observe the midnight sun, by taking into
consideration:
i) you are standing on a hill (height: 300 m, rising from a perfectly spherical Earth).
ii) the Earth has an atmosphere. For the purpose of this problem, assume that the
atmosphere is 3 km high and at standard pressure and 0¡ C, and that the Earth is a
perfect sphere. [James]
2) Making light of photons
Since light is a wave, two identical beams
can add together constructively or
destructively depending on their relative
phases. Interferometers, whether large
such as found in a satellite array
suspended high above the Earth or else
small like the one on my laboratory table,
make use of this fact to measure very small
distances.
Consider my simple
interferometer: the beamsplitter reflects
50% of the incident intensity and transmits mirror x1
mirror
x2 beamsplitter 1 1 the rest. All reflections involve a 180¡ phase change.
i) What is the electric field at the output (1) as a function of the relative displacement
(∆x=x1Ðx2). Consider the input field to be monochromatic and to have an electric field
of E=Eocos(2π v(x/c Ð t)) where Eo is the amplitude, t is time, ν is frequency (Greek
ÔnuÕ), and c is the speed of light.
ii) You will notice for certain values of ∆x, there is zero electric field at (1). Where has
the energy from the light gone?!
iii) Put a screen at (1). For a beam from my HeNe laser (frequency ν = c/632nm), what
is the smallest ∆x you could clearly distinguish?
iv) If you travelled back a century, you would find that many believed that light moved
in a medium, just as water waves travel in water or sound waves travel in air. This
supposed medium was dubbed the ÔetherÕ. If the ether were also moving, this could
change the effective speed of light, by a version of the Doppler shift. Pretend that you
are a scientist at this time and you want to establish the existence of the ether. How
might you use your interferometer to measure the speed of the ether relative to your
lab? What limit you can place on the etherÕs speed? (For simplicity, assume you still
had a monochromatic source of the same wavelength as in part iii). [James]
3) Imagining imaging
When you look at yourself in a mirror, you see yourself behind the mirror by the same
distance as you are in fact in front of it. Your mirror image might even appear far
enough from you to be in the next room! No light really passes from the place your
image seems to be, so the image is called ÔvirtualÕ. On the other hand, when you take a
photograph, you need real photons to hit the film; this kind of image is called ÔrealÕ.
i) Prove that any point A a distance d from a flat mirror has a virtual image AÕ which is
the same distance d on the other side of the mirror. What is the magnification of this
virtual image? [Hint: magnification M can
be defined as the apparent size of an image
divided by the size of the original object; it
can also be defined using the angle between
two rays of light leaving a point on the object,
and the corresponding angle the rays seem to
form when traced back to the virtual image.]
F
F'
ii) Any hyperbola has two foci, which we can
label F and FÕ. Show that one is the virtual
image of the other, reflected in the hyperbola.
Can you find a welldefined magnification?
[Robin]
2 4) Correct time and temperature, at the toneÉ
Consider a pendulum clock:
i) If the pendulum arm is made of a material with a linear coefficient of thermal
expansion α , determine an expression for the ratio of the final period over the initial
period, T ′ / T , if the change in temperature is ∆ Τ .
ii) If the suspension system is brass and the change in temperature is 20¡ C how much
time does the clock gain/lose in one day if Τ = 1.000 s?
A nickelsteel guitar string, initially of length lo , is stretched to a length L such that it
has a frequency f for its fundamental frequency.
iii) If the temperature is increased by ∆ T, determine an expression for the new
fundamental frequency in terms of the old one, i.e., f ′ / f .
Assume Young's modulus, Y, and the linear expansion coefficient, α , to be constant.
iv) If the initial frequency is 440 Hz (ÔAÕ above middle ÔCÕ), how Ôout of tuneÕ will the
guitar be, due to an increase in temperature of 20¡ C? Take lo = 0.795 m and L = 0.8 m.
[Simal]
5) Toys, Toys, Toys
What are the physics rules for oscillating molecules? In this question we will consider
different physical models of a linear molecule (like O2 , N2 , CO2 , or some longchain
polymers) which has just been hit by a singleatom molecule (like He or Ne). What
weÕll change is the model of the force between atoms, and what weÕll examine is their
motion after the collision.
Model a linear molecule as consisting of evenly
spaced particles (atoms, balls or whatever) of
mass m each. All collisions are perfectly elastic,
etc. Initially, the first (leftmost) particle of the
chain is hit by an object of mass m , and velocity
V [ to the right].
Consider first a chain consisting of just 1 particle (a m onomer ) Ñ i.e., we have an
ordinary 2particle collision. What is the speed and position, as a function of time, of
the 1atom chain afterwards?
Part I Ð stringy bonds
Consider now a 2atom chain (a dimer) having an interparticle spacing L. The particles
are connected by an initially extended ideal (massless, nondissipative and nonstretchable) string. The left ball [at (0,0) ] is hit by the incoming object. Find the motion
of the centre of mass of the lattice.
3 Generalize to an nparticle chain (a polymer); keep it a linear (straightline) molecule!
Describe (in words and equations) also the motion of the two balls as seen from both the
centre of mass frame, and an outside (laboratory) frame of reference. Sketch the
corresponding positiontime graphs for each ball (use one graph for each of the two
reference frames).
Part II Ð springy bonds
This is same as Part I, except that now the two atoms are connected by a spring (with
spring constant k). Describe (in words) the ensuing motion from both frames (laboratory
and centre of mass) of reference; find the motion of the centre of mass and sketch the
positiontime graph of the left ball in both frames. [HINT: consider the centre of mass
frame first].
Discuss the plausibility of the models used in Part I and Part II.
BONUS: Consider that there still is a spring, as in Part II, but there is now also a
dissipation mechanism. That is, there is some dissipative force of size Ð γ* (instantaneous
drag force on object) = Ðγ(dx/dt). What is the motion of the centre of mass now? Find
and draw a sketch of the position (as a function of time) of the left ball in both the centre
of mass and outside frame.
[HINT: a solution to the differential equation: md2x/dt2 = Ðkx Ð γ(dx/dt) is
2Ue − γt
m 1
k γ 2 4−
t
sin
2
m m 4 k γ 2
−
m m where the initial conditions are: x(0) = 0, dx/dt(0) = V(0) = U; also, 4k > γ2 ]
Can you suggest a realistic dissipation mechanism for molecules? [Peter]
6) Making a shoebox spectrograph, using a compact disc
Have you ever noticed the rainbow colours reflected from a compact disc? The tracks
on the CD are arranged with such mathematical regularity that a CD makes a nice
diffraction grating. This pretty easy experiment lets you make a spectrograph from a
shoebox and a CD, one which is good enough to reveal the individual lines contained in
the spectrum of light from a fluorescent light tube, streetlights, or other light sources
you may have. 4 You will need:
• • •
•
•
•
•
• peepcover
any compact disc (my
favorites are the goldcoloured CDR blank
u sed for recording
output slot
CDROMs; about $2slit
3. It saves my music
collection, too!)
a cardboard box the
size of a box for
input
hiking boots;
a
shoebox or small
corrugated cardboard
CD diffraction grating
box might do
aluminum foil
some duct tape, or other reasonably opaque tape
scissors
a pen or marker
a utility razorknife, or Exactoª knife (CAUTION: danger to fingers!)
a ruler To set up, read all these instructions before beginning. Refer to the illustration, too.
1. lay the box on its side, as sketched in the picture above. Measure and mark a
midline around the outside of the box, to use for alignment (up, across, and
down the middle of each face Ñ see dotted line in figure). Put a similar line
around the inside of the box, also at exactly the midpoint.
2. use scissors or razorknife to cut a lengthwise slot about 1 cm wide along the
top of the box, centred on the alignment line. Cut from about the midpoint
lengthwise, and go all the way to the end of the topmost panel.
3. cut a hole through the box, on the small endpanel of the box farthest from
the slot you cut in (2). It should be roughly 2 cm high by 3 cm wide, and
positioned toward the top of the boxend. It must be centred on your
alignment line.
4. cut a piece of aluminum foil about twice the size of this hole in (3). Put it on a
firm pad of paper to cut it, and along the middle of this foil make a 3cmlong
cut with your razorknife (hold the knife at a shallow angle, to pull the
cutting edge, and not the bladetip, across the foil). YouÕre making the input
slit Ñ you want the narrowest slit that will be sure to pass enough light: 0.1
mm is pretty good. When done, tape the foil over the rectangular hole in the
endface, with the slit horizontal and centred.
5. tape a compact disc, labelside down, inside the box on the side boxpanel.
Set it up so that the hole through the CDÕs middle is centred on the alignment
5 line you drew inside the box. Cover the front half of the disc with opaque
tape, so only the half farthest from the input slit will be used. The placement
of this disc depends on your placement of the input slit: a ray from the input
slit to the middle of the exposed section of the CD should make an angle of
about 30¡ from the CD surface. You may have to try moving the disc forward
and back in the box to get the input and output right, relative to the disc.
The spectrograph is now complete, except for figuring out how to use it! YouÕll need to
look down at the CD through the long slot in the top, and try different places along the
slot. But you want the only light inside the box to have to come through the slit, so try
cutting a piece of cardboard or dark cloth to cover the slot, with a 1 cm peephole to
look through. Then you can peep at different points along the slot, and still keep stray
light out by sliding the peephole card.
To use your spectrograph, you should point it so that light will pass through the slit and
onto the compact disc in the bottom. When this is exactly right, you can look at
different places along the output slot to find the best spreadout rainbow, from an
incandescent lamp, or many spectral lines from a fluorescent light, streetlamp, or neon
light.
The experimental assignment:
i) Write down your own description of how you made this spectrograph, and how you
made it work.
ii) Describe what you see, for different lightsources such as fluorescent lights (different
types?), incandescent lights, candleflames, neon lights, streetlights of different types, a
laserpointer (caution: never stare directly into a laser pointer, or HeNe laser, or the
sun. Even these simple sources can cause damage to your eyes, though you may not
recognize the damage immediately).
iii) Can you use the formula above, and your own observations, to figure out how many
total tracks there are on a CD? [Robin] 6 INFOBITSª Ñ Useful Bits of POPTOR Information
Remember to check the POPTOR webpage for hints and any necessary corrections!
www.physics.utoronto.ca/~poptor index of refraction for air at standard pressure and 0 deg C: n = 1.0003
6 radius of the earth: R = 6 × 10 m
nickel steel
10 YoungÕs modulus: Y = 18.2 × 10 Pa
Ð5
Ð1
coefficient of thermal expansion: α Å 1.4 × 10 K
brass
10 YoungÕs modulus: Y = 9.0 × 10 Pa
Ð5
Ð1
coefficient of thermal expansion: α Å 2.0 × 10 K
MINITUTORIAL: The physics of diffractiongrating spectrographs:
The input and output angles of light diffracted from a diffractiongrating depend on
wavelength, and on the diffraction grating itself:
n λ = 2d • (sin θi + sin θ d )
where θi is the input angle, measured from the perpendicular to the surface (the surface
normal), and θd is the angle of the diffracted light, measured similarly; d is the spacing
between grooves of the diffractiongrating, or tracks of the CD. The n is an integer
which gives the order of diffraction, which you can investigate.
The input light is scattered from very many regularly spaced grooves or tracks in the
compact disc. Only at very special angles are the conditions right for constructive
interference for waves coming from all of the grooves. At these special angles the light
of a certain wavelength is bright; at other angles there is only random or destructive
interference, and practically no intensity of light is sent on. 7 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 4: Optics and Waves
1) The land of the midnight sun (is where, exactly?)
a) At midnight on the summer solstice, the sun rays are tangent to the perfect sphere at
the Arctic Circle. The Ôside viewÕ looks like:
Axis for Earth rotation
observer on Arctic Circle 66.7deg
sun rays equator Once the picture is drawn it is obvious that the tilt of the sphereÕs axis to the plane it
sweeps out around the sun is 66.7¡.
b) i) Exaggerated for clarity:
s
sun rays
hill c Re Thus c=arccos(Re/(Re+300m)) and the subtended arc opposite from angle c is 60 km.
Thus you are 60 km south of the Arctic Circle.
ii) Refraction causes the rays to bend at the air/vacuum interface. Modelling this as a
single interface isnÕt exactly right (the interface is continuous and still causes refraction 1 due to the index gradient). SnellÕs law: sin(90¡)=n sin(a) as defined in the cartoons
below: s Ra sa
Re+Ra
Re
So:
Thus a=88.6¡. Solving for s gives an imaginary number (i.e., with − 1 ). What went
wrong? We assumed that the refracted ray that intersected the interface exactly over
the Arctic Circle would eventually intersect the ground, but it does not. If the interface
were lower, the index difference larger, or the Earth bigger, it could intersect the EarthÕs
surface. Thus the most southern ray to intersect the ground intersects the atmosphere at
a point towards the sun: s Ra
AC
b
a s
c b
As shown on the diagram, this refracted ray is tangent to the earth surface. We find
s=190 km. We still have to find out how far south we are. For this triangle, the
refracted angle of the ray in the atmosphere (angle: a) is 88.19¡ to the normal. The
incident angle (angle: b) is 88.85¡ to the normal. Thus we find that angle c is
(90Ð88.19Ð(90Ð88.85)) = 0.66¡. Using the value of Re, the corresponding arc length is 69
km. This is the distance you are south of the Arctic Circle. [James]
2 2) Making light of photons
a) The output at (1) is the sum of the two beams. Remember that though the intensity
after one reflection followed by transmission through the beam splitter is 0.25, the
2
electric field changes by 0.5 since I~E .
The math in this problem is easier if you replace 2*cos(a) by exp(ia)+c.c. where c.c.
indicates complex conjugate (i.e., in this case c.c.=exp(Ðia) ). Convince yourself that this
relationship is true, using the fact (you may not know) that exp(ia) = cos(a) + i sin(a),
where i = − 1 . The problem still can be done by writing cos() and sin() everywhere,
but this phasor notation is easier and extremely useful!
Thus the sum of the two beams at position (1) is: E 2 ⋅ x1 + x0 2 ⋅ x 2 + x 0 Etotal = 0 exp(iπ)exp i 2 πv
− t + exp i 2 π
− t + c.c. 4
c
c where x0 is the distance from the beam splitter to position (1).
Rewriting: E
2 π ⋅ v ⋅ x0 2 ⋅ x 2 + x0 Etotal = 0 exp i 2 πv
−t +π+ {exp(i 2 π ⋅ v 2 ⋅ Dx) + 1} + c.c. 4
c
c where Dx is (x1Ðx2).
This can be rewritten in terms of cosines but for the purposes of the rest of this question,
this form is easier to deal with.
ii) From the above expression, you can see that there is no electric field at position (1)
c
when exp(i 2πv 2Dx ) + 1 = 0 ie. when Dx = n
where n=..,Ð3, Ð1,1,3,.. Thus, if the
( 4v)
difference in distance is 1/4 of the wavelength (recall wavelength=c/ν), no electric field
is found at (1) and we have total destructive interference. Where does the energy go?
Consider the electric field that is reflected back in the direction of the original beam.
Due to fact that one beam undergoes three reflections and the other only one (instead of
two and two), you can show using the same method as above that the MAXIMUM
c
reflected occurs when Dx = n ( 4v) where n=..,Ð3, Ð1,1,3,.. . Thus if the beam does not
exist at position (1), it is reflected back the way it came. Energy is not lost.
iii) There is some subjectivity to what it means to Ôclearly distinguishÕ but I would argue
that you can clearly distinguish between a maximum and a minimum output. Maxima
c
at (1) occur when Dx = n ( 4v) where n=..,Ð2,0,2,.. So we can write
(Dxmax − Dxmin ) = 0−c
, i.e., 1/4 of the wavelength.
( 4 v)
3 The wavelength is 632 nm, so the smallest distance change you could clearly measure is
158 nm. That is pretty small. To contrast, atomic spacing in a solid is on the order of
5× 10Ð 10 m, so you could measure a distance corresponding to ~300 atoms.
iv) First assume the ether exists and is moving relative to your laboratory. This seems
to make sense unless the ether is held to the surface of the Earth due to gravity. That
would require the ether to have sizable mass. The ether must exist in space though since
light travels in space, and then you would get some interesting light bending effects due
to the concentration of ether around massive objects. Also you would still have ether
existing in vacuum, thus you would have something with mass existing in a vacuum
and now my head is starting to hurt. Even if the ether is not moving , you could put
your apparatus in a vehicle, such that you would be moving relative to the ether.
The trick is to rotate your interferometer and see if you can see a change in the light
output at (1). If one arm is parallel to the ether motion and the other is perpendicular to
it, you find that destructive interference occurs if x1
x1
2x 2 +
− exp i 2πv + 1 = 0
c (c + V ) (c − V ) where V is the speed of the ether relative to your interferometer.
interferometer by 90¼ and this term changes to: Rotate the 2x1
x2
x2 −
− exp i 2πv + 1
c + V (c − V ) c Define the minimum V such that by rotating your interferometer, you go from a case of
destructive interference to a constructive interference. (You might argue that this is too
much to require; the output intensity only needs to change by a small amount. This is
really determined by the noise in your system, which is a subject we are not going to
discuss here, so we take the strongest case required.). Thus: x1
x1
2x 2 1 (c + V ) + (c − V ) − c = ( 2 v) (Destructive) 2x1
x2
x2 c − c + V − (c − V ) = 0 (Constructive) Simplifying this gives:
V2 = c2
( 4 v) c x1 + x 2 + 4v 4 To make your scheme as sensitive as possible, you want to make x1+x2 as large as
possible. Say 19th century technology limits you to x1+x2=1m. For a source of
wavelength 632 nm, you would require V~1× 10 5 m/s. Increasing x1+x2 to 100m
reduces minimum V required to be observed to ~1× 104m/s. [James] 3) Imagining imaging
i) From the law of reflection,
θi = θr
Also θi + θr is the exterior angle of the triangle AAÕP, and is therefore equal to the sum
of the alternate interior angles ∠VAP and ∠VAÕP.
But ∠ VAP = θ i , and therefore ∠VAP = ∠VAÕP. This makes the triangles VPA and
VAÕP congruent, in which case d(object) = d(image).
As for the magnification, this can be determined using the angle between two rays of
light leaving a point on the object, and the corresponding angle the rays seem to form
when traced back to the virtual image. We have already shown that ∠ VAP = ∠VAÕP,
which is true of all object/image points. Thus the apparent size of an image divided by
the size of the original object, or the magnification is +1. The virtual image is lifesize
and erect.
ii) To demonstrate that one focus is the virtual image of the other we will prove that the
angles the incident ray and reflected ray make with the tangent are equal. This is most
easily done with the following geometry in mind.
On the diagram given in the question label the following:
5 the focus F → object point with coordinates (Ðc, o)
the focus F′ → image point with coordinates (c, o)
the incident ray (leaving F) → line l 2
the reflected ray (`leaving' F′) → line l 1
the point of reflection on the surface of the F′ branch of the hyperbola → point A
with coordinates (xo, yo)
Also draw in the tangent line at point A → label it line l
Then label
the angle between l 2 and l → angle α
the angle between l and l 1 → angle β
where both α and β are counter clockwise angles.
So we want to prove α = β . Suppose the curve is in `standard position' so that the
equation is,
x2
a2 y2 − 2 =1
b 2 x 2 yy ′
∴ 2 − 2 = 0 → y′ =
a
b b2 x
a2 y Substituting (xo, yo) yields the slope of the tangent line,
xx yy
b2 x
m = 2 o → gives the equation of the tangent → o2 − o2 = 1
a
b
a yo
Now we can use the formula for the tangent of the counterclockwise angle from one
line l 1 to another l in terms of their respective slopes m1 and m, namely
tan α = m − m1
1 + mm1 yo − 0
where the slope of the line l 1 is m1 = x − c
o plugging in the values for the slope you get something not as bad as it seems, 6 b 2 xo tan α = − yo
xo − c 2
b 2 x 2 − b 2 cxo − a2 yo
a yo
=2o b 2 x y a y o ( xo − c ) + b 2 xo y o
1+ 2 o o a y o xo − c 2 = 2
2
− b 2 x o c + (b 2 x o − a 2 y o ) ( a2 + b 2 )xo yo − a2 cyo − b 2 xo c + b 2 a 2 b 2 ( − xo c + a 2 ) −b 2
=2
=
=
c xo yo − a2 cyo cyo (cxo − a2 ) cyo
2
2
where we have used the fact that a 2 + b2 = c 2 and b 2 xo − a2 yo = a2b 2 for a hyperbola. The same calculation with Ðc replacing c gives
tan(−β) = b2
cyo so tan(β) = −b 2
cyo so tan( α ) = tan(β), and therefore α = β.
In other words reflected rays appear to come from F′ or the focus F′ is the virtual image
of F. The same argument can be made for the reverse situation (F ′ the object point and F
the virtual image). Thus each focus is the virtual image of the other.
The magnification is n ot well defined. This is clear when you compare the angle
between two rays of light leaving a point on the object with the corresponding angle the
rays seem to form when traced back to the virtual image.
If for example you consider the rays drawn in the diagram (given in the question) and
the corresponding angles they make with the xaxis (our second ray reflected back
along the xaxis) the magnification, M
M= tan( F ′)
→∞
tan( F ) Since the angle F ′ → 90o
If another two rays are taken an entirely different value for M can be obtained. Thus the
magnification depends on which pair of rays you are tracing out and thus cannot be
well defined. A proper image isnÕt really formed. [Carrie]
4) Correct time and temperature, at the toneÉ
i) Let the initial and final lengths, due to a temperature change of ∆ TK , be L and L′
respectively. The respective periods are then:
T= g
L and T′ = g
L′
7 Since LÕ = L(1 + α∆T), we have
T′ = g
L(1 + α∆T ) = (1 + α ∆ T ) − 1
2 T
1 Thus the required ratio is −
T′
= (1 + α ∆ T ) 2
T ii) Now for brass we have α = 1.9 × 10 −5 K −1 and we're told ∆T = 20°C = 20 K
Since T = 1.000 the new period is
T ′ = (1 + 1.9 × 10
= 0.9998 −5 (10)) − 1
2 Since the new period is shorter the clock will measure time faster and so gain time.
Since the now warmer clock measure is in just 0.9998 s. Therefore the clock gains time
1
by a factor of = 1.0002 per unit of real time.
T
Thus after one day (as measured by the clock initially cold) the warm clock gains 0.0002
days = 0.0002(24) (60) (60) s = 17.28 s.
Therefore the clock when warmer by 20 K gains approximately 17.28 s per day.
iii) In the case of a stretched string, it isnÕt exactly the change in lengths which change
the frequency. The string is stretched to a length L, no matter what. Instead, as the
string expands, it becomes less taut Ñ the tension in the string isnÕt as great, since the
string doesnÕt need to be stretched as much.
For transverse waves in a stretched string we use the equation c = f λ where c is the
speed of the wave, f the frequency and λ the wavelength. The first harmonic vibrates
with the only nodes at the end points of the string. This distance L is then equal to one
c
half the wavelength λ. Thus f = .
2L
τ µ where τ is the tension in the string and µ the mass
1τ
density per unit length. Thus f =
.
2L µ The wave speed c is given by τ/ A
= Y , where
∆l / l
A is the cross sectional area ∆ l is the change in length due to the applied tension, l is
the initial length and Y is ÔYoungÕs modulus,Õ a constant. We can determine τ from the formula for Young's modulus, namely 8 So we have f = 1 Y A ∆l
2L
lµ Initially we have stringlength lo,, and the string is stretched out by an amount
∆l = (L − lo )
µ =ρA and where ρ is the density of steel at the initial temperature. After the temperature increase
∆T we have
length lÕ prior to being stretched, and stretch amount ∆lÕ
l′ = lo (1 + α ∆T )
∆l′ = L − l′
= L − lo (1 + α ∆T )
The linear mass density doesnÕt change, because we always stretch to length L. The
crosssectional area A will increase as the metal expands, but our YoungÕs modulus,
assumed to be constant, is defined in terms of the initial A. (We would have been better
off to assume that Y¥A was constant).
Therefore
f= 1 YA ∆l
1 YA {L − lo }
1 YA L
=
= − 1
2 L µ lo
2L µ
2 L µ lo
lo f′= and L
1 YA ∆l© 1 YA {L − l©
} = 1 YA L − 1 = 1 YA − 1
= l©
2L
µ l© 2 L µ
2 L µ l© 2 L µ lo (1 + α∆T ) It follows:
L − lo (1 + α ∆T )
f′
=
f
(L − lo ) (1 + α ∆T )
and this is the required ratio.
Now, using α = 1.1 × 10 −5 K −1 , ∆T + 20 K , f = 440 HZ, lo = 0.795 m and L = 0.8 m
we have:
f′
=
440 0.8 − 0.795 (1 + 0.00022)
= 0.9822
(0.8 − 0.795) ⋅ (1 + 0.00022) f ′ = 432.2 Hz,
Therefore the final frequency of the nowwarmer guitar is 432.2 Hz, which is out of tune
by about 8 Hz. [Simal, James & Robin] 9 5) Toys, Toys, Toys
The masses are equal, so after the collision the ÒlatticeÓ will move at V m/s [right];
x(t)Ê=ÊVt is the position as a function of time.
First, note that momentum will be conserved, so that at any point in time we will
have:
mV = mL + mR
V =L+R [1] where L and R are the velocities of the left and right balls, respectively (they clearly
point in the xdirection, so we drop the vector sign).
The position of the centre of mass is given by:
xc.m. = mxL + mxR 1
= ( xL + x R )
2m
2 [2] where xL and xR are the positions of the left and right balls (in the laboratory frame),
respectively.
Hence,
Vc.m. = dxc.m. 1
V
= ( L + R) =
dt
2
2 using [1]. In the centre of mass frame then, which moves at V/2 m/s [right], the velocities of the
balls become:
L′ = V − VV
=;
2
2 R′ = 0 − V
V
=−
2
2 at t = 0. (the balls are thus approaching each other)
The situation is perfectly symmetric and we can right away conclude that after the two
balls collide, which will take a time of L/2/(V/2) = L/V their velocities will be:
L′′ = − V
;
2 R′′ = V
2 (the balls are going away from each other)
Next, the balls will be pulled by the string, and conserving momentum and energy
yields that after the collision we will have:
L′′′ = L′ = V
V
; R′′′ = R′ = −
2
2 The balls therefore oscillate with a period of 2(L/V). Note that the motion is not simple
harmonic, as acceleration = 0  Ðkx.
The motion may be written as
10 L′(t) = V
L
2L , t<
mod 2
V
V L′(t) = − 2L V
L
, t>
mod
2
V
V xL ′ (t) = 2L V
L
t, t <
mod 2
V
V xL ′ (t) = − 2L V
L
mod
t, t >
2
V
V the primes denote centre of mass frame; ÒmodÓ gives the remainder after division, so
using it we can figure out what part of the period t is in. If you have never seen ÒmodÓ
before, donÕt sweat it. I think it is (more or less) clear the graph looks something like
this: (the dashed line corresponds to the right ball, the solid corresponds to the left one)
Since we have the motion in the centre of mass frame, and we know its velocity, we
simply transform to the laboratory frame. This gives:
2L L
<
VV
2L L
L(t) = 0, t mod
>
VV
L(t) = V , t mod 2L L
<
VV
2L L
xL (t) = const , t mod
>
VV xL (t) = Vt , t mod Plotting this gives:
11 The solid line is the motion of the left ball; the dashed that of the right; the thick line is
the average of the two, which is also the motion of the centre of mass.
The system looks somewhat like a centipede Ð first the back moves, with the front being
stationary, then the front moves, and so on.
Note that if we had a body of mass 2m colliding with one of mass m we wouldnÕt get
the same results as here Ð our system loses energy (itÕs a soft object, like jello).
In an nparticle lattice, the speed of the centre of mass would be V/n.
It is instructive to try this by brute force Ð keeping the same symbols as in i) (xL, L
for left object, x R, R for right)
L+R= dxL dxR
+
=V
dt
dt
2 2 dxL dxR mV = m
+m
+ k(L − ( xR − xL ))2 dt dt 2 This is a system of differential (momentum and energy) equations that just screams out
Ògo awayÓ [NOTE: if you rearrange it Ð it takes lots of work Ð you end up with
something workable, but not at the high school level]
LetÕs take the hint and go to the centre of mass frame. From ii), the centre of mass still
moves at V/2 m/s [right]. In that frame we once again get the balls approaching at
equal speeds (V/2). Since the compression on each half of the spring will be equal, we
can treat the system as two springs of length L/2 (kÕ = 2k).
The motion is simple harmonic with a period:
T = 2π m
2k The motion of the left ball is
xL ′ (t) = V
2 2k m
t
sin
2k m [this gives the correct initial conditions Ð xLÕ(0) = 0, L(0) = V/2]
Thus, in the outside frame, the motion looks like this:
x L (t ) = V
V
t+
2
2 2k m
sin
t
2k m Plotting gives (not to scale!): 12 The black line is the right ball, green line is the left ball, straight line is the motion of
centre of mass. The motion looks once again like a centipede, but kind of weird (nonuniform).
This is very similar to ii), except that the solution is now xL ′ (t) = V−
2e
2 γt
m 2
1
k γ sin
4−
t
2
m m 4 k γ
−
m m 2 Plotting gives something like this: 13 Note that the particleÕs motion eventually decays and remains at x = 0.
In the outside frame, then:
Ve
x L (t ) = − γt
m V
t+
2 1
k γ 2 4−
sin
t
2
m m k γ 2
4−
m m which gives (not to scale!):
Here, the left ball is black,
right ball red and centre of
mass is green (straight line).
Clearly, the last model is most
realistic. After a body collides
with something, we might
expect its molecules to vibrate
for a little bit, but not forever
as in ii) or iii). ii) is most
unrealistic as it experiences
infinitely long vibrations and
the bond between molecules is
most likely not a rigid string
(this implies that the force
acting on the molecules occurs
14 over an infinitely short amount of time, leading to infinite accelerations, etc.).
The dissipation mechanism is due to radiation Ð an accelerating charge radiates away its
energy [see problem set 1]. [Peter] 6) Making a shoebox spectrograph, using a
compact disc
As given in the Ômini tutorialÕ under
InfoBitsª, the input and output angles of
light diffracted from a diffractiongrating
depend on wavelength, and on the diffraction
grating itself:
n λ = 2d • (sin θi + sin θ d )
where θ i is the input angle, measured from
the perpendicular to the surface (the surface
normal), and θ d is the angle of the diffracted
light, measured similarly. 2 d is the spacing
between grooves of the diffractiongrating, or tracks of the CD (rather
than d , as given Ñ this is
something of a convention). The n
is an integer which gives the order
of diffraction, which you can
investigate. Diffraction orders: This figure was made using a CD
and a laserpointer. You can see the laserpointer,
held at the right. The brightest spot of light is the
undiffracted specular reflection (sometimes called the
Ôzeroth order,Õ and then to the right are three orders or
diffracted light, labelled n = Ð1, Ð2, Ð3. Try the
equation yourself to see why theyÕre negative. We used a HeNe laser from the
lab, and a CD that was kicking
around, and found that with
normal incidence (θi =0), the 1st
order diffracted angle was found
to be 22 deg. Since the HeNe
wavelength is 632.8 nm, the
spacing between adjacent grooves
on the CD works out to be
approximately 1.7 µm.
If you look at fluorescent lights,
you can easily see the different
spectral lines, of different colours,
that make the fluorescent light
15 look white. The colours come from different phosphors coated on the inside of the glass
tube Ñ each spectral line comes from a quantum transition in the atoms making up the
phosphor. The atoms are excited by absorbing ultraviolet light, which comes from
mercury atoms which are themselves excited by electrons accelerated up and down the
tube. So fluorescent lights are doubly fluorescent Ñ first the mercury vapour inside, and
then the phosphor coating. Your school may have a wallchart that shows characteristic
spectral lines of different materials (these are available through Edmund Scientific,
among other places); a picture of part of one of ours is below. If you look at street lights, you may find that there are different kinds. Incandescent
lamps give a whole or continuous spectrum, but lowpressure and highpressure
sodium lamps are different: the lowpressure lamps give spectral lines, but in highpressure lamps the much more frequent collisions between atoms make the spectral
lines spread out or broaden. Astronomical observatories care very much which sort of
lights a city installs, since too much scattered light (light pollution) can wreck
observations. If lowpressure sodium lamps are used, the scientists can use the
frequencies between spectral lines to peer through to the deep sky. [Robin]
16 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 5: Electricity and Magnetism
Due March 5, 1999
1) Bully for you!
i) Back in kindergarten, the big bully challenges you to a game of ÔNomercy pink leg
wrestlingÕ. You decide that this endeavour is not in your best interest so you challenge
him back to a Ôreal toughÕ game of pointcharge placement. ÒHuh?Ó says the bully so
you explain: ÒPretend you are given a set of electrically charged balls, and you must
arrange them to be free, but static Ñ so that they do not fly together or apart.Ó ÒHey,
what about gravity or friction or redistribution of the charges on the balls?Ó the bully
demands. Unfortunately, you seem to have challenged the only bully in advancedplacement physics in the entire kindergarten class. ÒHmm, ignore them and pretend itÕs
a perfect insulator, so the charges cannot moveÓ you respond, hoping that if this doesnÕt
go well, the recess bell will ring before legwrestling becomes an option again. ÒOkay,
but I go firstÓ says the bully. ÒThree balls: two with charge +4q, the other with charge
Ðq. Hurry up twerp!Ó What is your answer?
ii) You got him on that one but now it is your turn. You ask him ÒThree balls, all with
charge q, placed on the points of an equilateral triangle. Two other balls with identical
charges on them, but you get to pick the amount of charge. What is the charge and in
what arrangement?Ó He screws up his face to think, exclaiming ÒThere better be a real
answer for this, tweek!Ó You assure him there is one. Better think quick, what is it?
[James]
2) Charged with resistingÉ
A current of 1 mA flows through a conductor made of
1 mA
two wires, one copper and the other iron. The crossCu
Fe
sections are identical, and the wires are buttwelded as
shown in the figure. What electric charge naturally
accumulates at the boundary between the two metals? How many elementary charges
does that correspond to? [Hint: GaussÕs law] [Gnädig/Honyek]
3) ÔAscending and Descending VoltagesÕ: a circular argument
ÒPsst!Ó, a man in a dark trench coat whispers to you, ÒPsst buddy, ya interested in an
almostnew physics equation?Ó Being the foolhardy type, you decide to check out his
1 wares. ÒOkay buddy, IÕll even throw in one for free to show you that IÕm legit. Say ya
got a wire carrying a current. That current induces a magnetic field. Take your right
hand like this ya see. Point your thumb out in the direction of the current and the
magnetic field curls around the wire like your fingers do.Ó
You are not impressed. Heck you learned the righthand rule in Grade 3!
ÒOkay, waitasecond. I got a real good one. Real hot!Ó Whipping out a small backboard
from within the long folds of his coat, he draws the following diagram while saying, Ó
Say ya got a closed circular circuit, with a magnetic field going through it. If that field
changes in time, there is going to be an electric field induced in the circuit, according to:
v dφ
vv
E=
where φ = B ⋅ A
dt
B
here, E is the induced electric field around the circuit, B is
the magnetic field, and A is the area enclosed by the circuit
(with its unit vector pointing perpendicular to the circle, in
the direction defined by the right hand rule). So waddya
think of them apples?Ó A E Wow! This is cool stuff, but something about it sounds fishy. ÒOkay, IÕll show you.
Take a look a this.Ó He motions you into the dimlylit alley behind him. Sure enough,
your new friend has an electromagnet stashed back there. Above the end of the magnet
is a circular loop of five 100 ohm light bulbs all wired in series. Even though there is no
battery in the circuit, the bulbs are all glowing weakly. You whip out your
magnetometer, and verify that yes, there is a uniform magnetic field running
2
perpendicular to the plane of the circuit (area=0.001 m ) and it seems to be ramping up
in time: B(t) = B0 ¥ t.
a) Using your friendÕs equation, what do you conjecture is the total current in the
circuit? And what is the voltage across one of the bulbs?
b) Darn, you think, I wish I had brought my voltmeter. So you say to your new friend:
ÒLet me get this straight. There is current running through each bulb. That means that
voltage must be decreasing across the bulb. Therefore as you go around the circuit the
voltage keeps dropping after each bulb. You end up back at where you started but
now it seems you are at a lower voltage. DoesnÕt this remind you of a print by some
Dutch guy named M.C. Eaker, Esther, or something like that?Ó
ÒNo, no, no, buddy. You got it wrong!Ó. What is wrong with your logic? Draw an
equivalent circuit. And what is title of the picture that you are mumbling about?
c) ÒOkay, now I understandÓ, you tell the huckster, Òbut there is something still wrong
with your equation. If your equation was really right, those bulbs would blow up.Ó
You got him there. Explain. [James]
2 4) Flux compression Ñ putting the squeeze on a Bfield
This is an actual experimental problem in practice Ñ a way in which very high
magnetic fields can be produced, using a currentcarrying circuit and high explosives.
i) Consider a circuit of total area A with some current I passing through it. This causes
a uniform magnetic field B perpendicular to the plane of the device. The sides of the
device are suddenly imploded so as to cause the area to shrink to AÕ. What is the
resulting magnetic field BÕ ? (total area A) (total area AÕ) ii) If the object is a solenoid shrinking from a radius 10rÕ to a radius rÕ, with I = 1A and
n = number of turns per unit length = 100 / 1 cm, what is the resulting field?
[HINT: one of MaxwellÕs equations will be useful] [Peter and Bryan]
5) Scratch and dent sale on capacitors
i) What is the capacitance of the earth, taking it as a perfect sphere?
ii) Consider the earth not as a perfect sphere Ñ say a comet smacks into the earth and
dents it, changing its volume by 3%. By what percentage does the earthÕs capacitance
change?
[B ONUS : extra points awarded for deriving the formula for capacitance of a sphere,
given in InfoBits below] [Robin & Gnädig/Honyek]
6) Mocking mirrors
The method of images is very useful for solving a variety of electrostatics problems.
Consider a point charge +q a distance d away from an infinite conducting plane.
Because the plane is a conductor, the charges on it will move so that the potential on the
surface of the plane is constant (until this happens, there is a field in the conductor; this
field makes the charges move). But you could also make an infinite constantpotential
surface if you had two opposite charges (+q and Ð q) a distance 2d apart Ð the potential on
their perpendicular bisector is a constant.
This wouldnÕt be very important, except there is a theorem which states that the
solutions to electrostatic problems are unique: if you find some electric field that satisfies
the boundary conditions, it is the o nly solution. The twinned charges give the right 3 potential for the infinite plane, so their field around +q is the
same as the field of a single +q charge above a conducting plane.
i) So, what is the field at any point y just outside our infinite
conducting plane? +q
d ii) Consider t wo infinite conducting planes, crossing at a 90¡
angle.
A charge is placed
symmetrically inside one quadrant,
a distance d f rom either plane.
d
What is the field just outside the conductors now? What
image charges can you use to produce the same field?
d
+q
iii) Consider now two infinite,
conducting, parallel planes
separated by a distance 2d,
with a charge +q midway. Where do you have to place
+q
image charges so that the planes will be equipotential
d
d
surfaces?
[HINT: in this case you can think of the planes as mirrors,
the charge +q as an object, and the image charges as images
of the object formed by the two mirrors] [Peter] INFOBITSª Ñ Useful Bits of POPTOR Information
Remember to check the POPTOR webpage for hints and any necessary corrections!
www.physics.utoronto.ca/~poptor
Csphere = R
,
k C = capacitance, R = radius
k = 8.9875 × 10 9 N m 2 C −2
= 8.9875 × 10 9 F m −1 Ecapacitor = Q2
, E = energy, Q = charge, C = capacitance
2C 4 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 5: Electricity and Magnetism
1) Bully for you!
i) The symmetry gives it away. Put the negative ball in the middle, and place the
other two to balance the Coulomb force. No matter how far apart are the positive
charges, the net force on each of the three balls will be zero.
ii) This is a classic lateral thinking problem. Trying to balance three balls in a triangle,
with two balls sounds impossible, UNLESS you think in 3D. Place each ball above and
below the plane holding the three balls. Symmetry says that each
negative ball must be an equal distance from each +q ball. Now balance
B
the Coulomb forces. Define the equilateral triangle to have sides of
length 1 unit. From the top this looks like the figure at right, from which
BB
1
we can find that B =
.
3
From the side, this looks like the figure at left, where
angle b is opposite side B. We find that we have to
satisfy two equations: n
b C A q B n2 3nq cos(b)
=
4A2
C2
2nq
sin(b) = 3 q 2
2
C (vertical forces acting on ball Ðn)
(horizontal on ball q) Since A = C * cos(b), the first equation simplifies to:
n A
= 12 C
q 3 The second equation simplfies to
n33
=C
q2
So 2 A = C 2
Recall that A 2 + B2 = C 2 . These two equations then give:
A = 1± 2
3
1 and the value for n=10.4q or 0.33q. You win! [James]
2) Charged with resistingÉ
Let I denote the current flowing in the wire, A the crosssection of the wire and ρ1 and
ρ 2 the resistivities of the materials. OhmÕs law for a wire of length l gives U = I Ê ρ l / A
which yields E = U / l = ρI/A for the electric field strength in the wire.
The resistivity of copper is lower than that of iron. Therefore to give the same current in
each part, the electric field strength has to be smaller in the copper than in the iron. Cu Fe According to GaussÕs law, the difference in the electric field strengths implies an
accumulation of charge at the boundary of the two metals (see figure). The total charge
accumulated at the interface is
Q = ε 0EA = ε 0 I (ρFe − ρCu )
It is interesting that this quantity depends purely on the current and material constants,
but not on the crosssectional area of the wire (i.e., of the interface).
Substituting the known data, the charge is found to be Q ≅ 5.10 −21 C, which is only
1/30th of an elementary charge! Though a measurable macroscopic current flows
through the wire, the accumulated charge is only a small proportion of the microscopic
elementary charge. This strange result shows that classical electrodynamics (imagining
charge carriers as small balls) cannot always correctly describe electrical phenomena.
Only application of the more sophisticated laws of quantum theory and statistical
physics can give a more accurate description. [You might be interested to note that the
1998 Nobel Prize in Physics was shared by Robert Laughlin for his theory of the
fractional quantum Hall effect, and fractional charge excitations.] [Gnädig/Honyek]
3) ÔAscending and Descending VoltagesÕ: a circular argument
a) Obviously, I did not miss my true calling as a screenwriter. The total induced
electric field, if the given equation is correct, is simply 0.001 Bo , over a distance of
2 0.001π , so the total voltage (E ¥d) is: Bo ⋅ 0.001 ⋅ 2 0.001π , and the total current is this
value divided by 500 ohm. The voltage across one bulb is 1/5 of the total.
b) The picture you are mumbling about is called ÒAscending and descendingÓ (note
hint in the title of the question). M.C. Escher makes use of a lovely illusion to draw a
staircase with no beginning or end, and with people on it constantly descending or
ascending. Actually, my colleague Julian (who is a fine violinist I might add) comments 2 that there is an additional Escher print on the same theme, with a water fall linked in a
similar way, endlessly producing power. V RV R R
V Of course, we have no such ÔtricksÕ in physics,
so what is going on in the circuit? The voltage
ÒsourceÓ is distributed through every bit of
wire, so the voltage drop across each resistor
matches the voltage increase across 1/5 of the
length of the wire. Thus you start out at the
same voltage you began with. This is similar to
the circuit at left. c) Consider the righthand rule in comparison
to the equation your new friend is trying to sell
you (another hint that I included in the
equation). The induced current will create a
magnetic field that will increase the flux in the
loop, this will increase the induced current that
will create more flux, you have a runaway effect! Any small perturbations in any small
magnetic field will cause it to increase dramatically and without end. Of course this is
not realistic so there must be a problem with your buddyÕs model. The real equation
has a negative sign. The induced current acts against changes in the magnetic field.
Mother Nature is truly a conservative force (no political insinuations intended) [James] V R V R 4) Flux compression Ñ putting the squeeze on a Bfield
This experiment is apparently done at Los Alamos National Laboratory, in New
Mexico, USA. The two sides of the device have explosives attached to them, so that
they start touching each other at the far end and finally end up being connected at the
low end. The area of the loop is actually much smaller than that of the sides. The current
is caused by a capacitor discharging.
As hinted, the Maxwell equation we
will use is that for induced emf. We
have:
dΦ B
= −Emf
dt
→ → Φ B = ∫ B • dA
So the change in the magnetic flux
causes and induced emf. In our case,
the initial flux is just B¥ A, since the
3 field is uniform and perpendicular to the plane of the device. As the area of the device is
shrinking, the flux is decreasing. However, we see from above that this change induces
an emf in the circuit, which in turn produces a current. Which direction is this current?
From the equation we see that it will oppose the decrease in the flux Ð i.e., the current
will flow in a direction to oppose the decrease in flux and will thus increase it. How
much does the flux grow by? It will keep growing until it equals the original flux,
because then ∆B = 0, emf induced = 0, and we have reached a steady state.
Thus, we see that flux will be conserved in this process and we will get
Φ B = const = BA = B′A′
A
B′ = B
A′
Field of solenoid is B = µ 0 In using AmpereÕs Law. A direct application of i) gives
2 2 r r
which gives 1.26 T in this case. Note that the field grows as the
B′ = B
= µ 0 In r′ r′ square of the radius, which means that we can achieve very large fields (~ 1000 T) with
this technique. [Peter and Bryan]
5) Scratch and dent sale on capacitors
i) Capacitance C is defined such that Q = C V, where Q is the charge on the capacitor
and V its electrostatic potential. Thus we can write C = Q/V. For a spherical
distribution of charge, the field and potential outside the sphere is exactly as it would
be if the entire charge were concentrated at a point at the centre of the sphere. Thus at
the sphereÕs surface or a vanishingly small distance beyond, V = k Q/R, where R is the
sphereÕs radius. From this it follows that the capacitance of an isolated sphere is:
Csphere = Q/V = Q/ (k Q/R,) = R/k
(the ÔgroundedÕ capacitor plate is at infinite distance, where the electrostatic potential
approaches zero)
6 9 2 Ð2 The earthÕs radius is 6.378 × 10 m, and k = 8.9875 × 10 N m C ; from this it follows
Ð4 2
Ð1
Ð1
that the capacitance of the earth is 7.1 × 10 C N m , or 710 µF. Seems kind of small
for such a big planet!
ii) The energy of a capacitor of charge Q and capacitance C is Q2/(2C). If the change in
energy of the capacitor can be found the change in its capacitance can also be calculated.
The energy of the capacitor is higher when it is indented, since the surface charges
move in a direction opposite to the force acting on them. Also an electrostatic field of
field strength E has an energy εE2/2 per unit volume and when the capacitor is
indented the electric field penetrates a volume where it was not previously present. 4 If the surface of the capacitor is only changed a little the electric field can be considered
as identical to the original one near the surface. Thus, the change in energy depends
purely on the change in volume and not on the actual shape of the indent.
Imagine that the capacitor is hammered so that its volume decreases by 3%, but its
shape remains spherical. Its radius is therefore reduced by 1% (as the volume of the
sphere is proportional to the cube of its radius). The ratio of the energy of such a
capacitor to the energy of the original one is the same as the ratio of the energy of the
indented capacitor of the problem to that of the original one. Thus, the relative change
in their capacitance is identical as well.
Further, the capacitance of a spherical capacitor is proportional to its radius. The
capacitance of the new capacitor (of reduced size) is therefore 1% smaller, and the
capacitance of the capacitor of the original question decreases by the same amount.
[Robin & Gnädig/Honyek] 6) Mocking mirrors
i) We have: So that the field at a point (0, y) is given by:
r
2 cos(θ)q )
2dq
)
E(0, y) = −
x=−
x
2
3/2
4 πε o R
4 πε o y 2 + d 2 ( ) And it has no ycomponent. Hence, the potential, which is just the integral of the field
(up to a constant) gives zero (since ∫ 0dy = 0 ), so the potential is indeed constant in the
planes.
ii) By comparing this to i), or noticing that the image charges correspond to images of
object as produced by a mirror, a selfconsistent solution is: 5 The field at any point (0, y) is
r
2dq
2dq
)
E(0, y) = −
x+
3/2
4 πε o ( y + d)2 + d 2
4 πε o ( y − d)2 + d 2 ( ) ( ) 3/2 )
x which is also the field at a point (Ðx, 0), by symmetry.
iii) Consider this reasoning: to make A equipotential we have to put a charge Ðq a distance
d left of it. To make A' equipotential, we put Ðq a distance d to the right of A'. But now
the two Ðq charges cause A and A' not to be equipotential again. So we put a charge +q
a distance 3d left / right of A / A' (respectively). And so on, ad infinitum.
This is equivalent to the forming of infinitely many images when you stand between
two mirrors Ñ the 1st seems a distance d away, the 2nd, a distance 3d = d + the distance
of the light ray bouncing from the opposite mirror.
We have: [Peter] 6 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 6: AC Circuits and Electronics
Due April 2, 1999
1) Thief of time
100V 33 kΩ
neon lamp 30 µF A chum of mine once upon a time had built this circuit,
and set it by the basement windows of his house as a
burglar deterrent Ñ it makes a flashing light, and gives
the impression it might be part of a burglar alarm.
The flashing light comes from a small neonfilled bulb,
with closely spaced electrodes. The bulb can support
about 80 V across it without problems, but at a higher
voltage the gas inside ionizes in the strong electric field,
the resistance drops drastically, and a discharge takes
place between the electrodes. The discharge electrons
excite the neon atoms, which produce the light. Once
electrons stop flowing, the ionized gas recovers and the
resistance rises again to the original value. i) How frequently does this particular design flash?
ii) If the neon lamp goes from 220 kΩ in its normal state to about 0.5 Ω during the
discharge, roughly how long does each flash of light last?
iii) During a flash, roughly how much electrical energy goes into the lamp?
iv) How could you easily modify this circuit to make the flash last longer? How long
would the flash then be? [Robin]
2) Logic rules (!)
For the questions below assume that a logical high (Ô1Õ) means a voltage of about 5 V, and
a logical low (Ô0Õ) means a voltage around 0 V. Straight lines are wires and things labeled
A, B, CÉ are inputs.
i) Consider this attempt at a logic gate. Make a
truthtable for this setup: show the output for each
possible combination of inputs (use 0s and 1s). What
kind of gate is it? Name at least one major problem A
B (output) 1 this gate would have, if used in some circuit. [HINT: does the gate have, say,
directionality?]
To avoid the problem you may have found above, we can use a wonderful device
known as a transistor. A transistor essentially functions as a switch. It has 3 electrical
connection points Ð the base, the collector and the emitter. When the base is at a higher
voltage than the emitter, the conduction band of the semiconductor inside shifts toward
the collector and then current can flow from the collector to the emitter. Otherwise, the
band is near the base and the device does not conduct between the collector and the
base. Semiconductor Ñ get it? Sometimes itÕs like a conductor and sometimes it isnÕt.
ii) Knowing this, can you figure out what the following circuits do? Give a basic
description of each and figure out the truth tables (as described above). V
b) a) V V
R
R R (output) (output)
collector A A base
emitter B [NOTE: in all circuits the resistance of the load on the output (e.g., voltmeter) is much
higher than R] [Peter]
3) Small, but wireyÉ
Your mission Jim, if you chose to accept it, is to connect the dots using the given
components to obtain the required currents or voltages (see figure on next page). The
wires and components are delicate so do not bend them. Connecting diagonal dots
does work though. Remember Jim, all the components are ideal (wires have no 2 resistance, there is no voltage drop across an ammter, etc.). You have 15 seconds before
this unit selfdestructs. 14, 13.... e.g.) Set voltage=5V, current: I=5A. 5V
wires 5V V A
1Ω 1Ω A V 10Ω a) Set I=2mA. 1kΩ 1kΩ A b) Set I2=18mA. What is I1? A1 1kΩ 9V 9V A2 12V 9V 1kΩ wires wires 0.5kΩ
c) The voltage across the unknown element AC changes as a function of time (see V1(t)). Set V2(t)
as shown.
V1(t)
V2(t)
t V1 V2 AC t
Recall Jim that the current across
a diode looks like:
I wires diodes V 13kΩ 1V 13kΩ A d) Set I=47.5 mA. (Be careful Jim, we think this one might be harder than it looks.) wires
connectors
13kΩ
unlimited supply of all three components [James] 3 4) Is resistance futile?
Given a resistance R (made of n smaller resistors of value r connected together any way
you want), is it possible to add two more rÕs so that the total resistance remains R? Note
that all the resistors must have both their ends connected Ð i.e., the following is illegal:
The resistance from a t o b obviously
remains at a value R, after connecting two
more resistors r, which makes the
question far too boringÉ [Peter] r r R a b 5) ÔE pluribus unumÕ
Given a large number of exactly 1 Ω resistors, describe using circuit diagrams how to
make a resistance of any (arbitrary) given value.
[HINT: the value of the new resistance is arbitrary, but it can only be measured to a
finite number of digits] [Peter]
6) Inductive reasoning S The circuit shown in the figure Ñ
composed of three identical lamps
and two coils Ñ is connected to a
DC power supply. Some long time
later, the switch S is opened. In
order of brightness, how do the
lamps rank? Explain.
[Gnädig/Honyek] C
L C
L L Remember to check the POPTOR webpage for hints and any necessary corrections!
www.physics.utoronto.ca/~poptor 4 19981999 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 6: AC Circuits and Electronics
1) Thief of time
i) LetÕs use these symbols for the various bits:
R = 33kΩ, the resistor after the battery
RB = 220kΩ, the normal resistance of the bulb
C = 30 µF, the capacitance of the capacitor in parallel with the bulb
Our basic equations are:
VC = VB (in parallel)
VR + VB = 100 V (the whole voltage drop around a closed loop)
I R = I B + IC (the current is conserved as it splits over the bulb and capacitor paths)
VR = I R ⋅ R (OhmÕs law)
VB = I B ⋅ RB
QC = C VC (charge on a capacitor), which gives:
˙
˙
⇒ IC ≡ QC = CVC (just taking the derivative)
˙
= CVB
Then this can go into OhmÕs law to give a voltage,
VR = I R ⋅ R = ( I B + IC ) ⋅ R ( ) ˙
= I B + CVB ⋅ R
V
˙
= B + CVB ⋅ R RB and then this can be used in the voltagesum expression
VR + VB = 100 V , to reduce just to an expression for the bulbvoltage: VB
˙
+ C VB ⋅ R + VB = 100 V RB 1 1
1 100 V
˙
VB + VB + −
=0
C RB R RC
which we can write in a simpler form for the constants,
1 ˙
VB + VB = b + c = 0
b= 1 1
1
+ C RB R 100 V
RC
We can solve this several ways Ð the most straightforward is just to integrate, using
˙ dV
VB = B , so that you get an expression d VB on one side and dt on the other, to
dt
integrate.
c=− A neater answer is to think about what this means Ñ what is going on. After a long
time, we can guess that if the bulb hasnÕt broken down, the capacitor will be all charged
up and that nothing will be changing (i.e., will have a Ôsteady stateÕ). Then there could
˙
be no changes in time, so VB (t ) = 0 , and therefore:
VB (∞) ⋅ b + c = 0
⇒ VB (∞) = − RB
c
= (100 V ) ⋅
R + R b B Thinking about the start, when the battery is connected but no charge has yet flowed
through R, we need to have VB(0) = 0. So the right solution starts at zero and
asymptotically (over a long time) goes to a constant as a limit. If you already know RC
circuits, or if you know 1st order differential equations, or if you integrate the equations
(or if youÕre lucky!), you might guess that the change is an exponential relaxation,
which if we start at zero and go to VB (∞) would have the form ( VB (t ) = VB (∞) ⋅ 1 − e − at ) If you put this into the equation above, youÕll see easily that it actually works: (− a) VB (∞) e − at + VB (∞) (1 − e − at ) ⋅ b + c = 0 {(a − b)⋅ VB (∞) e−at } + {VB (∞)⋅ b + c} = 0 This has to be true for all values of t, which means each parenthesis term must be zero
independently. Taking the second term to be zero gives VB (∞) , if we didnÕt already
guess it above. The first term to be zero gives us a=b, so the solution is:
1 1 1 −
+ t
RB C RB R VB (t ) = 100 V ⋅
⋅ 1− e R + RB or with our exact values put in: 2 ( VB (t ) = 87.0 V ⋅ 1 − e −1.16 t ) ( VB (t ) reaches breakdown for the bulb (80V) when 80 = 87¥ 1 − e −1.16t ) ⇒ t = 2.18 s ii) At breakdown, the capacitor discharges through the bulb, until the bulb ÔrecoversÕ.
V C = VB
IB+IC=0
QC = C VC
˙
⇒ IC = C VC
and with
VB=IB RB
and also (with current lefttoright taken positive)
0 = I B + IC
This then gives
VB
˙
+ C VB = 0
RB
1
˙
VB = 0
⇒ VB +
RC
which has a solution VB (t ) = VB (0) ⋅ e − 1
t
RC So a good characteristic time of the bulbÕs flash is RC = 0.5 Ω ⋅ 30 µF = 15 µs
iii) The energy stored in a capacitor charged to voltage V i s: = 30µF ⋅ (80 V ) = 96 mJ (not enough to make the bulb explode, I hope!) It is all
dumped through the neon lamp.
1
CV 2
2 1
2 2 iv) going back to (ii) we can make the flash last longer by adding a resistor R′ series
with the bulb. Then τ = RC = (0.5 Ω + R′) ⋅ 30 µF which you can control by the value of
R′ and we can make the bulb light up for longer. But since
˙
˙
I B = IC = − CVC = − CVB
˙
and VB (t ) will be smaller with R′ added, then the current through the bulb will be
smaller and it wonÕt be as bright. [Robin] 3 2) Logic rules (!)
i) This is an OR gate. When all of A, B, C are low (i.e., voltage is, say, 0 V), the output is
naturally low. But when either A or B or C are high, so will the output be.
Problem: when either A or B or C conduct, some current flows through the circuit. But
when more than one input is high, the currents actually add. If we have something
fairly sensitive connected to the output, we could fry it if the current is too large. WhatÕs
worse, in a computer many gates are connected together Ð but if we connect a few of
these gates to the inputs of another gate, the currents will add (if we have n inputs, final
current is nI. If we feed n of such gates into another similar gate, the current is now n2 I)
Ð this is clearly a problemÉ
ii) a) This is a NOT gate. When A is low (ca. 0 V), the base and the emitter are both at 0
Volts. The transistor is thus cutoff Ð no current flows from the collector to the emitter (it
acts like an open circuit). If we now connect a device with a large resistance (>> R), the
voltage drop across R is negligible and almost all voltage (Vo) goes through the device.
So the output is high.
[Voltage across R would be RI and that across the device rI, but since r >> R the first
may be ignored. Note that the current flowing through both is the same]
When A is high, the transistor conducts (with almost no resistance) Ð thus, all the
voltage drop will be across R [since it is smaller than r Ð the resistance of the output
device]. Hence, the output is low.
So this is a NOT gate.
b) The second gate (from the left) is a NOT, from above. The first gate is a NAND, by a
similar argument as above. Thus, the gate is a NOT (NAND) = AND.
Note that using transistors the output and current voltage are constant. There is still,
however another problem, but IÕll let you figure it outÉ [Peter] 4 3) Small, but wireyÉ b) Set I2=18mA. What is I1? 1kΩ A 1kΩ 1kΩ a) Set I=2mA. A1 9V 9V
12V
Be careful about the polarity of the 12V
source! A2 9V So what is I1? The normal
voltage/current laws do not
help you. The only answer
that I can argue would be
using symmetry, and thus
I1=9mA. c) V1 AC V2 This is known as a fullwave rectifier
and can be used to chance AC voltage
into DC voltage (with an extra
smoothing filter) d) Set I=47.5 mA.
total: 618 resistors
I had a more complicated answer but this one, handed in by a
few students, is more straightforward. 13kΩ 13kΩ 1V 13kΩ A 13kΩ [James] 4) Is resistance futile?
LetÕs try all 6 possibilities:
a) R r r 5 b)
r
R r c)
R
r
r d)
R
r r r R e) r f)
r r
R From a) we get: R = R + 2r, which is impossible (r > 0)
From b) we get: R = r/2 + R, which is again impossible
From c) we get: r = r + 2R Ð impossible.
1+ 5
From d) we get: R = r 2 −1 + 5 From e) we get: R = r 2 From f) we get: 2r = 2r + R Ð impossible.
So it is possible, using (d) or (e).
A sophisticated aside for question 4:
Note that we have a way of making resistances r that are rational. But note that both
the solutions (d) and (e) in question 4 are irrational Ñ if our arbitrary r is rational then
6 it follows that R is irrational. But to make R we only used rÕs in some combinations Ð we
only used rational numbers.
LetÕs recall what kinds of circuits we can make using resistors:
We could have resistors in parallel and series. But the equivalence formulas involve
additions (series), or divisions (parallel), so there is no way to obtain a rational number.
In the general case, we need to solve KirchoffÕs Laws, which are linear and create linear
systems of equations. Solving such a system involves adding / subtracting rows, and
multiplying and/or dividing numbers Ð once again there is no way to obtain an
irrational number.
And thus we must conclude that theoretically it is not possible to make the circuit from
question 4, unless perhaps we could make R from an infinite number of resistors r.
Experimentally, on the other hand it would be possible. For in experiments we can only
measure things to a finite precision. Thus, there are no irrational values Ð since we only
measure a certain amount (say 10) digits, every number is finite and hence rational.
And hence we could achieve the circuits from either (d) or (e). [Peter] 5) ÔE pluribus unumÕ
Within experimental error, the other resistor (call it R) can only be measured to a finite
number of digits. Given any such number, we can express it as a rational number Ð i.e.,
in the form p/q where both p and q are integers.
To create a circuit to express p/q we build the following circuit Ð we connect in series p
groups of q groups containing 1 Ω resistors in parallel.
q resistors p groups [solution from a Polish Olympiad book ÒOlimpiady Fizyczne XXI I XXIIÓ] [Peter] 7 6) Inductive reasoning
When the switch is closed,
currents as shown at left flow
S
round the circuit. (The value of
the current is determined by the
C
C
voltage of the battery and the
I
L
L
L
I
I resistances of the lamps.) A very
short time after turning the
switch off the current flowing in
the coils is practically
unchanged. (If this were not the case, there would be a rapid change in the magnetic
flux, which would induce a very
I
I
high voltage in the coil.)
S
Currents of 2I and I continue to
C
C
flow therefore in the coils, and
these determine the currents
L
L
L
flowing through the lamps
2I
I
I
(figure at right).
I I This means that the lamp closest
to the switch suddenly flashes
and the brightness of the two other lamps does not change. (This is only true for a short
time, later all three lamps fade and go out.) [Gnädig/Honyek] 8 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 1: General
1) Pumpkin paradox
a) Mbig = 10000*Msmall Therefore, for the same relative strength we need a cross sectional area 10000
times larger than the poles in the small model. So,
Abig = 10000 Asmall
2 2 (1/4) π (Dbig) = 10000 (1/4) π (Dsmall) or
Dbig = 100 Dsmall
So Farmer Joe is going to need poles that are 100 times thicker.
b) Mass is proportional to volume which in turn is proportional to area
(3/2)
Thus, mass is proportional to area
. So,
(Mbig) (2/3) (2/3) /(Msmall) = Abig/Asmall = (1000kg/0.1kg) (3/2) . (2/3) Therefore, Abig = 464 Asmall
Since the big pumpkin surface area is 464 times greater than the small,
Joe is going to need 4.64 L of paint.
c) Mweight/Mpumpkin = 450/1000 thus,
Mweight = 9/20 The sum of the moments is zero when balanced, thus,
Rweight Mweight g – Rpumpkin Mpumpkin g = 0
Therefore, Rweight = (20 Rpumpkin)/9 [Carrie] 2) Cubic quandry
a) There are 6×4 = 24 equal cubes that can perfectly surround the lamp.so we
can say that the fraction of power heating one cube is 1/24. 1
Qt = mc∆θ
24
24mc
[seconds]
? t=
Q b) A unit area of a blackbody radiator in temperature T, emits energy by the
4
2
4
power of σT Watts. so our cube is emitting 6L σ T J/s by the assumption that
our cube is in thermal equilibrium each time. Consider our case that any small
change in T, will affect the emission rate and the time needed to increase the
temperature accordingly. But, in this case, we are looking at changing the
temperature by 1K at room temperature, i.e., from 300K to 301K. The relative
4
change is on the order of 1/300. and the change in T is about 4 × (1/300). 3014 = (300 + 1)4 = 300 4 (1+ 14
1
) ≅ 300 4 (1+ 4 ⋅
)
300
300 To a good approximation, we can ignore the changes in dT/dt due to a small
change in T, and we can substitute the room temperature instead.
1
Q t = mc ∆θ
24
Q ⇒ t = {mc}/  6L2 σ T 4 24  6L2 σ T 4 t + c) In thermal equilibrium, the emission rate and heating rate are equal. so we
have :
 6L2 σ T 4 = 1
Q
24 Q
⇒T = 2 144σL 1 4 [Amir] 3) Poles apart, telling
First of all let us note that the NorthSouth (NS) designations on magnets are a
little messed up. You see, the N pole is actually the Northseeking pole, meaning
that it is actually the South pole (since opposites attract). We will follow these
conventions in this document (it doesn’t matter how you approach this, so long
as you’re clear what a North pole of a magnet means to you).
After this lengthy introduction, let us get right to the point. The magnetic field of
the magnet will bend the electron orbits into circles. We assume here that the
field is uniform throughout the face of the monitor (clearly not true, as the field gets smaller with distance; this solution is thus only valid for electron trajectories
not far from the magnet).
i) Suppose we put the Npole (really the South pole) of the magnet on the right
r
rr
side of the screen. The electrons will be bent according to F = − ev × B (e is the
size of the electron’s charge). From this we see that the direction of deflection will
r
r
be “DOWN” (looking directly at the monitor, v is out of the screen, B is from left
to right).
1
ii) From the voltage we can find the speed of each electron Energy = eV = 2 mv 2 ,
7
where V = 15 kV. From this, v = 7.3× 10 m/s (Note that this is actually
relativistic… we will ignore this, as the correction is small. Bonus marks will be
given to people who realize this). We also have mv 2
= e vB , where R is the radius of rotation of the electron.
R We find R = 0.41 cm. Now, each electron will actually only be deflected when it’s
in the field of the magnet. In this case, this would be roughly the size of the
magnet, which we cleverly enough didn’t tell you. (we assume that the magnetic
field drops off really quickly at the sides of the magnet). Seeing that the turning
radius is only 4.1 mm, the magnet really can’t be bigger, or there would be no
picture whatsoever (all electrons would turn back). Let’s assume that the width
of the magnet is indeed 4.1 mm. Then, electrons will turn by 90 degrees, and the
vertical distance traveled is clearly 4.1 mm.
DISCUSSION: This somewhat makes sense, but I would expect something
bigger. This is mostly due to the fact that the magnetic field actually decays
slowly with distance (and doesn’t become abruptly zero), whereby the force acts
over a larger distance and deflects the electrons more. y
x In the general case, we can resolve the motion into 2 directions: x (towards the
screen), y (down the screen). We have: x (t ) = − R sin(ϖt )
y (t ) = R cos(ϖt ) where v = ϖR . We want to find a time τ when the electron reaches the screen (solid line) so
–x(τ) = width of magnet, and calculate what the deflection y(τ) will be.
[Peter]
4) Physics Rocks!
a) To determine the critical height one must consider the equation of motion of
the stone in the water. The forces acting on the stone in water are,
(i) the downward force of gravity = –m*g
(ii) the upward drag on the stone = 0.5 b p A v 2 (b = drag coefficient, p = density of water)
and from Archimedes' principle,
(iii) the upward buoyant force = p V g
(V = volume of fluid displaced)
Summing the forces we get,
2 p V g + 0.5 b p A v – m*g = 0
The sum of the forces is zero at equilibrium, when the velocity is the terminal
velocity. Given that the diameter of the stone is 5 cm and that the density of
3
quartz is 2600 kg/m it is easy to find the following,
–3 m –5 m A = 1.96 × 10
V = 6.54 × 10 2 3 m = 0.17 kg
Plugging in the values yields a terminal velocity of 1.6 m/s (of the stone in
water). Thus if the stone is dropped in air with Vo = 0 m/s then under the
acceleration of gravity the distance it travels before reaching a velocity of 1.6 m/s
is,
2 D = Vt /(2 g) = 0.13 m
So the stone should be dropped from a height of 0.13 m above the surface of the
water.
b) At terminal velocity the waterdrag problem is trivial! After 6 seconds the
rock has fallen, 6*(1.6 m/s) = 9.6 m. [Carrie] 5) Light diversions
Consider a ray of light reflected from one of the mirrors. The reflected ray is a
mirror image of the continuation of the incident ray. Therefore, an easy way to
trace the ray path is to make another square beside it and follow the continuation
of the incident ray in that square.
For more than one reflection , more squares should be drawn as shown. It is
observed from the figure that as the continuation of the initial ray pass through
any corners, it means that the light have gone out from the square. Therefore, the
necessary condition would be tan(θ) = m /n which m and n are two arbitrary
integers.
Doing the first part, makes solving this part much easier. We have to build up
some triangles which are image of one another. Since the cone angle is 30°, last
triangle fir exactly the first one. Two regular polygons with 12 sides is produced.
Analogous to condition that the rays leave through the upper hole after many
reflections is that the continuation of the incident ray pass through the smaller
polygon. After some simple mathematics, we figure out that D must be smaller
than the width of the polygon which is d / tan 15°. [Yaser] 6) Physicists’ pipe dreams
a) The water column will get narrower and narrower because the flow is
constant and the speed is increasing, therefore the surface area of the water will
decrease until it breaks up into droplets.
b) If we consider low viscosity for water, molecules of water are freely falling
down.
2 v 2  v 0 = 2gh
vπr 2 = v 0 πR 2
r=R ν
ν
2 2 r = R { v 0 /( v 0 + 2gh ) } 1 4 c) The stream will break up into droplets. It cannot get arbitrarily skinny, the
way the formula gives, because:
1 . The radius can’t be less than the radius of a water molecule!
2 . Before you reach the limit (1) above, surface tension of water won't let you
make it as skinny as possible. In fact, surface tension could affect the result
which we obtained , but we are neglecting that!
Consider the droplets have radius a . What is the height of a samevolume
cylinder of radius r? The energy of surface tension for these two area ( as you
could see in many books & specially at the end off this problem set :)
4
π a 3 = πr 2 ∆x
3
4
? ∆x = a 3 /r 2
3 The surfacetension energy of these two states ( water tube & droplet ) are : σ 2 π r ∆x and σ 4 π a 2 substituting deltax, we have and comparing them:
a3
σ 8π
3r a 3 3r
and σ 8π 〈
3r 2a it means that, if 3r/2a is less than 1 , then the surface energy of droplets is less
than that of a water cylinder, so it’s energetically favourable for the water to
break into droplets instead of staying in a cylindrical shape. What does
determine the actual radius of a droplet? Several factors may affect that — for
example the circular motion of water flow. But the most important one is density
of water. It will break into droplets when the weight of the droplet has the same
order of magnitude as the surface tension. It means:
ρ = 4πa 3 / 3 g = 2πrσ using previous result, we will get: r= 4σ
= 2 ↔ −3 m = 2 mm
10
4ρg and then the height at which we reach this redious is:
2
h = v 0 / 2gr 4 (R 4 − r 4 ) d) For the experimental part, I just used a tap with diameter of 6 mm and tried
to see at what height the water breaks up. After that, I fixed the height and
measured the volume of the poured water and the time interval.
Here are my results: Height of break up Volume
(cm)
(ml) Time needed
(s) v 0 = V /( πR 2 T )
0 56 cm 100 ml 78 4.5 cm/s 710 cm 100 ml 62 5.7 cm/s 1620 cm 200 ml 50 14.7 cm/s 2630 cm 300 ml 61 17.4 cm/s Due to large error I have, the experimental behaviour is not exactly the same as
our expectation, but `h’ is still proportional to v 2 .
[Amir]
0 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 2: Mechanics
Due December 17, 1999 (revised date)
1) Going for a spin
Cally is playing on the whirlyride (like a merrygoround, but nonmotorised) in her
neighbourhood park. To make it go faster she pushes against the ground using one of
her feet.
a) If she accelerates the whirlyride from rest to a speed of 15 r.p.m. in 3 seconds, what
is the corresponding angular acceleration?
b) By what fraction does the angular velocity change when her 20 kg kid brother Matt
leaps radially onto the edge of the whirlyride? Assume the initial moment of inertia
2
(just Cally and the whirlyride) is 2000 kg m and that the whirlyride has diameter 5 m.
c) Since one of her alltimefavorite past times is tormenting her little brother, Cally
decides to speed up the merrygoround in an effort to fling her little bother from the
ride. Assume that Matt is sitting at a radius of 4.5 m, that the coefficient of static friction
between him and the floor of the whirlyride is 0.7 and that he is naively not holding
on! At what speed will Cally’s little brother start to slip? . [Carrie]
2) Bubble bonanza
a) What is the presure inside a soap bubble of radius ‘r’ if the surface tension of the
soap solution is σ?
b) What will happen if two bubbles are brought to touch each other? (assume that one
of them is bigger than the other, and there is still a bubblefilm between them
c) What will be the end result if the wall between the the two touching bubbles breaks,
and the two bubbles merge?
d) What if those two bubbles were attached to opposite ends of a tube (e.g., a drinking
straw) like a weightlifter’s dumbell? [Amir]
3) Siphoning cellar
My outside water faucets in my new house don’t work, and I’m reduced to running a
hose from the laundry tubs out my basement window to fill my aboveground
swimming pool. If I forget and leave the hoseend in the pool, then when I disconnect
1 my hose from the tap, water from the pool will siphon back into my basement. The
setup looks like this: 1.75 m 1.0 m 1.5 m 10 m Water in a 2 cmdiameter hose flows with a drag force Fdrag on the hose which is
proportional to the length of the hose and to the speed of water flow. This drag results
in a pressure drop along the hose. This pressure drop, per unit length, is proportional
to v by the formula:
∇p = ∆p 32 Uη
=
∆x
d2 which is described more fully in INFOBITS™ below.
a) If I forget the hose when I leave for work, and it siphons for 12 hours while I’m gone,
how much will the water level in the pool drop? Estimate, or calculate exactly.
b) When the last of the water drains out of the pool, then the water in the hose becomes
of shorter and shorter length. This changes the drag on the water. See if you can
approximately calculate the flow rate of water out of the hose as it empties. [Robin]
4) Gimme a brake...
The Bumble Bee ride at Super Fun Fun World consists of carts (painted yellow and
black of course), which seat 4 people, that traverse a windy, hilly track. The figure
below illustrates one section of the ride. On a particularly unusual Tuesday there is a
massive power
Aieeee!
failure at Super
Fun Fun World
(which
consequently isn’t
so fun fun). The
power fails just as
6m
one of the carts
4m
nears point A with
a velocity of 0.5
m/s.
Point A 5m Point B 2 a) If no emergency brakes are applied with what speed will the cart hit the bumper at
point B?
Take it that you have drum brakes, which press on a hub which is half the wheel
diameter. Each drum presses onto
a 30° arc of the hub. The thickness
of the hub is the same as the wheel.
For steel wheels on steel rails, the
30°
coefficient of static friction is 0.78, 30°
and the coefficient of kinetic
friction is 0.42. You can assume
that only about 3 mm of the rail is
in contact with the wheel.
b) At the bottom of the first hill the emergency brakes are applied radially to either side
of the wheels. If the brake pads have a static coefficient of friction 0.85 and kinetic
coefficient 0.65, what is the optimum force to apply to the brake pads such that the
braking force is a maximum?
c) With the brakes applied (assume with optimal constant force) how far up the second
hill will the cart go?
d) With the brakes held on, will the cart go back down the hill? [Carrie] 5) It’s a ball? What a gas!
This question is the beginning of a kinetic model for an ideal gas. It can lead you to
figure out the ideal gas law PV = nRT.
a) A ball with mass m and velocity v collides with a stationary wall. If friction is
negligible, and bouncing is perfectly elastic, the ball bounces back with the same speed.
But what happens if the wall, itself, moves toward the ball with velocity u? Consider all
velocities be in the same direction, perpendicular to the wall.
b) Going through (a) you can find out that the momentum has been changed. This
means that the wall did some work on the ball. Can you show explicity that the
difference in kinetic energy of the ball is exactly the same as the work done by the wall?
BONUS: Can you use this model to show how much the temperature of a gas of atoms
behaving like such elastic balls goes up as it is compressed — as long as no heat is
allowed in or out? [Yaser] 3 6) Dimensional thinking
A cylindrical vessel of water is leaking via a small hole of radius ‘r’. We want to keep
the height of the water constant. The best way is to add water at some rate R (kg per
second).
a) Can you guess what should be R in terms of ρ (the density of water), H (height of the
water level in the vessel), r (radius of the hole), g (acceleration due to gravity) and A
(the crosssectional area of the vessel). [HINT: Use your intuition, or common sense, and
–2
watch that you keep the correct units in each stage (i.e., acceleration has units L T ,
where L is distance and T is time)]
b) Can you prove your formula with physics theorems?. [Amir] 4 INFOBITS™ — Useful Bits of POPTOR Information
Coefficient of resistance for pipe flow:
There is a drag on the inside of a pipe, for a liquid flow through it. This drag means
that there’s a pressure drop as you go along the pipe — it is why small plumpingpipes
lead to poor ‘water pressure’ for a shower or similar. The pressure gradient (change in
pressure per unit length) is:
∆p λ 1
=
ρU 2
∆x d 2
where p is pressure, ρ is fluid density, d is pipe diameter, U the average speed of liquid
flow, and λ is a dimensionless constant which is equal to:
∇p = λ= 64
Re for laminar (smooth) flow Re is called the Reynold’s number. (usually a value of a few 1,000 for laminar flow) The
Reynolds number for a pipe can be found from:
Re = ρUd
η putting all this together, we see that:
∇p = ∆p 32 Uη
=
∆x
d2 C HECK THE POPTOR WEB PAGE for other hints, and any corrections we might post:
www.physics.utoronto.ca/~poptor 5 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 2: Mechanics
1) Going for a spin
α= ∆ω 15 ⋅ 2 ⋅ π / 60
=
= 0.5rad / s 2
∆t
3 Angular momentum is conserved (but energy is not!!)
I oω o = ( I o + mr 2 ) ω
ω
2000
=
= 0.9%
ω o 2000 + 20(2.5)2
Ignoring the fact that Matt is sitting off the whirlyride, what happens is that the force of
friction is providing the centripetal acceleration necessary to keep him on the ride. The
force is ≤ µmg = 137.2 N .
Thus, at maximum,
v
v2
= 1 rad s ±1
= 0.7 ⋅ 9.8 , v = 6 m s ±1 and ω =
4.5
4.5 [Carrie] 2) Bubble bonanza
a) Consider 2 hemispheres of a bubble of radius r. They are repelled by a force of
F = ∆PA = ∆Pπr 2 , where A is the area of the projection of half a sphere and ∆P = Po − P
(difference between outer (atmospheric) and inner pressure). The hemispheres are
pulled together by surface tension (which is given as force per unit length); moreover,
there are 2 surfaces to a bubble, so the total force here is F = 2(2πrσ ) . Solving yields
P = Po + 4σ
r b) Note that pressure is inversely proportional to the radius. Thus, the smaller bubble
will actually have a higher pressure inside! This means that the film between the 2
bubbles will be pushed into the larger bubble. 1 If the film breaks, the bubbles will either break (but that’s so boring), or merge. Using
the fact that the numbers of moles inside the bubble stays constant, assuming that
temperature stays constant and using the ideal gas law, one gets
PV = PV1 + P2V2
1 4σ + P r 3 = 4σ + P r 3 + 4σ + P r 3 o
o 1
o 2
r r1 r2 (P, V, r refer to the final bubble, the remaining variables refer to the 2 initial bubbles)
The equation could in theory be solved; note that for Po = 0, the final bubble will be
bigger than either of the 2 initial bubbles…
The system is obviously not in equilibrium. Since the smaller bubble is at a higher
pressure, it will in fact contract and force air into the bigger bubble. Thus, we will end
up with a membrane at one end and a big bubble at the other. [Peter, Amir] 3) Siphoning cellar
Solutions for this problem will be provided with Set #3 [Robin] 4) Gimme a break...
First I have to comment that this question was so hard that nobody got it right,
including some of us here at POPTOR…
a) Ok, that was easy. Conserving energy yields
1
2 mv 2 = 1 m(0.5)2 + mg(6 − 4)
2 From this, v = 6 m/s.
b) This is where things get difficult. Clearly we want to stop the cart using static
friction, as these coefficients are higher than those for kinetic friction, and thus the force
will also be larger. Something has to slip, however (you can’t just stop the cart dead) –
since the coefficient of kinetic friction between the wheels and the rail is lower than that
between the brakes and the wheel, we want to avoid letting the wheels slip (it will yield
the lowest force).
Now, if we have an object of mass m lying on a level plane, where the coefficient of
static friction is µ it certainly is true that the maximum force we can apply so that the
object doesn’t start moving (and kinetic friction doesn’t kick in) is FMAX = Ffriction = µmg .
But a rolling object is different!! For one, it is not stationary, it is already moving. So
what force can we apply? Well, we have to ensure that the cart keeps on rolling, i.e. V(t)
= ω(t)R, and not slipping… 2 I am also going to assume that there are 2 forces acting on the wheel – each of
magnitude F/2  pointing in different
directions.
We have to analyze the linear and
F/2
rolling motions (about the axle) while
F/2
the brakes are applied, knowing we
Friction
start with the cart rolling (m is the
mass of the cart, M the mass of the
wheel, I the moment of inertia of, F
the applied force, R the radius of the wheel and µ the coefficient of static friction
between the wheel and the track).
ma = −µmg
Iα = µmgR − FR
V From this we find
v(t) = vo + at = vo − µgt
ω(t) = ω o + αt = ω o − F / 2 ⋅ Rt µmgRt
+
I
I We have that vo = ω o R at t=0, but to ensure that this is true at later times, we clearly
need that
µg = F /2 ⋅ R µmgR −
R I
I The moment of inertia is that of the wheel, i.e., of 2 cylinders, one of radius R and the
2 R = 5 MR2 (I’ve assumed for simplicity
2
1
1
other of radius R/2. Thus, I = 2 MR + 2 M 2
8
that both sections of the wheel weigh the same amount; it would probably make more
sense to assume that their densities are the same, but oh well…)
F/2 = 5
8
8
µMg + µmg ≈ (0.78)mg
8
5
5 (assuming the wheels are not very heavy)
This is the force felt by the wheel; the actual applied force is
F = Fa (0.65) (the brakes have to be slipping) Solving for the applied force gives
Fa = 2 ⋅ 18.8m = 38m
NOTE: most people put that F = 2 µmg , which is less than the above. Plugging this into
the equations above we can see that V (t) ≠ ω(t)R , and so it seems that the cart slips…
but you can do some experiments at home to convince yourself otherwise. What’s going
3 on here? Well, the force of friction is actually Ffriction ≤ µmg , and what will in fact happen
is that it will adjust itself to a lower value so that the cart doesn’t slip… The above force
is the maximum that can be applied so that the cart doesn’t slip; a bigger force and it’s
all over…
c) The above force does work ( W = Fa d ) to slow the cart down. It starts with an energy
2
1
2 m( 0.5) + mg6 . After a distance d it will stop; here, d turns out to be 1.6 m. Thus, the
cart never even makes it to the hill…
d) If the cart was on the hill, we need it to be in static equilibrium. From the above
diagram we see that this amounts to mg sin θ ≤ 0.78mg cosθ , or θ ≤ 38o . Eyeballing the
diagram, this seems to be false, and thus the cart will probably fall… [Peter]
5) It’s a ball? What a gas!
a) Doing the question in the reference frame of the wall really simplifies things. In the
rest frame of the wall, the ball velocity is the same before and after the collision but
opposite in the direction. In this system, the ball’s initial velocity is v + u. So the ball
bounces back with v + u in reference to the wall. This velocity is v + 2u in reference to
the laboratory.
b) The change in kinetic energy = 1
2 1
m(v + 2u)2 − 2 mv 2 = 2mu(u + v) The work done by the wall = Fx
But F = ∆p m(v + 2u − ( − v)) 2m(u + v)
=
=
; x = ut
∆t
t
t Therefore work = Fx = 2 mu(u + v) , same as the change in kinetic energy above…
BONUS: (solution based on Feynman’s); “< >” denotes an averaging
Let us work in 1D (results in 3D are very similar); let ρ be the density (#) of atoms per
unit volume.
Assume the compression is slow (so the gas stays in equilibrium). Then, the wall is
practically stationary. The force exerted by the wall = change in momentum / ∆t is
F = 2ρmv 2 A
We want to average this over all the speeds, v. Notice that only _ of all atoms (balls) are
moving toward the wall – we do not want to average over them, as they do not
contribute to the force on the wall. This yields
P = F / A = ρ < mv 2 >
Multiplying by the volume, V gives,
PV = N < mv 2 >
4 1
1
In 1D we have that 2 kT = 2 < mv 2 > and thus PV = NkT
If you want to know how temperature changes with volume, you can note that this is an
adiabatic process and look up the formula in any book (or derive it using the above).
[Peter, Amir]
6) Dimensional thinking
a) First of all, figuring out how much water to add is equivalent to figuring how much
water is actually leaving. There a couple of ways to do this, but dimensional analysis is
perhaps the most instructive. First, since the hole is small, the amount of water flowing
out in a short time will change the volume in the vessel by very little; thus, the area A is
irrelevant.
Now, we need to combine the remaining variables into a flow rate R that has units
[kg/s]. We now assume that the formula is of the form
R = Cρx H y r z g w
The units must combine to [kg/s]. It’s pretty easy to see that x=1 and w=1/2. Now, it
would seem logical that the flow is proportional to the area of the hole, i.e., r2 . Since the
flow also grows as H to some power (not as inverse of H), the only logical choice is to
set z=2 and H=1/2.
Thus, R = Cρr 2 gH , where C is some constant.
b) Using Bernoulli’s Law to find the flow rate of the water leaving the vessel leads one
to
1
2 1
ρv 2 = ρgH + 2 ρU 2 Here, v is the speed of flow in the pipe and U is the speed of water flow in the tank. For
a large area A (compared to the area of the hole), the U term is negligible. However, we
don’t have to make this assumption. Since the flow is steady, the continuity equation
applies and we have UA = vπr 2 . The rate R is R = ρvπr 2 . Solving gives
R = ρπr 2 2 gH πr 1− A
2 2 . (for A >> πr 2 , this is indeed equivalent to a), if C = π 2 ) [Peter, Yaser] 5 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 3: Thermodynamics
Due January 14, 2000
1) Clocks a’ Rockin’
A friend of mine knows someone who overclocked his computer, that is, he made it run
faster than it was originally designed to. The entire system remains the same, except
now twice as much data is transferred per unit time, as before. Thus, in effect, the
current has doubled. My friend claims that the new system is so hot, that the cooling
system’s temperature has to be around freezing.
Consider a resistor R, placed in surroundings with room temperature 20 °C, which has
been overclocked as described above. If the original system causes a temperature rise of
13 °C, what temperature will the new system heat up to?
[Peter]
2) Literally linear
Consider a rod with the linear coefficient of expansion λ and initial length of L. It starts
out at temperature T 1 .
a) Heat the rod to a temperature T2 — what is its new length? Then cool it back to its
initial temperature T1 .— what is the length of the rod after this cycle, assuming the
usual formula for linear expansion and contraction? Go through the calculation in a
little detail — don’t just guess, or use common sense.
b) Do this heating and cooling, again and again for n times. What is the length of the
rod now? Does your answer seem strange? (say ‘yes!’) Can you explain (or guess!)
what has made the answer weird? You may think that this happens because we
assumed that the linear heat expansion is the same in all temperatures. This is not the
issue we’re looking for. Try to find another reason. [Yaser]
3) We will break dividing walls
A thermally insulated cylinder, like a vacuum flask, contains a
single species of ideal gas in two separate compartments,
divided by an insulating wall. The pressure, volume and
temperature of gas in each part are : P1, V1, T1 and P2, V2, T2.
What is the final P, V and T of the system, if we remove or
break the separating wall? [Amir] P1, V1, T1 P2, V2, T2 1 4) The nonaligned axis
There are special crystals which have different
coefficients of linear heat expansion along two
different directions, x and y. Say that the
coefficients have values Le 1 a nd Le 2 i n the
directions x and y. Such crystals do exist, and are
termed anisotropic .
Consider a squaresided rod cut from a block of
y
such a crystal, with the rod’s longitudinal axis
making an angle of 30° with the x axis. What is
the linear heat expansion of this rod along its
length? (Ignore the third dimension) Does this rod
expand properly, or does it somehow bend? [Yaser] 3 0°
x 5) Patent, or pshaw?
Here’s an idea: venetian blinds which are painted black on one side and silver on the
other. Assume that the blinds are installed between two panes of glass, with vacuum in
between (so air and its convection has no effect in this question), and assume that the
reflectivity of the silver side is 90%, whereas that of the black side is 10%.
a) How well does it work to turn the silver side of the blinds outward when the room
is already too hot, so as to reflect the light, while turning the black side outward to
–2
absorb heat when the room is too cold? Take it that sunlight is 1kW m , and find the
steadystate temperatures due just to thermal radiation (StefanBoltzmann law).
b) Is there any change you can see that would improve the blinds’ function to control
radiation of heat into the room? [Robin]
6) Tea? For who?
Berto used an electrical heater to heat 3 kg of an unknown liquid in a container (don’t
ever try heating any unknown liquids at home, though!). The heater delivers 1000
watts of power. The initial temperature of the liquid is 20 °C. As the liquid warmed up,
Berto wrote down its temperature in different times. In Tables 1 & 2 below, some of the
data collected by him are shown.
a) Neglecting the heat capacity of the container, find the specific heat of the liquid.
b) Berto turned off the electrical heater when its temperature reached 60 °C. Then, he
started writing down the liquid temperature as it was cooling . You can see in the
table, below, the liquid temperature for some different times. Using these data, try to
find the heat capacity of the liquid to a better approximation. 2 Hint: You should consider the loss which is the heat transfer between liquid and room
in your calculation in part (a).
c) Sir Isaac Newton, in the 17th century, showed that objects cool at a rate proportional
to the difference between their temperature and the ambient temperature, such as room
temperature. Try to find the proportionality constant for (b).
Table 1: Liquid temperature during heating
time(min) 0 3 4 5 6 temperature (°C) 20 34.1 38.5 42.9 47.2 Table 2: Liquid temperature during cooling
time(min) 0 8 13 19 26 temperature (°C) 60 50.2 45.1 39.8 35.2 [Yaser & Carolyn] INFOBITS™ — Useful Bits of POPTOR Information
Remember to check the POPTOR webpage for hints and any necessary corrections!
www.physics.utoronto.ca/~poptor
Specific heat of water: 4.2 kJ kg –1 K–1
–8
–2 –4
StefanBoltzmann constant σ = 5.7 × 10 W m K
–2
–2
Solar constant = 1.353 kW m ; at surface of the earth, about 1 kW m 3 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 3: Thermodynamics
1) Clocks a’ Rockin’
First, I have to say that this was not an easy question — probably way too hard for #1.
The key is to realize that the resistor is radiating energy with a temperature T, with a
background temperature 20°C. Most radiating objects behave more or less like black
bodies, and I will assume this to be the case here as well …
Running at input power P, the computer’s resistor reaches a steady state temperature T
at which the input power is balanced by thermal radiation power (from the StefanBoltzmann law). Since all objects radiate thermally (unless they’re at zero temperature),
the resistor also absorbs a little bit of thermal radiation from the roomtemperature
surroundings at To = 20° C or 293 K. So the power balance is:
P = σ{(T )4 − (T o )4 }
When the resistor is "overloaded", the power (P’) goes up by a factor of I2 = 4 (i.e., P =
I2R, where R is constant). The new temperature T’ will radiate more power to balance
this.
P ′ = 4 ⋅ P = σ{(T ′)4 − (T o )4 }
thus, dividing these equations:
(T ′)4 − (T o )4
=4
(T )4 − (T o )4
The question tells us that the temperature rises by 13° C, so we know:
T’ – T = 13 K
therefore,
(T + 13)4 − (293)4
=4
(T )4 − (293)4
One way to solve this is to do it graphically, plotting the left and right side of: { (T + 13)4 − (293)4 = 4 (T )4 − (293)4 } and seeing where they cross. This isn’t hard to do, with a computer. Or you can use a
spreadsheet program to try a whole lot of guesses. 1 2 10 10 ( T + 1 3 )4 T o 4 function values 1 10 4 ( T4  To 4 ) 10 5 10 9 0 10 0 5 109 1 101 0
280.0 290.0 300.0 310.0 320.0 Temperature
The solution is T = 297.8 K, or normally about 5° C above room temperature. With the
overclocking, the temperature is therefore about 18° C above room temperature, or
38° C.
If we wish to cool the system down to ∼ 20°C (room temperature), and our cooling
system is the same size as the resistor, the cooling system all by itself has to be at
Tc ~ 0°C, for then
0 + 38
≈ 20o C
2
so the cooling system is indeed at 0 ° C when the resistor (actually the processor) is off.
In case you are curious, this is a true story — you may even verify it yourself (I would
suggest using someone else's computer, though …). 2. Literally linear
a) Let T2 − T1 = ∆T > 0
(1st) heating:
2 L1 = L(1 + λ∆T)
(2nd ) cooling:
L 2 = [L(1 + λ∆T)][1 − λ∆T] = L(1 − λ2 ∆T 2 )
Every cycle (heating and then cooling the length of the rod is multiplied by (1 – λ 2∆T2)
∴ after n cycles L2n = L(1 – λ 2∆ T2)n
(this can be proved by induction, if you like)
λ∆T better be less than 1 (otherwise the length of the rod becomes negative), and hence: lim (L 2n ) = 0 n→∞ This clearly isn't right  whatever happened to conservation of mass???
The explanation is that this is actually a differential statement ( dL = L ⋅ λ ⋅ dT ) valid only
for tiny ∆L and ∆T. (otherwise "L" in the equation is not actually constant).
Because of this, only 1st order terms (i.e., ∆ L and ∆T) are important — 2nd order (∆ L2,
∆T2) should be ignored. Hence, 1 – λ2∆T2 ≈ 1 and there is no problem.
If λ were constant, one could write:
dL
= Lλ
dT
and solve this:
dL
= λ dT
L
dL
∫ L = ∫ λdT
ln(L) = λ (∆T) + C
λ∆T L = Lo e (you can check for yourselves that the length of the rod is indeed conserved after a cycle
when using this formula)
Expanding this yields:
L ≈ Lo (1 + λ∆T + (λ∆T)2 + …)
and ignoring 2nd order terms and higher gives:
L ≈ Lo (1 + λ∆T)
Unfortunately, λ varies with T (among other things), and this is not really true … 3 3) We will break dividing walls
Clearly, V = V1 + V2 (1) ("conservation of volume") and (2) ("conservation of mass") n = n1 + n2 The vessel is insulated, and hence energy has to be conserved.
E = α nkT
= E1 + E2
= αn1kT1 + αn2kT2
⇒ nT = n 1T1 + n2T2 (3) since PV = nRT, we get from (3):
PV = P1V1 + P 2V2
and P = P1V1 + P2 V2
V1 + V2 using 2 yields:
PV P1V1 P2 V2
=
+
T
T1
T2
(P1V1 + P2 V2 )
=T
P1V1 P2 V2
+
T1
T2 4) The nonaligned axis
The x and y directions expand independently. 4 x ′ = x(1 + Le1∆T)
y ′ = y(1 + Le 2 ∆T) y x = L cos(30°)
y = L sin(30°)
Assuming
straight: x the rod expands
DL = Sqrt ( Dx^2 + Dy^2)
Dy= Y*H2*(T2T1) L’ – L = L α∆T Dx = X*H1*(T2T1) and (L’ – L)2 = (x’ – x)2 + (y’ – y)2
∴ (L’ – L)2 = L 2 (Le12 cos2(30) + Le22
sin 2(30)) (∆T)
And hence α2 = Le12 cos2(30) + Le22
sin 2(30) L Y= LSin 30
30 X = LCos 30 In fact, from the diagram we can
see that the expanded part of the
rod is at a different angle to the
horizontal, and thus the rod must
bend to compensate. Dy
Tan Theta = Dy /Dx = Tan 30 * (H2 / H1)
Dx 5) Patent, or pshaw?
Here's the situation:
In steady state, nothing is either heating
up or cooling down. So the net power
being absorbed as radiated heat must
equal the power being emitted as
radiated heat (we’re ignoring
convection and conduction here).
The blinds absorb the sunlight,
certainly. More subtly, though, there is
radiated energy from the roomtemperature room inside, and also
radiated energy from outdoors —
because neither has zero temperature!
So all of these go into the absorbed power, keeping track of what the absorption
coefficient is on each face. 5 Let the ‘outside reflectivity’ be written as R, regardless of whether it’s the silver or black
side. Likewise, let the ‘inside reflectivity’ be r, regardless. The absorption is then (1–R)
or (1–r).
outside reflectivity = R
outside temperature = Tout
inside reflectivity = r
inside temperature = Tin
–2
incident sun power = 1000 W cm
area of each side of blinds = A
Absorbed power = absorbed sunlight + absorbed outdoor radiation + absorbed indoor
radiation
–2 4 4 = 1000 (1–R) W cm • A + (1–R) σ Tout • A + (1–r) σ Tin • A
Radiated power
4 4 4 4 = eout σ Tblinds • A + ein σ Tblinds • A = (1–R) σ Tblinds • A + (1–r) σ Tblinds • A
where e is the emissivity of the object, and equals the absorption a ( = 1– R or 1– r ). Both
sides of the blind radiate, of course.
These two powers must balance:
1000 (1–R) W cm –2 4 4 • A + (1–R) σ Tout • A + (1–r) σ Tin • A
4 4 = (1–R) σ Tblinds • A + (1–r) σ Tblinds • A
and we solve for the temperature of the blinds (when they’re hotter, they’re sending
more heat to the room, of course).
Tblinds 4
4 1000 (1 − R)/ σ + (1 − R) Tout + (1 − r ) Tin = (2  R  r) 1/ 4 Then when we have the black side outward R = 0.1, and r = 0.9; when the silver side is
outward, R = 0.9, and r = 0.1. So we get two different temperatures for the blinds. If we
take the indoor temperature to be 20° C, and outdoors to be 0° C, we find:
silver side out: Tblinds = 307.5 K or 34.4° C
black side out: Tblinds = 383.0 K or 109.9° C
The ratio of absolute temperatures is 1.25 — 25% hotter (Kelvin) with black out.
Interestingly, if we ignore the room and outside temperature, we get the answers we’d
have if both were zero — the answers we’d have basically in deep space!
silver side out: Tblinds = 204.7 K or –68.5° C
black side out: Tblinds = 354.5 K or 81.3° C 6 The ratio of absolute temperatures is 1.7 — 70% hotter (Kelvin) with black out.
However, the temperature is actually much lower in this case, just because the blinds
are sitting in ‘deep space at zero degrees’.
This isn’t the whole story, though! Even if the blinds are hotter with the black side
outwards, how efficient are the blinds at radiating into the room from the silver side of
them?
2 silver side out (black side in): power/m into room:
= (1–0.1) σTblinds = 0.9• 5.7 × 10 –8 Wm –2 K –4 4 –2 4 –2 • (307.5 K) = 459 W m black side out (silver side in): Tblinds = 383.0 K or 109.9° C
= (1–0.9) σTblinds = 0.1• 5.7 × 10 –8 Wm –2 K –4 • (383.0 K) = 122 W m Though the temperature is higher, the temperature factor of 1.25 leads to an increase of
4
2.44 in T . The emissivity factor, however, is 1/9th — more important than the
temperature. So, really, the silver side out leads to more power radiated into the room!
This offers plenty of control over the temperature – so the patent is indeed working, just
not the way I thought it might!
b) To make them work even better: you could see what happens when you make the
black even blacker and the silver more reflective; you could use another set of blinds
set up parallel, too — with both silver sides facing outwards, the effect would be even
greater (can you calculate this? What then do you do when you put the outside black
blinds outwards? Do you do the same with the inside blinds or do you leave them
open, neither side outwards?) You could also use a fan, so you could convectively cool
the blinds when they are black side out, and put the heat into the room more efficiently.
Lots of answers are possible… [Robin] 6) Tea? For who?
The best way to analyze data is to plot it, whenever possible.
a) Assume no power loss. Then,
dQ
dT
= 1000 = mc
dt
dt
Solving, 1000t = mc(T (t ) − 20) . Plotting T(t)20 versus t (the time in seconds) should
yield a straight line with slope 1000/(mc). Note that the line has to pass through (0, 0),
as when t=0 we know T(t) is exactly 20°C!
Power = Many commercial packages (e.g., Excel) can fit data, but they will not give you an errorestimate on the slope. Instead, it’s best to plot the data yourself, and then fit the
7 steepest and shallowest lines that still fit the data. The best fit is then the average of the
two slopes, and the error is the slopes of the steepest line minus the average slope. I’ve
done this in Excel; see below 30 y=(0.077 +/ 0.001) t 25
20
15 Series1 10 Series2 5 Series3 0
0 100 200 300 400 1000
= (0.077 ± 0.001) and hence c = ( 4330 ± 60) J /(kg oC ) . We have used
mc
∆c 0.001
the fact that
=
where ∆c is the error in c. Note that all errors are rounded off to
c
0.077
1 significant figure, and all results are rounded off to the same decimal place as their
errors. We thus have b) The system clearly loses power. This means that our value for c is too high, as we
assumed that the liquid absorbed all 1000 W of power, when it in fact did not. Let’s
assume that the system loses power at a steady rate. Our equations become
(1000 − L)t = mc(T (t) − 20)
and − Lt = mc(T ′(t) − 60) , where T’(t) is the temperature during cooling. Plotting this
data shows that it is not in fact a straight line (so L is not a constant), but we won’t
worry about this here. Plotting and fitting the data as in a) yields
0
5 0 500 1000 1500 Series1 10
15 Linear
(Series1) 20
25
30 2000 y =  .0171x
0
2
R = 0.9757 8 (I have only plotted the average slope above).
This yields a value c of (3600 ± 100) J /(kg oC)
∆T
= − k(T ′(t) − 20) . This may be solved, but calculus is required. The
∆t
easiest way to proceed is probably to plug in all values for T’(t), find a bunch of k’s (4, c) We have that in fact) and then average them. Doing this gives k ≈ 5 ⋅ 10 −4 1 / s . This value could be
used again to obtain a revised value for c; it will not be much different from the current
value, however.
For completeness, let us solve the above equation.
dT dx
dx
=
and our equation reads
= − kx . We also know that
Let x = T − 20 ; hence,
dt
dt
dt
x = 40° C at t = 0. Rewriting the above and integrating both sides gives ∫ dx
= − k ∫ dt or ln( x ) = − kt + C
x To make C fit our initial condition we finally arrive at
T − 20 = 40 exp( − kt )
This can be fitted e.g. by taking ln() of both sides – plotting ln(T20) versus t yields a
straight line with slope 0.0006 1/s…
Since the Carnot refrigerator is the most efficient, therefore we conclude that this value
is a minimum for the required power. [Peter & Yaser] 9 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 4: Optics and Waves
Due February 25, 2000 (revised date)
1) Bobbobbobbin’ along
A terrifically important concept in waves and oscillations is that of the harmonic oscillator.
We want to use some basic principles of the harmonic oscillator in later POPTOR
questions, so have a look here at the basics.
The ‘generic’ harmonic oscillator that you’ll see in university, and forever after that on
into graduate school and beyond, is a mass resting on a frictionless table and attached to
a spring. A property of springs, for small stretches or compressions, is that the force
exerted by the spring is proportional to the amount of stretch. This is called a linear
restoring force and the force is represented F = –k x, because the force tends always to
restore the position of the mass to its original neutral position (unstretched spring).
a) Write the force equation for such a massonaspring (inertial force and restoring
force). Since acceleration is the timederivative of velocity, and velocity in turn the
timederivative of position, this actually is a ‘secondorder linear differential equation’.
b) You can do pretty simple derivatives. Suppose that sin(ω t) is a solution of this
particular differential equation. For that to work, what must be necessary?
c) So, what is the rate of oscillation of a 100g mass on a spring that has a spring
–1
constant of 80 N m ? [Robin]
2) Keeping up an image
The distance between a screen and a light source lined up on a table is 120 cm. Moving
a lens between them, sharp images can be obtained at two different positions,
producing two different images. The ratio of sizes of these two images is 1:9.
a) What is the focal length of the lens?
b) Which image is the brighter one? Determine the ratio of the brightness of these two
images. [Gnädig/Honyek]
3) Shifty radar
Officer Smith and Officer Wesson are on patrol monitoring speeding along Hwy 401.
They are using a handheld radar gun which is set to detect the speed of approaching
1 cars, sending out waves at a frequency of 1000 MHz. After several hours of speed
monitoring Wesson turns to Smith and asks, "How does this thing work anyway?" To
which Smith replies, "Well now, are you familiar with the Doppler effect?"
a) Finish Smith's explanation.
Suddenly the two officers are distracted from the conversation, coffee and donuts by the
sounds of tires screeching and horns honking. They look up to see a milk truck racing
down the 401. Smith gives Wesson the eye and Wesson fires the radar gun.
b) If the observed frequency difference is 330 Hz, how fast is the milk truck going?
[Carrie]
4) To air is human
A cylinder of radius r contains n moles of a monatomic ideal gas. A tightfitting piston
of total mass M is fitted into the cylinder, sealing it. The room temperature is T .
Consider that the cylinder is really well thermally insulated, so all changes are
‘adiabatic’.
a) What is the new resting height h (height of the piston from the base of cylinder)?
Now, we attach or add a small mass m to the piston at rest. As you may guess, it'll start
to oscillate.
b) If the piston’s oscillations damp out (e.g.., due to friction) what would the final h be?
Suppose the motion is not damped, and that height from (b) is the equilibrium height of
piston and the piston will oscillate around that equilibrium point.
c) Each oscillator can be modelled by a system of mass and spring. What is the
corresponding mass and spring coefficient for this example?
d) Find the minimum amount of h in this oscillation. [Amir] 5) Bending the truth
Here’s an actual test used to find the index of
refraction of a material: Consider a piece of paper
with a drawing of a group of parallel lines.
Consider also a polished cylindrical rod made of
some unknown glass (for which we want to find
the index of refraction). Set the rod down on the
paper, at some angle θ such that 0 < θ <π/2. ? ? ? ? ? ? ? ? ? ? a) How will the lines appear as viewed through the glass rod? Make a sketch.
b) Find the index of refraction of the glass, using details from what you see in
part (a). [Amir]
2 6) Cool abrrrrations!
Consider a beam of light incident on a
convex lens. We expect all the parallel
rays to converge at a focal point. But, in
reality, the visible beam of light consists
of different wavelengths from violet to
red. Since the index of refraction of the
lens depends on the wavelength we are faced
with a defect called chromatic aberration —
different wavelengths will focus at
different places (see figure at right). Chromatic Abberation NOT TO SCALE
white Light Red light blue light a) Using the lensmaker’s formula, find a
formula that gives the change in the focal distance of a lens due to a small change in the
index of refraction.
b) One way to minimize the chromatic aberration is to put two lenses of different
materials together, to make one new lens. Suppose one part is made of Light Crown
Glass and the other from Heavy Crown Glass. Below is a table showing the index of
refraction vs. wavelength for these two different kinds of glass. We are looking for a
lens with f = 10 cm. What should be the focal lengths of the two Light and Heavy
Crown glass lenses so as to minimize the chromatic aberration?
Index of refraction of two kinds of glass in terms of wavelength
Wavelength (nanometer) 488 520 568 632 694 890 Light Crown Glass 1.4877 1.4859 1.4836 1.4813 1.4796 1.4758 Heavy Crown Glass 1.5793 1.5766 1.5735 1.5704 1.5681 1.5634 Hint : Lensmaker formula 1/f = ( n–1) * ( 1/R1 + 1/R2)
f : focal length of the lens
n : index of refraction of the lens material
R1 & R2 : The radii of curvature of the two sides of the lens [Yaser] Remember to check the POPTOR webpage for hints and any possible corrections!
www.physics.utoronto.ca/~poptor 3 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 4: Optics and Waves
1) Bobbobbobbin’ along
This question was a sneaky way to get you solve a second order differential equation
quite painlessly.
a) The forces involved included the restoring force, –kx and the inertial force, ma,
d2x
where a = 2 . These two balance each other (the net force is zero). Thus the force
dt
equation for our system is, d2x m 2 = − kx dt Which is a secondorder ordinary differential equation.
b) We are given x = sin(ω t) as a solution to our force equation. Taking the first time
derivative we obtain,
dx
= ω cos (ωt)
dt
Differentiating again with respect to time yields,
d2x
= − ω 2 sin (ωt)
2
dt
Plugging this expression for the acceleration into our force equation, along with
x=sin(ωt) yields ( ) m −ω 2 sin (ωt) = − kx = − k sin (ωt)
2
Therefore, solving for ω we get ω = k
.
m So this oscillation x = sin( ωt) isn’t always a solution — only for a certain ω will there be
a solution. So we find that the frequency of oscillation depends this way on both the
spring constant of the spring (its strength) and the mass of the bob attached.
c) Putting the values k = 80 N/m and m = 100 g into our equation for ω, gives us a value
ω= 80
= 28 /sec, or 4.5 Hz
0.1 [Carrie] 1 2) Keeping up an image
Since the object distance u and the image distance v can be exchanged in the lens law
2
and their ratio shows the magnification to be (v/u) = 9 (or 1/9), then ν / u = 3 (or 1 / 3) .
Thus, the object distance is 30 cm (or 90 cm) and the new image distance is 90 cm (or 30
cm). The focal length can be calculated from the lens law: ƒ = 22.5 cm.
If the same amount of light passed through the lens in both cases then the 9 times
smaller image would be 81 times brighter, as the smaller image occupies a surface 81
times smaller on the screen than the larger one. However, the lens placed at a greater
distance receives only one ninth of the light reaching the nearby lens, therefore the small
image is only nine times brighter that the large one.
It can be shown in general that in such cases the small image is as many times brighter
than the large one as the large one is larger.
[Gnädig/Honyek] 3) Shifty radar
a) The radar gun uses the Doppler shift to detect the speed of cars. For light the correct
doppler effect involves relativistic considerations. For cars, which have relatively small
source velocities, we skip those. Thus we have for the officers viewing at rest,
fL ≈ (1 + v/c)f S
Where fL is the frequency perceived by the officers (or listeners at rest), v is the velocity
of the source, c the speed of light, and fS the source frequency (1000 MHz in this case). If
waves are reflected from a moving car the reflected frequency, fR, is that of the a source
moving with twice the car velocity. So,
fR ≈ (1 + 2v/c)f S.
The waves which are reflected from the moving cars are beat against the transmitted
waves, yielding a frequency difference given by,
F = fR  fS ≈ 2vfS/c
Solving for the velocity of the car we get, v ≈ Fc/2fS.
b) Plugging in the numbers to our velocity equation we get,
v = [(330 Hz)(3 × 108 m/s) / (2 ( 109 Hz))](1 km/1000m)(3600 s /1 hr)
= 178 km /hr !!!!
Book him, Dano! [Carrie] 2 4) To air is human
a) We know that in an adiabatic process PVγ is constant.
P0 = nRT0 nRT0
=
V0
Ah 0 P1 = P0 + Mg
A γ γ P1V1 = P0 V0 γ γ ⇒ P1 h1 = P0 h 0 P ⇒ h1 = h 0 0 P1 1/ γ P0 A = h0 P0 A + M g nRT0 P0 A h1 = P A + Mg P0 A 0 1/ γ 1/ γ b) Exactly the same as part (a) you can see that: nRT0 P0 A
h2 = P A + (M + m)g P0 A 0 P2 = P0 + 1/ γ (M + m)g
A c) The force acting in a massspring system is –k ∆ x. Consider that we change the
equilibrium state by moving the piston x away from equilibrium. The force acting on
the piston to push it back will be:
F = P0 A – PA + (M + m) g
Since P(h2 + x)γ = P2h2γ = P1h1γ = P0h0γ
P = P2 γ h2 (h 2 + x) γ and since x is quite small compared to h2, x
P = P2 1 + h2 −γ γ x
≈ P2 1 − h2 γ x
⇒ F = P0 A + (M + m) g − P2 A 1 − h2 ⇒ F = γ P2 A
x= kx
h2 3 the spring coefficient of the system will be
(M + m). γ P2 A
and the corresponding mass will be
h2 d) Since the harmonic oscillator goes up and down with the same distance, and the
upper level of this oscillator is (h 2 – h1) above the equilibrium, the lower level will be
h2 – h1 under the equilibrium.
h min = h1 − (h 2 − h1 ) = 2h1 − h 2 5) Bending the truth
a) A rod is a cylindrical lens which
basically magnifies in one direction
only, and leaves the other direction
unchanged, so the line will be magnified
in the direction perpendicular to the
rod, and won't change in the direction of
parallel to the rod.
b) Since we are looking far from the rod, the
outgoing beams (going to your eyes) are
parallel. The point A is emitting light in all
directions, but we are concerned about the
beam that will go out vertically. The beams
are sketched at right.
The point A will be seen as A’ and we can
AH′
find the ratio of
by trigonometry. First
AH
we assume that AH is approximately equal
to the BH' line:
HA = BH′ = R sin(θ − (ϕ − θ)) = R sin(2θ ϕ)
HA ′ = R sin ϕ
Magnification = HA ′
sin ϕ
=
HA sin(2θ − ϕ) We neglect rays which pass either far from the centre, or at large angles from the axis
sin θ ≅ θ
.
(the paraxial approximation). In this case, θ and ϕ are small and sin ϕ ≅ ϕ sin ϕ= n sin θ ⇒ ϕ = nθ ⇒ Magnification = nθ
n
=
2θ − nθ
2−n
4 In practice:
The rod is magnifying in the direction
of the line perpendicular to the axis of
rod. The point A will be seen as A′ .
HA ′
= m (Magnification)
HA
HA ′ = OH tan(α + β)
HA = OH tan(α) ⇒ HA ′ tan(α + β)
n
=
=
tan(α)
2−n
HA ⇒ n= 2 tan(α + β)
tan(α) + tan(α + β) Where α is the angle between rod and parallel lines and β is the angle between bent
lines and parallel lines. One can find n by measuring α and β. [Amir]
6) Cool abrrrrations!
a) 1
1
1
= (n − 1) • +
f R1 R2 ∆ 1
1 1
= ∆(n) • + f R1 R2 = 1 ∆(n)
• f (n − 1) b) When two different lenses are placed together, their inverse focal lengths add up:
111
=+
f f1 f2
We want the aberration to be minimum. So
∆
=> 1
=0 f 1 ∆(n1 ) 1 ∆(n2 )
+•
=0
•
f1 n1 − 1 f2 n2 − 1 Consider two wavelength 488 nm and 890 nm (nanometers).
∆(n1 ) = 1.4877 − 1.4758 = 0.0119
∆(n2 ) = 1.5793 − 1.5634 = 0.0159
Now we have two equations, two unknowns
5 1111
+==
f1 f2 f 10
1 0.0119
1 0.0159
•
+•
=0
f1 (1.48  1) f2 (1.57  1)
Therefore
f1 = 2.5 cm
f2 = – 3.3 cm [Yaser] 6 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 5: Electricity and Magnetism
Due March 17, 2000 (revised date)
1) Vile away the hours
To complete his giant, superevildestructor ray gun the great, evil scientist Dr. Vile
requires a copper sphere with a charge of exactly +3 µC. He has a copper sphere of
mass 2 g.
a) What fraction of the electrons must be removed from his copper sphere to give it a
charge of 3 µC ?
Having removed the correct number of electrons Dr. Vile places the 3 µC copper sphere
near two other spheres of charges +7 µC and +5 µC. The three spheres are arranged in
an equilateral triangle formation, with sides of 9 cm.
b) What is the resultant force (and its direction) on the 3 µC copper sphere? [Carrie]
2) Capacity for thought
Two parallelplate capacitors with equal capacitance C are charged to voltage V. The
distance between the plates is d and their area is A . There is no dielectric between the
plates.
Connect the two capacitors by two wires so as to make a circuit. Obviously, there is no
current in the circuit. Now, imagine that the plates of one capacitor move apart with
velocity v and the plates of the other capacitor move toward each other with the same
velocity v . At this time, there must be some current in the circuit. Find its value.
[Yaser]
3) Dielectric City
Consider two parallelplate capacitors C1 and C2 which carry corresponding charges
Q1 and Q2.
a) What is the total energy of the two capacitors, taken as a system?
Now, we connect the two positive plates to each other and two negative plates to each
other.
b) What will be the new energy of the system?
1 c) What happened to the energy of the system when the
plates were connected?
d) What is the energy difference between an empty
capacitor and a capacitor full of water, if each carry the
same charge?
e) What will happen if I put a chargedup capacitor on the
surface of water, as shown in the figure? Why? How
high will the column of water be when it rises into the
space between capacitor plates?..... [Amir] 4) Super loopy
Two identical superconducting circular loops are positioned
coaxially (a line passes through both their centres, perpendicular to
each) and far apart. The loops have selfinductance L. Equal
currents I are passing through each of these loops in the same
direction. Now, bring the loops together.
a) What is the current in each loop?
b) Find the difference between initial and final energy of the system?
[Yaser]
5) Keeping it together...
A narrow beam of electrons, of radius r and all moving at the same
velocity v much less than speed of light (c), produces a charge current I . We assume
that the beam has cylindrical symmetry.
a) What is the electric field at the edge of the beam?
b) What is the magnetic field at the edge of the beam?
c) What is the radial (diverging) velocity of electrons at the border of the beam, after
the beam has traveled a longitudinal distance 100 times r?
d) What is the diverging angle for I = 1 mA and r = 1 cm and v = 1000 m/s?
[BONUS: Since the B field depends on electron speed, can you find the speed for which
the force due to Bfield entirely cancels the force due to Efield? Hint: think ‘frame of
reference’] [Amir] 2 6) A feel for fields
a) Prove that it’s impossible to have a magnetic field which increases along the zaxis but
which has only a zcomponent. This field must have a radial component! By choosing a
short cylinder, show that for a cylindrically symmetric Bfield:
Br = r dB
2 dz A loop with electrical resistance R is falling in a cylindrically symmetric magnetic field.
The centre of the loop as it falls is exactly aligned with the axis of the cylindrically
symmetric field, and perpendicular to it. Along this axis, the field zcomponent is
changing as dBz/dz. The radius of the loop is r and its mass is m.
b) Write the equation of motion governing the fall of the loop, and sketch the graph of
its velocity vs. time.
c) What is the terminal velocity of the loop? [Yaser]
[Hint: you may want to use the fact that magnetic field lines are always closed loops —
they cannot start or end at a point, as Efields do...]
INFOBITS™ — Useful Bits of POPTOR Information
Remember to check the POPTOR webpage for hints and any necessary corrections!
www.physics.utoronto.ca/~poptor 3 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 5: Electricity and Magnetism
1) Vile away the hours
22 i) The 2.0 g sphere contains 1.99 × 10 atoms. The charge on the nucleus of each atom
22
23
is 29 e. Thus the sphere contains (1.99 × 10 )•29 = 5.77 × 10 electrons. One electron
–19
–6
–19
has a charge of 1.6 × 10
C. The electrons removed = (3 × 10 C)/(1.6 × 10
C per
13
electron) = 1.875 × 10 electrons. So the fraction of electrons removed is
(1.875 × 10 13 23 –11 ) / ( 5.77 × 10 ) = 3.25 × 10 b) The figure at right illustrates the configuration of
the charged spheres with q1 = 7 µC, q2 = 5 µC and q3
= 3 µC. Let the positive x direction be as shown, and
the y direction perpendicular to it, towards the upper
left. The net force on q3 is the vector sum of the
repulsive forces F 1 and F2 . The magnitude of F1 is
given by:
F1 = k q1 q3
r2 (9 × 109 N m2 C −2 ) • 7 × 10−6 C • 3 × 10−6 C
=
(0.09 m)2 = 23.3 N
Similarly F2 is:
F2 = k q2 q3
r2 (9 × 109 N m2 C −2 ) • 5 × 10−6 C • 3 × 10−6 C
=
(0.09 m)2 = 16.7 N
rrr
The vector sum F = F1 + F2 has components along x and along y. In the x direction:
Fx = F1x + F2 x
= 23.3 N + 16.7 N • cos(60°)
= 31.7 N
and in the y direction,
1 Fy = F1y + F2 y
= 0 N + 16.7 N • sin(60°)
= 14.5 N
Thus the magnitude of the resultant force is: (31.7 N )2 + (14.5 N )2 F= = 34.9 N
The direction is at an angle
arctan(14.5/31.7) = 25° from the x axis. [Carrie] 2) Capacity for thought
The charge of each capacitor is Q=CV. When the capacitors are connected to each other
and the separation between the plates change, the total charge is still conserved so
2Q=q1+q2 . Since the capacitors are parallel to each other, the voltage across them is
q
q
equal to V = 1 = 2 . We know that the capacitance is inversely proportional to the
C1 C2
separation of the plates. Therefore C1d1 = C2 d2 . Since the plates are moving together or
apart by speed v for the two capacitors, we have
d1 = d + vt
Therefore d2 = d − vt C1 d2 d − vt
=
=
C2 d1 d + vt On the other hand, q1 = C1
d − vt
q2 =
q2 .
C2
d + vt Substitute the value of q2 = 2Q − q1 in the above formula we have
q1 = Q d − vt
d and q1 = Q d + vt
d The current in the circuit is the rate of change of charge, which is equal to
I= dq2
dq
Qv
=− 1 =
.
dt
dt
d [Yaser] 3) Dielectric City
a) The energy stored in a capacitor with capacitance C and charge Q is
TOTAL energy of a system consist of two capacitors is:
E1 = Q2
. So the
2C 2
Q1
Q2
+2
2C1 2C2 2 b) At first, I’ll find the equivalent capacitor and the total
charge on it.
Those two capacitors are
connected as parallel
capacitors and the total
charge on the righthandside
plates is conserved (there is
no external current) we have
q 1 + q2 = Q1 + Q 2 (or similarly
for the left plates: (–q1 ) + (–q2 )
= (–Q 1 ) + (–Q 2 ))
Cequivalent = C1 + C2
Qtotal = Q1 + Q2
Now I can find the new energy of the system of one equivalent capacitor.
(Q1 + Q2 ) 2
Qtotal 2
E2 =
=
2Cequivalent 2(C1 + C2 )
c) One may find the difference between E2 and E1 as:
2
2
(Q1 + Q2 ) 2
Q1
Q2
E1 − E2 =
+
−
2C1 2C2 2(C1 + C2 ) = 2
2
2
2
C2 (C1 + C2 )Q1 + C1(C1 + C2 )Q2 − C1C2 (Q1 + Q2 + 2Q1Q2 )
2C1C2 (C1 + C2 ) 22
22
C2 Q1 + C1 Q2 − 2(C1Q2 )(C2Q1 )
=
2C1C2 (C1 + C2 ) = (C2Q1 − C1Q2 ) 2
>0
2C1C2 (C1 + C2 ) which is always equal or larger than zero.
The energy loss is because of the emission of accelerated particles due to connection of
two wires with different voltages. When you connect the first two wires (say negative
plates) nothing will happen and you will just have one ground to measure voltage on
each part of the two circuits. You can then calculate the voltage difference of two
positive plates.
Connecting the other two wires will create a large electric field between two wires
(when they are separated by a small distance) and then the electrons will be accelerated.
If we could put a resistor between two capacitors, this energy could have been changed 3 into heat. Often, the energy loss can be found in the form of electromagnetic waves in
visual range (you’ll see it as a spark!).
d) Capacitance of a capacitor full of water is ε times capacitance of an empty capacitor.
A
(Recall the formula for capacitance C = ε ε o , where ε is susceptibility of water.)
c
Energy of full and empty capacitor can be found by using that formula:
Eempty =
E full = Q2
2Cempty Q2
Q2
=
2C full 2 ε Cempty and the energy difference is:
∆E = Eempty − E full
Cf = ε εo Q2
1
=
( 1− )
Ce
ε A1
A
; Ce = ε ε o 2
d
d Ecapacitor = (total charge)2
2(equivalent capacitance) in which I've assumed that an equivalent capacitor have the total charge (initial charge).
Q2 ⇒E=
2 ε0
( ε A1 + A2 )
d Since A1 = wh and A2 = w(Hh) we can write:
Q 2d
Ec =
2 ε 0 w( ε h + H − h)
On the other hand, gravitational energy of water is:
Eg = mg × h
h ρw d g 2
= ρgwhd × =
h
2
2
2 And total energy of the system is:
Etotal = Ec + Eg = ρw d g 2
Q 2d
+
h
2 ε o w(H + ( ε − 1)h)
2 Since the energy tends to be minimized in physics world (and defines the equilibrium
state): 4 −Q 2d(ε − 1)
dEtotal
=0⇒
+ ρw d g h = 0
dh
2ε o w(H + (ε − 1)h)2
⇒ Q 2 (ε − 1) = 2ε oρw 2 gh(H + (ε − 1)h)2
3 2 Q2
2 h
1 h h
⇒
+
+
−
=0 H
ε − 1 H
(ε − 1)2 H 2ε o (ε − 1) ρ w 2 g H 3
*Imaginaryvalued solutions have no physical meaning, here.
*Answers with h
> 1 mean that the capacitor will be completely filled.
H h
< 0 mean that the water won't rise at all. (in our equation this will
H
h
could be negative).
not happen, but if one could find a medium with ε <1,
H *Answers with The condition that capacitor is completely filled with water is:
1+ 2
1
Q2
+
−
≥0
ε − 1 (ε − 1)2 2ε 0 (ε − 1)ρw 2 gH 3 [Amir] 4) Super loopy
The main point of this question is that the loops are superconductors. Superconductors
have no resistance. Therefore, when the loops are brought together, the magnetic flux,
which passes through each of them, must not change. Otherwise there would be a
voltage across each loop, which would cause an infinite current! To avoid infinite
current, the currents in the loops change so as to keep the flux fixed.
When the loops are far apart the flux passing through each loop is equal to
Φ i = LIi
which is because of the selfinductance of each loop. When the two loops are brought
close together, the mutual inductance is equal to the selfinductance. The flux passing
through each loop is combined in equal parts. One is from the selfinductance and the
other is from the mutual inductance. Therefore
Φ f = LI f + LI f = 2LI f
To avoid infinite current,
Φi = Φ f ⇒ I f = Ii
2 The energy of an inductance is equal to 5 U= 12
LI
2 Therefore the difference in the energies of the system is equal to
∆U = U f − U i = 12121212
3
LI f + LI f − LIi − LIi = − LIi2
2
2
2
2
4 It can be shown that this is equal to the work needed to bring the loops together. Here
the work is negative because the loops attract each other.. [Yaser]
5) Keeping it together...
i) I'll find the density of electrons
inside the beam.
I = ρv (πr 2 ) ⇒ ρ = −I
π r 2v The electric field at the edge of beam can
be found by using Gauss’s law.
E= ρ ⋅ π r2 ρr
λ
−I ^
=
=
=
r
2πrε o 2πε o r 2ε o 2π ε o r v Using Ampere's law, I can find B:
B= µo I
2πr The radial force on electrons on the
edge is
F = − e(E + v × B) µ I v
I
⇒ F = e
− o ^
r
2πr 2πε o r v
⇒F= Ie
(1 − µ o ε o v 2 )
2πε o r v as you may know, the quantity µ o ε o =
⇒F= 1
where c is the speed of light.
c2 Ie
Ie
v2
(1 − 2 ) ≈
2πr ε o v
2πr ε o v
c The change in radial momentum can be found by using 6 ∆Pr = Fr ∆t = Ie
100 r
•
2πε o r v
v ∆Pr = 100 I e
2πε o v 2 ∆vr = 100 I e
∆Pr
me 2πε o me v 2 in which I've assumed that Fr , r and v won't change too much during this process. One
may find the ratio:
100 • 10 −3 • 1.6 × 10 −19
∆vr
100 I e
=
=
v
2πε o me v 3 2π • 8.85 × 10 −12 • 9.1 × 10 −31 • (107 ) 3
∆v
tanϑ = r ≈ 0.3
v
⇒ ϑ = 0.3 radians ≈ 17°
(The speed in the problem set was wrongly written as 1000 m/s. It is 10,000 km/s.
Sorry!).
BONUS :
In part (c) we found that the force acting on electrons at the border is
F= Ie
2πε o v r v2 1− 2 c putting this force to be zero will result:
v2
=0 ⇒ v=c
c2 1− [Amir] 6) A feel for fields
a) Try to draw field lines, which have to be denser as the z
changes. The field lines must bend, which shows the presence of
a radial component. Use Gauss’s Law for this part. The
magnetic field lines that enter a cylinder must go out so as to
make a closed loop.
B z (z+∆z )
r ∆z B z (z) Br The flux from the top and bottom
surfaces is
Φ1 = πr 2 (B( z + ∆z) − B( z) = πr 2 dBz
∆z
dz The flux from the sides is
7 Φ 2 = 2πr∆zBr
For the total flux to be zero we should have
Φ1 = Φ 2 ⇒ Br = r dBz
2 dz b) There are two forces acting on the loop, gravitational force and the magnetic force.
The rate of change of the magnetic flux respect to time is
dΦ
πr 2 dBz
2 dBz
= πr
=
dt
dt
v dz
where v is the velocity of the loop falling down. For the current in the loop we have
V dΦ / dt πr 2 v dBz
i= =
=
R
R
R dt
The force exerted by the magnetic field, which is due to radial component of the
magnetic field and upward, is equal to
F = i l Br = i (2πr) Br
Substitute the value of current and magnetic field we already derived we have
F= 2 π 2 r 4 dBz v
R dz Therefore the equation of the motion of the loop is
2 d2v
π 2 r 4 dBz m 2 = mg −
v
R dz dt
The magnetic force is negative because of Lenz’s law. The
magnetic force is proportional to v and negative, causes
damping. Initially, the velocity increases but reaches a
terminal velocity which is when the acceleration is zero. v
vf t mgR
d2v
= 0 ⇒ vf = 2 4
2
π r (dBz / dz)2
dt
The velocity of the loop versus time is plotted. [Yaser] 8 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 6: AC Circuits and Electronics
Due April 3, 2000
1) Ohm my!
You are given the circuit shown, which starts with a battery of
voltage V and a resistor R across the battery. If I add another
resistor r in series with R, what should be the value of r so that the
power dissipated across it should be a minimum? Similarly, what
should be the value if r is added in parallel with R? [Robin] R r 2) A current affair
There are two circuits shown below, with the bottom part of the diagram held at
ground. If the input is a current turned on, then off, as graphed, what is the output
voltage vs. time? R I I Vo C
t
I
C R Vo [Yaser]
3) Pandora’s box o’ electronics
Consider a black box with 3 terminals (A, B, C). Only batteries & resistors are present
inside. Given these measurements below (taken with a multimeter, between pairs of
1 Voltage:
A Resistance:
C 0V 1.6 V A 1.6 V A B B B A C B C 99 Ohms nonsense (< 0)
nonsense (< 0) C Current:
A
A B C 0A 0.015 A B
C 0.031 A terminals), what is inside? Draw a circuit diagram,
and specify the values of components.
[HINT: the above values, since they are
experimental, have an errorbar in the last digit]
[HINT 2: measuring resistance across a battery will
result in an error] [Peter] 4) Only logical
Logic gates are used to electronically perform Boolean or Logical operations on input
signals. The simplest is the NOT gate which is symbolised by the
diagram at left. The response of the NOT
A
C
gate can be summarized in the following
1
0
truth table, at right. So if the input signal
(A) is true or 1 the NOT gate response (C) is
0
1
false or 0 (and vice versa). A B C 1 1 1 1 0 0 0 1 0 0 Another type of logic gate is the AND
gate with symbol shown at left and truth
table at right. Thus the output C is true
only when both A and B are true,
otherwise C is false. 0 0 The OR gate has symbol shown at left.
a) What do you suppose is the truth table for this gate?
b) Build an OR gate using NOT and AND gates.
c) Build an AND gate from NOT and OR gates.
BONUS
d) Build an EXCLUSIVE OR gate (where C is true if one of A or B is true but not both),
using AND, OR and NOT gates. … [Carrie] 2 5) Net worth
i) Consider an infinite square
array of resistors r, sketched
conceptually at left. Consider any 2
neighbouring points (separated by
a resistance r) — the equivalent
resistance is r /2. Show why.
ii) Consider now an array of
equilateral triangles (arranged so
that at any junction there are 6
resistors r coming off). What is the
equivalent resistance across any
arm?
iii) Consider an array of hexagons
(arranged so that at any junction
there are 3 resistors r coming off). Now what is the equivalent resistance across any
arm?
iv) Consider a network of regular polygons in which, at any point, N resistors r are
coming off. What is the equivalent resistance?
[BONUS: can you draw such a figure in 2 dimensions for N = 5 ?] [Peter]
6) My current analysis...
Roughly sketch the impedance of this circuit, at right, as a
function of the frequency ω of the sinusoidal driving source (the
circle with the squiggly bit). A qualitative description will do.
NOTE: 1
1
= α2 ≠
= β2
L1C1
L2C2 [BONUS: find the impedance as a function of frequency exactly]
[Peter] Remember to check the POPTOR webpage for hints and any necessary corrections!
www.physics.utoronto.ca/~poptor 3 19992000 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 6: AC Circuits and Electronics
1) Ohm my!
You might be able to do this one by inspection: if r
∞ then no current flows in the
circuit, so no power is dissipated through r. On the other hand, as r
∞ power
dissipation increases for a constant current. So a few equations are in order.
In the series circuit
V = I (R + r) (voltage V across resistors in series, current I through resistors)
2 P = I • Vr = I r (power P dissipated in r, current I through r same as R) 2 V
=
r R + r
For an extremum:
0= 2 dP V 2r =
1−
dr R + r R + r which happens at r = R. However, this
is a maximum, not a minimum.
This is easy to see by graphing it, by
finding the second derivative, or by
simply noting:
2
lim V r = 0 r → 0 R + r 2 lim V lim 2
r
lim 1 r =
= V2
=0
V 2
r → ∞ R + r r → ∞ (R + r) r → ∞ 2(R + r ) …so if it is zero at the origin and at infinity, has one extremum, and cannot be negativevalued, it must be a maximum. The minima you’re asked for correspond to r = 0 and
r = ∞.
2 How about in parallel? It’s even easier to see: Pr = Vr I r = (Vr ) /r. The voltage Vr = Vo
the battery voltage. Clearly this power decreases monotonically as r increases — the
miminum is at r = ∞. [Robin] 1 2) A current affair
Usually questions like this one are given with a specified v oltage applied — which
would be easier! Here it is a c urrent that is specified (without telling you how they
made sure such a current was produced, at whatever voltage needed). This makes
some parts tricky, unfortunately.
For the first case, all the current must go through the resistor, and onward to charge the
capacitor. Therefore, the capacitor charge goes up linearly respect to time, and the
q It
voltage across it would be V = = 0 . After time τ no current flows, so the charge
CC
remains on the the capacitor (except perhaps for leakage due to some large resistance
through the capacitor which ideally should be infinite. Leakage can happen through
moisture in the air, too). I(t)
I0
τ t V
I0 τ /C
τ t For the case where the capacitor and resistor are connected in parallel, the current Io is
divided through both of them. Initially most of the current will pass into the capacitor
and increase the capacitor voltage as it charges up (since we’ve assumed a constant
current flow, this the source voltage will have to increase to match the capacitor voltage,
to keep current constant). As this voltage increases, the current through the resistor in
parallel also will increase, according to V = IR, and this will make the current through
the resistor increase — a bigger share of the current flows through the resistor.
Even if the current were constant for all time, the capacitor cannot charge up forever: at
some voltage the current drain through the resistor would be equal to the whole current
supply, and the capacitor will not take any of the current. Then the circuit will reach
‘steady state’. This helps us figure out how to solve the following equation which
describes the circuit (a famous physicist once said: “Never write down an equation until
you’ve figured out what the answer should be.”)
2 I o = I capacitor + I resistor
dQ V
+
dt R
dV 1
=C
+V
dt R
= Where V is the voltage across both the capacitor and the resistor (in parallel), and Q is
the charge on the capacitor. The current Io is constant, so we can solve this equation
either systematically or by guessing from what we figured out about voltages, above.
We try a solution which has exponentials in it (because the derivative of the exponential
is again an exponential, which maybe would cancel with the original). In steadystate,
the dV/dt term will be zero (i.e., nothing is changing), so finally V(t ∞ ) = Vo = Io R. If
the voltage approaches this value asymptotically, then maybe it should have the form: ( V(t) = Vo 1 − e αt ) Try it in the differential equation: it works, as long as α = –1/ RC. Sounds OK, because
if R
0 (short circuit) or C
0 (no capacitor), then it
V
takes no time at all to reach the final state.
In fact what happens is that the voltage follows this
curve V(t) until the current is turned off. Then the
charge on the capacitor leaks away through the resistor
until it goes back to zero, exponentially with the same
timeconstant α = 1/ RC. [Yaser & Robin] I0 R
t τ 3) Pandora’s box o’ electronics
There is no unique solution to this problem (for
example, a 100 Ω resistor could also be drawn as
two 50 Ω resistors, etc.).
Trial and error always works, and here’s a
possible solution. The battery has a value of
1.5V, resistance at A is 75 Ω , resistance at B is
25 Ω and the resistance at C is also 25 Ω.
Using these values we first of all see that the
resistance from B to C is 100 Ω (within error of
the 99 Ω given); measuring resistance from A to
C or B to C yields an error, as we’re going across the battery [an aside: the way a
multimeter measures resistance is to send a small current across the circuit, measure the
potential difference created and find the resistance from Ohm’s Law. However, when a 3 battery is in the way it will change the current, and perhaps even reverse it, which will
yield in very strange values shown by the multimeter].
The current from A to B is 0, and the voltages from A to C and B to C will be 1.5 V
(within error of 1.6V), as given [the internal resistance of the multimeter is extremely
high when measuring voltage]. The current from A to C (remember that the resistance
of the multimeter is very small when measuring current) is 1.5 / 100 = 0.015 A and the
current from B to C will be 1.5 / 50 = 0.030 A, again within error of the given values.
If you like a more systematic approach, try this:
• First, try to keep things as simple as possible (i.e., start with the simplest
components, like resistors and batteries, adding more complicated ones later).
• Next, the most complex circuit possible is when all terminals are interconnected.
Now, it may be shown that such an ‘outer’ triangle is
equivalent to an ‘inner’ triangle, just as I’ve drawn before.
So, we redraw the diagram to the one I had first (this is
not necessary; I’m only doing it so that this solution is the
same as the one above). Now, it is pretty clear that there
has to be at least 1 battery between AC and BC. The
easiest way to satisfy this is to put one at C.
• As to the remaining resistors, you can either guess the
values or solve 3 equations and 3 unknowns, as below:
R A + R B = 100
R A + R C = 1.5 / 0.015 = 100
RB + R C = 1.5 / 0.030 = 50 (RA, RB and RC are the resistances in the arms A, B and C, respectively, in the 1st
diagram). And this indeed does it… [Peter]
4) Only logical
The truth table for the OR gate is shown at right: If either
A or B is true (1) then the output is true (1).
b) Using NOT and AND gates the following OR gate can
be constructed: A B C 1 1 1 1 0 1 0 1 1 0 0 0 4 c) Using NOT and OR gates the following AND gate can be constructed: BONUS
d) One possible EXCLUSIVE OR gate built out of AND, OR and NOT gates: [Carrie]
5) Net worth
iv) Let me do the general problem first. Consider an
infinite, homogenous grid where at any junction N
resistors are connected together. Note that a
voltage source is really equivalent to a current
source, i .e. , something which injects & draws
current.
Consider injecting a current I into the junction.
Because the
whole system is Nsymmetric,
the current
will split evenly and I/N will flow into every
resistor. See the example, above left.
Consider now drawing a current I from a
junction. Again, a current I/N will be drawn
from every resistor. Now s uperimpose the 2
current flows, offset by 1 arm (see right). We
5 will have a current I coming in, and a current I coming out one arm away — equivalent
to a battery. The current in the given arm will be I/N + I/N = 2I/N. The resistance is r.
Thus, the voltage drop is 2Ir/N which must also equal RI, where R is the equivalent
resistance. From this we see that R = 2r/N. This also solves the remainder of the
questions…
This question is inspired by a Polish Olympiad question, from way back when…
Answers:
i) r/2
ii) r/3
iii) 2r/3
iv) 2r/N
BONUS: it is not possible to draw such a figure in 2D; in 2 dimensions, the only possible
figures are those for which N = 3, 4 or 6. To convince yourself of this you can note that
at each junction the sum of all the internal angles has to be 360 degrees (of course, you
don’t have to be able to draw this figure for part (iv) to hold…). [Peter]
6) My current analysis...
Here’s what you have to know:
i) when the driving frequency ω is small, the current is almost steady (DC). The
inductor behaves like a piece of wire, with zero resistance. The capacitor, on the other
hand, doesn’t let any current through and behaves like a break in the circuit.
ii) when ω is very large, the opposite is true: the capacitor lets current through but the
inductor doesn’t (the e.m.f. that builds up inside is very large)
iii) an LC or RLC circuit will resonate at a frequency of ω 2 =
power dissipated is maximum, and the impedance is minimum. 1
. At this point the
LC Now, the circuit in question contains 2 circuits (the R, L2 , C 2 circuit and the C1 , L1
circuit), each with their own resonant frequencies. We can thus expect that the overall
impedance will have minima near the resonant frequencies. In fact, the impedance of
the R, L 2 , C2 circuit will be higher than that of the other circuit, due to the resistor R.
Also, the impedance will never quite reach zero, as the two resonant frequencies are not
equal ( α ≠ β ).
From all this, we ‘guesstimate’ the curve on the following page (note that I have
arbitrarily assumed a < b).
For an analytical treatment I will employ the complex number technique. What happens
here is that a sinusoidal driving source (e.g., cos( ωt) ) can be conveniently represented
6 as a complex number
(cos(ω t) = Re(exp(iωt)),
where Re denotes the real
part). This is useful as
derivatives of the exponential are really easy to
calculate. This, in the end,
is useful to solve the
differential equations present when dealing with
capacitors and/or inductors.
The gist of all this is the fact that the resistor, capacitor and inductor, in a circuit driven
by a sinusoidal source, behave as resistors whose generalized ‘resistance’ (called
impedance) is a complex number (the number has to be complex because capacitors and
inductors change the phase of the current; the phase is represented by the angle the
impedance makes with the real axis in the complex plane).
The values turn out to be:
Resistor R: Z = R
1
iωC
Inductor L: Z = iωL Capacitor C: Z = 2 (Z is the impedance; i = –1)
Finally, the standard Kirchoff rules apply to impedances (i.e., they’re added when in
series and the reciprocals of the sum of their reciprocals is taken in parallel).
After this crash course, let’s apply this to our circuit. The total impedance will be
1
Z = (iωL1 ) + + iωC1 α2 = iωL1 1 − 2 +
ω 1+ R 1 1 1 +
1 R iωL2 +
iωC2 1
1 β2 iωL2 1 − 2 ω Real nasty… now, what we really want to plot is the magnitude of this thing, Z. This
yields some pretty ugly expressions, and not much can be done so simplify them.
7 You can take my word for the fact that this yields very similar results to the graph
above; the second minimum isn’t really a minimum, more of an inflection point, but it’s
close enough for this rough analysis. [Peter] 8 20002001 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 1: General
Due October 30, 2000
1) The path of least resistance (...and highest voltage)
You are given 32 identical 3volt batteries, each of which has an internal resistance 2
ohms (i.e., even if you shortcircuit them with a wire you still have 2 ohms appearing in
such a circuit). How can you connect all these batteries together so as to get highest
possible current through a 4ohm resistor?
[HINT: Think of the batteries set up in n rows and m columns...] [Yaser] 2) The last straw...
a) When you suck on a straw, you create a partial vacuum in your mouth. If a person
could make a partial vacuum in their mouth equal to 50% of atmospheric pressure,
what’s the farthest possible vertical distance above their drink of water?
b) Is it reasonable that your lungs might produce a vacuum 50% that of the
atmosphere, by inhaling? What else is at work?
c) Does the size of the straw (its diameter) have any effect on the maximum height?
To answer parts b) and c) it is suggested that you test the effects on height using
differentsized straws. Once you have collected the data you can determine the
approximate pressure — this is the empirical approach.
Materials: – a bucket or pitcher of water
– a few ~1.5 m lengths of plastic tubing of different diameters
(Canadian Tire stores sell this, for example)
– a measuring tape The plan — If you suck up water from the bucket up the tube, like a long straw, at some
point you can no longer raise the water level within the tubing. Measure the height of
the water level within the tubing above the water level in the bucket. Try it also by
inhaling, to draw up the water, instead of sucking.
Record your result and repeat the experiment for tubes of different diameter. Once you
have collected the data you can determine the approximate vacuum pressure your
lungs can achieve by themselves, and compare that to the best overall partial vacuum
you are able to produce otherwise. [Sal]
1 3) Fermi, fer you
Enrico Fermi, a famous physicist of the 20th century, was wellknown for asking
peculiar questions of his students, intended to develop their practical ability to figure
things out from reasonable assumptions and common knowledge (or good guesses!). In
one of these ‘Fermi Questions’ he’s reported to have asked “How many piano tuners are
there in Chicago?” The point was to figure it out on the spot, not to contact the Tuners’
Guild, and the reasoning might have started with a guess of what fraction of
households even owned a piano — based, for instance, on thinking about your own
friends’ households. Then for a typical pianoowner, how often might one hire a tuner
— maybe every year or two on average? In the end, one could figure out how many
workerhours of piano tuning per year there would be in a city the size of Chicago, and
thereby figure out how many tuners could be supported by that amount of work.
Try these yourselves! You’ll have to make a number of assumptions or approximations
— please describe each on a separate line, with the numerical estimate or guess you
make. Any final answer within a factor of three of the ‘real’ answer is considered an
excellent success, for others a factor of ten is good. Your thinking is the main thing of
interest to us — please don’t go and measure anything, or look anything up.
a) How many piano tuners do you figure there are in Toronto?
b) I saw lots of meteors streak across the sky this summer — they moved through an
arc of about 45 degrees of the sky in a little less than a second. How fast do meteors go?
c) How many kilograms of toothbrushes do the people of Canada throw away each
year?
d) If icebergs never broke off from the icecaps at the poles of the earth, roughly how
long would it take to tie up all the water on earth as ice at the poles?
e) How many atoms scuff off your sneakers onto the sidewalk with each step you take?
[Robin]
4) Number four with a bullet
–1 A 2 g spherical bullet 9 mm in diameter is fired from a gun at 250 m s into one end of a
thermally insulated cylinder. The cylinder, 15 m long and with a diameter of 1 m, is
pressurized to 20 atm with Argon gas at room temperature. While passing through the
cylinder, the bullet experiences a drag force which slows the bullet:
1
Fdrag = − ερAv 2
2
where ε = coefficient of drag (a constant)
ρ = density of gas
A = crosssectional area (area of an object’s shadow)
v = speed
2 Ignore gravity, and assume ideal gas behaviour.
Eventually, the bullet hits the opposite end of the cylinder and stops.
a) How long does it take for the bullet to reach the opposite end of the cylinder?
b) Calculate the bullet’s velocity just before hitting the cylinder wall.
c) Calculate the change in temperature of the Ar gas in the cylinder due to the bullet’s
motion before impact.
[HINT: If you can integrate, you can do this; if you get stuck, check the POPTOR
webpage for ways to solve the quite simple differential equation you will get]. [Brian]
5) Bubbling ideas
Suppose we have a perfectly round soap bubble of initial radius Ro, with internal air
pressure p, and external air pressure po.
a) What is the change of pressure inside the bubble if the bubble size is increased by a
small amount?
Now put a charge Q on the bubble, spread evenly over the surface.
b) What is the new radius of the bubble after charging it up?
[HINT: perhaps you know or can find the field outside a spherical charge distribution,
and figure out the field inside too. But what about the field right on the bubble wall,
neither inside nor outside? The right answer is a kind of averaging... [Peter]
6) Magnus matters
What does it take to pitch a curveball in baseball? Besides some talent, it takes just a bit
of physics — the Magnus effect.
The drag force on a ball depends on how quickly it is moving through the air. When a
ball is moving through the air and also spinning, the airspeed on one side of the ball is
greater than on the other. So it isn’t strange that the force on one side of a spinning ball
can be greater than on the other, and the ball can be pushed sideways — curve in its
path.
Here’s a simple experiment to find out about this. You need:
a cardboard tube of the type that is found at the centre of a roll of paper towels
a skinny elastic band, at least 5 cm long
a sticky note, like PostIt™ brand
We want to drop the tube through the air while it is spinning. To do this, cut the skinny
elastic band so that it’s now just an elastic string. Attach one end of the elastic to the
3 stickynote, using tape (duct tape works well for me). Then stick the sticky note onto
the cardboard tube right in the middle (as a kind of anchor) and wind the elastic around
the tube. Now hold the tube horizontally while holding the free end of the elastic. Let
go of the tube while keeping hold of the elastic — before it really falls, the elastic will
pull on the tube and spin it up, and the stickynote will peel off at the last.. It works
best if you can release it over a balcony or from a ladder.
a) What happens to the tube as it falls? Does it make a difference which way the tube is
spinning as you drop it? How does this compare to the way the tube falls if it isn’t
spinning at all? How do you explain the effect? How do you imagine the spinning
motion changes the forces on the tube?
b) Measure how far the tube falls, and how much it moves sideways. You may need to
make several trials, measuring each time, and average your answers together to get a
reliable result. If you can figure out the error in your measurement, that’s worth extra
points. Can you tell what p ath or trajectory the tube takes as it falls — curved or
straight? You might want to use a video camera and then look at the tape framebyframe to analyse things.
c) Use the results from (b) to figure out what the angle of descent is for the tube,
relative to the vertical. You should probably find that the falling tube pretty quickly
reaches terminal velocity , when the air dragforces exactly match the force of gravity (for
a parachutist in freefall this is roughly 200 km per hour!). At what speed does the
cylinder fall? Can you use this to figure out roughly how quickly the tube moves
sideways?
d) Theory — this just takes some algebra, though it looks a little hairy. Bonus points!
Use the Fdrag formula from question 4; for anything shaped like a c ylinder, the
coefficient of drag is ε = 1.0. Can you use this to predict the terminal velocity of the
falling spinning cylinder (you may also have to weigh your tube)? How does this
compare to what you measured? You can likewise look at the sidewaysmotion
terminal velocity. What must be the sideways force, to give the sideways terminal
velocity you found above?
e) BIG BONUS: For a given cylinder and spinrate, can you roughly predict the angle,
relative to the vertical, at which the spinning tube should fall? [Robin] 4 POPBits™ — Useful bits of information
Drag force:
Fdrag = ε ρ A v 2 ε = coefficient of drag (a constant)
ρ = density of fluid
A = crosssectional area (area of an object’s shadow)
v = speed where εsphere = 0.5
εcylinder = 1.0
C v (Ar) = 12.5 J mole –1 K –1 –1 Mm (Ar) = 39.948 g mole (heat capacity at constant volume)
(density of argon, per mole; molar density) ρair = 1.2928 g L–1 5 20002001 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 1: General
1) The path of least resistance (...and highest voltage)
If two identical batteries are connected to each other in series, the equivalent resistance
is the sum of their individual resistances. The voltage across them would also be the
sum of the individual electromotive forces.
When they are connected in parallel, the total resistance is half the individual resistance
and the voltage would be the same as individual voltages.
If we connect all the batteries in series, we could get a high voltage but at the same time
the total resistance is big and lowers the current. We could reduce the resistance by
connecting all in parallel but then the total voltage is small.
The optimum case is to set the batteries in n rows and m columns. If the voltage of each
battery is V and its resistance is r then the voltage in each row is mV and the resistance
is mr. So we have n equivalent batteries each having voltage mV and internal resistance
mr . Therefore the total votage is m V and the total resistance is mr/n. Then the current
that passes through a resistor R is equal to
I= mV
V
=
mr R r
+
R+
mn
n We have to maximize the above equation with the condition N=nm. Therefore we
should minimize the denominator, which is the sum of two terms having the product of
Rr/mn or, equivalently, Rr/N. We know that the sum of two numbers which have a
constant product is minimized when they are equal.
Therefore R/m = r/n. It means mn= 32 and m/n =2. So m=8 and n=4. [Yaser] 2) The last straw...
a) By creating a partial vacuum in your mouth, atmospheric pressure forces the liquid
up the straw. The difference in pressures will work only to a certain height h until the
weight of the water in the straw will balance the differentialpressure force:
P/Patm = 0.5, P = 0.5Patm = 5.05 x 104Pa
Patm = P + ρ gh
h = (P atm –P) / ρ g
= (1.01 x 105 Pa – 5.01 x 104 Pa) / {(1000 kg/m3 )(9.8 m/s2 )}
1 h = 5.15 m
Experimental Results
Note: Inhaling is different than sucking By sucking, resting, then continuing to suck,
your tongue acting as a piston can produce a vacuum effect capable of raising the liquid
in the tubing higher than a height your lungs are capable of sustaining. I did the
experiment both ways.
Experiment by inhalation
At the hardware store I picked up 3 1.5 meter tubes with diameters 1.58 cm, 1.90 cm,
2.54 cm. The tubing was taped along side my washing machine so as to get the tubing
as straight as possible. One end of the tubing was submerged in a bucket of water .
The other end was brought to my mouth. I exhaled, and then I inhaled as much as
possible. As the water level rose in the tubing, it became more and more difficult to
breathe in . This was not the result of me running out of lung capacity, because I was
able to continue breathing in if I took my mouth away from the tubing.
Once the water level stopped rising I quickly plugged the hole of the tubing with my
tongue. The level of the water in the tubing was then recorded. My lungs, I felt, where
getting tired after a few attempts so I rested every 3 to 4 attempts. 5 attempts for each
tube were performed. Below are the results.
Diameter of
tubing (cm): 1.58 1.9 2.54 Attempt 1 112 99 106 Attempt 2 105 114 105 Attempt 3 112 115 111 Attempt 4 106 108 117 Attempt 5 100 109 103 107 ± 2 109 ± 3 108 ± 2 Average Table 1. The heights of water within a tube achieved by inhaling on differentsized tubing.
b) This result would suggest that the you could not raise the water level within the
tubing to 5m simply by inhalation. 2 The results above show that the different average heights for each tubediameter are
within each other’s error. The diameter of the tube had no bearing on the height
attainable, as one expects. The weight of the water within a larger tube will exert
greater force against the atmosphere, but this force is distributed over a greater surface
area due to the increased tube diameter. The resulting pressure exerted by the weight
of the water, Force/Area, is therefore not affected.
These results suggest that your lungs are only capable of producing a partial vacuum of
0.895 atmospheres — 0.105 ATM below atmospheric:
1.08 = (P – Patm) / ρg = (1.01 x 105 Pa (1 – C) / {(1000kg/m3)(9.8m/s2 )}
solving for C,
C = 0.895
Why does this differ from our earlier calculations? Because experiment shows that a
typical person’s lungs cannot deliver 0.5 atmospheres of partial vacuum!
Experiment by sucking on tube
It would seem that the vacuum your mouth is capable of producing relies on another
mechanism besides lungs capabilities. As mentioned earlier your tongue acting as a
piston may be able to produce a greater partial vacuum in your mouth. To test this idea
we repeated the experiment by sucking. (Unfortunately we initially misguided you by
suggesting a short length of the tubing to use — sorry about that!)
What I did was I found a tube longer than 5 m and repeated the experiment, in a
stairwell this time. With a bucket of water on the ground floor I placed the tubing in the
bucket and walked up a few flights of stairs. I then began to suck on the tubing,
pausing occasionally to catch my breath. By placing my thumb over the opening of the
tubing I maintained the water level in the tubing while resting. I continued this pattern
of sucking and resting until I could no longer raise the water level. Measurements are
shown below.
1.3 cm diameter tubing Height in meters Attempt 1 4.5 Attempt 2 4.2 The results are close to those calculated earlier! With each suck I was able to draw the
water higher and higher. Not being restricted to one breath I was able to take
advantage of the mechanisms my tongue is responsible for. There finally reached a
point where the partial vacuum, created in my mouth, could no longer raise the water
level beyond the 4.5 meter. This is the height that determines the vacuum capabilities of 3 my mouth. Lungs & diaphragm muscles, in fact, are not nearly as sturdy as mouth
muscles, though a little baby crying might leave you wondering... [Sal]
3) Fermi, fer you
Folks don’t have to get the following answers – it only matters that you reason well, and
work with whatever scraps of information you already know. It isn’t OK to look up
significant answers for things. You’re supposed to work like on a desert island, with
what’s in their heads. Clever resourcefulness, inventiveness but mostly reasoning from
common ideas and facts is what gets marks here.
a) How many piano tuners do you figure there are in Toronto?
• about 3.5 million people in Toronto
people: 3.5 M
• about 20% of households have a piano
pianos: 700 k
• about 10% of these get their pianos tuned regularly, say 2x per year, others maybe
once every two years, (2 x 70 k + 0.5 x 630 k) =
tuning vists/year: 455 k
• say ~1.5 hours to tune a piano + 0.5 hours driving time ⇒ in an 8hour day maybe 4
visits. If a tuner works 280 days/year
tuning visits/year/worker 1,120
Total tuners in whole Toronto area: about 400 Concert pianos probably are the specialty of a small handful of tuners; likewise some
people have relatives who make a hobby of it, or have their pianos tuned by people ‘off
the books’. So I think this number is probably good within a factor of two.
How else can one get an idea? Try the Yellow Pages, which lists the fairly large tuning
services, some of which employ more than one person; then figure that a certain
number don’t pay for ads, or list in regional Yellow Pages, and it probably agrees with
this number.
b) I saw lots of meteors streak across the sky this summer — they moved through an arc of
about 45 degrees of the sky in a little less than a second. How fast do meteors go?
Well, angular speed is only good if we can guess the height of the meteors. That can’t
be any higher than the height of the atmosphere, which isn’t of course a sharp edge. I
think the International Space Station orbits at something like 240 miles, if I recollect
right. Also that the Russian space station Mir is going to be in trouble when it drops to
about 120 miles. Now, a person can still breathe on top of Everest at 10,000 m, about the
same height as airliners fly through the air, so there’s a reasonable amount of air at 6
miles up (maybe roughly 5070% of an atmosphere?) A meteor would burn in a lot less
air than that. I think the space shuttle is in a fireball, with protection from its heattiles,
at 40 miles up, so I think meteor showers must show up at heights of roughly 2040
miles up. 4 In that case, 45 degrees per second is 2 π (30) / 8 miles per second, about 25 miles per
second. [is that reasonable? I remember that the space shuttle orbits at about 90
minutes per orbit, and the earth radius is about 8,000 miles, so about 25,000 miles in 90
minutes, or 17.000 mph, or almost 5 miles per second, so this seems possible].
Well, meteor showers come often (like the Leonid shower at just this time of year) from
comet dust, as the earth moves in its orbit. So if the earth orbits through the dust, the
speed above ought to be roughly the orbital speed: 93 million miles radius, in one year,
and the earth is moving at almost 20 miles per second. So I think this probably is a
pretty good fit, for a sight at night!
c) How many kilograms of toothbrushes do the people of Canada throw away each year?
If we followed the dentist’s advice, it would be a new brush each 36 months, if I
remember right. Do you do that? Let’s say we buy 1.5 toothbrushes per year for each
Canadian (about 30 million); it might only be one, but some people buy new
toothbrushes for guests, and others treat their toothbrush like an old friend... Say 5
million are either kids or folks with dentures who soak ‘em. That would then mean 37.5
million toothbrushes per year that get bought. Unless we store up toothbrushes, they
all eventually get thrown out (even after being used for cleaning bathroom grout or
polishing fussy silverware), so we ditch 37.5 million toothbrushes per year. How much
do they each weigh? I’ve never weighed one, but I think it would take at least 20
toothbrushes to weight the same as a quarterstick of butter — 0.25 lbs /(2.2 kg/lb) / 20
= 10 g. That might be too low: my toothbrush is about 20 cm long, and very roughly 1
3
cm across, in average width, about 20 cm . If it has about the same density as water,
then that should be about 20 g.
So 37.5 million x 0.020 kg = 750,000 kg of toothbrushes each year.
I wonder if any of this plastic could be made recyclable...
d) If icebergs never broke off from the icecaps at the poles of the earth, roughly how long would
it take to tie up all the water on earth as ice at the poles?
This is the toughest one. It’s like the problem in wintertime that the air moisture in
your room ends up frozen on your windows – it’s a kind of dehumidifier or moisturepump, and your room can get a bit drier. Mostly winter dryness is because the frigid
outside air, even if it’s at 80% relative humidity, holds little moisture; if you only warm
up outside air, the relative humidity goes down quickly, because your warm houseair
could have held a lot more water, but it still has only the little amount of water in frigid
air.
Let’s see: air doesn’t circulate very well north and south, but it would eventually pass
over the polar icecaps. So, if it falls as snow on the icecaps and never again leaves,
eventually all moisture will get trapped in the icecaps. People suspect this may be an
5 issue for places like Mars. However, the polar regions actually get very little snow –
you may remember that people take icecores in order to look back at pollen and spores
that fell hundreds or thousands of years ago. It’s one way that people test for global
warming, because that biological record, frozen at different depths, indicates ancient
temperatures. Now, to go back 1,000 years in the Antarctic ice cap, one doesn’t have to
drill 1 km – it’s much less than 1 m of snow per year.
In fact, the poles are both desert regions, which by definition means less that 10 inches
of precipitation falls (measured as water). So, say that exactly 10 inches falls each year
over the area of each icecap. How big is each icecap? Smaller than the arctic/antarctic
circles which are at 22.5 degrees, which is the tilt of the earth on its axis. You could
figure out the area of a sphere within 20 degrees of the pole, but that’s too picky for
such a rough approximation. Good enough (with 20% probably!) to take the area of a
circle at the radius of the earth, 8000 miles, and say at 15 degrees away from the pole.
That means a radius about 15/360 th of the earth circumference, 25,000*(15/360) =
about 1,000 miles. That area is then about 6 million square miles at each pole, for 12
million square miles total.
If 10 inches falls each year there and is locked up, and the total surface area of the earth
2
is 4 π r = 780 million square miles, then the rest of the earth loses 10 * (12/780) = 0.15
inches per year. If 70% of the earth’s surface is water, then the oceans and lakes would
lose about 0.2 inches per year as an equivalent amount is locked up each year in the ice
caps. If the icecaps never calved icebergs, or evaporated or whatever, then in a lifetime
the oceans would drop by about 16 inches, which would be noticeable. In about 300,000
years the oceans might drop by a mile. If the oceans are in only a few places a few miles
deep, then the average depth might even be less than this – in less than a million years
the oceans might all be sitting as ice at the polar caps.
Except for icebergs.
e) How many atoms scuff off your sneakers onto the sidewalk with each step you take?
Let’s see: my sneakers pretty much wear through through the sole in about 18 months,
if I wear them to walk on sidewalks outdoors and such. That sole is about 1 cm thick,
but only about 1/3 the area of the sole has to wear down before they wear through
under the balls of my feet. So, for my size12s, that’s like 1 foot long and perhaps 4
inches wide, or 25.4 cm x 10 cm for a net wear of 1 cm x (25.4 cm x 10 cm) / 3 = roughly
3
80 cm of rubber & plastic. Plastic is a hydrocarbon, so it’s mostly C and H in very
roughly even proportions, with an average mass per atom of about 6.5 a.m.u. It has a
3
density a little above water, I think, so say 1.5 g/cm . So that’s about 120 g of rubber.
At 6.5 a.m.u. average, for the atoms, that means 6.5 g per mole of sole (per mole of atoms
in the soles of my shoes) – so I lose a little less than 20 moles of atoms from my soles in
one year.
6 And for one step? Well, how many steps in a year? Hmmm, well I walk about 600 m to
the streetcar from home, and about 500 m at the other end – 2.2 km just getting to work.
My steps are about 1 m per pace (I’m tall), so that’s 2,200 steps then. I walk for about 1
hour net, between classes and offices, during the day while I’m at work, and another 20
minutes at lunch. I figure I take about 1 step per second (very roughly), so that’s 1 1/3
hours x 3600 seconds/hour x 1 step/second = 4,800 steps. I take my sneakers off at
home. So I walk about 7,000 sneakersteps each day, roughly, 300 days per year
because I don’t always wear sneakers. So a little over 3 million steps in 18 months to
22
remove about 20 moles (20 x 6 x 10 ) of atoms.
17 So with each step I must be leaving behind roughly 4 x 10 atoms! Yikes, one day
people ought to be able to figure out where I walked! They just need to make a
combination upright vacuum cleaner and mass spectrometer!
REMINDER: all these values are very rough – within a factor of 2 or 3 either way is
reasonable. The important thing is the reasoning, as Sherlock Holmes might once have
said. [Robin]
4) Number four with a bullet
The Physics – The key to this problem is realizing that the acceleration of the bullet is
not constant while traveling though the cylinder of gas: From the drag equation
1
Fdrag = − 2 ερAv 2 we know that bullet experiences a force that is a function of its
velocity. Since F = ma and the mass of the bullet is constant, the acceleration has to be a
function of velocity. Therefore, the following constant acceleration equations cannot be
used:
1
x(t) = x(0) + v(0)t + 2 at 2
(1)
v(t) = v(0) + at
(2) including the variations on equations (1) and (2):
v(t)2 = v(0)2 + 2 a( x(t) − x(0))
1
x(t) = x(0) + 2 (v(t) + v(0))
1
x(t) = x(0) + v(t)t − 2 at 2 In this question, only the following fundamental definitions apply:
dx
v=
dt
dv
a=
dt
F = ma (3)
(4)
(5) (6)
(7)
(8) 7 The Solution – Given quantities converted to SI units where appropriate:
Variable Name Definition Value v(0)
m
r
L
R
T
P
ε
Mm
C Initial velocity of bullet
Mass of Bullet
Radius of bullet
Length of cylinder
Radius of cylinder
Initial temperature of gas
Pressure of argon gas in cylinder
Drag coefficient for a sphere
Molar mass of argon gas
Specific heat of argon gas 250 m/s
2.00 x 103 kg
4.500 x 103 m
15.0 m
0.500 m
300 K
2.027 x 106 Pa
0.500
39.948 g/mol
312.5 J/kg/K Quantities to be calculated:
Variable Name
v(t)
t
V
A
∆T
M
•
•
• Definition
Final velocity of bullet after traversing the cylinder
Time it takes for the bullet to traverse the cylinder
Volume of cylinder
Maximum crosssectional area of bullet
Change in gas temperature
Mass of argon gas in cylinder Calculate maximum crosssectional area of bullet:
A = π r 2 = π (4.50 × 10 3 )2 = 6.362 × 10 5 m/s
Calculate cylinder volume:
V = π LR 2 = π (15.0)(0.500)2 = 11.78 m 3
Calculate density of argon gas in cylinder:
n
P
=
V RT
n
PMm (2.027 × 106 )(39.948)
ρ = Mm =
=
= 3.246 × 10 4 g/m 3 = 32.46 kg/m 3
V
RT
(8.31451)(300)
Calculate the mass of argon gas in the cylinder:
M = ρ V = ( 32.46)(11.78) = 382.4 kg
PV = nRT (Ideal Gas Law) ⇒ • Part A
• Calculate the time it takes for the bullet to transverse the cylinder:
1
ma = Fdrag = − ερAv 2
2
dv
since a =
, then
dt 8 m dv
1
= − ερAv 2
dt
2 dv
ερA 2
=−
v = −β v2
dt
2m where β = this is a differential equation – what function v(t) can be
differentiated once to equal itselfsquared, multiplied by a
constant – β ? ε ρ A (0.5)(32.46)(6.362 × 10 −5 )
=
= 0.2581 m 1
−3
2m
2(2.00 × 10 ) (see A Guide to Solving Simple Ordinary Differential Equations (ODE’s) on the POPTOR
webpage for hints etc. for this problem set to understand the following steps)
dv
v2
v( t ) = − β dt
t dv ∫ 2 = −β ∫ dt
v( 0 ) v
0
1
1
−
= βt
v(t) v(0)
v(0) β t + 1
1
1
= βt +
=
v(t)
v(0)
v(0)
v(t) = v(0)
v(0) β t + 1 equation for the velocity of the bullet as a function of time We can’t solve for time t since we don’t know what v(t) is. However, we do know
what the distance is so use the definition for velocity:
since v = dx
, then
dt dx
v(0)
=
dt v(0) β t + 1 this is another differential equation (again see A Guide to Solving Simple Ordinary Differential Equations (ODE’s) )
dx = v(t)dt =
L t 0 v(0)
dt
v(0) β t + 1 0 1 ∫ dx = v(0)∫ v(0)β t + 1 dt L= v(0)
1
ln(v(0)β t + 1) = ln(v(0)β t + 1)
v(0)β
β ln( v(0)β t + 1) = Lβ 9 v(0)β t + 1 = e Lβ
t= e Lβ − 1
v(0)β equation for the time it takes the bullet to transverse the cylinder substitute in the numbers:
t= e(0.2581)(15.0) − 1
= 0.726 s
(250)(0.2581) Part B
Using the equation for velocity and the value for time found in Part A:
v(t) = v(0)
(250)
=
= 5.23 m/s
v(0)β t + 1 (250)(0.2581)(0.726) + 1 Part C
the change in the bullet’s kinetic energy is converted to heat in the gas:
∆EK = ∆Q ( ) 1
1
1
m v(t)2 − m v(0)2 = m v(t)2 − v(0)2 = MC∆T
2
2
2
∆T = ( m v(t)2 − v(0)2
2 MC ) equation for change in temperature of argon gas substitute in the numbers:
(2.00 × 10 −3 )((250)2 − ( 5.23)2 )
∆T =
= 5.23 × 10 −4 K
2(382.4)(312.5)
FINAL ANSWERS
Question Answer Part A
Part B
Part C t = 0.726 s
v(t) = 5.23 m/s
–4 ∆T= 5.23 × 10 K [Brian]
5) Bubbling ideas
a) The temperature of the air inside the bubble is always the ambient temperature.
Therefore, pV = constant, where V is the volume of the bubble. So if the volume
changes by a small amount ∆V then (using the product rule) we have
10 ∆( pV ) = ∆p ⋅ V + p ⋅ ∆V = 0 ⇒ ∆p = − p ∆V
V b) If we put a charge on the bubble, the bubble size increases as each part of the bubble
is repelled from every other part. This will reduce the internal pressure.
2
The electric field outside the bubble at the surface is Q ( 4πεR0 ) where R0 is the radius of
the bubble. The field inside is zero. We could say that the field exactly on the bubble is
2
the average of these two fields Q (8πεR0 ) . A fulltheory calculation gives exactly this
answer. Therefore, the electrostatic pressure is
PE = Eσ∆A
Q
Q
Q
=
=
2
2
∆A
8πεR0 4πR0 32επ 2 R04 where σ is the surface charge density, E is the electric field on the bubble and ∆A is the
area of a small piece on the bubble.
This pressure should be equal to the change in the internal gas pressure, following its
change in size
PE = − ∆p → ∆V = V
Q
Q
→ ∆R =
24
3
P 32επ R0
96επ 2 R0 P [Peter & Yaser] 6) Magnus matters
a) If the tube is not spinning, it just falls straight down. If it is spinning, then it doesn’t
fall straight, it scoots out to one side or the other, depending on which way it was
spinning. If the tube is spinning so one side moves backwards with the passing air, and
the other side moves forwards against the passing air, the tube swings away from the
side that moves forward against the passing air (greatest relative speed).
When the tube falls relatively slowly, as it does for us, the flow around the tube is pretty
wellbehaved — no very serious turbulence (which would make the problem tougher).
We call the flow laminar, which means it flows in smooth layers. Now, the air is very
slightly viscous — as syrup is more viscous than water, and water is more viscous than
alcohol. That means that as the tube moves, it carries a layer of air along with it, close
to its surface.
It also means that when the tube rotates, it carries the air with it then. Imagine the tube
holding still, and the air rushing past (as in a wind tunnel). The side of the tube
rotating forward against the oncoming air will actually slow the air slightly; the side of
the tube rotation with the oncoming air will carry the airflow a little faster. This means
the air is slower on the side of the tube that rotates against the flow, and faster on the
other side. Bernouilli’s principle (which really is just a conservation of energy formula
for fluids!) means that the air pressure will be higher on the side moving against the
11 airflow, and less on the other side. The net pressure difference is a force, which pushes
the tube sideways — the Magnus Force.
Bernouilli’s principle is a big part of how an airplane wing generates lift out of the
forward thrust of the engines. It’s also the reason why a strip of paper, held below your
lower lip, will rise as you blow air across the top surface.
[EXTRAS: When the object moves very quickly (like a fast baseball, or many real
airplane wings), it isn’t any more exactly the Bernouilli effect, but a bit more
complicated. What happens there has to do with the transition that airflow makes
when it isn’t smooth, but becomes turbulent and begins to detach from the flow around
the tube. When the relative airspeed is higher, the flow detaches earlier in the stream
around an object; where the surface moves back with the airflow, it delays this
boundarylayer separation . The result is that the air flowing past a spinning body is tilted,
in the stream past the object. So the air in the wake of the spinning tube actually kicks
over to one side. The momentum change of that air results in the force pulshing
sideways. (Look for a streamingvideo description of this on the archive of the Science
of Baseball website, from a show on the Discovery Canada: www.exn.ca)]
b) I recorded this on my computer, using a cheap CCD webcamera and shareware
NIH Image 1.62 [http://rsb.info.nih.gov/nihimage], which can put a timestamp on
each frame to help analyse motion. Here are a few frames in the sequence: You don’t have to do this, though — you can get very nearly as much by finding where
it hits the ground, measured from where you let it go, and using a stopwatch to time it.
b) I measured from the video frames on my computer. If you use a video camera, you
can measure from the TV screen. You might want to use a dryerase felt pen (but
remember to erase it afterwards!); but be careful about using any markers at all on a
12 computer monitor, since they are sometimes antireflection coated with a material that
dissolves (they used to often dissolve with ammoniabased cleaners like Windex too!
I’m not sure this still is true, so check the manual for your display for warnings!)
In the graph on this page, each marker on the trajectory is a measurement at a different
time (the blue ones are a little more than 0.1 second apart). The path is curved: it starts
off going to the right, because of the elastic band I used to spin it, I think. Then it drops,
but as it gains speed it begins to push to the left.
c) The angle of descent was about 45°, so the tube moves at the same speeds in the
downwards direction and the sideways direction. It almost reaches a constant speed
(terminal velocity) after 1 second, when it hits the floor, but doesn’t quite get there. If
you used a balcony in your school, you probably did better than I did. I was able to
–1
extrapolate that the terminal velocity would be around 3.8 m s .
d) At terminal velocity, there is no acceleration — all forces balance. Then the 45°
trajectory indicates that the net force is at 45° down and to the left, so in this case the
Magnus force is about the same as gravity, at terminal velocity. People have patented
‘sail’steamships with tall rotating cylinders for
0.5
sails; they could use windpower to make the
ship tack into the wind at 45° this way.
0 From POPBits™, at the end of questionset #1:
2 0.5 The mass of my cardboard tube was 25 g, and
2
its crosssectional area was 125 cm ( see
comments below). Balancing forces, and
therefore setting ma = F drag a t terminal
velocity, we can solve simply for vterminal:
m a = ε ρ A vterminal 1.5 2 –2 2 0.025 kg • 9.8 m s = 1.0 • 1.2928 g L
2
2
• 0.0125 m • vterminal
Thus vterminal = 3.9 m s 1 y (m) Fdrag = ε ρ A v –1 3 –3 • (10 L m )
2.5 –1
–1 3 as compared to 3.8 m s measured above (this
1.5
1.0
0.5
0.0
0.5
worked unusually well for such a formula).
x (m)
The sideways Magnus force is comparable to
the force of gravity, so the xcomponent of velocity is about the same as this value, and
the final terminal velocity is the resultant, about √2 times larger. 13 e) B IG B ONUS : You can come fairly close to the right answer with this slightly handwaving reasoning (you can also have a look in The Physics of Baseball (2nd Ed.), Robert
K. Adair, Harper Collins, NY, 1994, ISBN 006—950471).
The drag force is:
Fdrag = ε ρ A v 2 which can be described as a pressuredifferential ∆P (front – rear) multiplied by the area
presented to the wind or rushing air. Since this pressuredifferential depends on the
relative speed, if an object spins there will be a difference in the force from sidetoside
across the ball. Then
2 2 FMagnus = ∆Pleftright A = ε ρ A (vright – vleft )
now,
vright = v + 2πr • f
= v – 2πr • f vleft where f is the rotation frequency (revolutions per second) of the object. Substituting:
2 2 2 2 (vright – vleft ) = (v + 2πr • f) – (v – 2π r • f)
= 8π r v f
so
FMagnus = ε ρ A (8π r v f ) For our case at 45° falling angle, Fdrag = FMagnus so we can figure out the spinning
frequency f:
(8π r v f ) = v 2
–1 which we solve to find f = 6.3 rev s I do think the elastic band setup I used could provide this spinrate. But, in fact, this
handwaving argument fudged a little on the appropriate use of areas and direction of
airflow. Even so, any error is by a constant factor: the formula shows the appropriate
dependence on speed v and spinrate f.
Comments on my own setup
Cardboard tube:
circumference 14 cm > diameter d = 4.5 cm
length 28 cm
2
crosssectional area presented to wind (rectangle) A = 125 cm
mass 25 g 14 (I didn’t have a weighscale at home, so I took a 25 cm rod, a handle from a cattoy, and
hung a plastic grocery bag tied to a string on one end, and just a piece of string on the
other. The rod had a piece of string tied around the middle to suspend the whole thing
to tilt freely. I slid the string along the middle of the rod until the grocery bag and loose
string balanced each other evenly. Then I taped the cardboard tube to the loose string,
which tipped the balance over. I then dripped water dropbydrop into the plastic bag
until it balanced the cardboard tube again — it took 25 ml, which weighs 25 g.) [Robin] 15 2000–2001 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 2: Mechanics
Due November 27, 2000
1) Take your best shot
Evangeline, an Olympic biathalon competitorintraining (who hopes to use her keen
grasp of kinematics to win gold), keeps a water supply on her training course: on top of
a 10 meterhigh water tower platform sits a 10 metertall cylindrical drum of water, 3
meters in diameter. Unfortunately, the pipes to the drinking fountain below have
frozen solid. Evangeline skis up to a point 5 meters away from the base of the tower,
and with her last round she uses her target rifle to shoot a small hole in the wall of the
water drum.
Where, precisely, on the drum should the hole be so that she can have a drink without
moving from the place where she stands? [Sal]
2) The fountain of (misspent) youth
Sukumar has accidentally backed his pickup truck into a fire hydrant, and broken it off
–1
at ground level (oops!). Water gushes straight upward at a rate of 150 kg s , and with a
–1
speed of 20 m s . Suku tries to fix things by holding a nearby garbage can over the
flow, but finds himself and the can lofted into the air on the water flow.
If Suku weighs 60 kg, and the garbage can weighs 2.5 kg, what is the height that he’s
waving to his friends from, when the police get there? [Yaser]
3) Huh? Hmmm…
Consider three containers, as illustrated at right, each with the same area A at its base,
and each with height H. The first one is a cylinder, and the other two are tapered. The
area of the tops of the other two vessels are 2A and A/2.
a) If all three containers are
filled with water, what's the
water pressure at the bottom of
each? What total force does
this then exert over the whole
bottom plate of the container?
b) If the containers of water are
1 each put on a weighscale, individually, what weight will the scale show for each?
Treat the containers as being very thinwalled, and essentially massless. Only the
bottom plate touches the weighscale, so why is the measured weight different from the
force found as area × pressure you calculated in part (a) ?
Remember that balance shows the force which is exerted on it. [Yaser]
4) All aboard!
–1 a) A freight train traveling with an initial speed of 65 m s , cuts its engine and coasts 8
kilometers along a flat track before finally coming to a stop. Calculate the train's
coefficient of rolling friction. Is this a value for the whole train, or for each railroad car?
b) A locomotive is pulling an empty 15.0mlong hopperloader car at a constant speed
–1
of 1.00 m s . The mass of the locomotive and empty car is about 75,000 kg. As the
hoppercar moves along the track it passes under a grain dispenser that pours grain into
–1
the car, at a rate of 1000 kg s . The grain dispenser only operates when the car is
beneath it. How much work does the locomotive do while the hoppercar is being
filled? Assume that the coefficient of rolling friction is the same as calculated in part (a).
[Brian]
5) Java jumpup (eeyow!)
A few years ago McDonald’s™ was sued by a woman who was badly scalded when the
cup of hot coffee she was holding between her thighs — while driving — spilled. The
woman won the suit, and since then their coffee cups include a printed warning on
them that the presumably hot coffee inside is really hot.
So when the makers of the Java Jumpstart Commuter Mug* recently came out with a new
model, we at POPTOR suggested they have it evaluated by DeepeeDeetee Consultants
Inc ., a group of OAC Physics students, who evaluated the mug’s stability on the
dashboard of a moving car.
The physicsconsultants found that the mug's stability depends significantly on how
much coffee is inside. Find the height or coffee level in the mug for which the mug of
coffee is least likely to tip when the car accelerates. The mug is 20 cm tall, has a radius
of 6 cm, and weighs 200 grams (1/4 of which is the weight contribution of the bottom of
the mug) [Sal]
[*a fictitious company. If you find the first true storey unlikely, compare with other true stories
of human fallibility at the irreverant Darwin Awards homepage, http://www.darwinawards.com] 2 6) Weight! Bob!
In science, either experiment or theory can lead the way in discovery. But it pretty
much isn’t finished science until both experiment and theory come together. Here’s one
example where a simple experiment can tell you the
answer you might not be able to guess theoretically —
string and steer you in the right way to think, in order to
understand. pulley Two identical weights are connected by a 1 m string.
To begin, one weight is set on a frictionless table, and
the second is held as in the picture, with the midpoint
of the string between them draped over a massless
frictionless pulley. If the second weight is dropped, which happens first:
• does the first weight slide along until it hits the pulley, or
• does the second one swing around until it hits the side of the table?
Try this in an experiment, to find the true answer, and then see how well you can explain
it by theory. Try whatever you like to make things practically frictionless as given —
but tell us what you did in your experiment, to approximate these ideal conditions, and
then carefully describe what you measured.
HINTS: To make the first weight practically frictionless on the table, you might try two
child’s wooden cars with wheels. Or you might be able to use an airrail at school for
the table and tie together two identical airrail cars for the weights. Or perhaps a dryice puck on the table would be practically frictionless, and you could tie it to an equal
weight to drop at the side. The frictionless pulley can be nearly any small lightweight
wheel (no inertia!), or perhaps teflon tubing rotating on a nail — or you might use
tefloncoated electronics wire to connect the weights, and then a smooth round edge on
the table to slide over with little friction. [Yaser, Robin] C HECK THE POPTOR WEB PAGE for other hints, and any corrections we might post:
www.physics.utoronto.ca/~poptor 3 2000–2001 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 2: Mechanics
1) Take your best shot
Due to water pressure in the tank,
water squirts out horizontally from the
bullet hole at some speed, depending
on the ‘head’ or height of water above
where the hole is made. The water
then falls a certain vertical distance.
The horizontal distance the water
reaches depends on the squirtspeed
and the time it takes the water to fall
— a time that only depends on height
above the ground.
Let the hole the rifle makes in the tank
be the origin of a vertical axis which
we will set to zero. Let h be the height
above the hole to the top of the tank.
As water flows out through the bullethole, the height of the water in the tank must drop. In a time ∆t, the volume of water
flowing through the bullethole is:
Aflow × distanceflowed = Aflow (vflow × ∆t)
where Aflow is the area of the hole, and vflow is the flowspeed of water at the hole.
Likewise the water in the tank will drop by a volume:
Atank × distancedropped = Atank (vtank × ∆t)
Equating these, we get:
vtank = vflow × (Aflow /Atank )
As the water drains from the tank due to the bullet hole, the kinetic energy put into the
squirting water simply equals the potential energy of lowering all the water in the tank
from the hole up. In a small time ∆t this is:
1⁄2 Mflow vflow 2 = Μtank g h 1 2 1⁄2 (ρ Aflow (vflow ∆t)) vflow = (ρ Atank h) g (vtank ∆t)
Where ρ is the density of water. We can cancel terms, and use vtank we found above, to
get:
2 vflow =2gh This is the initial horizontal speed. The initial vertical speed is zero. The time of flight
for the water leaving the tank before it hits the ground 5 meters away from the base of
the tank is given by:
2 t = 2 (20 – h) ⁄ g
This time needs to be exactly the value that lets the water squirt 5 m sideways, at
constant speed vflow , before hitting the ground.
vflow t = 5 m
therefore,
4 h (20 – h) = 25
Solving for h we get the result that in order for the water to reach the athlete 5 m away
from the base of the tank, she must shoot a hole approximately 32 cm down from the
top of the tank. [Sal]
2) The fountain of (misspent) youth
Assume that Suku is at height H. The speed of the water at this height, before the
collision with the garbage can, is
2 2 v = vo – 2 g H
–1 where vo = 20 m s is the initial speed. After the collision, the water stops its motion,
since the collision between a liquid and any surface is always inelastic. The change in
momentum of the water during the collision acts provides the a force and compensates
for gravity in order to hold the garbage can and Suku up. This force is equal to:
F= ∆m * (v – 0)/∆t = 150 (kg/s) * v
where ∆ m is the mass of the water which has a collision at a certain time. ∆ t is the
collision time. Both ∆ m and ∆ t are supposed to be infinitessimals (vanishingly small
numbers) but their ratio is actually equal to the rate of the water flow, which is 150 kg/s.
This force is equal to the weight of the garbage can plus Suku.
F = (60 + 2.5) • 9.8 = 150 v = 150 •
From which we can find H = 19.5 m. Yikes! 400  2 • 9.8 • H
[Yaser] 2 3) Huh? Hmmm…
a) The water pressure at the bottom of all the containers is the same and equal to (ρ*g*
H) where ρ is the density of the water. The pressure is dependent on the height of the
water and not on the amount of the water. The total force which is exerted by the water
on the bottom plate, by this pressure, is equal to the product of the pressure and area of
the plate and is the same for all.
b) The weighbalance shows three different values for the containers full of water. For
the noncylindrical containers, this value is not equal to the product of the water
pressure and the bottomsurface area. For the cylindrical container, the water pressure
on the side walls is everywhere sideways, and contributes nothing to the weight. This
isn’t true for the other containers.
For the tapered containers, not only water pressure exerts a force over the bottom plate:
the rigid walls of the container also exert force. For the one in which the area of the top is
bigger, this force is downward and therefore the total force on the bottom plate is more
than the product of pressure and area. For the other container, in which the area of the
top is smaller, this force is upward, and therefore the total force is less than than the
product of the pressure and the area.
The easiest way to find the right answer is to imagine the sloped wall in a series of stairsteps, and then take the limit of the steps becoming vanishingly tiny.
Each step has a vertical part and a horizontal part. Pressure on the vertical part, which
is a tiny cylinder, gives an outward force everywhere; these radial forces add up to zero
net force. Pressure on the horizontal part, like a shelf around the bottom of the little
cylinder, gives a net downward force, and these sidewall forces add up to exactly the
weight of the water which lies above the sloped walls (i.e., outside the cylinder lying
above the circular base).
The net force can be found by adding all the contributions. Each horizontal step is
actually an annulus — a flat ring lying between two circles of radius r and r + dr.
For a wall sloped at an angle θ from the vertical, the radius r of the vessel as a function
of vertical position z from the base is:
r = ro + z ⋅ tan θ
It’s easy to find dr then (as the differential of r):
dr = dz ⋅ tan θ
The area of the annulus is then:
A = 2πr ⋅ dr = 2πr ⋅ dz ⋅ tan θ
= 2π(ro + z tan θ) ⋅ dz ⋅ tan θ
The downward force on this small annulus is then:
3 dFy = p( z) ⋅ A
where p (z) is the pressure, which is a function of z because z gives the depth. The
pressure is simply the weight per unit area of all the water lying above, and so
p( z) = ρg( H − z)
Thus
dFy = p( z) ⋅ A
= ρg( H − z) ⋅ 2π(ro + z tan θ) ⋅ dz ⋅ tan θ
The total force down on the side walls is then the sum of all these tiny contributions, for
all values of z from z=0 to z=H. This is just the integral (siimpler than it seems — the
integrand is only a polynomial, actually):
z= H ∫ dF Fy = y z=0
z= H = ∫ ρg(H − z) ⋅ 2π(r + z tan θ) ⋅ dz ⋅ tan θ
o z=0 z= H = 2πρg tan θ ∫ (H − z) ⋅ (r + z tan θ) ⋅ dz
o z=0
z= H = 2πρg tan θ ∫ {Hr + (H tan θ − r ) ⋅ z − tan θ ⋅ z } ⋅ dz
o 2 o z=0 H
= 2πρg tan θHro z 0 + { = 2πρg tan θ 1
2 1
2 H ( H tan θ − ro ) ⋅ z 2 0 } 1
− 3 tan θ ⋅ z 3 H
0 1
H 2 ro + 6 H 3 tan θ ⋅ Naturally, as θ goes to zero (as for a cylinder), this vertical wallforce contribution goes
to zero.
It is obvious that this downforce on the sidewalls is the same as the weight of the water
in the ‘extra’ region outside the cylinder — going back to the term for pressure, all we
have done is simply to add the weight of the water lying above every little horizontal
annular step. Since we considered every little horizontal step, we’ve considered the
whole volume of water in the extra region. [Yaser & Robin] 4 4) All aboard!
a) Assume the train experiences constant acceleration (deceleration).
Known quantities converted to SI units where appropriate:
Variable Symbol Description Numerical Value X(0) Initial displacement of train 0.00 m X(t) Final displacement of train 8000 m V(0) Initial velocity of train 65.0 m/s V(t) Final velocity of train 0.00 m/s Unknown quantities:
Variable Symbol Description µ Coefficient of rolling friction the train experiences a rolling frictional force (proportional to its normal force) that is
opposite to its direction of motion:
F = −µmg
F = ma
ma = −µmg
µ=− a
g (1) constant acceleration displacement equation:
X(t) = X(0) + V(0)t + 12
at
2 since X(0) = 0.00 m, then
X(t) = V(0)t + 12
at
2 (2) constant acceleration velocity equation:
V(t) = V(0) + at
since v(t) = 0.00 m/s, then
V(0) + at = 0
t= −V(0)
a (3) 5 substitute equation (3) into equation (2) to find the train’s constant acceleration:
X(t) = − V(0)2 V(0)2
+
a
2a 2 aX(t) = V(0)2 − 2V(0)2
a=− V(0)2
2X(t) (4) substitute equation (4) into equation (1) to find the coefficient of rolling friction:
V ( 0) 2
(65.0)2
µ=
=
2 gX(t) 2(9.81)(8000)
µ = 0.0269
SOLUTION FOR PART (B):
Note that although the rolling frictional acceleration µg is constant, since F = µgm, the
rolling frictional force will not be constant if the train’s mass is changing – that is: F(t) =
µgm(t).
Known quantities:
Variable Symbol Description Numerical Value V Constant velocity of train 1.00 m/s m Initial mass of train and hopper
car 75000 kg l Length of hopper car 15.0 m r Mass transfer rate of grain 1000 kg/s µ Coefficient of rolling friction 0.0269 Unknown quantities:
Variable Symbol Description m(t) Mass of train and hopper car as
a function of time m(x) Mass of train and hopper car as
a function of displacement W Work done by the train
6 mass of train and hopper car as a function of time:
m(t) = m + r t
convert m(t) to m(x) using the fact that t =
m(x) = m + r x
:
V x
V in order to maintain a constant velocity the train must exert a force that balances the
rolling frictional force: ∑ F = Ffriction + Ftrain = 0
Ftrain = − Ffriction = −(−µ mg) = µ mg
since m is a function of displacement:
x F(x) = Ftrain = µ g m(x) = µ g m + r V
the definition of work is:
l W = ∫ F(x)dx
0 l
l
x W = µg ∫ m(x)dx = µg ∫ m + r
dx V
0
0
l x2 W = µg mx + r
2V 0 (15.0)2 l2 W = µg ml + r
= (0.0269)(9.81) (75000)(15.0) + (1000)
2V
2(1.00) W = 3.27 × 10 5 J [Brian] 5) Java jumpup (eeyow!)
Given that the weight of the mug is M and that the weight of the bottom of the mug is
M
–3
the height of the centre of mass (yc.m.) of the mug with coffee of density ρ (~1 g cm )
4
filled to height h is given by, yc.m. = 2 3 MH
2h 4 2 + ρπR 2 ( M + ρπR 2 h) Define a constant a = ρπR 2 such that the mass of the coffee contained is m = ρπR 2 h ≡ a h
7 ∴ yc.m. 3
ah 2 MH +
8
2 =
( M + a h) The maximum stability of the mug of coffee is achieved for the lowest centre of mass.
To find the minimum of yc.m. we need
d( yc.m. )
=
dh 3
ah 2 a h( M + a h) − ( a) MH +
2
8 ( M + ah)2 =0 3
ah 2
MH +
8
2
3
ah 2 + Mh − MH = 0
8 ∴ h( M + ah) = Solving for h we get
1 1
23 2
h=
− M ± M + MHa 4
a the negative solution can be ignored.
1 1
23 2
h=
− M + M + MHa 4
a Given:
M = 200 g, H = 20 cm a = ρπ R 2 = 36π g / cm Solving for h we get,
1 1
3 2
h= −200 + 40000 + (200)(20)(36π) 36π 4
h = 3.6 cm
The mug is most stable with coffee filled to a level h = 3.6 cm. [Sal] 8 6) Weight! Bob!
In an experiment, the tablebased weight hits the pulley
before the free one smacks into the side of the table. string How to explain this?
• It’s easiest to analyse the forces in x (horizontal) and
y (vertical) components. pulley • The force on either block comes from tension in the
string
• There’s only one tension, so the forces on the block are equal in magnitude, though
they may change in time
• The difference is that the force on the lefthand block is always directed along the
string, towards the pulley. The force on the righthand block is also always directed
along the string, but this vector changes as the angle of the string changes.
• The lefthand block moves horizontally towards the pulley with an acceleration:
aleft = T/m where m is the mass of the block, and T is the tension in the string
• The righthand block moves horizontally towards the wall with an acceleration:
aright = T cos(θ) /m
where m: mass of the block, T: tension in the string, θ: the angle the string makes with
the horizontal
• Before the righthand block falls, there is no tension in the string. Once it does begin
to move, then it makes an angle θ >0, so aright < aleft
• The lefthand block is always accelerated to its collision with the pulley faster than
the right hand block to its collision with the wall.
The lefthand block hits first. [Yaser, Robin] 9 20002001 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 3: Thermodynamics
Due January 15, 2001
1) Bye, metallic bar!
Mechanical thermostats for home heating, and flasher units for car turnsignals (and
Christmas tree lights!) often use this principle: Two strips, copper and iron, are bonded
together, side by side. They share the same elastic properties (e.g., Young’s modulus),
and the same length l, and diameter d . They d on't have the same linear thermalexpansion coefficient, though — if we heat them both to a higher temperature, one
expands more than the other, and they bend. If the change in temperature is ∆T and the
linear thermalexpansion coefficients are λFe and λ Cu, find the angle of deflection.
[Yaser]
2) Maxing Maxwell
Have you ever wondered how fast molecules are flying about, maybe air molecules in
your lungs? The atmosphere we breathe is composed of particles of different gases, the
kinetic energy of these particles being related to their thermal motion. These gas
particles collide with one another constantly, affecting their velocity. To speak of the
velocity of a particle is therefore not very practical. What we can speak of though is a
probability distribution of velocities — how likely it is to have one velocity or another.
A result known as Maxwell's Distribution describes how particles are distributed over
different speedranges:
M
P(v) = 4π 2π RT 3/2 v2 e−M v 2 /2RT where M is the molar mass of the gas, T is the absolute temperature of the gas and R is a
constant. The distribution gives the probability (maximum 1.0 = 100%) that a particle in
the gas will have a velocity that falls between v and v+dv, where dv is an incremental.
a) Calculate the most probable velocity for molecules of oxygen at T = 300 Kelvin.
–1
–1 –1
Oxygen has a molar mass M = 0.0320 kg mole . R = 8.31 J mL K .
b) Now graph the Maxwell Distribution . [Sal] 1 3) A pressing concern
There is cylinder filled by a gas, with area A. In the cylinder, there is piston with mass
m on top which makes the height of the piston from the bottom of the cylinder a
distance h. How much weight should we add onto the piston to reduce the height to
half of its initial value? Assume the air pressure to be Po and the temperature of the gas
is always room temperature. [Yaser]
4) Ringing the sphere
An aluminum sphere is heated to 100 °C. The diameter of the sphere, at this
temperature, is 25.4508 mm. The sphere rests on top of a 20 g copper ring which has
been cooled to 0 °C. At this temperature the inside diameter of the ring is 25.4 mm, so
the sphere cannot pass through the hole.
A long time later, when the system reaches equilibrium (when the temperature of the
sphere and ring are equal), the aluminum sphere is barely able to slip through the
copper ring. Assuming there’s no loss of heat to the surroundings, only conduction
between the sphere and ring, calculate the mass of the aluminum sphere.
Linear coefficients of expansion
–1 Cu = 431.8 nm K
–1
Al = 584.2 nm K
Specific Heats
–1 –1 Cu = 386 J kg K
–1 –1
Al = 900 J kg K [Sal] 5) Reality Check
A 1liter thinwalled stainless steel container is filled with helium gas at standard
temperature and pressure. The container sits on a hotplate so that temperature of the
container and gas can be controlled.
a) Not only the gas, but the container too, will expand when heated! Calculate the
maximum pressure of the gas in the container, and the temperature at which it occurs,
assuming ideal gas behaviour and linear thermal expansion of steel. What is the
volume of the container at this temperature? The equation for thermal expansion in one
dimension is: ∆x = α x ∆T
In this equation ∆x is the amount of expansion, α is the expansion coefficient, x is the
initial length and ∆T is the change in temperature.
–5 1 α (steel) = 1.1x10 K 2 b) Do the values you calculated in part (a) make sense? If not, explain possible reasons
why not. [Brian]
6) Chillin’
Sam likes to enjoy a tall glass of cold lemonade on a hot summer day. He would like to
know in advance how much ice he’ll need to put into a pitcher of lemonade in order to
lower its temperature by the desired amount.
a) Calculate the volume of water Sam needs to freeze in order to make ice enough to
lower the temperature of a litre of roomtemperature lemonade by 10 ° C. The
temperature of the freezer Sam uses to make the ice is –10 °C. Assume the lemonade is
initially at room temperature (20 °C), and is basically water, and that all the ice has
melted before the temperature of the lemonade is measured.
b) Now try it! As an experiment, measure the change in temperature of a litre of water
caused by a small ice cube. You may want to use a vacuum flask (e.g., Thermos™
bottle) to do this experiment, so that heat energy is conserved. Compare your
measurements to theory, using the equations from part (a). Comment on your results.
density ρ (water) = 998 kg m
ρ (ice) = 917 kg m –3 –3 specific heat C(water) = 4190 J kg
C(ice) = 2220 J kg
latent heat –1 –1 –1 K –1 K Hf(ice) = 333000 J kg –1 [Brian] Remember to check the POPTOR webpage for hints and any necessary corrections!
www.physics.utoronto.ca/~poptor 3 20002001 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 3: Thermodynamics
1) Bye, metallic bar!
1) If the change in temperature is ∆T then the final lengths of the two strips are
lFe = l(1 + λ Fe ∆T ) lCu = l(1 + λ Cu ∆T )
Since they are bonded together, they have to bend in order to have different length. If
they are assumed as an arc in which has a (subtended) angle of α , the angle of
deflection from a straight line, measured at either end, would be α/2.
If the radius of the curvature, measured to the midpoint between the two strips, is R
then we have (α in radians):
d lFe = R + α 2
d lCu = R − α 2
Combining all the formulas above, and eliminating R from them, we obtain the angle of
deflection in radians to be
α /2 = l∆T (λ Fe − λ Cu )
2d [Yaser] 2) Maxing Maxwell
The most probable velocity is the one for which
there is a p eak in the velocity distribution curve.
To find the maximum of this distribution we
find the local maximum: take its derivative
with respect to velocity and set that equal to 0.
Solving for the velocity we get,
1/2 v = (2RT/M) Oxygen gas at a temperature of T = 300 Kelvin
–1
with a molar mass of 0.0320 kg mol has a most probable velocity of,
v ~ 395 m s –1 1 A plot of the velocity distribution of oxygen at a temperature T = 300 Kelvin is shown
above.
[Sal]
3) A pressing concern
If the initial pressure inside the cylinder is P i, and the piston is at its equilibrium
position, we must have a balance of forces:
Po A + mg = Pi A
When we put an additional mass M on top of the piston then we have
Po A + (m + M ) g = Pf A
where Pf is the final pressure.
Since the temperature inside the cylinder is constant and equal to room temperature,
the product PV of pressure and volume must finally turn out the same before and after
putting the weight. Knowing that the height is reduced to half its initial value, we get
h
PV = Pi ⋅ h A = Pf ⋅ A
2 ⇒ Pf = 2 Pi Substituting Pf in above formulas we get the weight needed to be
Mg = Po A + mg [Yaser] 4) Ringing the sphere
The relative change (per degree Celsius) in the diameter of the copper ring and
aluminum sphere is associated with their respective linear coefficients of expansion.
After some time the copper ring, initially at a temperature of 0 °C, will reach
equilibrium with the aluminum sphere at some mutual temperature Teq. The changes
in temperature for each will result in changes ∆D in diameter D:
∆D = (Tf –Ti) D α
where Tf and Ti are the final and initial temperatures of the copper ring or aluminum
sphere. α is the linear coefficient of expansion of each metal.
We are told that when the system reaches equilibrium the diameter of the Al sphere
equals the inner diameter of the Cu ring.
DCu = DAl
Therefore,
DCu(initial) + ∆DCu = DAl (initial) + ∆DAl
Solving for the shared temperature Tf, we find that the system reaches equilibrium at a
temperature Tf = 50.38 °C. 2 We are also given that the system reaches equilibrium with no loss of energy to the
surroundings. This means that heat is only transferred between the metals — the heat
lost by the aluminum sphere as it cools is absorbed by the copper ring as it warms.
Therefore,
CCu MCu (∆TCu) = CAl MAl (∆TAl)
Where CCu , CAl and Mcu , MAl are the specific heats and masses of the two metals.
We are given the mass of the copper ring, and the changes in temperature of both
metals were calculated above. We can therefore solve the equation above for the mass
3
of the aluminum sphere: MAl = 8.71 x 10 kg.
[Sal]
5) Reality Check
It is very important to f ully understand where the equations we use come from.
Sometimes students are tempted to use the “equation hunting” method when solving
problems. This is a very dangerous practice since many of the equations we use (such
as those found in textbooks) can be:
(1)
(2)
(3)
(4) A limited case of a more general equation
An approximation to a more complicated expression
Derived under certain assumptions
All of the above The “equation hunting” method always generates interesting solutions, but not always
correct ones – in fact, some pretty crazy answers can result as you will soon find out!
Always ask yourself if the answers you calculate make sense.
Known quantities converted to SI units where appropriate:
Variable Name Variable Definition Variable Value To Initial temperature of gas 300 K Po Initial pressure of gas 101325 Pa Vo Initial volume of gas 1.00 × 10 −3 m 3 α Linear expansion coefficient for steel 1.10 × 10 −5 K1 Unknown quantities:
Variable Name Variable Definition T Temperature of gas at maximum pressure P Maximum pressure of gas n Number of moles of gas in the container
3 In this question we will assume ideal gas behavior (the “ideal” in ideal gas should raise
some red flags here) and linear thermal expansion of an object in one dimension.
• Generate an expression for the volume of the container as a function of temperature:
change of length in one dimension as a function of the change in temperature
∆x = α xo ∆T
total length in one dimension as a function of the change in temperature [ x = xo + ∆x = xo (1 + α∆T ) = xo 1 + α (T − T0 ) (1) definition of volume
V = x3 (2) substitute (1) into (2): [ 3 = Vo[1 + α(T − To )] 3 3
V = x 3 = xo 1 + α(T − To ) (3) where
3
Vo = xo • Generate an expression for the pressure inside the container as a function of
temperature:
ideal gas law
P= nRT
V (4) substitute (3) into (4)
P=
• nRT [ Vo 1 + α (T − To ) (5) 3 Solve for the temperature of the gas at maximum pressure:
at maximum pressure dP
=0
dT [
[ nR 1 − α (2T + To )
dP
nR
3αnRT
=
−
=
3
4
4
dT V 1 + α (T − T )
Vo 1 + α (T − To )
Vo 1 + α (T − To )
o
o [ [ solve for T: [=0
4
Vo [1 + α (T − To )] nR 1 − α (2T + To ) 4 T=
• 1 1
1 1
− To = − 300 = 4.53 × 10 4 K
−5 2 (1.10 × 10 )
2 α (6) Calculate the number of moles of gas in the container (this is a constant):
PV
(101325)(1.00 × 10 −3 )
n= o o =
= 0.0406 mol
RTo
(8.31451)(300) • Calculate the maximum pressure – substitute (6) into (5):
P= nRT [ =
3 (0.0406)(8.31451)(4.53 × 10 4 ) )] 3 3 = (1.00 × 10−3 )[1 + (1.10 × 10−5 )(4.53 × 104 − 300)] 3 Vo 1 + α(T − To ) [ ( (1.00 × 10 −3 ) 1 + (1.10 × 10 −5 ) 4.53 × 10 4 − 300 P = 4.58 × 106 Pa = 45.2 atm
• Calculate the volume at maximum pressure – from (3) [ V = Vo 1 + α (T − To ) V = 0.00334 m 3 = 3.34 L
So the temperature of the gas and container would be roughly 40,000K, pressure is
around 40 atmospheres and the volume of the container has tripled. Note that the
melting point of steel is about 2,000K. Clearly the container would either melt or rupture
before reaching the temperature, pressure and volume calculated above. Note, too, that
for large changes in temperature, it is usually incorrect to assume a linear change in
length.
[Brian]
6) Chillin’
The physics: the ice absorbs heat (thermal energy) from the lemonade until the
temperatures of the ice and lemonade become equal.
There are two types of energy involved in solving this problem: (1) t hermalmotion
energy and (2) crystallization energy. The first is the kinetic energy associated with the
movement of molecules or atoms in liquids, solids and gases. Crystallization energy is
the bond energy required to take apart a crystalline solid molecule by molecule or atom
by atom – in this case, it is the energy required to convert ice (a solid) to water at
constant temperature. Crystalline energy is also called latent heat of melting. 5 Known quantities converted to SI units where appropriate:
Variable Name
C(water ) Variable Definition Variable Value Specific heat of water 4190 J/kg/K Specific heat of ice 2220 J/kg/K H f ( H 2 0) Heat of formation for H20 333000 J/kg V (water ) Volume of lemonade ρ(water ) Density of water T (water ) Initial temperature of
lemonade 293 K T (ice) Initial temperature of ice 263 K Final temperature
lemonade and “ice” 283 K C(ice) Tf 1.00 × 10 −3 m 3
998 kg/m3 of Unknown quantities:
Variable Name
m(water )
m(ice)
V Variable Definition
Mass of lemonade
Mass of ice
Volume of water needed
to freeze in order to make
ice Solution:
• Calculate mass of lemonade:
m(water ) = ρ(water )V (water ) = (998)(1.00 × 10 −3 ) = 0.998 kg • Calculate mass of ice:
As with all thermodynamic problems we need to begin our solution with a
conservation of energy equation (energy cannot be created or destroyed). If we
consider the ice and lemonade as a closed system (that is, no heat can get in or out of
the container holding the lemonade and ice), then the change in internal energy of
the ice plus the change in internal energy of the lemonade must equal zero:
∆E(water ) + ∆E(ice ) = 0 • (1) The heat absorbed by the ice from the lemonade takes place in three steps:
6 (1) Heat is absorbed to take the ice from its initial temperature up to 0 °C
(273 K) (2) Heat is absorbed to convert the ice to water at constant temperature 0 °C
(this is the crystalline energy) (3) Heat is absorbed to raise the resulting icewater at 0 °C to its final
temperature from equation (1) m(ice)C(ice)(273 − T (ice)) + =0
m(water )C(water ) Tf − T (water ) + m(ice)H f ( H 2 0) + m(ice)C(water ) Tf − 273 ( ) ( ) solving for m(ice):
m(ice) = )
C(ice)(T (ice) − 273) − H f ( H 2 0) − C(water )(273 − Tf ) m(ice) =
• ( (0.998)( 4190)(283 − 293)
= 0.105 kg
(2220)(263 − 273) − ( 333000) + ( 4190)(273 − 283) m(water )C(water ) Tf − T (water ) Calculate the volume of water Sam needs to freeze in order to make enough ice to
lower the temperature of the lemonade by 10 degrees:
V= V= m(ice)
ρ(water ) (0.105)
= 1.05 × 10 −4 m 3 = 0.105 L .
(998) [Brian] 7 20002001 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 4: Optics and Waves
Due February 16, 2001 (revised date)
1) Sub hubbub
In our deepsea research submarine
POPSub there is a frosted (not clear) light
bulb of radius R. With you in it, the sub is
at a height H from the bottom of the sea,
examining life on the ocean floor. The light
bulb is at a height h from the square (L × L)
window which is d thick and made of a
glass with index of refraction n. h
d
H L a) Calculate the area at the bottom of the
sea which is illuminated.
b) As you look out, how big does the spot look? [Yaser] 2) “Out, out damned spot!”
Each time there’s a solar eclipse, a certain number of people suffer eye damage, or even
are blinded, because they look directly at the sun. It isn’t even necessarily safe at
maximum, when the sun is blocked. To be safe, one needs to use the sort of goggles
that prevent the same kind of eyedamage in welders, whose torches can be very very
bright; there are other special filters and glasses, but some you might think are good
aren’t actually enough, since, e.g., invisible infrared or ultraviolet may pass through and
cause damage .
When you think about using a magnifying glass, in the sun, to burn wood or set fire to
paper, it really isn’t very surprising your eyes are in danger. Direct sunlight is about 1
–2
kW m , at the earth — how about calculating the intensity of light focussed from the sun
using a lens?
Your eyeball is roughly the size of the circle you make with thumb and forefinger
placed tiptotip, as when you make an ‘OK’ sign. Make a guess, from that, of roughly
what the focal length is of the lenssystem of your eye, and let’s figure the numbers:
a) Estimate the size of your eye’s p upil when it’s at its smallest, and figure out the
power in watts that would shine through if you looked straight at the sun (don’t do this
1 by experimenting!). Now, using the focal length of the lens of your eye, determine the
size of the image of the sun formed at the back of your eye. Then find the intensity of
light that this gives.
It could be worse: if the sun gave the same intensity of light but was much smaller or
much farther away, the focussed image would be much smaller and the intensity much
higher. That’s because all the rays of sunlight would be practically parallel, and focus
nearly in the same place. But lasers give nearly parallel rays, so they will focus nearly to
a spot…
b) Consider a laserpointer with 5 mW total output and a beam about 1 mm across. It
doesn’t focus to a point, quite, because of diffraction. So, if all the light goes into your
eye, what is the focussed intensity at the back of your eye?
This value is about the maximum intensity for a Class IIIa laser. This can cause
temporary loss of vision, like a cameraflash going off in your face; beyond this power
eyedamage can result. [Robin]
Laser Safety: http://www.adm.uwaterloo.ca/infohs/lasermanual/documents/tblcont.html
Classifications: http://www.adm.uwaterloo.ca/infohs/lasermanual/documents/section8.html
3) Musical buildings
I used to enjoy watching fireworks on the 4th of July in Rochester, NY, where I went to
graduate school. I noticed something quite interesting, though: when the loud ‘bomb’
fireworks went off, making a noise like a cannon, the buildings would play music.
Reflected from several different
buildings, the noise sounded something
like the grating on a bridge when your car
drives over — a brief rough note of
music.
This was because several buildings in
Rochester had long vertical ribs in their
architecture, running the height of the
buildings, separating sets of windows.
The ‘bang’ sound would reflect off each
rib, and the reflections coming to me
would make a note like running a pencil
along a comb.
a) What frequency should you hear
reflected? Consider a building with ribspacing D at a distance L1 from the ‘bomb’ and
L2 from the spectator. You may need to know also the angles θ1 and θ 2 from the
building of these two paths. You may need to make L1 and L2 large, to simplify.
2 b) There’s a different ‘picture’ of how this works: that the single bang of a firework
‘bomb’ is a sound with many many different frequencies in it, rather than a single note
— something like white light, as opposed to coloured light. What frequency would you
expect to hear, standing in the same place as (a), if all the different wavelengths of
sound simply diffracted from the ‘grating’ made of buildingribs? [Robin]
4) A hightech diet, high in optical fiber...
The internet and most telephone communication depends most heavily not on copper
wire, nowadays, but on optical fiber . Rather than electrical pulses in wire, for signals,
optical fiber communications use laser pulses propagating in tiny flexible glass strands.
Think of the following model of a laser pulse propagating in an optical fiber. Take a
glass rod with a circular crosssection of radius r. Say that light is launched into the
fiber through the end, and that all rays lie within a cone of angle θ measured from the
axis of the rod. The index of refraction of the fiber is n.
a) If the input pulse is very shortduration, what is the duration of the pulse of light
coming out at the end of the fiber? Assume the length of the fiber to be L.
b) If the input light is a train (series) of pulses with duration τ, and timeseparation ∆t,
what should be the maximum length of the fiber in order that the output pulses are still
distinguishable. This is a real issue for communications, since we need to know the
fastest digital signal that can be sent.
c) If the peak intensity of a single input pulse is I, what is its output peak intensity.
[Yaser]
[check the POPTOR website for an example of sound in a tube doing the same thing as this...]
5) Wave goodbye
Hong and her brother Liang have a clothesline in their backyard. They mentioned to
me their experiments with their giant onestring guitar... If they hit the cable sharply
with a stick, what they see is two waves that run away from the impactplace, one in
either direction. They want to know why — can you explain for them?
What they see is called a travelling wave. The shape of the wave stays the same, as it
simply moves along the wire as if it were sliding. A travelling wave has the form:
f ( x , t) = f ( x + ct)
or
f ( x , t) = f ( x − ct)
If you look closely, you’ll see that the top one is a wave with shape f(x) travelling to the
right ( increasing x with time), and the lower one is the same shape, but travelling to the
left (decreasing x with time). In fact, almost any shape will work.
3 There are some very interesting properties of such travelling waves:
1) if you have a shape that works as a wave, then the same shape would work as a
wave if it were bigger or smaller by any factor a.
2) if you have two different shapes that work as travelling waves, then if you add the
two together, the sum will also work as a wave.
These are properties of what we call a linear system. They’re true, for instance, for
sound waves in air (except for explosions and such). If they weren’t true, we couldn’t
listen to music (think about that...).
a) Say that Hong and Liang suddenly put a ‘dent’ f(x) in the shape of the clothesline by
quickly striking with the stick. How do you figure out the two waves that run away in
either direction? Look for a trick or idea that uses properties (1) and (2) above to help
prove what happens; you don’t have to use much math.
b) Say that the clothesline is tied firmly at both ends. Then the position of the cable at
the end cannot move. What happens when a wave travels right to the end of the cable?
Try the same kind of ideas that work for (a).
c) Say that at one end the clothesline is tied to a ring that slides up and down an
upright pole; say that this slides perfectly without friction., and the ring has negligible
mass. This is a different boundary condition. What happens when a wave travels right to
the end of this cable? [Sal/Robin]
6) Ten yards for interference…
Many people have home theatres these days and although technology makes having a
home theatre relatively simple, in order to get the maximum effect one still has to take
into account the geometry of the setup. Consider a basic home theatre setup, with
distance 2d between the speakers, and a distance L from the speakers to the opposite
wall of the room. Assume the speakers are point sources, that each puts out exactly the
same singlefrequency pure sine wave (this must be a pretty boring movie, or a test of
the Emergency Measures Systems…) and that the walls do not reflect any sound.
a) Find a general expression giving the position(s) P of maximum intensity along the
opposite wall as a function of speaker spacing d and room length L. Give the position(s)
P in terms of distance y from the centreline between the speakers. Do not assume d<<
L.
b) If L = 5 m, d = 3 m and both speakers are producing an inphase 320 Hz sound
wave, calculate the position in one direction of the first three points of maximum
intensity along the opposite wall, as measured from the centreline between the
speakers. The speed of sound in air is 331 m/s. 4 HINT: This question involves solutions to an equation that can’t easily be written as a
tidy equation y = f(x), but more like 0 = f(y) + g(y) . You’re welcome to solve the
equations graphically — plotting, and then looking for roots. You’re also allowed to
solve by hand iteratively, or get a computer to solve the equation numerically.
[Brian] INFOBITS™ — Useful Bits of POPTOR Information
You don’t need this to solve question 5, but it’s interesting.
Waves satisfy a differential equation (an equation of derivatives) called the wave
equation :
∂ 2 f ( x , t)
∂x 2 1 ∂ 2 f ( x , t)
−2
=0
∂t 2
c It is actually a very simple equation — you just need to take the derivative twice. The
unusual notation is because these are partial derivatives. The p artial derviative with
respect to x means that you take the derivative with respect to the xvariable (totally
ignoring the tvariable). So, for instance,
∂ sin( x + 3t)
= cos( x + 3t)
∂x
but
∂ sin( x + 3t)
= 3 cos( x + 3t)
∂t (using the chain rule) You just pretend the other variable doesn’t even exist.
A travelling wave can always be described by f(x,t) = f(x ± ct), if f is any function that can
be differentiated properly. It gives the shape f(x) in space, travelling in the direction of
increasing x for f(x – ct) and decreasing x for f(x + ct). Remember to check the POPTOR webpage for hints and any possible corrections!
www.physics.utoronto.ca/~poptor 5 20002001 Physics Olympiad Preparation Program
— University of Toronto —
Solution Set 4: Optics and Waves
1) Sub hubbub
This question is supposed to be the easiest one in the set. If you want to consider the
size of the window and the light bulb, you’ll find the geometry a bit fussy without
teaching very much. The reasonable approximation is that the light bulb is tiny
compared to other sizes, like the distance from window and therefore its size should be
neglected. A thick window also can be solved as we do below (with an extra refraction
surface), but let’s take the window to be thin compared to other distances, like h and the
height H from the bottom of the sea.
The area at the bottom of the sea
which is illuminated, is square in
shape, each side is equal to an
unknown, X. The angles that the
rays traveling in the sub and sea
makes with the normal are α and γ .
These ray which hit the edge of the
window determine the size of the
illuminated area at the bottom of the
sea. X is equal to
X = L + 2( H + d)tan γ
From Snell’s law
sin γ = sin α
nw ⇒ tan γ = sin α
2
nw − sin 2 α where nw is the index of refraction of the water. From the geometry of the system we
have
sin α = L
L2 + 4 h 2 ⇒ tan γ = L
2
2
(nw − 1)L2 + 4nw h 2 1 Thus, 2( H + d)
X = L 1 + 2
2
(nw − 1)L2 + 4nw h 2 If you were in the sub you would see the illuminated area formed by the continuation
of the ray, which is in the sub. Therefore, the length of any side of the square is
X ′ = 2( H + d + h)tan α = L( H + d + h)
h [Yaser] 2) “Out, out damned spot!”
Most of the focussing comes from the curved shape of your eyeball at the cornea , and
the lens of the eye mostly ‘touches up’ the focussing. Makes little difference for this
question.
a) The focal plane of the eye is located approximately 22 mm behind the cornea or front
of the eye. The iris controls the size of the pupil and on a bright day a typical pupil size
is 2 mm in diameter.
Approximating the eye
size of
focal length of
sun
lens as a “thin lens”
eye (3 cm)
we can use Newton’s size of
distance to sun
image
form of the thin lens
equation (a similartriangles law, actually):
Hi / Ho = f / x
Where: Hi
Ho
f
x height of the image;
8
height of the object (radius of Sun ≅ 7 x 10 m)
focal length of the lens
11
distance of the object (distance SunEarth ≅ 1.5 x 10 m) The size of the image on the focal plane of the eye is;
–4 Hi = (Ho • f) / x ≅ 10 m
The area of the image on the focal plane is therefore;
A ≅ π • 10 –8 m 2 The pupil of the eye, approximately 2 mm in diameter under bright conditions will
allow into the eye approximately;
P 3 2 ≅ ( 3 • 10 W/m ) • (π • 10 –8 2 –3 m ) ≅ 3 • 10 W 2 This is about the power of a highschool HeNe laser. The resulting intensity in the focus
on the retina of the eye is:
I ≅ P/A ≅ 5 2 3 • 10 W/m About 300 times concentrated from direct sunlight. If your pupil has not closed down
–2
fast enough, this can be much higher — more like 1000 kW m .
b) In the problem above, the image size was determined by geometry — actually by the
angle the sun subtends at your eye, which is the angular size, responsible for parallax.
When you consider a laser with practically parallel light rays, the focus should be
essentially a point, except for the fact that there is diffraction of the light, which makes
the final focal spot slightly blurry.
The formula for this is:
d=2λf/a
where d is the focal spot diameter, λ is the wavelength, f is the focal length, and a is the
beam diameter at the lens. Often the quantity (f/a) is called the fnumber. For our case,
λ ~ 700 nm (red light), f = 3 cm, a = 1 mm; then d = 42 µm. The laser power of 5 mW all
6
–2
goes into this spot, for an average intensity of 3.6 × 10 W m .
[EXTRAS : How do you understand this formula for diffraction? Only an infinite plane
wave propagates without any change at all — but it extends infinitely! For any real
wave, there is a general relationship:
∆x ∆k ~ 1
where ∆ x is the uncertainty in the position, and ∆ k is the uncertainty in the wavevector,
related to the momentum; it’s direction is the direction of propagation. For a plane
wave, ∆x is infinite, so ∆k can be zero — you know exactly where the wave is going (but
not at all any special place where it is). When a plane wave passes through a slit, or the
pupil of your eye, the hole size sets a limit on ∆x, and so ∆k cannot be zero. The wave
doesn’t lose any momentum, going through the hole, so what happens is the direction of
k spreads out — the wave propagates not just in a straight line, but it diffracts from the
aperture. The same thing applies for the focal spot, if you trace backwards: light from a
tiny focal spot spreads out faster than from a big focal spot. So tiny focal spots go
together with smaller fnumbers.
tanθ = (a/2)/f
where θ is the halfangle of the spreading out of the light. Then the formula for d can be
written as
d • tanθ ~ 1
It makes sense, when you realize that d = ∆x (the width) and tanθ = ∆k (the spreading).]
[Sal & Robin]
3 3) Musical buildings θ2 a) The question is pretty easy if you let the distances be
large. Then the rays are parallel, and the solution is much
like the way you solve for interference in a thin layer, like
oil on water — you need to
examine
the
pathD
differences. θ1 The figure at left shows the
basic setup of angles for
incoming and outgoing
sound waves; I draw them
as rays here. The figure at
right
illustrates
the
pathlength differences: the straight lines across the path
mark where paths are equal, and the extra distance one
ray must travel is marked by the fat line (in red, if you
have a colour version of this). This extra pathlength is
from two parts: extra distance coming in, and extra
distance coming out from the buildinggrating. ∆l1 D D ∆l2 θ2 The figure at left shows the details for the incoming, and the
figure at right shows the detail for the outgoing distance.
The total extra pathlength is ∆l1 + ∆l2 :
∆l1
∆l 2
= sin θ1 and
= sin θ 2
D
D
∆l = ∆l1 + ∆l 2 = D ⋅ (sin θ1 + sin θ 2 ) θ1 At the speed of sound Cs, one reflection will be later than the
other by a time T:
T= ∆l D
=
⋅ (sin θ1 + sin θ 2 )
Cs Cs In fact, all reflections are separated in time by this — the ‘bang’ of the fireworks burst
comes to the listener as a series of pulses, and T is the p eriod of this repetitive signal.
The frequency is then:
ν= 1
Cs
=
T D ⋅ (sin θ1 + sin θ 2 ) 4 For a building with ribs about 2 m apart (roughly the size of a window), with Cs = 300
–1
m s , and θ1 = 22° and θ2 = 45°, this gives ν = 140 Hz. That’s a fairly low note, but not
as low as the famous 40 Hz organ note in Bach’s Tocatta and Fugue in Dminor which
isn’t actually present on many recordings…
It isn’t necessary to assume that the distances L1 and L2 are very large. If they aren’t,
and interesting thing happens: not all the periods T between notes are exactly the same,
but they increase or decrease as the ‘bang’ ripples off all the ribs of the building. (This is
not difficult to show — just redraw the diagrams above, with rays coming from a
point.) As a result, the note that you hear actually shifts a little while you’re listening.
The technical term for this is ‘frequency chirp’, like the whistle of a bird may rise or fall.
b) The fireworks ‘bang’ has many frequencies in it, so another way to imagine this
same problem is to compare it to white light hitting a diffraction grating and spreading
out — like a room light diffracting from a compact disc. In that case, it’s easy to find
what frequency you’ll hear, in a given place. It’s just the same wavelength formula as
for a diffraction grating:
nλ = D ⋅ (sin θ1 + sin θ 2 )
where n is the o rder of diffraction (check this out on a CD, and you’ll see several
rainbows in the colours from a pointsource light).
In terms of sound, the note you hear is usually determined by the lowest frequency, or
longest wavelenth — the fundamental. The other wavelengths are shorter; these higher
frequencies are then multiples of the fundamental, called the overtones. These overtones
determine the timbre or basic soundquality of an instrument, like flute vs. oboe.
For the fireworks, the fundamental frequency is:
C
Cs
ν= s =
λ D ⋅ (sin θ1 + sin θ 2 )
which is the same formula we found before. [Robin] 4) A hightech diet, high in optical fiber...
a) There are many paths light can take down the fiber. That ray of light which moves
in a straight line parallel to the axis of the fiber reaches the other end of the fiber earlier
that some ray which has an angle θ with respect to the axis, and which travels more in a
zigzag pattern to get to the end of the fiber. The time it takes for the straight ray to
travel a distance L along the fiber is
t1 = nL
c where c is the speed of light. The other ray has to travel a longer distance:
5 L cos θ
and it gets to the other end at a time
t2 = nL
c cos θ The duration of the pulse at the output of the fiber is usually measured as the width of
the pulse measured halfway down from the peak of the pulse — the fullwidth at halfmax (FWHM). To find
this exactly, one needs to
know how the input light
goes into the different
L
rays, but it will be roughly
Td
half of the time delay
between the first rays to
arrive and the last rays to
arrive:
t −t
nL 1 −1
τd = 2 1 = cos θ 2
2c
where τ d is the pulse duration. This effect is usually called modal dispersion. We
assumed in our calculation that the initial duration of the input pulse is much smaller
that the output pulse, i.e., much smaller than this spreading–out by different rays.
Otherwise the timestretching adds to the original pulse duration the way that two sides
of a rightangle triangle give the hypotenuse:
t τ = τp2 + τd2
b) Here we again assume that the initial pulse duration is much
smaller than the dispersion broadening. In order that we can
tell the pulses apart, after they broaden, they shouldn’t overlap
with each other much. Therefore, the duration of the output
pulses should be smaller than roughly half of the time
separation between them.
τ d ≤ ∆t 2 ⇒ L ≤ c∆t
n(1 cos θ − 1) c) The energy of a pulse is equal to the integral of the intensity
in terms of the time. If the pulse looks similar to a Gaussian pulse, we can approximate
the energy of the pulse by the product of the peak intensity and the pulse duration. If
the fiber is lossless, we expect the energy of the output pulse to be equal to the input
pulse. Therefore, 6 Iτ
in
out
out
I peak τ = I peak τ d ⇒ I peak =
∆t
in
out
where I peak = I and I peak is the peak intensity of the input and output pulses. We assume
that the length of the fiber is the maximum length such that the output pulses are
distinguishable.
[Yaser] 5) Wave goodbye
a) Let us assume
that the shape of
the “dent” after
the stick strikes
the clothesline is
given by f(x, t=0).
An example of such a “dent,” frozen in time, is shown in the figure above.
From the properties of waves given in the question, such a f(x, t=0) can be the
instantaneous result of the addition of two identically shaped pulses, each half the size of
the original “dent” — one pulse travelling to the left and the other to the right with the
same speed.
f(x, t=0) = 1/2 f1 (x)+ 1/2 f2 (x)
where f1 (x, t =0) = f(x, t= 0) and f1(x, t ) = f1(x +vt)
f2 (x, t =0) = f(x, t= 0) and f2(x, t ) = f2(x –vt) Once a solution, always a solution: the dent created instantaneously immediately
decomposes into these two travelling waves after the line is struck. This explains the
two waves you will see emerge, after impact, travelling in opposite directions.
b) The fixed ends of the clothesline impose some conditions on the evolution of the
waves. For all times t, the actual wave functions f1 , f2 must be constant at the fixed ends.
f1 (x = L1 , t ) = 0 [1] f2 (x = –L2 , t ) = 0 [2] where L 1 and –L2 are the positions of the fixed ends relative to the position of impact.
The following trick can satisfy conditions (1) and (2) while revealing something
familiar from experience.
For any wave f 2 ,
imagine a matching
wave which is a
negative version,
and travelling in
7 the opposite direction toward the fixed end in some imaginary extension of the wire
(see the figure below). There is no fixed point in this picture. As the wave and its
negative counterpart approach the spot which ought to be fixed, the waves will always
add up to zero — satisfying the condition for the fixed point!
What you observe
mathematically is the
two waves passing
each other and
leaving the tiedpoint fixed. Since
the point in the middle never moved, we could have nailed it down — it’s our fixed
end. So when a real wave goes into a fixed end, the resulting wave is the negative wave
coming back from the fixed point, just the solution you’d have from the doppleganger.
c) Using a similar trick as in part (b), we consider superimposing a waves f2 with a
positive clone coming the other way. This will ‘explain’ or model for us reflections at a
sliding ring on a pole. No longer do we have restrictions (1) and (2): at the sliding ring,
they’re replaced by the condition that there can be no vertical forces — only horizontal
ones (or the massless ring would immediately move in response and the vertical force
would vanish).
With only a h orizontal
force possible, the
tension in the string,
for the right solution,
must give a horizontal
force: so the string
itself
must
be
horizontal. That’s fine
— our doppleganger
wave now is always a
mirror image of the real
one, and therefore
symmetric. When these two add, the tangent to the string at the place of the pole is
always horizontal; this is true for any smooth wave adding with its mirror image.
The waves, as they add, rise to a height of twice the amplitude of each. So the ring at
the pole rises to twice the height of the wave that travels to it, and then the wave
coming back will be identical to the one going in.
[Sal] 8 6) Ten yards for interference…
The Physics: In this question, we do not assume the spacing 2d between the speakers is
very small when compared to the
sound wave
paths
distance L from the speakers to the back
wall. If the difference in path length
speaker 1
from speaker 1 to point P and speaker 2
y
to point P is an integral multiple of the
d
sound wavelength, there will be
big
center of constructive interference.
d speaker 2 screen
TV
L room
back
wall Speaker 1 to point P: d1 = Speaker 2 to point P: d2 = a) Path lengths from speaker 1 to point P
and speaker 2 to point P c an be
determined using the Pythagorean
theorem: (y − d)2 + L2
(y + d)2 + L2 Therefore, the difference in path length is:
y ≥ 0: δ = d2 − d1 = y ≤ 0: δ = d1 − d2 = (y + d)2 + L2 − (y − d)2 + L2
(y − d)2 + L2 − (y + d)2 + L2 For constructive interference, the difference in path length must be an integral multiple
of the wavelength (δ = mλ ):
y ≥ 0: (y + d)2 + L2 − (y − d)2 + L2 = mλ = m v
f (1) y ≤ 0: (y − d)2 + L2 − (y + d)2 + L2 = mλ = m v
f (2) where m = (0,1,2,K , ∞ )
In equations (1) and (2), m is the order of the interference (i.e., n=2 means they interfere
by being two waves out of phase, f is the frequency of the sound wave, v is the speed of
sound the medium and λ is the wavelength of sound in the medium.
b) We need to solve for y using equations (1) and (2) when L = 5 m, d = 3 m, f = 320 Hz, v =
331 m/s:
y ≥ 0:
y ≤ 0: 331
320
331
(y − 3)2 + 52 − (y + 3)2 + 52 = m 320 (y + 3)2 + 52 − (y − 3)2 + 52 =m 9 The first maximum occurs when m = 0 :
y ≥ 0: 331
(y + 3)2 + 52 − (y − 3)2 + 52 = (0) 320 = 0 y ≤ 0: 331
(y − 3)2 + 52 − (y + 3)2 + 52 = (0) 320 = 0 by inspection, both equation yield y = 0 at the first maximum. The second and third
maximum occur when m = 1:
y ≥ 0: 331 331
(y + 3)2 + 52 − (y − 3)2 + 52 = (1) 320 = 320 (3) y ≤ 0: 331 331
(y − 3)2 + 52 − (y + 3)2 + 52 = (1) 320 = 320 (4) Not everyone will find it easy to determine a regular closed form solution for y in
equation (3) and equation (4) (some did!). There are, however, a number of easy ways
to solve these equations:
• plug in values for y in the equation and iteratively converge to zero in on a solution • graph the left hand side and right hand side the equation to see where the curves
intersect • use a math program to solve the equation numerically. I have solved equation (3) and equation (4) numerically using a math program called
Maple. The answers I got are:
y ≥ 0: y = 1.02 m y ≤ 0: y = –1.02 m Therefore, the three points of maximum intensity closest to the centerline are:
y = {–1.02 m, 0 m, 1.02 m} [Brian] 10 20002001 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 5: Electricity and Magnetism
Due March 9, 2001 (revised date)
1) Bridging the gaps
Consider an insulated wire of length L wound on a metal cylinder. The insulation
breaks at some hidden place, and the wire inside touches the cylinder. Using a battery,
an ammeter and two variable resistors, figure out where along the wire is the break.
[Peter]
2) Roadkill this question
In the circuit drawn at right, all circuit elements
lie on the edges of a cube. All resistors have
identical resistance R , all capacitors have
identical capacitance C . The voltage between
points A and B is V. Figure out what charge is
produced on the capacitor next to B. [Yaser] B
A 3) Loopy circuits A
B C D A wire loop is laid out on a plane to make the closed circuit
shown at right. A uniform magnetic field B is oriented
perpendicular to this plane — sticking directly out of the
page. This magnetic field changes at the rate dB/dt.
What is the induced electromotive force in the wire? Find
the answer in terms of the area of each section, A, B, C, and
D. [Yaser] 4) Use the Force, Luke!
A mass spectrometer is a device that measures the mass of charged particles (called ions)
by steering particles with the same mass to the same point in space. Mass
spectrometers are analogous to prisms, for light: a prism spreads a beam of white light
1 into its component
colors
(or
B
into
page
wavelengths); in a
particle
detector
mass spectrometer
a beam of ions with
different masses is
spread
out
according to mass.
There are many
uses for mass
y
spectrometers such
as in radiocarbon
accelerator
dating, where mass
E
spectrometers
d
measure the ratio of
numbers of carbon
13
14
isotopes C and C. The figure at right shows the geometry of a simple mass spectrometer: A given ion is accelerated by an electric field to a specific energy — and hence
velocity. As the ion leaves the accelerator, it enters a region containing a uniform
magnetic field that is perpendicular to its direction of motion. As the ion moves
through the magnetic field, it experiences a centripetal force that causes its path to
curve along a circle. A detector set along the yaxis, in this figure, can determine the
intercept of the ionpath and that axis.
a) Derive an expression relating the energy, charge and mass of the ion to its location
(along the yaxis) on the detector. Assume nonrelativistic energies.
b) What minimum pathradius can this apparatus accommodate.
c) If B = 2 Tesla, what is the minimum measurable mass for a ion with charge q = +1 e
(where e is the magnitude of charge on an electron), accelerated to an energy 25 eV?
What is the maximum mass? Do you think there may be any difficulty in trying to
detect extremely heavy ions? If so, why? Note: an electron volt (eV) is another unit for
energy. [Brian]
v
vv
F = qv × B is the force experienced by a charged particle moving through a
HINTS:
magnetic field. In this equation, q is the charge of the particle, v is the velocity of the
particle and B is the magnetic field. The righthand rule for cross products may be
useful in determining which way the particle curves. 2 5) A ringing charge
A thin ring of radius R
positioned in the xy plane has a
positive linear charge density + ρ
[C/m] on the top half and an
equalmagnitude negative linear
charge density – ρ [C/m] on the
bottom half.
Derive an
expression for the magnitude and
direction of the electric field at
point P , a distance d away from
the center of the ring. y + + + + + +
+ + x R P –
– –
– z d –
– –
– – HINT: what is contribution to the
electric field at point P from any small segment of the ring? [Brian]
6) Scoping out electrical charge
Two identical spheres of mass m are identically charged, then each suspended from the
same point by two nonconducting threads. Take each thread to have zero mass and
fixed length L. The electrostatic forces acting on each object will act to push them apart.
THE THEORY
a) Derive an expression for the charge q on the spheres, found from their mass, the
length of strings, and the angle θ between the threads.
THE EXPERIMENT
You will need the following:
1) 2 pieces of light string or sewing thread, and sewing needle
2) Styrofoam™ (or similar) packing material — packing ‘peanuts’ or blocks
3) utility knife or paring knife or razor blade
4) A protractor or ruler
5) A bit of tape, or pushpin
Procedure:
From the Styrofoam block cut out two cubes and try to cut the corners off to make
something nearly spherical. Try fairly small cubes — about a centimeter or less. Using
the sewing needle and thread, thread equal lengths of thread through the centers of the
spheres. Using the tape or pushpin, attach the free ends of the thread together and 3 suspend the whole thing freely — I tied the two ends together and then taped the end to
the edge of my bookshelf.
Charge the spheres up — I rubbed mine against my hair, but you could rub a balloon
against a sweater and then touch it for a few seconds to the styrofoam. The suspended
charged balls will now hang with some space between them.
b) Now use the results of your measurements, and part (a) above, to determine the
charge on the Styrofoam spheres. [ Brian] INFOBITS™ — Useful Bits of POPTOR Information
Remember to check the POPTOR webpage for hints and any necessary corrections!
www.physics.utoronto.ca/~poptor 4 20002001 Physics Olympiad Preparation Program
— University of Toronto —
Problem Set 6: AC Circuits and Electronics
Due April 6, 2001 (revised date)
1) Inducting Henry
A circuit contains an inductor L 1 = 70 mH in series with
a resistor R1 = 50 Ω, both of which are in parallel with a
R2 = 350 Ω resistor and a 90 V battery. The switch is
closed for 110 µs and then opened. L1 R1 R2 What is the value of the current in both resistors at the
moment the switch is opened at 110 µs? [Sal] 90 V
+ 2) Inducing Resistance
a) Graph the current that would flow from a power
supply in the circuit at right (across terminals marked
with circles) if the following voltage were applied:
0
V(t) = Vo C t<0
t≥0 b) What condition on L, R, and C will make the
impedance seen by the power supply one which is
purely resistive? (You may want to see the POPTOR
Primer™ at the back of this set). [Robin & Cambridge
Tripos] R L R 3) Rectifying a situation
In its simplest description, a diode is an electrical component that permits current to flow
in only one direction. It can be
D
used as a r ectifier t o change
alternating current (AC) to direct
C
current (DC).
RL
upply
a) In the circuit shown at right, a
stepdown transformer converts
110V AC to 6.3V AC. D is a diode Transformer 1 and the load resistor R L is 100 Ω. If the frequency of the AC power supply is 60 Hz, plot
the voltage across R L versus time.
b) In order to get a smoother
output voltage, capacitors can be
used as shown in the circuit at
right. If Cs is 330 µF, plot the
voltage across RL for three values
of R L, so that D1
AC
Supply Cs RL Transformer RLCs infinitely large
RLCs = 1/6 sec
RLCs = 1/60 s
[Yaser] 4) Zeners and the art of electronics maintenance
A Z ener diode maintains a more or less constant
voltage across it in the reverse direction independent
of current passing through it for a wide range of
currents. This property may be used to stabilize a
power supply against variations in input supply
voltage and in load current drawn.
Assume that there is a Zener diode, which the IV
curve is as below. This diode is placed in the circuit,
which is shown. Transformer gives an output voltage
reduced from the AC supply by a factor of 11.
Rectifier is used to change
the AC voltage to DC.
a) If Rs is 100 Ω , for RL =
100 Ω , what is the voltage
across it if the voltage of
AC supply is varied from
85 volts to 130 volts? C
upply I 5V
V 0.7V Vs
IL Rectifier Rs RL Iz b) Assume that the voltage of AC supply is 110 volts. What is the voltage across RL if
we vary its resistance from 50 Ω to 150 Ω?
[Yaser] 2 VL 5) Active courage
a) By replacing an opamp with its ideal model (see the POPTOR Primer™ below),
derive an expression for Vout as a function of Vin and use this expression to calculate the
V
ratio out .
Vin
V
b) What is the limit of the equation out you derived in part (a) as the gain A
Vin
approaches infinity? What effect does the opamp have on the output signal with
respect to the input signal?
c) Is the equation in part (b) a good approximation to the equation in part (a) if the gain
A is equal to 10,000? (less than 1% is good)
d) Why doesn’t the opamp amplify the input voltage by the opamp gain A?
e) BONUS (Worth an insane number of marks): When the opamp gain A is very large,
what does the voltage on the negative terminal V 1 become? What happens to the
voltage V1 when V2 is nonzero (not grounded)?
[The result of this bonus question illustrates the real usefulness of opamps: opamps in
negative feedback mode allow one to arbitrarily set the voltage of any point in a circuit
without sinking current (i.e., with infinite effective resistance). This is the basis for circuits
such as the Miller Integrator.]
[Brian] 6) Freaky Filters for “Phreaky” Physicists
Various types of frequency filter appear in many devices. Here are some examples:
Electrical – radio antenna
Mechanical – automotive shock absorbers
Optical – sunglasses
Acoustical – earplugs
Data – NetNanny
This problem will concentrate on understanding how simple firstorder electrical filters
work. But before getting into solving the actual problem, you’ll need to understand the
useful concept of impedance: See the POPTOR Primer™ below, for a briefing, if you
don’t know about impedances very well.
For the two filters on the next page, answer the following questions. 3 jwC R
Vout
Vin +
– jwC Vout
Vin +
– R Filter A Filter B a) Derive expressions for Vout
(these should complex functions of ω).
Vin b) From part (a) produce expressions for the magnitude and phase of
function of frequency. Vout
as a
Vin Vout
when ω = 0 and as ω approaches infinity
Vin
(HINT: you may want to use l’Hôpital’s Rule)?
c) What is the magnitude and phase of d) Twoway speakers usually have a tweeter for highfrequency sounds and a woofer for
low to midfrequency sounds. The reason why twoway speakers are used is because a
single speaker cannot efficiently produce both high and low frequency sounds (FYI:
speakers act like filters too!). The tweeter and woofer each have a separate filter that
filters the incoming signal from the amplifier. If you were building your own twoway
speaker, which filter (A or B) would you use on the tweeter and which filter would you
use on the woofer? Since the magnitude of the filter responses do not cut off sharply,
the tweeter and woofer will have some overlap in the sound frequencies they produce.
Use the equations for magnitude developed in part (b) to calculate the c rossover
frequency for the two filters if R = 10,000 Ω and C = 0.02 µF.
BONUS (Worth a reasonable number of marks): What effect will the phase have on the
music played through our twoway speaker?
[Brian] Remember to check the POPTOR webpage for hints and any necessary corrections!
www.physics.utoronto.ca/~poptor 4 POPTOR Primer™ in AC Electronics
Active Circuits
For many of you this may be your first encounter with an active circuit element. Unlike
resistors, capacitors, inductors and diodes which are passive circuit elements, active
components like opamps and transistors have the ability to put energy into a circuit in
a controlled manner based on electrical input from the circuit itself!
Opamp stands for “operational amplifier” and is one
of the simplest of all the active elements. The figure at V 1
–
right shows the circuit symbol for an opamp. An opVout
amp is a threeterminal device that amplifies the
difference in voltages between the “+” and “–” input V 2
+
terminals by a factor A and outputs the result (“A” is
called the g ain or amplification factor). The gain for a
typical opamp is usually between 10,000 and 1,000,000. Electrical characteristics of a
“reallife” opamp are very complex. However, for most applications it is more than
reasonable to represent the opamp by a simple model.
The circuit at left shows a model
V1
–
Vout
of an ideal opamp – that is, if we
could build a perfect opamp, this
+ A(V2 – V1 )
model would completely describe
–
its electrical response. Circuits
V2
containing opamps can be
+
analyzed using standard circuit
techniques (such as Kirchoff’s
Voltage and Current Laws and Ohm’s Law) by replacing all the opamp symbols with
its model.
R2
The circuit on the right shows an
opamp in a n egative feedback
R1
configuration.
All negative
–
feedback means is the output is fed
V1
back into the “–” input terminal; V
+
Vout
in
–
in this case, the negative feedback
V2
occurs through resistor R2 .
+
If you go on to take electronics
courses at university or college,
you will learn more about negative
feedback. In the meantime, for a motivating explanation, negative feedback is used to
5 eliminate nonlinearity in circuit response and increase stability – but you do not need
to know this in order to solve this question!
AC Impedance
Ohm’s Law V = IR relates the voltage V across a resistor R to the current I through the
resistor. But circuits can include capacitors and coils in addition to resistors. How then
can we analyze such circuits? Is there an Ohm’s Law for capacitors & coils?
The answer is “yes” – it is the impedance version of Ohm’s Law, V = IZ, where Z is called
the impedance (boldface denotes a complex number, here). The impedance Z can be
thought of as a complexvalued “resistance” having the form a + jb where a is the “real”
part and b is the “imaginary” part [1]. Don’t get scared of complexvalued impedances.
Complex numbers are only used to simplify the bookkeeping when determining things
like magnitude and phase as you will soon find out [2]. The great thing about
impedances is that the familiar techniques used to analyze DC circuits carry over!
When we want to extract a physically meaningful number from the impedance version
of Ohm’s Law, we just have to take the magnitude and phase of the complex
expression. There are a few things about impedances you need to know about: The
phase of a positive imaginary impedance is 90° and the phase of negative imaginary
impedance is –90°. Similarly, the phase of a positive real impedance is 0° and the phase
of a negative real impedance is 180°. The phase of an impedance with real and complex
parts can be anywhere between 0° and 360°.
Experiments show that when a sinusoidal (sinewave) voltage is applied across a
capacitor, the current through the capacitor is 90° out of phase with respect to the
applied voltage and increases proportionally with frequency. Similar experiments show
that when a sinusoidal voltage is applied across an inductor (coil), the current through
the coil is –90° out of phase with respect the applied voltage and decreases inversely with
frequency. (N.B.: the frequency of the applied voltage and current are the same for both
the capacitor and coil). Without proof (or you can take this to be an experimental result
−j
[3]), the impedance of a capacitor is jϖ C and the impedance of a coil is
where C is
ϖL
the capacitance, L is the inductance and ω is the frequency.
1. In circuit analysis the engineer’s “ j” is used to denote √(–1), instead of using the mathematician’s “i”
which can be confused with current.
2. In principle, we can solve these problems without complex numbers. However, doing so would
induce a really bad case of trigonometric diarrhea.
3. The impedances of a capacitor and coil can be derived from Maxwell’s equations. However,
Maxwell’s equations themselves cannot be derived, but are based upon empirical results. 6 20012002 Physics Olympiad Preparation Program
– University of Toronto – Problem Set 1: General
Due 9 November 2001
Since this is the first problem set, a few comments before starting:
• If you want to know what kind of physics you might need to solve these problems, look at
the Physics Olympiad syllabus at http://www.jyu.fi/ipho/syllabus.html. You don’t need
to know the syllabus by heart to do the problems, but you do need to be able to
recognize what you need to know so you can look it up.
• Don’t forget to look at the information given in the POPBits™ section at the end of the
problem set, it sometimes has information helpful or necessary for particular problems.
• Pay attention to words like “estimate” or “about”. They indicate that the expected answer
is not exact because either the input data is not precisely known or because
approximations or simplifying assumptions are necessary. Much of real physics is
learning how to turn insoluble exact problems into soluble approximations.
• “Nothing ventured, nothing gained”: Whether you finish a problem or not, please make
sure your reasoning and analysis are clear. If you write down nothing, it is easy for us to
mark – we just give you zero – but pretty boring. Your basic ideas may be right even if
you make a mistake or get stuck.
• Now, on with the show! 1) Let the sun shine in!
Hold a small, flat, square mirror toward the sun and send the reflected light onto a wall. Describe
how the shape of the bright spot on wall changes as you move the mirror closer to and farther
from the wall. What information about the sun can you obtain with this experiment? Write us
your measured value and its error.
(Hints: I made my mirror by masking a small round mirror with masking tape and paper so
that only a 0.5 cm square was exposed; this way you can also easily experiment with different
size or shape mirrors by just changing the tape. The mirror does not have to be an exact square,
but it does have to be flat, so don’t use a mirror which magnifies or reduces, as do many round
makeup or hand mirrors. If it is more convenient, you can leave the mirror fixed and look at the
reflection on something like a piece of flat cardboard which you move closer to and farther from
the mirror.)
[Yaser] Page 1 of 4 2) PingPong!
A ball will bounce back from a wall with the same speed it hits, if the collision is elastic. What
happens if the ball (which is moving with a speed v) hits a “wall” (e.g. a table tennis paddle) that
is moving (with a speed u) toward the ball?
(a) What is the velocity of the ball after it bounces back from the wall?
(Hint: Think of a frame in which the speed is the same before and after the bounce.)
(b) What is the change in the kinetic energy of the ball if its mass is m?
(c) Try to show that your answer to part (b) is equal to the work done by the wall on the ball.
(Hint: Assuming the duration of the collision is ∆T, find the force exerted on the ball and
its displacement.)
[Yaser]
3) Don’t break the window!
You have a ball of mass m attached to a massless string. The maximum tension the string can
handle before breaking is T. (What we are actually talking about is a yoyo that has been left
outside until its string starts to rot, but we have translated this into “textbook physics”.)
(a) If you hold one end of the string at a height h above the ground, what is the maximum
length the string can have so that you can still swing the ball in circles around you without
the ball touching the ground or the string breaking?
(b) If you do swing it around in circles slightly too hard and the string breaks, for what length
string will the ball fly the longest horizontal distance before hitting the ground?
[Alex]
4) Cowabunga!
First a well known joke: Farmer Smith was not satisfied with the yield of his milk cows, so he
decided to called in an animal psychologist, an engineer and a physicist to try to improve matters.
All three inspected the farm and the cows and made their recommendations. The animal
psychologist went first, “If you paint the milking shed green the cows will be happier and happy
cows will give more milk.” Then came the turn of the engineer. “If you narrow the milking stalls
by 10 centimeters you will be able to add an extra stall and thus be able to milk an extra cow in
the same time.” Farmer Smith was very happy so far, now it came to the turn of the physicist.
She got out a black board, drew a circle and said: “First, we assume a spherical cow.”
(a) Estimate the capacitance of a cow.
(Hint: To help any cityslickers, I’ll point out that a typical runofthemill Holstein1 dairy
cow weighs about 700kg and is more or less neutrally buoyant in water. )
(b) Estimate the resistance of a cow.
(Hints: People and cows have similar resistivity.
Sometimes different approximations are necessary for different parts of a problem.)
(c) When a cow walks across a carpet in winter, it can easily charge itself up to 10KV, and it
can get a nasty shock if its nose touches a doorknob. Estimate the peak power in such a
shock.
(Hint: The total resistance of a mammal in an electrical circuit is normally dominated by
1 http://www.holstein.ca/ Page 2 of 4 skin/contact resistance which can be quite variable, but cows tend to lick their noses
which keeps the skin conductivity up, so let’s ignore the skin/contact resistance.)
[David]
5) Symmetry Cubed!
Here’s is a great project for your art class: an electrified hanging mobile made from wire models
of the 5 platonic solids.
Consider a wire tetrahedron where each edge has a resistance R.
(a) What is the resistance between any two corners?
B
A
What if we replace the tetrahedron with a cube?
(b) What is the resistance between two opposite corners (e.g. AD)?
(c) What is the resistance between two diagonally opposite corners on the
same side (e.g. AC)?
(d) What is the resistance between two adjacent corners (e.g. AB)?
D
C
We’ll leave calculating the resistance of the various vertices of the octahedron,
the dodecahedron, and the icosahedron for your own amusement.
[Alex]
6) Is it hot enough for you?
The Mars Pathfinder mission’s Sojourner rover1 carried 3 small radioactive plutonium heaters to
keep its electronics warm. How warm would your radioactivity keep you?
The two largest sources of natural radioactivity in your body are the radioactive isotopes
Carbon14 and Potassium40. (Carbon14 is continually produced in the earth’s atmosphere by
cosmic rays, while the Potassium40 is part of the primordial composition of the solar system. )
About 1012 of the carbon in your body is Carbon14, and 0.0117% by weight of all potassium is
Potassium40. The decay of a Carbon14 atom releases an energy of 0.16×106eV, while the
decay of a Potassium40 atom releases 1.3×106eV. Both these decays are beta decays, so on
average about half the energy released in a decay is carried out of your body by a neutrino, but
the other half of the energy is deposited in your body as ionizing radiation. A typical human is
about 19% carbon by weight, and about 0.3% potassium,
(a) What is the total activity (i.e. decays per second) of the Carbon and Potassium in your
body?
(b) What is the total power (in Watts) deposited in your body by these decays?
(c) If you were put into suspended animation and fired into interstellar space, what would
your equilibrium surface temperature be if your internal Carbon14 and Potassium40
were the only source of heat keeping your body warm?
[David] 1 http://mpfwww.jpl.nasa.gov/MPF/index1.html Page 3 of 4 POPBits™ – Possibly useful bits of information
Constants and units1,2
astronomical unit (mean earthsun distance) electron volt
Newtonian gravitational constant
solar luminosity
speed of light in vacuum
standard acceleration of gravity
at the earth’s surface
StephanBoltzmann radiation constant
tropical year (2001) 149 597 870 660±20 m
(1.66053873±0.00000013)×1027 kg
(1.602176462±0.000000063)×1019 C eV
GN
LÍ atomic mass unit: (mass
atom)/12
elementary (i.e. electron) charge au
u
e 12C (1.602176462±0.000000063)×1019 J c
g σ
yr (6.673±0.010)×1011 m3/kg/s2
(3.846 ± 0.008)×1026W
299 792 458 m/s
9.80665 m/s2
(5.670400±0.000040)×108 W/m2/K4
31556925.2 s Radioactive halflives of some isotopes3
14C (Carbon14)
(Potassium40)
238Pu (Plutonium238)
40K 5730±40 years
1.277±0.001 × 109 years
87.7±0.1 years Resistivity of human tissue4
blood
bones
fat
muscle
nerve 1.7 Ω·m
160 Ω·m
27 Ω·m
7.0 Ω·m
2.5 Ω·m Great excuse for a party
Birthday of Marie Curie5 (1867) and Lise Meitner6 (1878) November 7 1 http://physics.nist.gov/cuu/Constants/index.html
http://pdg.lbl.gov/2000/contents_sports.html
3
http://ie.lbl.gov/education/isotopes.htm
4
http://www.acusd.edu/~mmorse/BMES2000.shtml
5
http://www.mariecurie.org/mariecurie
6
http://www.users.bigpond.com/Sinclair/fission/LiseMeitner.html
2 Page 4 of 4 20012002 Physics Olympiad Preparation Program
– University of Toronto – Solution Set 1: General
1) Let the sun shine in!
Let’s say the mirror is 0.5*0.5 cm2 square. It is important that your mirror shouldn’t be round to
observe the effect. When the wall is close to the mirror the bright spot has the same shape as the
mirror, in this case square. When you go farther and farther it changes gradually to the circle.
This effect comes from the fact that the rays coming from the sun are not exactly parallel due to
the size of the sun.
We can solve this question by considering that the square shape mirror consists of
infinitely large number of very tiny mirrors. Each tiny mirror makes a circular image of the sun
on the wall. If the distance between wall and the mirror is small all these small circular images
will form together the shape of the mirror. At large distances, these circular images are big and
they all overlap with each other form a big circle. By geometry, the ratio between the radius of
the image to the distance between the wall and the mirror is equal to the ratio between the radius
of the sun and the distance of sun to earth. For 0.5*0.5 cm2 mirror, up to about 50 cm the spot on
the wall looks square. As we go farther the spot changes to circle and at about 1.5 meter away,
the circle has the radius of about 0.75 cm. Therefore the ratio between diameter of the sun and its
distance to the earth is about 0.01.
The distance x where the image shape starts changing from square to the circle, is about where
you get the smallest image and by geometry
x
Distance − between − sun − and − earth
=
Diameter − of − the − mirror
Diameter − of − the − sun
which is about half a meter in this case. (Right) At small distance between the mirror and the wall in which the circles form the square .
(Left) At large distance between the mirror and the wall in which the circles overlap and the
whole image looks like a circle. Page 1 of 12 2) PingPong!
(a) A table tennis paddle is much heavier than the pingpong ball. Therefore we can assume
that it’s mass is infinity compared to the ball. Therefore, in the wall’s frame (or that of a
table tennis paddle!), the velocity of the ball is the same before and after it hits the wall,
but in the opposite direction. The velocity of the ball in this frame is v+u and so it
bounces back with v+u relative to the wall. Therefore its velocity relative to the ground is
v+2u.
(b) The change in the kinetic energy is
1
1
∆K = K f − Ki = m(v + 2 u) 2 − mv 2 = 2 mu( u + v )
2
2
where m is mass of the ball and K is the kinetic energy.
(c) The change in the linear momentum of the ball is
∆P = Pf  Pi = m(v + 2u)  (mv) = 2m(v + u)
Therefore if the collision time is ∆T, then the force is
∆P 2 m( u + v )
F=
=
∆T
∆T
The distance that the ball moves during the collision is equal to the distance that the wall
moves which is u∆T . Therefore the work done by the wall on the ball is
W = Fu∆T = 2 mu( u + v )
which is equal to the change in the kinetic energy of the ball as we expect.
3) Don’t break the window! r L h string d Page 2 of 12 FT
Fc
Fg The faster you spin the ball around you, the higher it will go, but the more stress we put on the
string. Looking at the above diagrams, we see that
h − d Fg
mg
g
=
=
⇒
d=h− 2
2
r
Fc mω r
ω
This is independent of the length of the string, and as we intuitively feel, d→h as ω→∞. The
tension in the string is, however, proportional to the length of the string:
L
F
F
mgL
FT =
= T= T
⇒
= mω 2 L
h − d Fg mg
h−d
(a) Rearranging the above equation, the length of the string
F (h − d)
L= T
mg
So L is a maximum when FT is a maximum and d is a minimum, i.e. when FT=T and d=0.
Th
∴ Lmax =
mg
(b) The distance the ball flies is equal to its horizontal velocity, v=rω, multiplied by the time
it takes to fall to the ground, t. If the starts to fall at a height, d, then the time it takes to
hit the ground is
1
2d
d = gt 2
⇒
t=
2
g
What we need to do is express the distance of travel, s=vt, in terms of L, and find the value of
L which maximizes s.
2d
2d
s = vt = rω
= r 2ω 2
g
g
The maximum distance will occur for the maximum tension, FT=T, and by geometry r 2 = L2 − ( h − d ) 2
and using previous results
F
T
ω2 = T =
mL mL and d=h− g ω 2 =h− g
mgL
=h− FT T mL we have Page 3 of 12 2 L2 − h − h − mgL T 2( h − mgL T )
s= T mL g 2 mg
hT
= 2 L1 − − L T mg Looking at this we easily see that the maximum distance will be when we maximize hT
L
− L mg
This is just a parabola in L with an obvious maximum at
hT
Lmax =
2 mg
For our interest, we can now calculate the maximum distance smax = T2 h
2 m 2 g2 −1 Note
Some people assumed the motion of the ball was vertical instead of horizontal. You can’t
actually “swing the ball in circles around you” in a vertical plane, but semantics aside, let’s see
what happens if you swing the ball in a vertical plane beside you. v
ball L θ string h
path of ball
for part(b) d
s
• The ball must orbit fast enough so the centrifugal force can overcome gravity at the top of its
orbit, so the velocity of the ball at the top of the orbit must be such that v2/L>g.
• If the string breaks, it will break at the bottom of the orbit since the tension must be a
maximum since gravity and the centrifugal force add together at that point.
To figure this out in detail, let’s assume you have got the ball going and it is now orbiting with no
further energy input from you. The constant total energy of the ball is Page 4 of 12 E tot = E potential + E kinetic
1
= mgL cosθ + mv 2
2
where we have set the gravitational potential energy zero at the height h. The velocity of the ball
thus depends on where in the orbit it is
E − mgL cosθ
v 2 = 2 tot
m
Not surprisingly, the maximum velocity is at the bottom of the orbit (θ=180°).
The tension in the string must balance the sum of the centrifugal force plus the component of the
gravitational force along the string:
v2
− mg cosθ
L
2( E tot − mgL cosθ )
− mg cosθ
=
L
2 E tot
=
− 3mg cosθ
L
As expected, this shows the tension is a maximum at the bottom of the orbit and a minimum at
the top; to minimize the tension we want Etot to also be a minimum. The tension at the top
(θ=0°) cannot be less than zero or the string will not be taut and the ball will fall. So we want
3
E tot = mgL
2
∴ FT = 3mg − 3mg cosθ = 3mg(1 − cosθ )
The tension at the bottom must be less than the breaking tension, T, so we must have
T > 6 mg
This is independent of the length of the string, so as long as the string is strong enough to support
6mg, the maximum length of the string is just set by the height you hold the end of the string
above the ground, i.e.
Lmax = h
If we take the same string and swing it horizontally, we have Lmax=(T/mg)h=(6mg/mg)h=6h, so
you can swing a much longer string horizontally instead of vertically. Even the weakest string
that can support the weight of the ball, i.e. T>mg, has Lmax>h if swung horizontally.
FT = m Finally, let’s check how far the ball will fly if the string breaks when swinging vertically. Since
the tension is maximum at the bottom, the ball will fly horizontally from the bottom of the orbit
and the time to hit the ground is
t= 2( h − L)
2d
=
g
g Again we want to maximize s=vt, Page 5 of 12 s = vt
=2 E tot − mgL cos(180°) 2( h − L)
m
g 3
mgL + mgL h − L
= 42 m
g
= 10 L( h − L) So the maximum value is L=h/2, and the maximum distance the ball will fly in the vertical orbit
case is
5
smax =
h
2
Comparing this with the horizontal orbit, we see that the minimum strength string strong enough
to be swung1 vertically (i.e. T>6mg) will fly exactly the same distance if swung horizontally or
vertically. In the vertical case, however, the maximum distance is independent of the string
strength, but in the horizontal case the maximum distance increases with the strength of the string
without limit (except for mundane factors such as air resistance and the size of the earth).
4) Cowabunga!
(a) For this question, it is convenient to assume that the cow is a sphere, since that is a shape
we know how to deal with and we only want an estimate. The capacitance of an
conducting object is the ratio of its electric charge to its electric potential or “voltage”, i.e.
Q
C=
φ
By Gauss’s Law and symmetry, the potential at the surface of a conducting sphere with
charge Q and radius r is the same as the potential a distance r from of a point charge Q:
q
φ=
4 πε0 r
(Where, by convention, V=0 at infinity defines the zero potential.) The capacitance of a
sphere of radius r is therefore:
Q
C = = 4 πε0 r
φ
The radius of a sphere with the same volume, V, as a Holstein2 cow (mass M=700kg,
density ρ=1000kg/m3) is 1 / 3 1 / 3 1 / 3
3
V M /ρ 700 kg /1000 kg / m = = = 0.55 m
r =
4
4π 4π π 3 3 3
The cowpacitance is thus
1
2 Try saying “the minimum strength string strong enough to be swung” 5 times fast.
http://www.holstein.ca/ Page 6 of 12 ( ) C = 4 πε0 r = 4 π 8.854187817 × 1012 F/m 0.55 m = 60 pF
(While writing this problem, I discovered that the standard capacitance of an isolated cow
– I am not making this up – is 0.1nF, in good agreement with our estimate. If the cow is
standing in a field, the earth acts as a conducting plane and this increases the capacitance
to about 0.2nF, but we don’t care about a factor of 2 in this problem.)
Note that you had to provide the value of ε0; we do not always give the values of basic
constants available in any appropriate physics textbook.
(b) A sphere is not a useful approximation for resistance, since it is hard to calculate the
resistance of a sphere. We’ll assume a cube, since the resistance of a cube of side length
L and resistivity ρ is very simple:
L
Lρ
R=ρ =ρ 2 =
L
A
L
For cowsize cube
M
1 / 3
cow L =
= 0.88 m ρ cow The capacitance of a cow just depends on the cow being a conductor of a certain shape,
but resistivity depends on the internal structure of a cow. My rough estimate of the
average resistivity is based on the values given for people’s insides in the POPBits™ is
ρcow~5Ω·m. The resistance is dominated by the low resistance pathways (e.g. the blood
vessels), so the high resistivity of bones and fat are largely irrelevant. So my estimate of
the resistance of the cow is just
5Ω / m
R=
= 6Ω ~ 10Ω
0.88 m
This is much cruder than our capacitance estimate. This is because parallel capacitances
add, so the size of the object pretty much determines the total capacitance; parallel
resistances add inversely, so any small scale structure (e.g. blood vessels, noses) can have
a big effect. The resistance is also sensitive to how the cow is connected in a circuit. In
this case the resistance of the cow is probably dominated by the cow’s nose and is
probably an order of magnitude higher than our simple estimate since the cow’s nose is an
order of magnitude smaller than the whole cow. (Note: This seems consistent with the
150Ω I get when I hold one probe in each hand and measure my resistance with a
multimeter, I have not yet tried to measure the resistance of my nose.)
(c) The peak power is just the initial power, since the cow’s voltage drops as it discharges:
2
2
V peak Vinitial (10 kV ) 2
Ppeak = I peakV peak =
=
=
~ 10 MW
10Ω
R
R
No wonder static hurts!
Note: By mistake, a draft pdf version of this problem set was originally posted with 600kg
cows at 20KV, instead of 700kg cows at 10KV as in the print version. This is an
approximate problem so either set of numbers are equally good.
5) Symmetry Cubed!
(a) By symmetry, the resistance between any two vertices must be the same, so let’s choose
vertices to measure the resistance between A and B. Page 7 of 12 A
C
B D By symmetry, C and D will have the same potential so no current will flow between them, so
that resistor can be ignored and the tetrahedron is equivalent to A
C
B D So the total resistant between A and B can be calculated by adding the 3 paths in parallel, i.e.
1
1
1
12
=+
+=
RAB R 2 R R R
R
∴ RAB =
2
(b) First we calculate the resistance between A and D. A B E F
H G C D There is lots of symmetry, so it easy to see that G, E, & B will have the same potential and
can treated as the same point, and similarly C, H, & F are at the same potential and can be
treated as the same point, so the circuit is equivalent to Page 8 of 12 A B E
G F
H C D So we see that R R R 5R
++=
3636
(c) Now we want the resistance between A and C. By symmetry, E, F, G, & H are all at the
same potential. As with part (a), the circuit is not affected by ignoring the resistors
connecting points at the same potential, so our equivalent circuit is:
RAD = A B E F
H G C D So 1
RAC 1 RAEC RAGC −1
R + 1 + 1 + R CD AB RBFD RBHD 1
1
1
=
+
+ 2R 2R
−1 R + 1 + 1 + R 2R 2R 1
1
=+
R ( R + R + R)
3
∴ RAC = R
4
= 1 + 1 + Page 9 of 12 (d) Now we want the resistance between A and B. This is the toughest one yet, but there is
still some symmetry, and we see that vertices E & G are at the same potential and can be
connected without changing the circuit, and F & H are at the same potential and be
connected. Our new equivalent circuit is A B E, G F, H C D A B
R
R/2 R/2 A B R/2 R E,G F,H R/2 R/2
R/2 R/2 R/2 E,G F, H R
C D 2R A B
R
R/2 R/2
2R/5 G F, H Page 10 of 12 R
A B 5R/12 A 7R/5
∴ RAB = B 7
R
12 Note: You can, of course, solve all these systems using Kirchoff’s rules and Ohm’s law to
write down systems of linear equations which can be solved by substitution or matrix algebra.
This is fine, but we thought using symmetry was more fun.
6) Is it hot enough for you?
(a) The activity A is
N
τ1 / 2 /ln 2
where N is the number of atoms and τ1/2 is the halflife of the radioactive isotope. The
number of atoms for each isotope is simply its mass divide by its atomic weight: 15 atomic
mass units (amu) for Carbon14 and 40 amu for Potassium40. For a 70kg person the masses
of each isotope are:
A= mC14 ~ 70 kg × 19% × 10−12 ~ 1.33 × 10−11 kg =13ng
mK 40 ~ 70 kg × 0.3% × 0.0117% ~ 2.46 × 10−5 kg
Therefore the activities are AC14 ~
AK 40 ~ 1.33 × 10−11 kg
1
= 2193 / s
(5730 yr /ln 2) (14 amu) 1.66 × 10−27 kg / amu ( 2.46 × 10−5 kg ) ( 1 ) (1.277 × 10 yr /ln 2) ( 40 amu) 1.66 × 10−27 kg / amu
9 = 6363 / s So the total activity is Atotal=8600 decays per second.
(b) About half the energy is deposited in the person’s body, therefore the power deposited by
each isotope is:
1
PC14 = AC14 0.16 × 10 6 eV 1.60 × 10−19 J / eV = 2.8 × 10−11W
2
1
PK 40 = AK 40 1.3 × 10 6 eV 1.60 × 10−19 J / eV = 6.63 × 10−10 W
2
So the total activity is about Ptotal=0.7nW. (
( )(
)( ) ) (c) In thermal equilibrium you will radiate as much heat, P, as you produce, so your
temperature, T, will be P 1 / 4
T = εAσ Page 11 of 12 where A is your surface area (about 2m2), and σ is the StefanBoltzmann constant. Your
emissivity, ε, is less than the black body value of 1, but it doesn’t matter too much since the
temperature only varies as the 4th root, so we’ll assume it is 1. So 1 / 4
0.7 nW T ~
= 0.3K
2
8
2 4 2 m 5.670400 × 10 W/m /K The cosmic microwave background radiation
(http://background.uchicago.edu/~whu/beginners/introduction.html) is 2.7K so it will keep
you warmer than than your internal radiation. ( ) Page 12 of 12 University of Toronto
20012002 Physics Olympiad Preparation Program
Problem Set 2: Mechanics
Due Monday 17 December 2001
(Note small extension to due date!)
Welcome back.
• This and the next 4 problem sets each concentrate on different areas of physics, but this
does not mean that you do not need to use a broad range of physics. For example, every
question on this problem set requires knowledge of mechanics1, but knowledge of
mechanics alone may not be enough to solve every problem.
• Each problem set has an experimental question. They typically require a few common
items, but you should look at them early in case you have to find something you don’t
immediately have. Your Physics teacher may be able to help.
• If you can’t do all of a question, do the parts you can do.
• Unless otherwise specified, assume everything takes place on the earth’s surface and
gravity points down in all our diagrams.
• We often provide links to websites, but the problems do not require any information from
these websites. Often the links are just for fun.
1) Salty log!
Salt water is denser than fresh water, and in the ocean you will sometimes find a sharp vertical
discontinuity in salinity (known as a “halocline”) between fresher water on top and saltier water
underneath. This often happens near coasts where fresh water runs into the sea or where glaciers
or sea ice are melting. Variations in seawater salinity and temperature drive the circulation of
deep ocean waters2 and have a major impact on climate.
Consider a log washed down a river and out to sea. Eventually the log becomes saturated
with water and it starts to sink, but if it reaches a halocline it may float on the boundary. If the
log has a uniform density ρL, and the (assumed) uniform densities of the surface and deep water
are ρS and ρD, what fraction of the volume, V, of the log will be above the halocline in the fresher
water.
[Isamu]
1
2 At least “mechanics” as defined by us.
http://www.oceansonline.com/watermasses.htm Page 1 of 4 2) Don’t leave me!
(a) Show that the Sun’s gravitational force on the Moon is more than twice the force of the
Earth on the Moon. (The distance from the moon to the earth is about 400000 km.)
(b) So why doesn’t the Sun pull the Moon away from the Earth?
[David]
3) Singing in the rain!
If you are suddenly caught out in the rain, is it better to run or walk to the nearest shelter?
(a) First consider the case where there is no wind and you have no umbrella. Assume it is a
distance L to the shelter, the raindrops are falling with a constant velocity u, and the rain
is falling at a constant rate of n mm/hour. Calculate how wet will you get (i.e. how mach
rain will hit you) if you walk with a constant speed v to the shelter. What speed, v, should
you walk or run to minimize how wet you get? (Your maximum speed is vmax.)
(b) What is your answer if you have an umbrella?
(c) What is your answer if you have an umbrella and the rain is failing at an angle θ in to the
direction you are walking? i.e. θ=0° is the rain falling vertically, θ=90° is the rain blowing
horizontally into your face, θ=–90° is the rain blowing horizontally into your back.
(Assume the vertical component of the velocity of the rain is still u.)
Hints: Simplify a person as a rectangular cube. An umbrella prevents water from hitting
the top of the cube, but not the front, back, or sides. By “rate”, I mean that after one hour an open
container left out in the rain will have a layer of water n mm deep.
[Yaser]
4) Sproing!
A block of mass m sliding on a ramp is connected to a spring that has a negligible mass and a
force constant of k, as shown in the figure. The spring is unstretched when the system is as
shown in the figure, the block has an initial velocity v down the ramp, and there is friction
between the incline and the block. The static and kinetic coefficients of friction are µs and µk;
you can assume the friction is small but not zero. k v (a) Estimate how much smaller the kinetic energy of the block will be the next time it passes
α
through the initial point.
(b) About how often is the block’s velocity zero? (i.e. What is the frequency that v=0?)
(c) Plot qualitatively the position of the block as a function of time until it permanently stops.
(d) Where is it more likely for the block to permanently stop: at a point higher or at a point
lower than the initial point (where the spring is unstretched)? Explain your answer.
[Isamu] Page 2 of 4 5) Fun before television!
Professor Bailey lived on a steep hill when he was a kid, and he and his friends used to throw
balls down the hill to try to get them into the woods at the bottom with only one bounce.
(Fortunately they had some helpful dogs who thought it was great fun to do the hard part of
finding the balls and – usually – bringing them back.) S1 S2
If you are standing on top of a smooth linear hill with aθ
slope of tanθ and throw a ball out
horizontally with an initial velocity v,
(a) how far down the hill, S1, will the ball hit on its first bounce?
(b) What is the ratio, S2/S1, of the distance (S2) travelled by the ball on its second bounce
over the distance (S1) travelled on the first bounce?
Hints: Assume the bounces are perfectly elastic. Ignore air resistance.
[Alex]
6) Soup’s on! Roll over Galileo!
Dr. Zed loves science, but even he sometimes gets misquoted in the press. As reported in the
Summer 1996 issue of the popular Canadian children’s magazine “Chickadee”3, Dr. Zed likes to
race empty and full soup cans against each other down a ramp. The magazine said that full cans
always won because they were heavier.
(a) Is it true that full cans roll faster than empty cans? Was the magazine’s explanation
correct?
Note: The empty and full cans in each race are identical except – Duh! – the empty cans are
empty and the full cans are full, and both the top and bottom of the empty cans are removed by a
traditional can opener that leaves the rims on the can. The ramp is not too steep, so the cans roll
on the rims without slipping. Dr. Zed used condensed cream soup so the contents of the full cans
are jellylike and act more or less like a solid as long as they are not subject to too high a stress.
Dr. Zed likes soup a lot, so he has a big supply of empty and full soup cans. He also has cans full
of all sorts of other stuff – beans, tomato paste, pumpkin pie filling – if he wants some variety for
his can races or for his dinner. 3 http://www.owlkids.com/ Page 3 of 4 (b) What is the relative acceleration, afull/aempty, of the two cans?
i)
Measure it experimentally. (Explain your methods and present your data.)
ii)
Calculate it theoretically.
iii)
Discuss any differences and give possible qualitative explanations for any
discrepancies (larger than your uncertainties) between your experimental data and
your theoretical calculation.
(c) What happens if you race a can of condensed Cream of Mushroom (or any other soup
which is a nonfluid jelly in the can) against the same size can of chicken or vegetable
broth (or consommé or any other soup with a waterlike consistency in the can)?
i)
Which can is first? What is the experimentally observed relative acceleration,
acream/abroth, of the two cans?
ii)
Give a qualitative theoretical explanation for your results.
Hints: Measure the relative accelerations either by timing each can separately or by racing
them and seeing where the second can is when the first can reaches the bottom – whatever works
best for you. Many cheap digital wristwatches have stopwatch functions. The second method
works best if you have two people. For a ramp I just used a shelf propped on some books. Don’t
forget to tell us any possibly relevant experimental parameters, e.g. the size of the cans you used
and the slope and length of your ramp.
[David] POPBits™ – Possibly useful bits of information
Constants and units4,5
astronomical unit (mean earthsun distance)
Earth mass
Newtonian gravitational constant
Solar mass
standard acceleration of gravity
at the earth’s surface
tropical year (2001) au
149 597 870 660±20 m
M ⊕ (5.974±0.009)×1024 kg
GN
(6.673±0.010)×1011 m3/kg/s2
MÍ (1.98892 ± 0.00025)×1030kg g 9.80665 m/s2 yr 31556925.2 s David’s Maxim of the Month
“A measurement without an uncertainty6 is just a number;
a measurement with an uncertainty is data.”
Great excuse for a party
Nobel Prize awards7 (Eric Cornell, Wolfgang Ketterle, Carl Wieman) December 10 4 http://physics.nist.gov/cuu/Constants/index.html
http://pdg.lbl.gov/2000/contents_sports.html
6
Also known as an “error bar”.
7
http://www.nobel.se/physics/laureates/2001/index.html
5 Page 4 of 4 20012002 Physics Olympiad Preparation Program
– University of Toronto – Solution Set 2: Mechanics
1) Salty log!
If the log has a density greater than the fresher surface water but less than the deeper saltier
water, it will float at the halocline such that the net sinking force on the fraction, f, of the volume
of the log above the halocline will be equal and opposite to the upward buoyancy force on the
fraction, 1f, below the halocline, i.e.
fV ( ρ L − ρ S ) = (1 − f )V ( ρ D − ρ L ) ∴ fρ L − fρ S = ρ D − ρ L − fρ D + fρ L
∴− fρ S = ρ D − ρ L − fρ D
ρ − ρL
∴f = D
ρD − ρS
As a quick check of our calculation, we can consider the case before the log begins to sink and it
is less dense than the fresh water. In this case it will float on the surface with a fraction in the
water
ρ − ρL
ρ
f = 1− S
=L
ρ S − ρ Air ρ S
which is exactly what we expect.
2) Don’t leave me!
(a) The ratio of the forces of the earth and the sun on the moon are
GN mMoon mSun
2
FSun
rSun
=
FEarth GN mMoon mEarth
2
rEarth
mSun
= mEarth 2.0 × 10 30 kg
2
rSun
2
rEarth = (1.5 × 1011 m) 6.0 × 10 24 kg 2 = 2.4 (400000km) 2 (b) This question is not trivial, since the threebody problem is not exactly soluble for a
inversesquare law force. We cannot write down the exact general equations of motion for
an arbitrary system of three (or more) bodies interacting by gravity, so we cannot prove
that the earth and the moon will never part company, even if we assume they are ideal
point masses. But if they do part company, it certainly won’t be for a very, very, very,
very long time. This is because the acceleration of the moon due to the sun is the same,
on average, as the acceleration of the earth due to the sun, so once they are moving Page 1 of 15 together around the sun (as they do) they will continue to move together. The maximum
difference in acceleration between the earth and the moon towards the sun is much
smaller than the acceleration of the moon due to the earth.
GN mSun
GN mSun
2−
rSun
aEarth − Sun − aMoon − Sun
(rSun − rEarth ) 2
=
GN mEarth
aMoon − Earth
2
rEarth mSun 1 2 − 1
r
(rSun − rEarth ) 2 Sun =
mEarth
2
rEarth
≈
= 3
2 mSun rEarth
3
mEarth rSun 2 × 2.0 × 10 30 kg( 400000 km) ( 6.0 × 10 24 kg 1.5 × 1011 m ) 3 3 = 0.013 For the moon to separate from the earth, it would have to gain or loose angular momentum
from the sun or the earth and there is no simple mechanism to do so.
3) Singing in the rain!
(a) We make the approximation that we are vertical rectangular cube with a top surface area
T and a front (or back) surface area F. The time it takes us to reach the shelter is t=L/v, so
our “top wetness” (i.e. the amount of rain landing on our top) is
WT = nTt = nTL/v
and the rain we sweep out of the air onto our front is
WF = ρFL = (n/u)FL
where, ρ, is the volume fraction of the air taken up by rain drops.
Our total wetness is
W = WT+WF = nTL/v + (n/u)FL = nL(T/v + F/u)
So our minimum wetness is when we run as fast as we can, i.e.
v= vmax
(b) If we have an umbrella, then WT=0 and our total wetness will just be
W = WF = (n/u)FL
So wetness is independent of our speed, v.
(c) If we have an umbrella, and the rain is falling at an angle (along our Front/Back plane)
then the amount of rain reaching our front or back is not just the amount we sweep out as
we walk, but also includes a contribution due to the horizontal movement of the rain,
W = WF = (n/u)FL + (n tanθ)Ft = (n/u)FL + (n tanθ)FL/v
= nFL (1/u + tanθ/v ) Page 2 of 15 where n tanθ is the rate at which the rain is falling horizontally. If θ = 0 then we just
recover our answer to part (b) as expected.
If θ > 0, then the rain is blowing in our face, (1/u + tanθ/v )>0, so the best we can
do is run as fast as we can, i.e. v= vmax, as for part (a), and our total wetness will be
W = nFL (1/u + tanθ/vmax )
If θ < 0, then the rain is blowing from behind, (1/u + tanθ/v )>0, so we can keep
dry if we just walk at the horizontal speed of the rain so that it is falling vertically in our
reference frame, i.e. our optimum speed is v= –u tanθ = u tanθ and our total wetness is
W=0.
(Note: We want to get to the shelter so we assume v is positive. If our goal is just
not to get wet, an alternate forθ > 0 is just to walk away from the shelter with speed
v= –u tanθ = –u tanθ)
4) Sproing!
The total force on the block along the ramp is the sum of the spring, gravitational, and
frictional forces:
2 dx
F = m 2 = kx + mg sinα  µ k mg cosα
dt
where x is the displacement of the block along the ramp from the position where the spring
dx
dt is the unit direction of velocity of the block along the ramp. i.e.
ˆ
force is zero, and v =
dx
dt
The frictional force always opposes the direction of motion, and we have made the usual (but
not always accurate) textbook assumption that sliding friction is independent of speed.
Without the frictional damping term, it is just an harmonic oscillator d2x = kx + mg sin α
dt 2
which you are expected to recognize after some rearrangement mg sin α d 2 x + mg sin α k k
= x + 2
m
k
dt
and know the solution to be mg sin α x + = x 0 sin(ωt), ω = k m k
mg sin α
∴ x = x 0 sin(ωt) −
k
where x0 is a (as yet unknown) constant. The block’s velocity (in the zero friction limit) is
thus
dx
= ωx 0 cos(ωt)
dt
m Page 3 of 15 v
. The
ω
minimum (i.e. lowest) and maximum positions of the block (in the zero friction limit) are
mg sin α
mg sin α
xmin = − x 0 −
xmax = x 0 −
,
k
k
which will occur at times
π
3π
ωtmin = + 2 nπ ,
ωtmin =
+ 2 nπ ,
n = 0,1, 2,…
2
2
The total kinetic (T) and potential (U) energy of the block are:
1 dx
T = 2 m dt 2 We know the initial velocity at t=0 is v (I probably should have said v0), so x 0 = 1
U = 2 kx2 + mg sinα x
where we have chosen the (arbitrary) zero for the gravitation potential energy to be the initial
point x=0 so that the total potential energy is zero at the initial point.. The initial kinetic
energy is
1
T0 = 2 mv2
and we can find the maximum extension points where the block’s motion reverses itself since
the velocity and hence kinetic energy at these points is zero so all the energy must in the form
of potential energy, i.e.
12
12
mv = kxmin + mg sin α xmin
2
2 1 1 2
−mg sin α ± ( mg sin α ) − 4 k − mv 2 2 2 ∴ xmin =
1 2 k 2 = −mg sin α ± (mg sin α ) 2 + kmv 2 k
By definition, xmin must be less than zero, so only one of the two solutions is possible, i.e.
xmin = −mg sin α − (mg sin α ) 2 + kmv 2 k
(a) The frictional force is small, so the motion of the block is damped only very slowly. The
kinetic energy lost is just the integrated frictional energy loss. On the way down to its
minimum position the frictional energy loss is W= ∫ 0xmin µk mg cosαdx = µk mg cosαxmin −mg sin α −
= µk mg cosα (mg sin α ) 2 + kmv 2 k Page 4 of 15 On the way back up to the initial point the block will lose the same amount of energy
since it travels the same distance, so the kinetic energy will be smaller by an amount 2
2 −mg sin α − ( mg sin α ) + kmv ∆T = 2W = 2µk mg cosα k (b) The block’s velocity is zero twice every oscillation cycle, i.e. at a frequency
1k
2ω/(2π)=ω/π=
.
πm
(c)
0.8
0.6 Relative x amplitude 0.4
0.2
0
0 5 10 15 20 25 30 0.2
0.4
0.6
0.8
1
1.2
Time (in oscillation cycles) (d) Because of gravity, the minimum potential energy position of the block is below the
initial position (which is the position where the spring force and potential energy are
zero). The springs will oscillate about a position below the initial position (i.e. from our
mg sin α
solutions about the middle point of the oscillation is x middle = −
), so the block as
k
it slows down will be more likely to stop below its initial position.
5) Fun before television!
(a) Lets choose an xy coordinate system in which the positive x direction points vertically
down and the positive y direction points horizontally away from the hill. Before the first
bounce has a constant horizontal velocity and constant vertical acceleration and the path
of the ball as a function of time is Page 5 of 15 y=vt , 1
x = 2 g t2 The parameterization for the slope of the hill is
x/y = tanθ
and the ball will bounce when its coordinates intercept those of the hill, i.e. when
2
2
1 y 1 x x1 = g 1 = g 1 2 v
2 v tan θ 1
∴ gx12 = x1v 2 tan 2 θ
2 v 2 tan 2 θ
g
The distance down the hill for the first bounce is
∴ x1 = 2 S1 = x1 + y1 = x1 +
2 2 2 x12
tan 2 θ = 1+ v 2 tan 2 θ
v 2 tan 2 θ
=2
g
g sin θ
tan 2 θ
1 2 v 2 tan θ
g cosθ
(b) This is pretty much the same as part (a), except a bit messier. First we need the ball’s
˙˙
vector velocity ( x, y ) before the bounce.
˙
˙
y1 = v ,
x1 = gt
To get the x component we need the time of the bounce:
=2 t1 = 2 x1
2 v 2 tan 2 θ 2v tan θ
=
2
=
g
g
g
g ˙
∴ x1 = gt1 = 2v tan θ
˙
y1
By energy conservation in an elastic collision
the speed ( v1 ) of the ball must be the same
˙
x1
before and after the bounce, so looking at the
θ
φin
angles in the bounce, we can see that the
˙
xb
α
˙˙
velocity components before ( x, y ) and after
φout
˙b
y
˙˙
( x b , y b ) the bounce are related by
θ
˙
˙
y1 = v1 cos φin , x1 = v1 sin φin , φin = θ + α
˙
˙
y b = v1 cos φout , x b = v1 sin φout , φout = −θ + α
These can be solved in various ways, some of
them involving messy trigonometry, but I
think the easiest is to think about how each
˙˙
component of ( x, y ) “bounces” separately and then adding them back together, i.e.
˙
˙
˙
˙
˙
˙
y b = x1 sin(2θ ) + y1 cos(2θ ), x b = − x1 cos(2θ ) + y1 sin(2θ ) So the path of the ball after the collision is given by
1
˙
˙
y = y b t + y1, x = gt 2 + x b t + x1
2 Page 6 of 15 α We can make things easier for ourselves by choosing to redefine our coordinate system
such that x1 = y1 = 0 , since the distance bounced can only depend on the initial vector
velocity , the acceleration due to gravity, and the angle of the slope. So
y
˙
y = ybt
⇒
t=
˙
yb
2
y
1
1 y
˙
˙
x = gt 2 + x b t = g + x b
˙
˙
yb
2
2 yb Once again the bounce will occur when the ball’s coordinates intersect that of the hill
(using our redefined coordinate system)
2
y
1 y
˙
∴ y tan θ = g + x b
˙
˙
yb
2 yb ∴y = ˙
˙
x
2 yb2 tan θ − b ˙
g
yb ˙
˙
2( x1 sin(2θ ) + y1 cos(2θ ))
=
g
= ˙
˙
− x cos(2θ ) + y1 sin(2θ ) tan θ − 1 ˙
˙
x1 sin(2θ ) + y1 cos(2θ ) 2 2
2(2v tan θ sin(2θ ) + v cos(2θ )) −2v tan θ cos(2θ ) + v sin(2θ ) tan θ − 2v tan θ sin(2θ ) + v cos(2θ ) g ( ) 2v 2
tan θ 1 + 2 sin 2 θ
g
So distance travelled along the slope on the second bounce is
y
S 2 = x 2 + y 2 = tan 2 θ + 1y =
cosθ
and the ratio compared to the first bounce is
= ( S2
y
g
g
2v 2
=
=2
y= 2
tan θ 1 + 2 sin 2 θ
S1 S1cosθ 2v tan θ
2v tan θ g ( = 1 + 2 sin 2 θ ) ) This does not depend on v or g; it is 1 when the slope is near horizontal and goes to
infinity when the slope is near vertical, both of which make sense since in the former case
the ball picks up very little gravitational energy so every bounce should be the same
length, and in the latter case the ball picks up a lot of gravitational energy before
bouncing.
6) Soup’s on! Roll over Galileo!
(a) I practically choked on my breakfast cereal when I read the magazine’s explanation in
1996. One of the most famous experiment’s in all of science is Galileo dropping balls of
different weights from the Leaning Tower of Pisa to show that gravitational acceleration
is independent of mass. (For a virtual version of this experiment, see Page 7 of 15 http://www.pbs.org/wgbh/nova/pisa/galileo.html. For a more modern version, check out
the video of David Scott dropping a hammer and a feather on the moon at
http://cass.jsc.nasa.gov/expmoon/Apollo15/apo15g.avi.) Whether Galileo actually
dropped balls from the Tower is somewhat controversial, but we know for sure that
Galileo studied acceleration of gravity using balls rolling down inclined planes. From
Newton’s Laws (of Gravity and Motion) I knew heavier cans should not roll faster, except
for effects such a friction and air resistance.
Jumping up from the table, I immediately took a bookshelf (825±10 mm long) and
put one end on some books. I took some full cans and empty cans (from our recycling
bin) and quickly established that full cans do roll faster than empty cans, but it has
nothing to do with how heavy they are. A full 284ml can rolls just as fast as a full 796ml
can; similarly a small empty can rolls at about the same speed as a large empty can. The
difference must have to do with how the mass is distributed, not how much their is.
I must admit I find it hard to believe that Aristotle actually thought acceleration was
proportional to mass. Yes, a flat piece of paper and a rock do fall at very different speeds,
but all you need to do is drop a small pebble and a large rock to see that acceleration isn’t
even close to being proportional to mass. (Even a crumpled up piece of paper hits the
ground at almost the same time as a rock if dropped from chest height.) Aristotle was not
an idiot (e.g. see
http://www.batesville.k12.in.us/Physics/PhyNet/AboutScience/WasAristotle.html), but in
this case he seems to have created a classic example of how “common sense” can
produce spectacularly wrong but long lasting dogma.
Note: The answers I give below is far longer and more detailed than we expect from you,
and even so I have left out much of the details of what I did.
(b) For this problem set, I redid my 5 year old experiment more carefully.
i)
In order to measure the acceleration I used 796mL cans of Campbell’s Cream of
x
ω
Mushroom and Consommé soup; the cans were
D=2R=101±1mm in diameter and 118±1mm in
length. (All length measurements with a Canadian
v
Tire Mastercraft 7m long tape measure.) I
removed one end from some cans, made soup,
then removed the other end. The empty
L
h
cans weighed mempty=70±5g
(on a kitchen scale) and
the full cans
weighed
θ
mfull=0.9±0.1kg. (The kitchen scale only went to 250g, so I had to empty the can and
measure the contents in several portions which is why the error is large. I roughly
checked the calibration of the scale using some water, but the quoted uncertainties are
based on how reproducible my measurements were.)
After playing around with various cans and ramps (“playing around” is an
essential part of good experimental technique – it is how you get a feel for how to do
things), I decided to use a ramp made from a L=825±10mm long Ikea wooden Page 8 of 15 bookshelf resting on VHS video cassettes. I used the cassettes instead of books since
they were all the same size (each about 30mm thick). I used this ramp because it was
straight, wide, not as slippery like laminate surfaces (so the cans always rolled instead
of starting to slide), and it was short enough so I could release the can with one hand
and catch with the other. (This last point saved me a lot of time since I didn’t have to
get up to fetch the can after each run.)
When I raced two cans, the full Cream of Mushroom can beat the empty can by
10±2cm, 8±3cm, 13±3cm for ramp heights of h = 120±5mm, 60±5mm, and
182±5mm respectively. I found I could not release the cans simultaneously and see
their separation accurately at the bottom (I was alone sine my kids lost interest after
about 30minutes), so I decided it was better for me to time the cans one at a time with
the stop watch function on my $30 Casio AQ140 wristwatch. After more playing
around the best method seemed for me to hold the watch in my right hand which also
blocked the can at the top of the ramp. I would start the watch with my thumb while
raising my hand which released the can. I would stop the watch when the can hit my
hand at the bottom of the ramp. I hoped that if I was consistent my reaction times
would mostly cancel out.
In order to have a consistent set of data (e.g. what would happen if my reaction
times changed) did runs alternating an empty can, a full Cream of Mushroom can,
and a full Consommé can so I would have the data for part (c) as well. I rolled each
can about 10 times for a given ramp height. I did this for 3 ramp heights (h=120±5,
60±5, and 182±5mm), before realizing that my timings depended on where I was
looking when I released and caught the cans, and also that by lifting my hand up to
release the can, I might sometimes give the can a little push. I decided to
continuously look at my catching hand from release to catch, and I pulled my hand
away from the can and up to avoid pushing the can (although there is still the
possibility of air movements “pulling” the empty can after my hand).
I started again and did a consistent set of measurements (“Can Runs”) which are
given in a table at the end of this solution set. The times (in seconds) are given for 5
runs for each type of can (Empty, Cream of Mushroom, Full Consommé). The data
are presented in the order I took them, but notice that I did not just steadily increase
(or decrease) the height of the ramp, to avoid the possibility of a correlated time bias
(e.g. maybe I got tired and this affected how I measured the time). I was actually
quite pleasantly surprised by how reproducible my measurements were; I didn’t
realize my reflexes are consistent to 0.1s.
If constant forces (e.g. gravity) on the cans are dominant, then the mean
acceleration of the cans is just given by
a= 2L
t2 So since L is a fixed distance,
a full
aempty t
2
empty =
t full Page 9 of 15 The date for the full and empty cream of mushroom soup is summarized in the table and
chart below:
a full
h
a
(in
h
a empty σ( full ) Correct
a empty
videos (in mm) σ(h) tempty σ(tempty) tfull σ(tfull)
ed
0
10
2
4.72
0.07
8.35 1.33
0.32
0.05
0.30
1
29
2
3.02
0.05
3.22 0.13
0.88
0.04
0.87
2
60
2
2.17
0.03
2.05 0.05
1.11
0.03
1.13
3
90
3
1.83
0.06
1.66 0.03
1.21
0.05
1.26
4
120
3
1.57
0.03
1.43 0.05
1.20
0.05
1.25
5
152
4
1.43
0.04
1.30 0.03
1.22
0.04
1.28
6
184
4
1.31
0.04
1.17 0.04
1.25
0.06
1.33
8
243
5
1.20
0.03
1.04 0.05
1.32
0.08
1.44
11
337
5
1.02
0.04
0.94 0.03
1.17
0.06
1.25
2.00 a(full)/a(empty) 1.67
1.33
1.00
0.67
0.33
0.00
0 100 200 300 Height of ramp (cm) For shallow ramp heights the full can is slower than the empty can, but for steeper ramps
the relative acceleration seems to flatten out approaching about 4/3. If I average the last
5 points, I get
a full
aempty
ii) = 1.23 ± 0.05 The empty can is a hollow cylindrical (radius R) with a moment of inertia:
Page 10 of 15 I empty = mempty R 2 If the cream soup contents rotate uniformly with the can, then the contents of the can
are a solid cylinder with a moment of inertia 1
I contents = mcontentsR 2
2
The total moment of inertial of the full cream of soup can is the sum of the moments
of the metal shell and the contents, i.e. ( ) ( ) 1
1
1
I full = mcontentsR 2 + mempty R 2 = mcontents + 2 mempty R 2 = m full + mempty R 2
2
2
2
where m full = mcontens + mempty .
There are various ways of calculating the acceleration, but I like to follow the
energy. When the can rolls down the ramp (height h, length L, angle with the floor θ),
the can gains both translational and rotational kinetic energy, i.e. 1
1
T = mv 2 + Iω 2
2
2
Since the can rolls without slipping, the angular velocity is ω=v/R, so I
1
1 v2 v2 T = mv 2 + I 2 = m + 2 2
2R
2
R
The can’s kinetic energy after it has rolled a distance x down the ramp must equal
the gravitational potential energy it has lost, i.e.
v2 I m + 2 = mgx sin θ
2
R v2
I
∴x =
m + 2 2 mg sin θ R
∴ but v ≡ dx
v
d2x I
=
m + 2 dt mg sin θ dt 2 R dx
, so
dt
a= d2x
dt 2 = dv
g sin θ
= I
dt
1 + mR 2 Page 11 of 15 a full
aempty 1 + I empty 2
m
empty R =
= I full 1 + m R 2 1 +
full 2
m
1 + empty R 2
m
4
empty R = 1
m full + mempty R 2 3 + mempty 2
m full 2 m full R ( ) So in the limit where we ignore the weight of the empty can compared to the full can,
we have a full
aempty m → empty m full →0 4
= 1.333
3 or using my measurements for my cans
a full
aempty = 4
= 1.300 ± 0.004 70 ± 5 g 3 +
0.9 ± 0.1kg (Note: If you solve the torques with respect to the point of contact and you use the
wrong moment of inertia  the one with respect to the center of the can you get a
fraction for the two accelerations that is really close to my experimental result: 1.23.
This is a coincidence.)
iii) Our theoretical calculation is very close to the experimentally observed value for
larger slopes, but several factors could be responsible for the difference and the odd
results for small slopes: friction, a constant offset in my timings, air resistance.
Air resistance would have a larger affect at higher speeds, so we would expect it
to affect the higher ramp data more, which is not what we see.
I investigated the vary large variations or the shallowest slope (h=10mm, which
had no videos and only a thin book propping up the ramp), by trying another can.
This can rolled slowly in jerks and almost always stopped and did not reach the
bottom of the slope. A little playing around showed that this can preferred to rest in
one orientation, apparently because its centre of gravity did not lie on its axis. Since
mushroom soup has chunks, this is easy to understand if the mushroom chunks were
resting on one side inside the can. The original can probably had a smaller imbalance
which was enough to affect its results. (c) The data at the end of the solution set also includes data for a consommé can. Page 12 of 15 The consommé was clearly the fastest. The plot below shows the data for the
relative accelerations of the consommé compared to both the Cream of Mushroom
and the empty can. a(consommé)/a(X) i) 6.00
5.67
5.33
5.00
4.67
4.33
4.00
3.67
3.33
3.00
2.67
2.33
2.00
1.67
1.33
1.00
0.67
0.33
0.00 X=empty
X=cream 0 100 200 300 Height of ramp (cm)
If we again only consider the last 5 data points, then the average measured ratios are aconsommé
= 1.66 ± 0.09
aempty
aconsommé = 1.35 ± 0.08
acream of mushroom
ii)
The consommé flows freely inside the can. If friction and viscosity were zero, the
can would roll down the slope but the consommé would not rotate within the can.
This means that almost all the gravitational potential energy gained by the can would
go into translational, not rotational, motion so it would roll with maximum possible
acceleration.
aconsommé = g sin θ I
1 + consommé m
2 consommé R = g sin θ 2 1 + mempty R 2
m
consommé R = g sin θ mempty 1 + mconsommé Page 13 of 15 aconsommé
2
= → 2 mempty mempty aempty
=0
1 +
m
consommé mconsommé 1
m full + mempty 1 + 2
m full = mempty 1 + mconsommé ( aconsommé
acream ) 3 m + empty 2 m 3
full = → mempty mempty
=0 2
1 +
m
consommé mconsommé So data indicate the consommé is not quite as fast as our model, which is reasonable
since the consommé undoubtedly does rotate somewhat.. Page 14 of 15 Data from Can Runs for Problem 6 (19 December 2001)
height (in videos)
5
height in mm
152
±
4 Empty (s)
1.45
1.4
1.43
1.48
1.39 Cream (s)
Consommé (s)
1.31
1.11
1.33
1.09
1.28
1.06
1.26
1.07
1.3
1.08 3
height in mm
90
±
3 1.9
1.82
1.76
1.78
1.88 1.67
1.68
1.63
1.62
1.7 1.39
1.45
1.4
1.36
1.37 1
height in mm
29
±
2 3.04
3
3.02
2.95
3.1 3.39
3.15
3.34
3.09
3.15 2.4
2.5
2.45
2.4
2.29 6
height in mm
184
±
4 1.36
1.32
1.27
1.31
1.27 1.13
1.21
1.19
1.12
1.19 0.95
0.97
0.97
1.05
1.07 2
height in mm
60
±
2 2.15
2.13
2.18
2.21
2.16 2.05
2.03
2.09
1.98
2.11 1.68
1.67
1.64
1.69
1.7 4
height in mm
120
±
3 1.57
1.53
1.62
1.59
1.56 1.43
1.41
1.37
1.49
1.47 1.19
1.23
1.22
1.19
1.24 8
height in mm
243
±
5 1.25
1.17
1.18
1.19
1.2 1.01
1.05
0.99
1.03
1.13 0.93
0.94
0.92
0.94
0.93 11
height in mm
337
±
5 0.99
1.03
1.06
1.01
0.96
1.05 0.91
0.95
0.91
0.91
0.99
0.96 0.79
0.82
0.78
0.84
0.9
0.88 0
height in mm
10
±
2 4.8
4.78
4.66
4.71
4.66 8.4
7.99
6.85
10.49
8.01 3.77
3.84
3.93
3.95
3.73 Page 15 of 15 University of Toronto
20012002 Physics Olympiad Preparation Program
Problem Set 3: Thermodynamics
Due Monday 14 January 2002 1)
How much heat could a heat pump pump if a heat pump could pump heat?
The heat in my house (and many other old houses in Toronto) is provided by an ancient coal
boiler converted to burn oil. It is starting to leak water so we are considering what to replace it
with, and one possibility is an air heat pump. (Geothermal heat pumps1 work better, but our back
yard is too small to fit the pipes.)
As its name implies, an air heat pump pumps heat by absorbing some heat (Qout) from the
outside air (temperature Tout) and supplying some heat (Qin) to the air inside the house
(temperature Tin). The heatpump works cyclically and for each cycle the heat pump motor does
an amount of work (W). Assume the heat pump has the maxiumum possible efficiency.
(a) What is the relationship between Qin, Qout, and W?
(b) How much heat will be pumped into the house for every joule of work done by the
motor if the inside temperature is 20°C and the outside temperature is –20°C?
(c) If the outside temperature was absolute zero, the heat pump would still pump heat into
the house. How is this possible and where does the heat come from?
[David]
2)
Relatively heavy
When I have a fever, I sometimes feel lethargic and heavy. Estimate the fractional amount I am
heavier if my temperature increases from 37°C to 39°C?
Hints: In addition to writing down the most famous equation in all of science, did you
know that Einstein had over 40 patents on refrigerators?2
All experimental evidence is consistent with General Relativity’s principle that inertial and
gravitational mass are equivalent.
Assume I neither gain nor lose any molecules as I heat up, and you may make the crude
approximation that I am a bag of ideal gas water molecules. (My kids sometimes think I am full
of gas, but not ideal.)
[David] 1
2 http://www.earthenergy.ca/tech.html
http://gtalumni.org/news/magazine/sum98/einsrefr.html Page 1 of 5 3)
Jet cooled!
The New Zealand physicist Ernest Rutherford1, who discovered the atomic nucleus in 1911, is
often described as the greatest experimental physicist of the 20th century. His great experimental
tradition still lives on in New Zealand and manifests itself in such recent accomplishments as the
jet powered beverage cooler2. This cooler consists of a tank of liquid petroleum gas (LPG) in a
container of water. When gas is released from the tank, the water is cooled very quickly. To
avoid accidental and explosive accidental ignition of the very flammable gas, the released LPG is
burned in a jet engine with a very satisfying 125dB roar.
The specifications of the cooler are a bit sparse, but it appears to be able to cool about
10 litres of water from 14°C to 2°C in about 5 minutes.
(a) What is the average rate (in Watts) that the cooler removes heat from the water?
I don’t have any LPG in my lab (and I have enough sense not to risk explosion, fire, and
deafness just to cool a drink), but I do have a 50 litre bottle of helium gas (atomic weight 4.003)
at a pressure of 170 Atmospheres and a temperature of 21°C.
(b) If I release gas from my bottle and allow it to expand adiabatically to atmospheric
pressure, estimate how cold (in °C) will it get?
(c) If I then use this cold gas to cool water, how much gas (in litres measured at 170
Atmospheres and a temperature of 21°C) do I need to release in order to cool 10 litres
of water from 21°C to 9°C?
Hint: Assume the water and gas are brought into thermal contact and left to
adiabatically reach equilibrium.
[David]
4)
Perpetual motion?
Of the many fascinating perpetual machines proposed through history, that proposed Johannes
Taisnierus in 1570 and discussed by Bishop John Wilkins of Chester3 in his book “Mathematical
Magick, or the wonders that may be performed by mechanical geometry” is one of the simplest. The magnet fixed on the top of the column was to draw the iron ball up the ramp until the ball
reached the hole in the ramp. The ball would then fall and be returned to the ramp’s base, and the
procedure would begin again.4
1 http://www.nobel.se/chemistry/laureates/1908/
http://www.asciimation.co.nz/beer/
3
http://wwwgroups.dcs.standrews.ac.uk/~history/Mathematicians/Wilkins.html
4
A modern “perpetual motion machine” involving rolling balls and magnets can be found at
http://www.theverylastpageoftheinternet.com/magneticDev/finsrud/finsrud.htm.
2 Page 2 of 5 Let’s consider an modern electrostatic version of this device, as shown in the diagram
below. The electric dipole (point charges ±Q, separated by a distance d) at the upper left is
aligned with the path of the ball (charge –q, mass m) rolling up the ramp. The length L (>>d) is
the distance from the centre of the charged ball to the centre of the dipole. When the ball is over
the hole, its centre is a distance d from the +Q point charge.
d
Q +Q d L
?
? q θ m trap door
(lets ball move up
through hole,
but not down)
If you really made such a device, describe and discuss what would happen when the ball is
released (from the initial position shown in the figure) until it either stops or it finishes one
complete cycle of an infinitely repeated loop. Does your answer depend on the ramp angle, θ, or
any other parameters (possibly including parameters not listed here)? Assume that the device is
as well made as is humanly possible, and feel free to make any realistic minor design changes
that might improve the performance of the device.
[Isamu]
? ? 5)
A slow day on the Bruce Trail!1
If all else fails, a pleasant walk is better than moping in bed, and random walks are particularly
important in physics. A onedimensional random walk is when you walk along a line, and for
each step the direction – forwards or backwards – is determined randomly. If the probabilities for
going forwards or backwards are equal, then after N steps you will find yourself on (rootmeansquare) average a distance of N steps away from where you started. This is only
recommended for hikers who aren’t in a hurry to get anywhere.
A classic example of a random walk process is diffusion, and the classic elementary school
demonstration of diffusion is dye2 dropped into water. I think these demonstrations are great, but
it is not obvious to me that diffusion is the dominant process distributing the dye in most such
demonstrations. One website3 I looked at said it took 7 minutes for dye to diffuse throughout a
glass of cold water, but only 4 minutes for a samesized glass of hot water.
(a) Show that such a result cannot be due to random walk diffusion. Suggest possible
explanations for the reported result (or convince me that I am wrong in saying that the
result cannot be due to random walk diffusion).
1 http://www.brucetrail.org/
The “dye” is usually food colouring, but you can use ink, juice, pop, ….
3
http://hastings.ci.lexington.ma.us/staff/SLee/science/molecules.html
2 Page 3 of 5 Hints: Calculate the ratio of temperatures, Thot/Tcold, of the water in the two glasses if it
the results are due to random walk diffusion. The mean speed of the molecules depends on the
molecular mass and on the temperature, and the mean step size is roughly given by the mean
spacing between water molecules. Diffusion in 3dimensions can be treated as 3 independent
onedimensional random walks.
(b) Do the experiment yourself by dropping some dye into hot and cold glasses of water.
Do your results agree with the results I report from the elementary school website?
(c) Design, do, and describe an improved experiment to study the diffusion of dye in
room temperature water. Is the movement of the colour in your container in your new
experiment consistent with random walk diffusion?
Hints: Here are some of the “improvements” I tried: I found a spot where my water
containers were be safe from being moved or shaken or subject to sudden temperature changes.
(I discovered the top of my fridge or a south facing window are NOT good places.) I waited for
the water to be motionless and in thermal equilibrium before adding the dye. I tried adding the
dye to the bottom, not the top. Chunks of leftover Hallowe’en lollipops gently dropped into the
containers initially seemed to work well (and were used to bribe young children not to mess with
my experiments). The candy dissolved in the water and released its colouring, but a dense
bottom layer of sugar water was also created which seemed to affect the diffusion.1 Next I tried
injecting food colouring into the bottom of a container with an eyedropper2, but often the dye
leaked out before I reached the bottom. I also tried putting a few drops of food colouring into a
tiny paper pocket which I weighted with a nickel and taped the edges shut before dropping into
the container.
[David] 1 Note: A good experimenter always reports any interesting effects observed, even if they don’t match
initial expectations.
2
Actually a plastic straw with the top taped shut since I didn’t have an eyedropper. Page 4 of 5 POPBits™ – Possibly useful bits of information
Constants and units1,2
astronomical unit (mean earthsun distance)
atomic mass unit: (mass 12C atom)/12
Boltzmann constant au
u 149 597 870 660±20 m
(1.66053873±0.00000013)×1027 kg k (1.3806503±0.0000024)×1023 J/K elementary (i.e. electron) charge e (1.602176462±0.000000063)×1019 C electron volt eV
GN
ε0 (1.602176462±0.000000063)×1019 J Newtonian gravitational constant
permittivity of free space
Planck constant
solar luminosity
speed of light in vacuum
standard acceleration of gravity
at the earth’s surface
standard atmospheric pressure
at the earth’s surface
StephanBoltzmann radiation constant
tropical year (2001) h
LÍ
c
g (6.673±0.010)×1011 m3/kg/s2
8.854187817×1012 F/m
(6.6260755±0.000004)×1034 J·s
(3.846 ± 0.008)×1026W
299 792 458 m/s
9.80665 m/s2 atm 101325 Pa σ
yr (5.670400±0.000040)×108 W/m2/K4
31556925.2 s Interesting facts about water
Specific heat = 4186J/kg/K
Freezing temperature at standard atmospheric pressure = 273.15K
Elephants can smell water more than 5 km away.
Great excuses for a party
Birthday of the quantum3 (1900)
Isaac Newton’s birthday (1642) December 14
December 25 1 http://physics.nist.gov/cuu/Constants/index.html
http://pdg.lbl.gov/2000/contents_sports.html
3
http://www.nobel.se/physics/educational/tools/quantum/energy1.html
2 Page 5 of 5 20012002 Physics Olympiad Preparation Program
– University of Toronto – Solution Set 3: Thermodynamics
1) How much heat could a heat pump pump if a heat pump could pump heat?
(a) By energy conservation, the work must provide the difference between the heat pumped
from the outside and the heat delivered to the inside, i.e. W=QinQout.
(b) The maximum efficiency possible is for a Carnot engine,
∴ Qhot
T
= hot
Qcold Tcold ⇒ Qin
T
= in
Qout Tout To convert from °C to absolute temperature in °K, we just add 273°. The heat pumped per
joule of work is Q −1 T −1 273° − 20° −1
Qin
Qin
=
= 1 − out = 1 − out = 1 − = 7.3 273° + 20° W
Qin − Qout Qin Tin (c) The motor’s work is converted to heat, so the heat pump always provides at least W heat.
Rereading the question, one could argue that pump does not “pump heat into the house”
(especially if the motor is inside the house), so maybe “provide heat to the house” would
be better wording.
2) Relatively heavy
Another easy one, as long as you remember E=mc2. Some students (equally correctly) used
the heat capacity of an ideal gas or water, but this is how I did it. The total energy of a
molecule is the sum of its rest mass and its thermal energy. (We can define its potential
energy to be zero.) An ideal gas molecule has no internal or rotational degrees of freedom, so
its only thermal energy is its kinetic energy and the total energy is
3
E = mc 2 = m0c 2 + kT
2
3kT
∴ m = m0 + 2
2c
Where m0 is the rest mass of the molecule at T=0. In this approximation, my mass is just
proportional to my molecules masses, so the fractional increase in my mass going from
T1=37°C to T2=39°C is 3kT2 3kT 3k
− m + 21 m0 +
(T2 − T1 )
2 0
T2 − T1
m(T2 ) − m(T1 ) 2c 2c = 2c 2
=
=
3kT
3kT
m(T1 )
2 m0c 2
m0 + 21
m0 + 21
+ T1
2c
2c
3k Page 1 of 8 As is well known (i.e. you are supposed to know it) water is H2O and hence has an atomic
mass of 18, so m0=18u, where u is the atomic mass unit. Thus
m( 39°C ) − m( 37°C )
39°C − 37°C
=
2
−27
m( 37°C )
2 18 × 1.661 × 10 kg (299792458 m / s) ( ( ) 3 1.3807 × 10−23 J / K ) + ( 37°C + 273°K ) = 1.5 × 10−14 So the fractional mass increase is about 1014. (You get the same order of magnitude no
matter which way you do it.) 3) Jet cooled!
(a) The average cooling rate is just
dQ (10 l × 1kg / l)(14° − 2°)K
J
=
(4186J / kg / K ) = 1674 = 1.7kW
dt
s
(5 min × 60s /min)
where we have used the well known density of water (1kg/litre).
(b) The reason I chose helium1 for this problem is that it is a noble gas, and noble gases2 act
almost like an ideal gas at normal temperatures. For an ideal gas undergoing adiabatic
expansion, P(1−γ)Tγ is a constant, where P is the pressure, T is the absolute temperature,
and γ=5/3 is a constant.
(1−γ )
3 γ
P 170 atm 5 −1 ∴ Tf = Ti i = (273° + 21°)K = 38K = −235°C P 1atm f
(c) To cool 10 litres of water by 12°, we need
∆Q = (10 l × 1kg / l)(21° − 9°)K ( 4186 J / kg / K ) = 502320 J
of heat removed. The thermal energy of an ideal gas molecule is just 3kT/2, so using the
ideal gas law, our cold expanded helium gas has a thermal energy
3
2
Q = N kT = PV .
2
3
To provide the required ∆Q, we must have
2∆Q
.
N=
3k∆T
where ∆T =9°C(235°C)=244K is the temperature increase of the gas as it reaches
equilibrium with the water. Using the ideal gas law the volume of gas (at 170 Atmospheres
and 21°C) required is
2(21K + 273K )(502320 J )
NkT 2T∆Q
V=
=
=
= 23 × 10−3 m 3 = 23l .
P
3P∆T 3(170 atm × 101325Pa/atm)244 K 1
2 http://www.webelements.com/webelements/elements/text/He/key.html
http://www.xrefer.com/entry.jsp?xrefid=490629&secid=.&hh=1 Page 2 of 8 That corresponds to about 4 cubic metres at atmospheric temperature  enough helium to
inflate about 300 11" balloons.
Once again, many students used (equally correctly) the heat capacity of an ideal gas.
My answer corresponds to using the heat capacity at constant volume, but you get 14l if you
use the heat capacity at constant pressure. Rereading how I worded the question, assuming
constant pressure probably does make more sense (since the gas is being released and
expands in air at atmospheric pressure), but we accepted both answers.
4) Perpetual motion?
When the ball is released, it will not move unless the attractive force of the dipole can
overcome gravity and static friction. We will assume that we can make it such that the static
friction is very small, so we must have
Qq
Qq
−
> mg sin θ
2
2
4 πε0 ( L − d 2)
4 πε0 ( L + d 2) 2 Ld
Qq > mg sin θ
∴
2 4 πε0
2 L2 − d 4 Since L>>d, this requirement becomes
Qq d
> mg sin θ
2πε0 L3
This is possible for appropriate charges and masses, although it may not be easy. (For
example, everything would have to be constructed from insulators and the experiment done in
a vacuum to prevent charged particles in the air being attracted to our charges.)
Once the ball starts to move, it will roll up the ramp until it hits the barrier where it will
be deflected down. (If there is just an wall perpendicular to the direction of the ball, the ball
will bounce backwards with a strength proportional to the elasticity of the collision, so I
added a deflector to make sure the ball goes down in the right direction.) The ball will roll (if
the deflector is just right, otherwise it will bounce) around the circular return path. Because
of the 2nd Law of Thermodynamics (e.g. friction, inelasticity, …) the ball will not quite get
back to its initial position. (For example, even if friction is essentially zero, the ball will not
have quite enough energy to raise the flap since in the initial position the flap was down, and
the flap has some mass which must be raised against gravity.)
Since the ball can’t quite get back to its starting point, it will roll back down the circular
return path and end up oscillating back and forth with small and smaller amplitude until it
finally stops. Page 3 of 8 Note that the oscillations may include bouncing back down the ramp, but because of the
energy losses, the ball never can get back to its start. The exact path depends on the amount
of friction and the inelasticity of any bounces.
The equilibrium position depends on the strength of the dipole. If it is strong enough to
overcome gravity, the ball will stick at the top of the ramp. (It will stick
the first time it reaches the top of the ramp if the collision with the wall
Fdipole
θ
is inelastic enough, but we have assumed we are doing a good job in
designing this device.) If we want the ball end up with the ball finally
Fgravity
stopping on the circular path, we must build the dipole such that
Fdipole sin θ < Fgravity 1
Qq 1
sin θ < mg
i.e.
−
4 πε0 d 2 (2 d ) 2 3Qq
sin θ < mg
∴
16πε0 d 2
This requirement combined with our earlier requirement to get the ball started gives
Qq d
3Qq
sin 2 θ
> mg sin θ >
3
2
2πε0 L
16πε0 d ∴d > ( 3 sin 2 θ
8 ) 1 3 L Since we want L>>d, this means the angle θ must be pretty small. When the ball stops on the
circular path, it will stop slightly to the left of the bottom of the ramp.
5) A slow day on the Bruce Trail!
A couple of comments about experimental questions on POPTOR. First, the important
thing is make measurements and then to interpret them to the best of your ability. Second, in
many ways, one never “completes” a real experiment, since one can always improve the
experiment in some way.3 Third, it is traditional in schools and universities – and this is also
the case for POPTOR that experiments do not usually receive a weight in marking
proportional to the effort required to do them well.4
(a) The density of liquid water doesn’t change very much between freezing and boiling
temperatures, so the number of steps to diffuse over the whole glass is pretty much
3 An interesting book discussing these kinds of issues is “How Experiments End”, see
http://www.fas.harvard.edu/~hsdept/faculty/galison/how_experiments_end.html.
4
For POPTOR, we “correct” for this by keeping a separate record of how people do on experiments. Page 4 of 8 constant independent of temperature. i.e. The mean free path that a diffusing dye
molecule travels before colliding with a water molecule is about the same for both
temperatures. The change in diffusion times is dominated by the change in mean
molecular velocities at different temperatures. The diffusion time, τ, is inversely
proportional to the mean velocity, i.e.
1
τ∝
v
Using the relationship between kinetic energy and velocity, and temperature and kinetic
energy for an ideal gas we have
1
τ∝
⇐ E kinetic = 1 mv 2
2
E kinetic
1
3
⇐
E kinetic = 2 kT
T
So the diffusion time is inversely proportional to the squareroot of the absolute temperature.
The maximum temperature range of liquid water is from freezing (273K) to boiling (373K),
so the maximum ratio of diffusion times is
T τ 373
max cold = max hot =
= 1.17
T 273 τ hot cold So the maximum ratio of diffusion times in water is about 1.2, which is much less than the
reported ratio of 7min/4min=1.75.
One obvious possible reason for the reported results is that hot water was moving more
which mixes the dye faster. One might expect convection in the hot water driven by
temperature differences at the water surfaces. Another possibility is that the measurement
errors were very large, i.e. people aren’t very good at judging when the dye has completely
diffused, especially if they have prior expectations about what should happen. The whole
point of the demo was to show that diffusion is faster in hot water, and a 17% (or less)
difference is not subjectively “large enough”.
(b) I reproduced the experiment using identical cylindrical glass glasses 10±0.2cm tall and
6.3±0.2cm in diameter (measured with a plastic ruler). I used water from my taps and the
hot water was 50±1°C and the cold water was 15±1°C, where the uncertainties are just my
reading errors since I didn’t calibrate my rather cheap outdoor thermometer.
7:35pm
Food colouring added to each glass. The hot water clearly had more
swirling. In the cold glass the dye when mostly to the bottom and then
swirled slowly. Clearly the residual motion from adding the water to
the glasses had not stopped.
7:38pm
Movement is clearly visible in the hot water, consistent with
convection. The is only slow movement in the cold water, and the dye
is in thin strands..
7:40
The dye is almost uniformly distributed in the hot glass, but
it is clearly slightly darker at the top than the bottom. More
clear is a dark vertical swirl along the axis of the glass,
which is wider and darker at the bottom. The cold glass is
still has clearly visible strands of colour.
∝ Page 5 of 8 7:51 Can still see dark central region in hot glass, and strands in cold, but
mixing is steadily progressing.
8:02
Dark central region in hot glass is fading, but still just visible. The
strands in the cold water are still visible, but fading.
8:12
Both hot and cold are still fading and mixing, but otherwise the same
8:31
Both hot and cold look uniform to me.
It took about an hour before both hot and cold glasses looked uniform to me, and there was
clearly nondiffusive mixing at the beginning. There was no obvious difference in the
diffusion.
(c) I used the same glasses as in part (a), and as mentioned in the question, I let the water
reach equilibrium before adding the dye.
The candy dropped to the bottom was interesting in that after dissolving the dye slowly
diffused up a couple of centimetres, darker at the bottom than the top, but then all movement
stopped until I threw it out after a week. There was a constant 2 cm layer of coloured water at
the bottom, and no sign of movement. (Within the 2cm layer it was darker near the bottom.)
When I swirled the glass before throwing out the water I could clearly see the sugar (because
of the change in index of refraction). My guess (and it is only a guess) is that so the candy
formed a network of molecules which stopped the diffusion.
The thing that worked best was a better eyedropper than the one I mentioned in the
question. This one had a narrow tip and very little food colouring (my preferred dye) leaked
out while I gently inserted it, and I could deposit a nice layer of dye at the bottom. The first
time I did this, I put it in a window and I noticed a beautiful (but very slow) convection cycle
starting with water on the warm (room) side moving up about 1.5cm after an hour, but no
movement on the cold (window) side.
Once put in a uniform temperature position, I observed the following:
11:30pm
Food colouring (red and blue) added to two glasses. It formed a thin
layer on the bottom of each glass, but initially not all the bottom was
covered.
11:39pm
Dye has almost covered the bottom of each glass.
11:49pm
Dye covers bottom of both glasses uniformly with a layer 12mm thick.
(The range is because it is hard to determine the top of the layer.)
12:20am
Dye layers now 23mm thick
12:43am
45mm thick.
1:15am
56mm thick
2:03am
68mm thick
5:03am
There is a light greenblue colour throughout the blue glass, but also a
dark 510mm thick layer at the bottom. The red glass has a uniform
pink colour througout, with a similar darker red layer on bottom.
7:15am
The bulk colour is a bit darker maybe, but the dark 510mm thick layer
at the bottom is still there.
8:00pm
The liquid is uniform blue in the “blue” glass, but the red still has a
darker layer on the bottom.
8:00am
One glass is a unifrom blue, the other a uniform red. Page 6 of 8 It is very hard to measure the progress, to see if it follows a squareroot time spread.
This is because diffusion does not have a sharp edge. Nevertheless the time scale for diffusion
of the food colouring throughout my 10cm tall glasses is about a day, with the red taking a bit
longer than the blue.
We can’t confirm the square root time dependence, but we can estimate the diffusion
time to see if it is consistent. The molecular weight of water is 18, so the density of water
molecules is
ρ H 2O
1kg / m 3
n=
=
= 3.35 × 10 25 m−3
27
mmolecule(H 2O ) 18(1.66053873 × 10 kg)
and the mean free path between collisions should be about λ= ( mmolecule(H 2O )
1
=3
= 3.35 × 10 25 m−3
3
ρ H 2O
n ) −1 3 = 3.1 × 109 m = 3.1nm The mean velocity of the water molecules is calculated from the kinetic thermal energy, i.e.
1m
v2
2 molecule(dye) ∴v = 3
= 2 kT 3kT
mmolecule(dye) For water molecules diffusing at room temperature ∴ vH 2 0 = 3kT
mmolecule(H 2 0 ) = ( ) 3 1.3807 × 10−23 J / K (273 + 20K )
18(1.66053873 × 1027 kg) = 637 m / s and the mean time between collisions is τ= λ
v The mean number of steps to diffuse a distance L is about L 2
steps = λ
and the mean time to diffuse a distance L is about ρ H 2O L 2
L2
t = τ =
= L2 3 λ
mmolecule(H 2O )
λv mmolecule(dye)
3kT For water this is L 2
L2
L2
s
t = τ =
=
= 5.1 × 10 5 2 L2 λ
λv
3.1nm(637 m / s)
m For our 10cm tall glass we expect
t = 5 × 10 5 s
2 (0.1m) 2 = 5 × 10 3 s m
So the water molecules take well over an hour to diffuse through the glass, and the
heavier dye molecules should take even longer. A quick search on the web and we can find Page 7 of 8 that most common food colourings have molecular weights5 of 4001000, with most around
500, so scaling by the square root of the molecular mass (from the above formulae) we expect
their diffusion times to be about
500
tdye ≈ 5 × 10 3 s
≈ 7 hr
18
which is quite consistent with what I saw. (Note that the diffusion time tells us the average
time for a molecule to travel the 10cm, but some molecules will be slower and some faster, so
it will take longer than the diffusion time to get a perfectly uniform colour.) 5 http://www.kiriyachem.co.jp/eng/syokuyou/chem2.html Page 8 of 8 University of Toronto
20012002 Physics Olympiad Preparation Program
Problem Set 4: Optics and Waves
Due Monday 11 February 2002 1) Yes, Officer!
A car driven by a physicist is stopped by a police officer for running a red light. The physicist
explains that the light appeared yellow because of the Doppler shift. (The wavelengths of red and
yellow light are about 690 nm and 600 nm respectively.) Does the physicist deserve the ticket?
[David]
2) You push, I’ll pull.
Radiation pressure is a very important factor for interstellar dust clouds1, comet tails2, and stellar
and planetary system formation3. Once a cloud of gas and dust collapses and a star ignites, its
radiation blows away smaller particles.
How big must a spherical dust particle be to be currently attracted by the sun instead of
being repelled. i.e. What is its minimum radius, R?
Hints: Assume the dust has a typical mineral density of about density 3 g/cm3 and that the
dust totally absorbs the sunlight.
[David]
3) 500 channels  but nothing to watch!
For years I have been waiting for my favourite TV show, Fraggle Rock4, to be available again in
Toronto. (That part isn’t made up, but from here on I may wander just a teeny bit from reality.) I
recently learned that the show is playing on a small station broadcasting on Channel 2
(57 Mhz).which is not on Toronto cable. The station is 50 km away, but fortunately my house is
on a hill and with a good antenna I can pick up the station by direct line of sight.
One day in the middle of the show I notice that the signal is fading in and out, going from
maximum to minimum and back again 8 times per minute. I then learn from the radio that one of
my atmospheric physics colleagues has lost control5 of a humongous balloon6 and it is currently
just halfway between me and the TV station. The balloon is 20 km up and slowly rising at a
1 http://www.seds.org/messier/more/m045_i349.html
http://antwrp.gsfc.nasa.gov/apod/ap001227.html
3
http://www.seds.org/messier/more/m042_hst2.html
4
http://www.henson.com/television/series/fraggle_home.html, http://home.no.net/fraggel/fragglerock.htm
5
http://www.atmosp.physics.utoronto.ca/MANTRA/press.html
6
http://www.mscsmc.ec.gc.ca/events/balloon/launch/launch17_e.cfm
2 Page 1 of 4 constant speed, and the signal I am receiving is the sum of the direct line of sight signal plus the
reflected signal from the balloon. How fast is the balloon rising?
[David]
4) An intense experience!
Nonlinear optical materials have an index of refraction that is dependent on the intensity of the
incident beam. For a linear nonlinear optical material, n=n1+n2I, where I is the intensity of the
beam, n is the index of refraction, and n1 & n2 are positive, real constants.
Most laser beams have a Gaussian distribution of intensity in the xy plane normal to the
direction of the beam, z. If for simplicity we only consider one axis we have
I ( x) = I 0 e − x2
w2 (a) Plot the intensity versus x. What parameter determines the size of the beam in this
formula?
(b) If a Gaussian beam hits a nonlinear material, how does it travel inside it? Show
qualitatively that the beam has to focus down. How does the beam continue after it
focuses?
(c) Estimate the distance from the surface where the beam focuses. Use the fact that different
rays which meet each other at the focal point have to travel the same optical distance
(Optical distance=Index of refraction * distance). If it helps simplify your answer, you
may assume n1>>n2I0.
(d) How do your answers change if n2 is negative?
[Yaser]
5) All the world in a drop of water
A microscope can be made using a drop of water as a lens.1
(a) What is the focal length of roughly 5mm diameter drop of water supported by a horizontal
piece of transparent plastic?
(b) If you put your eye very close to the water drop and look through it you can see quite
impressive magnification of objects. (I thought the ridges on my fingers were fascinating,
but I am known to be easily entertained. By “very close” I mean your eyelashes may be
touching the plastic.) Estimate the magnification of your water drop microscope.
If you can calculate the answer to these questions I’ll be very impressed, since the only way I
know to get the answer is to do the experiment. (To calculate the answer you’d have to calculate
the shape of the drop which depends on its mass, the acceleration due to gravity, the surface
tension of water, the molecular forces at the water/plastic boundary, …)
(c) Play around with different drops. Very roughly, discuss how the optical parameters (e.g.
the focal length, magnification, and distortion) depend on the physical parameters (e.g.
diameter) of your droplets?
Make sure you describe what you did to get your answers. 1 http://www.microscopyuk.org.uk/mag/art98/watermic.html Page 2 of 4 Hints: The drop should be more or less circular as seen from the top. I have used shiny
transparent tape, transparent food wrap, transparent CD or audio cassette cases…. (Plastic seems
to work better than glass.) You want the drop to be as rounded as possible, not smeared out, e.g.
like the drop on the left, not like the one on the right. I use a drinking straw to gently deposit my drops on the plastic. Your droplet lens is fairly
sensitive to vibration, so it is better to support your piece of plastic so you don’t have to hold it
and you can put things underneath your lens without touching it, but this isn’t necessary. Make
sure you have good lighting; I used either bright daylight or a flashlight. With a little practice, I
could put samples (e.g. salt crystals) on the bottom of an audio cassette case and consistently put
the right sized drop on the top which gave the right focal length.
The magnification is conventionally defined as the apparent size compared to the size seen
at the naked eye at a distance of 25cm from the eye. I found pieces of linear graph paper very
useful for studying magnification and distortion; I darkened some of the lines to make them more
visible if looked at out of focus while looking through a droplet at another piece of graph paper in
focus. Overhead lights provided a convenient light source for the focal length measurement.
[David] Page 3 of 4 POPBits™ – Possibly useful bits of information
Constants and units1,2
astronomical unit (mean earthsun distance) au
u 149 597 870 660±20 m
(1.66053873±0.00000013)×1027 kg k (1.3806503±0.0000024)×1023 J/K elementary (i.e. electron) charge e (1.602176462±0.000000063)×1019 C electron volt
index of refraction of water
Newtonian gravitational constant
solar luminosity (i.e. light output)
solar mass eV
nH20
GN
LÍ
MÍ (1.602176462±0.000000063)×1019 J
1.33
(6.673±0.010)×1011 m3/kg/s2
(3.846 ± 0.008)×1026W speed of light in vacuum
standard acceleration of gravity
at the earth’s surface
StephanBoltzmann radiation constant
tropical year (2001) c
g atomic mass unit: (mass
Boltzmann constant 12C atom)/12 σ
yr (1.98892 ± 0.00025)×1030kg
299 792 458 m/s
9.80665 m/s2
(5.670400±0.000040)×108 W/m2/K4
31556925.2 s Great excuses for a party
High energy Nobel birthdays:
Hideki Yukawa3 (1907)
Samuel Chao Chung Ting4 (1931)
Abdus Salam5 (1926) January 23
January 27
January 29 1 http://physics.nist.gov/cuu/Constants/index.html
http://pdg.lbl.gov/2000/contents_sports.html
3
http://www.nobel.se/physics/laureates/1949/press.html
4
http://www.nobel.se/physics/laureates/1976/press.html
5
http://www.nobel.se/physics/laureates/1979/salambio.html
2 Page 4 of 4 20012002 Physics Olympiad Preparation Program
– University of Toronto – Solution Set 4: Optics
1) Yes, Officer!
If the driver is relativistic, the police officer clearly should issue a ticket, so we can just use the
nonrelativistic Doppler formula for frequency (f ) or wavelength (λ):
f ′ v
λ v
= 1 + or
= 1 + f c
λ′ c 690 nm
v
∴
−1 =
600 nm
c
m
km
v = 0.15c = 4.5 × 10 7 = 1.6 × 10 8
∴
s
hour
So the driver is clearly speeding and deserves a ticket. The exact relativistic Doppler shift
differs by only a factor of 1 − (v c ) = 0.99 .
2 2) You push, I’ll pull.
The momentum and energy of the photons are
h
hc
pγ =
and
Eγ =
λ
λ
The number of photons produced per second by the sun are
L
Lλ
Nγ = Í = Í
Eγ
hc
A dust particle absorbs the momentum of any photon hitting it, so the pressure on a dust
particle at a distance r from the sun is the total momentum of the photons per unit area per unit
time, i.e.
N γ pγ
LÍ hλ
L
PÍ =
=
= Í2
2
2
4 πr
hc 4 πr λ 4 πcr
The total light force on a spherical dust grain of radius R is FL = PÍ × Area = PÍπR 2 = LÍπR 2 = LÍ R 2 4 πcr 2
4 cr 2
The mass of a spherical dust grain of radius R and density ρ is m = 4 πR 3ρ
3
The dust grains will be attracted to the sun when the light pressure force is weaker than the
gravitational attraction of the sun, i.e. when Page 1 of 5 ∴ FL < FG
∴ LÍ R 2 ∴ LÍ R 2 4 cr
4 cr 2 < GN mMÍ
r2 4 πR 3ρM
Í
< GN 3
2
2 r
3LÍ
∴R >
= 1.9 × 10−7 m
16πcGN ρMÍ
So the dust grains must be larger than 0.2µm to fall into the sun. This is why the clouds of gas
and small dust are blown away when a star ignites. Another nice example of the effect of
radiation pressure on interstellar dust is the Trifid Nebula
(http://antwrp.gsfc.nasa.gov/apod/ap990608.html).
3) 500 channels  but nothing to watch! r r
2 2 x d d
2
2
The distance
between me and the TV transmitter are fixed, so the interference indicates that the signal path
length is increasing by 8 wavelengths per minute. The path length of the reflected signal is
r = 2 x 2 + ( d / 2) = 4 x 2 + d 2 = 4 (vt + x 0 ) + d 2
2 2 where t=0 is defined to be when the balloon has a height x 0 = 20 km . The balloon rises with a
constant velocity, v=dx/dt, so
4 (vt + x 0 )v
dr d
2
=
4 (vt + x 0 ) + d 2 =
2
dt dt
4 (vt + x ) + d 2
0 We are interested in the time t=0, when the rate of change of pathlength is
dr
2v
=
dt t = 0 2
1+ d 2 x0 The wavelength of the signal is c 3 × 10 8 m / s 100
=
=
m
f 57 × 10 6 / s 19
So the balloon is rising with a velocity of λ= 2 2 1 8λ
1 dr
1 100 25
= 0.56 m / s
m 1 +
1+ d 2x =
1 + 50 km = (2 × 20km) 15s 19 0 2 dt t = 0
2 60 s
16
Note: Some people used differences instead of derivatives to get almost the same answer. This
works fine in this case since the change in distance is so small compared to the initial distance.
v= Page 2 of 5 4) An intense experience!
(a)
1.2
1 I/I0 0.8
0.6
0.4
0.2
0
4 2 0 2 4 x/w
The width is determined by the parameter w.
(b) The beam focuses down because the index of refraction is proportional to the intensity
and so is higher nearer to the axis. When passing through varying indices of refraction,
light always bends in the direction of higher index of refraction. This follows from
Snell’s Law or the law of optical distance given as a hint in part (c). In this case there is
positive feedback (i.e. the light bends towards the region of high index of refraction which
increases the intensity there which increases the index of refraction, …), so the light will
form a thin beam. After it focuses, it diverges and again focuses and the focusing and
diverging will be repeated in the medium, but the net result is a beam. (See, for example
http://focus.aps.org/v8/st26.html.)
(c) The variation in n drops over a distance
w, so we can make the approximation
of triangular focusing
L2 = w 2 + f 2
At the edge the index of refraction is n1,
while at the centre it is n1+n2I0, so
using the hint that all the light meeting
at a focus will have travelled the same
optical distance, we have n
Ln1 ~ f n1 + 2 I 0 2
where we use the average index of
refraction between the centre and the
edge. w
L f Page 3 of 5 n
∴ w 2 + f 2 n1 ~ f n1 + 2 I 0 2 2
n
n2 2
2
2
∴ w 2 + f 2 n1 ~ f 2 n1 + 2 I 0 = f 2 n1 + n1n 2 I 0 + 2 I 0 2
4 ( ) ∴f ~ wn1 → n1
w
n2 I 0 n1>> n 2 I 0 n n1 + 2 I 0 n 2 I 0 4
(d) If n2<0, then the beam will blow up instead of focusing, since the index of refraction will
now be lower in the high intensity region. 5) All the world in a drop of water
I used the top part of a CD case to support my drops. The overhead lights in my office
provided a nice light source. I had some linear graph paper which had 1mm squares; this
made it easy to measure the diameter of the droplets by simply resting the plastic on the graph
paper and counting the squares under the drops.
I measured the focal length by seeing how how high above my desk I had to hold the
drop to get a sharp image. I initially used a ruler, but then I realized that since I had 2 long
parallel fluorescent lights separated by 128±1cm at a distance 203±1cm above my desk, I
could get the focal length by measuring the separation of the images on the graph paper and
multiplying by 203/128. The drops were rather thick (1.5 mm for the smallest drops, almost
3mm for the biggest diameters), so the thin lens approximation was not very accurate, but
probably good enough.
Measuring the magnification was a bit tricky, but first you need to make your drop work
as a microscope. Since it is a microscope you have to put your eye very close to the droplet
(as the question says quite emphatically). If you don’t put your eye close, all you have is a
poor magnifying glass. When you do put your eye close, you have to move what your looking
at (e.g. your finger nail) nearer and farther until it is in focus. (When it does come into focus
the typical reaction, at least for Grade 5 students in my kids’ school, is to say “Wow!”.)
To measure the magnification for a drop, I first measured the diameter and focal length
of a drop using my graph paper. I then rested the CD case on a box of facial tissue which was
resting on its end. The box was 23cm long, which is very close to the 25cm standard reference
distance, so the drop was held at (approximately) the right distance above the table. I then put
a piece of graph paper on the table underneath the drop, and looked at another small piece of
graph paper held close underneath the drop. With my eyes very close to the drop I would
adjust the small graph paper and move my head until the small graph paper was in focus (often
with my eyelashes touching the CD case). I could then see both the graph paper on the table
and the magnified small graph paper, and could easily (if not terribly accurately) measure the
magnification by comparing the magnified and nonmagnified squares. This was hard to do
with the smallest drops. It also required a steady hand holding the small piece of graph paper. Page 4 of 5 Diameter (mm)
3.1±0.3
3.9±0.3
5±1
6±0.4
7.7±0.5
12.5±0.5
18±1
5±1
3±0.3 Focal Length (mm)
4.4±0.6
4.8±0.6
6.0±0.8
8.2±0.8
13.5±1
29±3
54±6
6.0±0.8
3.5±0.5 Magnification
40±5
32±4
30±3
22±3
17±2
9±1
4±1
30±3
40±3 (a) As you see from the above table, I found a 5 mm drop to have a focal length of 6±1mm.
(b) A 5mm drop has a magnification of 30±3.
(c) In general, the larger diameters gave larger focal lengths, larger field of view, and smaller
magnification. There are several different kinds of distortion. When looking at graph
paper, the lines were more curved near the edges for small drops, but for large drops it
was less likely all the field of view would be in focus and the drop was less likely to be
circular which lead to more distortion.
One thing I noticed was that difference support plastics made different height drops.
CD cases made very high (i.e. thick) drops, while the drops were lower on other plastics
(e.g. overhead transparencies, name tag holders, …). I assume this depended on how
hydrophobic the plastic is. I could also make the drops lower by waiting for them to
evaporate or poking them with a paper clip. Shorter drops had longer focal lengths and
less magnification.
You can see a nice example of a water droplet microscope in the journal Physics Education 36
(March 2001) 97101 (http://www.iop.org/EJ/S/1/NTO803278/abstract/00319120/36/2/301). If
you cannot access the IOP website, it is also available (although the beautiful colour photos are
only in black and white) from the Applied Spectroscopy Laboratory at the University of
Indonesia at http://www.geocities.com/spectrochemical/int_papers/ip34.pdf. Page 5 of 5 University of Toronto
20012002 Physics Olympiad Preparation Program
Problem Set 5: Electricity and Magnetism
Due Monday 11 March 2002
1) Purple Haze!
My old friend Steve studies top quarks using the CDF1 experiment which has a very large
superconducting solenoidal magnet with a uniform horizontal field of 1.5 Tesla. In addition to
physics, Steve also loves guitars.2 Is it safe for Steve to play his guitar inside the CDF magnet? In
particular, is he likely to shock himself or burn his fingers on the E4 string if it is vibrating with a
peaktopeak amplitude of 1mm?
Hints: The E4 string (330 Hz) on Steve’s guitar is 66 cm long, 0.23mm in diameter, and has
an electrical resistivity of 12 µΩ cm, a thermal conductivity of 80 W/m/K, a linear mass density of
0.3g/m, and a specific heat of 500 J/kg/K. (You may not need all this data, but we didn’t want to
leave out anything potentially useful.) The string vibrates sinusoidally in its lowest mode, and both
ends of the string are connected to pins at constant temperature (21°C). Assume that any heat
produced in the string can only escape through the ends of the string or through Steve’s fingers, i.e.
thermal radiation and convection are negligible. The time scale for heat flow out of the wire is
much longer than the oscillation period of the wire. Thermal conductivity and electrical
conductivity are completely analogous and the equations have identical form, i.e. electrical
conductivity relates charge flow and voltage differences, thermal conductivity relates heat flow and
temperature differences.
[David]
2) Relatively electric!
Near the surface of the earth there is typically a vertical electric field of about 120 V/m. (The
field can reach 20 kV/m underneath a thundercloud  this is why your hair may stand on end just
before you get hit by lightning  but we’ll ignore this interesting but irrelevant safety tip for the rest
of the problem.)
(a) Why aren’t people electrocuted by the typical 200 V potential difference between the air at
their feet and the air at their head?
(b) What is the typical surface electric charge density on the earth’s surface? Give your answer
in units of electron charges per metre squared.
(c) What is the typical surface magnetic field (magnitude and direction) at the equator produced
by this surface electric charge density, as measured by a magnetic field meter just above the
surface of the earth
1
2 http://wwwcdf.fnal.gov/cdf.html
http://webug.physics.uiuc.edu/courses/phys398/ Page 1 of 3 i)
if it is attached to the earth and rotates with it?
ii)
just above the surface of the earth, letting the earth rotate under it?
(d) Are either of your answers to the two parts of (c) consistent with the observed magnitude
(typically 30µT) or observed direction of the surface magnetic field at the earth’s equator?
Hints: My coordinates are such that an electric field pointing up is positive, pointing down is
negative. The surface of the earth, cows, and people, all have similar (noninfinite) resistivities.
Don’t worry about the fact that the meter sitting above the surface of the earth would be moving
supersonically, e.g. assume it is in a jet flying in an eternal sunrise, or sunset, or noon. The north
pole of a magnet is the one which is attracted to the north pole of the earth.
[David]
3) Spheres within spheres
An insulating sphere of radius R with uniform volume charge density of r has an electric field inside
r
r ρr
where r is the distance from the center of the sphere (r<R ). Note that this is a
it equal to E =
3ε0
vector formula.
(a) Consider two overlapping spheres of radii R1 and R2 with uniform charge densities ρ and ρ
and at a distance d from each other (d<R1+R2). Show that the electric field in the region of
overlap is uniform and calculate its magnitude and direction.
(b) If the two spheres have the same radii R and the distance between them is much smaller than
the radius (d<<R), you can consider the whole thing as one sphere with a surface charge
density σ.
i)
Find σ on the surface in terms of the angle that it makes with the line connecting the
two centres. What is the maximum surface charge density σ0? What is the electric field
inside the sphere in terms of σ0?
ii)
What is the electric potential, V, on the surface of the sphere in terms of the angle
that it makes with the line connecting the two centres? (The potential is defined to be
zero at infinite distance.)
[Yaser]
4) Does it (anti)matter?
When I was a kid watching Star Trek (during its original run), I sometimes wondered if antimatter
falls up. According to General Relativity it should fall down at the same rate as ordinary matter,
but some exotic theories predict it will fall with a different acceleration. Experiments trying to
observe the gravitational force on antimatter usually use conductors to shield the antiparticles from
external electromagnetic forces. Imagine a positron inside a solid piece of copper on the surface of
the earth. What is the vertical acceleration of the positron?
Hints: Assume General Relativity is correct. For the purposes of this question, a positron is
just like an electron, but with opposite charge. You can ignore the fact that the positron will
quickly annihilate with an electron, and assume it can move freely through the copper just like
conduction electrons. Before thinking about positrons, you may first want to think about the net
force on the free conduction electrons inside the copper.
[David] Page 2 of 3 5) Refrigerator Art
Experimentally and quantitatively determine the arrangement of north and south poles on a flat,
flexible refrigerator magnet. Describe what you did and what your conclusions are.
Hints: We are talking about the flexible magnetic cards which usually have advertising on
one side and a dark smooth surface on the back. We don’t care which poles are north and which
poles are south, but what is the arrangement of the poles. I figured it out just by playing around
with two similar fridge magnets and making a measurement or two with a ruler, but there may be
other methods.
[David] POPBits™ – Possibly useful bits of information
Constants and units3,4
Boltzmann constant
Earth equatorial radius
elementary (i.e. electron) charge
electron (or positron) inertial mass
electron volt
Newtonian gravitational constant
permittivity of free space
permeability of free space
speed of light in vacuum
standard acceleration of gravity
at the earth’s surface
StephanBoltzmann radiation constant k
R⊕ (1.3806503±0.0000024)×1023 J/K e
me (1.602176462±0.000000063)×1019 C eV
GN
ε0 µ0
c
g σ 6.378140×106 m
(510998.902±0.021) eV/c2
(1.602176462±0.000000063)×1019 J
(6.673±0.010)×1011 m3/kg/s2
8.854187817×1012 F/m
4π×107 N/A2
299 792 458 m/s
9.80665 m/s2
(5.670400±0.000040)×108 W/m2/K4 Interesting (?) Integrals ( a is a constant) ∫ sin(ax )dx = − 1 cos(ax )
a ∫ cos(ax )dx = 1 sin(ax )
a x
∫ sin2 (ax )dx = 2 − 41a sin(2ax ) x
∫ cos2 (ax )dx = 2 + 41a sin(2ax ) Great excuses for a party
Galileo’s birthday (1564)
Discovery of radioactivity (1896)
3
4 February 14
March 1 http://physics.nist.gov/cuu/Constants/index.html
http://pdg.lbl.gov/2000/contents_sports.html Page 3 of 3 20012002 Physics Olympiad Preparation Program
– University of Toronto – Solution Set 5: Electricity and Magnetism
dx 1) Purple Haze!
y
This can be done using the Lorentz force
x
law, but we’ll use Faraday’s Law. Assuming
the string is oscillating sinusoidally, the
induced electromotive force produced by the
string sweeping out an area A in the magnetic field B is πx dA
d L a pp
ϕ=B
=B ∫
sin(ωt) sin dx
L
dt
dt 0 2 app
L L
L πx ω cos(ωt) cos = −2 BLa ppν cos(ωt)
=B
π L 0
2
where ω=2πν is the radial frequency and ν=330Hz is the frequency of oscillation. The
magnitude of the peak e.m.f. is then
ϕ p = −2 BLa ppν = 2(1.5T )(0.66 m)(0.001m) 330 s−1 = 0.7V
This is not enough voltage to shock Steve, but is the string temperature safe? a pp By Joule’s and Ohm’s Laws, the power produced in a segment, dx, of the string with voltage
drop dV across its resistance dR is
2 a pp πx ω cos(ωt) sin dx B
AB 2 a 2 ω 2 πx L (dV ) 2 = 2
pp
cos2 (ωt) sin 2 dx
dP =
=
ρ
L
4ρ
dR
dx
A
where A is now the crosssectional area of the string and ρ is it’s resistivity. The average
power produced over one complete oscillation of the string is
dP = AB 2 a 2 ω 2
pp
4ρ 2 πx sin dx cos (ωt) =
L
2 AB 2 a 2 (2πν )
pp
4ρ 2 sin ∫ 02 π cos2 (ωt)dt dx
L
∫ 02 π dt 2 πx 1 1
2 π
−
sin(2ωt)
AB 2 a 2 π 2ν 2 2 πx 2 4ω
AB 2 a 2 π 2ν 2 2 πx 0
pp
pp
=
sin dx
=
sin dx
L
L
2π
2ρ
ρ
Since heat is being produced, the string will heat up. By symmetry, the middle of the string
must be the hottest point, and the net heat flow in the right half of the string (in the figure) is to
the right, and the flow in the left half is too the left. The string will heat up until it reaches a
steady state where the heat flowing through each segment just matches the joule heating of the Page 1 of 8 part of the string whose heat must flow out through the segment (i.e. integrated from the
midpoint to the segment):
x
x AB 2 a 2 π 2ν 2 πx ′ dQ
dT
pp
= −kA
= ∫ dP = ∫
sin 2 dx ′
L
dt
dx L / 2
2ρ
L /2 AB 2 a 2 π 2ν 2 x L L
AB 2 a 2 π 2ν 2 x ′ L 2πx ′ x 2πx pp
pp
=
=
sin
sin
−
−− 2ρ
2ρ 2 4 π L L / 2 2 4 4 π L (k is the thermal conductivity of the string.)
22 22
dT B a ppπ ν L 2πx ∴
=
L − 2 x + sin dx
8 kρ
π L The temperature profile of the string is thus
x
x B 2 a 2 π 2ν 2 dT
L 2πx ′ pp
T ( x ) − T (0) = ∫
dx ′ = ∫
L − 2 x ′ + sin
dx ′
dx ′
8 kρ
π L 0
0
2
B 2 a 2 π 2ν 2 pp
Lx − x 2 − L cos 2πx − 1
=
8 kρ 2π 2 L The peak temperature of the string is for x=L/2 B 2 a 2 ν 2 L2 π 2 pp T ( x) =
+ 1 + T (0)
8 kρ
4 (1.5T )2 (0.001m)2 (330 Hz )2 (0.66 m)2 π 2 + 1 + (21K + 273K )
= 8(12 × 10 −6 Ω ⋅ 10 −2 m )(80W / m / K ) 4 = 5100 K This means that Steve will burn his fingers, and if he tries to keep on playing with an insulated
pick the string will melt (the element with the highest melting point, 3800K, is carbon, but a
carbon fibre or diamond string would burst into flames first; the metal with the highest melting
point, 3695K, is Tungsten, see http://www.webelements.com). I think Steve should keep his
guitar out of the CDF magnet.
Alternate shorter approach from Alex:
The area swept out by the string is A= 660mm πx
∫ (1mm) sin 660mm dx = 420mm 2 0 and this area is swept out twice per cycle, so the average e.m.f. is ϕ ≅B dA
= 1.5T ⋅ 2 ⋅ 330 Hz ⋅ 420 mm 2 = 0.42V
dt The resistance is Page 2 of 8 ( ) −6
ρL 12 × 10 Ω ⋅ cm 66cm
ϕ ≅ BR =
=
= 1.9Ω
A
π (0.023cm ) So the total power is V2
ϕ ≅P=
= 93mW
R
Half the power is lost through the end of the string (with most power being produced in the
middle). Thus P σπd 2
∆T
=
2 4 L /2
LP
∴ ∆T =
= 4600 K
σπd 2
∴Tmiddle = Tend + ∆T ≈ 4900 K
2) Relatively electric!
(a) People are much better conductors than air, so they short out the field so their whole body
is at a constant potential. The energy density of the atmospheric “battery” is so low that
only an infinitesimal surface current flows through the body.
(b) The earth is a conductor, so it will have
zero field inside it and a charge density on
E=200V/m
its surface which can be calculated from
the electric field at the surface using
σ
Gauss’s Law. Near the surface of the earth
it looks like a plane, so we can just use a
E=0
Gaussian box with top (and bottom) area
A, i.e.
r
r
r
r
r
qinsidebox r
= E top ⋅ Atop + E bottom ⋅ Abottom + E sides ⋅ Asides
ε0 σA
= EA + 0 ⋅ A + E sides Asides cos π
2
ε0
σ
∴ =E
ε0
So the surface charge density is
∴ (8.854 × 10−12 F / m)(−120V / m) = −6.6 × 106 electron ch arg es / m 2
σ =ε E =
0 1.6 × 10−19 C / e
Note that the surface charge density is negative.
(c)
i)
In the rest frame of the earth’s surface, the magnetic field generated by the surface
charge density is zero, since it isn’t moving in this reference frame.
ii)
In this frame, the charge is moving with a velocity Page 3 of 8 2πR⊕
2π 6.378 × 10 6 m
m
=
= 464
1day 24 hours × 3600 s / hour
s
underneath the observer. This corresponds to a current sheet with a current density σv. By
symmetry and the BiotSavart Law, the magnetic field must run at right angles to the
velocity and parallel t to
B
North
the surface. The
v
Surface charges
⊗I
direction of the magnetic
up
field can be determined
L
to be from south to north
using the right hand rule. so we can use Ampere’s Law to calculate the magnetic field:
rr
∫ B ⋅ ds = 1 µ0 Ienclosed
2
v= loop ∴ BL = 1 µ0 Lσv
2
∴ B = 1 µ0σv = 1 µ0ε0 Ev =
2
2 Ev
2c 2 = 1β
2 E
=
c 1
2 (120V / m)(464 m / s) = 0.3 pT (3 × 108 m / s) 2 (d) So the magnetic field is the right direction, but much too weak.
3) Spheres within spheres
(a) The electric field in the region of overlap is
ρr −ρr2
ρ
E = E1 + E 2 = 1 +
=
(r1 − r2 )
3ε0
3ε0
3ε0
Choosing sphere 1 as the origin, and defining the x axis
to be the direction from the centre of sphere 1 to the
centre of sphere 2, then
ρ
E=
( x1 − x 2 , 0, 0)
3ε0
ρ
√
∴E =
dx
3ε0
(b)
i) The charges cancel where the spheres
overlap, so the surface charge density is
just σ(θ )= ρ t(θ ), where t(θ ) is the
thickness of the uncancelled charge layer
at the surface. The maximum charge
thicness is obviously at t(0 )=d, σ0=ρd. y
+ρ ρ
x
r2 r1 R1 R2 L2 L1 R θ +
d Using the cosine rule and the fact that the centre
of each sphere is ofset by ±d/2 from their combined
centre, we can calculate the thickness in the limit where d<<R, i.e. Page 4 of 8 t(θ ) = L2 (θ ) − L1 (θ )
= (L2 (θ ))2 − (L1 (θ ))2 d 2 2 Rd d 2 2 Rd
cosθ − R 2 + −
cos(π − θ )
= R2 + −
2
2
2
2
d
d ≅ R1 + cosθ − R1 − cosθ R
R = d cosθ
where we have used the approximation that 1 + ε ≅ 1 + 1 ε when ε<<1. So the surface
2 charge density is σ = ρ d cosθ = σ 0cosθ and the electric field strength inside the (overlapped) sphere is
σ
√
∴E = 0 x
3ε0
ii) Outside the spheres, this charge configuration is just an electric dipole with dipole
moment
p = Qsphered = 4 πR 3ρd = 4 πR 3σ 0
3
3
The electric potential outside the spheres is just the dipole potential
p⋅ r
p
V ( r) =
=
cosθ
3
4 πε0 r
4 πε0 r 2
so at the surface of the sphere
4 πR 3σ
Rσ 0
Rρd
Rσ
0
V ( r) = 3
cosθ =
cosθ =
cosθ =
3ε0
3ε0
3ε0
4 πε0 R 2 4) Does it (anti)matter?
Conduction electrons are free to move, so if they are at rest they must feel no net force. In a
piece of copper on the surface of the earth, the conduction electrons feel the force of gravity,
but they don’t all fall out the bottom of the copper, so there must be a force balancing gravity.
This force is an equal and opposite electrostatic force generated by the electrons themselves
sagging under the force of gravity. i.e. They start to fall but this very quickly causes the bottom
of the copper to become slightly negatively charged relative to the top, and the electrons will
stop moving when the charge imbalance is such that it exactly balances gravity everywhere
inside the conductor. The electrons natural arrange themselves for this to happen, and this is
just the same explanation why the electric field inside a conductor is always zero (except they
never mention that the effect of gravity is to create a tiny nonzero electrostatic field).
For a positron, the electrostatic force is the equal and opposite to that on a electron, but
the gravitational force is equal and in the same direction (according to General Relativity), so
if the net force on a conduction electron is zero, the net force on a positron must be Page 5 of 8 apositron = 2g =19.6m/s/s straight down.
(g is the standard acceleration of gravity at the earth’s surface.)
5) Refrigerator Art
Playing around with two fridge magnets, I noticed that if you stick them together
like this (where the white area is the
printed face and the dark area is the magnetic part), and then gently pull them like this, they
make a
“clack,
clack, clack” sound. Studying this effect more closely it is clear that what is happening is that
they start to repel each other as I pull them and then they are attracted, and then repelled, and so on, ,, , and each time they are attracted together they make a nice “clack!”. This only happens when I
pull them along their long axis, i.e. like this not when I pull them like this Page 6 of 8 This must mean that fridge magnet must consist of many magnetic domains (essentially little
dipole magnets) whose North and South poles must be arranged like NS NSNSNSN (I can’t tell which is north and which is south, but they must be alternating.). The positions
where the fridge magnets were attracted together were at 5, 9, 13, and 17 mm, so the width of
the domains is about 4 1/2 mm. (Your magnets may differ, although all the ones I tried seemed
similar.)
Looking from the side, the domains are probably arranged like this
NSNSNSNSNSNSNSNS
SNSNSNSNSNSNSNSN (A)
or less probably like this
(B) N SS NN SS NN SS NN SS N (C) N SN SN SN SN SN SN SN S . It cannot be like this since that us equivalent to this N S Page 7 of 8 (assuming all domains are equal strength magnets) which would not give the correct “clack,
clack behaviour.
To check that arrangement (A) is correct, I held the ends of the fridge magnets close to
each other. They either attracted or repelled, e.g.
NSNSNSNSNSNSNSNS
SNSNSNSNSNSNSNSN NSNSNSNSNSNSNSN
SNSNSNSNSNSNSNS When I flipped them over, they did the opposite e.g.
NSNSNSNSNSNSNSNS
SNSNSNSNSNSNSNSN (This took some care to see since the forces are weak and the main observable affect is not the
repulsion but the fact that the magnets tend to push to one side to try to align North to South
poles.) These observations are not consistent with arrangement (B), in which case the force
should not change direction. Page 8 of 8 University of Toronto
20012002 Physics Olympiad Preparation Program
Problem Set 6: Circuits and (a bit of) Modern Physics
Due Monday 8 April 2002
• You need a multimeter or ohmmeter for the last problem. If you don’t have one, ask your
Physics Teacher or a friend.
• Please be prompt with this problem set; we will be selecting students for our exciting
POPTOR training weekend right after the due date. 1) Everyone has the capacity to learn the quadratic formula!
Calculate the equivalent capacitance between points A and B of the following infinite circuit, if
all the capacitors are equal and of value C. A B
Hint: If you got rid of the two leftmost capacitors (one horizontal and one vertical), what
would the capacitance be of the remaining infinite chain?
[Isamu]
2) I always resist getting a shock
You may have noticed warnings on television sets or computer monitors not to mess about inside
even if the power is unplugged. This is because there are some large capacitors inside which are
charged up to high voltages – a potentially lethal combination. Large inductors can be equally
lethal but in a different way. If you have a circuit with a current flowing through a large inductor,
if you try to turn it off by opening a switch there can be spectacular sparks generated by huge
voltages across the switch. I work on the ATLAS1 experiment at the CERN2 laboratory and we
have a very large inductor. It is a superconducting solenoidal magnet 5.3m long and 1.2m in
radius. The solenoid is wrapped uniformly with about 1200 turns of superconducting wire which
carry a current of 8000A.
(a) Estimate the magnetic field at the centre of the solenoid.
1
2 http://pdg.lbl.gov/atlas/atlas.html
http://public.web.cern.ch/Public/, “Where the Web was born!” Page 1 of 4 (b) When we talk about “how big” a resistor is, we need to specify both its resistance (in
Ohms) and its power rating (in Watts). (The power rating tells us how much power the
resistor can dissipate continuously without melting, blowing up, bursting into flames, …
whatever.) How big a resistor would you need to put in series with the solenoid so that if
you had to switch the magnet off suddenly, the voltage across the switch would not
exceed 80V?
(c) With the resistor in parallel, how long will it take for the voltage across the open switch to
drop to a safe value of 8V?
[David]
3) Cracked!
After buying a new TV, I connect it to an satellite dish with a 10m long rubber coated copper
wire. I then discover that I have trouble picking up any channel below Channel 19 (500MHz). (I
switched to satellite after dealing with that pesky balloon in Problem Set 4, but I seem to be
doomed to never manage to get Fraggle Rock.) When I measure the resistance of the wire with a
DC ohmmeter I discover it has infinite resistance so I decide it must have a hairline crack of
width d (see figure, not to scale). The reason the low frequency channels can’t get through is that
d 10m
the wire has an RC time constant of about 1/(500MHz)=2×109s. What is the width of the crack,
d?
Hints: Assume the sides of the crack are parallel and straight, and that the crack is filled
with rubber. The resistivity of copper is 1.7×10−8Ω·m; the dielectric constant of rubber is κ=7.
Assume all resistances and capacitances other than the wire’s are negligible.
[David]
4) Gravity isn’t Bohring
Some cosmologies allow the possibility that other universes exist in which the forces and
constituents differ from our universe. Imagine a universe in which there is no long range
electromagnetic force, so the only long range force is gravity. Assume the proton, neutron
and the electron exist with exactly the same masses as in our universe, but there are no
electromagnetic forces between them. A hydrogen atom would still consist of an electron
bound to a proton, but it would only be bound by gravity.
(a) What would be the energy difference (in electron volts) between the ground state and the
first excited state of a hydrogen atom in the imaginary universe without
electromagnetism? Assume the value of Planck’s and Newton’s constants (and all the
other laws of physics except for electromagnetism) are the same in the imaginary universe
as in our universe.
Page 2 of 4 (b) Assume the current size of the imaginary universe is about the same as ours (radius ~
1010 light years). Can gravitationally bound hydrogen atoms exist in the imaginary
universe?
[David]
5) Resisting the light
When a light bulb heats up, does its electrical resistance go up or down?
Find an unbroken 120V incandescent light bulb; any typical size (15W, 25W, 40W, 60W,
75W, 100W) will do. (If you – temporarily – take a light bulb from a lamp, make sure that
the lamp is turned off, unplugged, the bulb has cooled down before removing it, and you have
permission from the owner of the lamp.)
(a) Measure the resistance of the light bulb when it is at room temperature. (Just use an
ohmmeter or multimeter; ask your physics teacher for help if you don’t have one at
home.)
(b) Calculate the resistance of the light bulb when it is turned on and hot under normal
operating conditions. (Do not try to measure the resistance while the light bulb is
plugged into a circuit! Just calculate the resistance from the bulb’s voltage – 120V  and
power rating, e.g. 60W.)
(c) Assuming your measurement and calculation are accurate, answer our original question:
When a light bulb heats up, does its electrical resistance go up or down?
[David] Page 3 of 4 POPBits™ – Possibly useful bits of information
Constants and units3,4
astronomical unit (mean earthsun distance) au
u 149597870660±20 m
(1.66053873± 0.00000013)×1027 kg k (1.3806503 ± 0.0000024)×1023 J/K e
me (1.602176462 ± 0.000000063)×1019 C (1.602176462 ± 0.000000063)×1019 J permittivity of free space eV
GN
ε0 permeability of free space µ0 4π×107 N/A2 Planck constant h (6.6260755 ± 0.000004)×1034 J·s atomic mass unit: (mass 12C atom)/12
Boltzmann constant
elementary (i.e. electron) charge
electron (or positron) mass
electron volt (eV)
Newtonian gravitational constant (510998.902 ± 0.021) eV/c2
(9.10938188 ± 0.00000072)×1031 kg
(6.673±0.010)×1011 m3/kg/s2
8.854187817×1012 F/m (4.13566727 ± 0.00000016)×1015 eV·s
proton mass
solar luminosity
speed of light in vacuum
standard acceleration of gravity
at the earth’s surface
StephanBoltzmann radiation constant
tropical year (2001) me (938271998 ± 38) eV/c2 LÍ (1.67262158 ± 0.00000013)×1027 kg
(3.846 ± 0.008)×1026W
299 792 458 m/s c
g σ
yr 9.80665 m/s2
(5.670400 ± 0.000040)×108 W/m2/K4
31556925.2 s Great excuses for a party
Most famous physics class demonstration ever
(Oersted discovers that electric currents generate magnetic fields) 3
4 April 1820 http://physics.nist.gov/cuu/Constants/index.html
http://pdg.lbl.gov/2000/contents_sports.html Page 4 of 4 20012002 Physics Olympiad Preparation Program
– University of Toronto – Solution Set 6: Electronics and Modern Physics
1) Everyone has the capacity to learn the quadratic formula!
Since this is an infinite chain, removing the two leftmost capacitors still leaves the same chain
with the same total capacitance CAB. A
C C CAB B
From this we see that we must have
1
1
1
=+
C AB C C + C AB ∴ C (C + C AB ) = C AB (C + C AB + C ) 2
∴ C AB + C AB C − C 2 = 0 −C ± C 2 + 4 C 2
−1 ± 5
=C
2
2
We can’t have a negative capacitance, so
5 −1
C AB =
C
2
∴ C AB = 2) I always resist getting a shock
(a) We can estimate the magnetic field using the formula for an infinitely long solenoid:
B = µ0 nI
N
N 1200
∴ B ≈ µ0 I = 4 π × 107 2
8000 A = 2.3T
L
A 5.3m
So the magnetic field will be about 2 Tesla.
(b) Since the resistance is in parallel, no current flows in it as long as current can flow
throught the superconducting (and hence zero resistance) solenoid, but when the solenoid
is switched off, all the current flows through the resistor and the voltage across the
resistor is the voltage across the switch. The initial current is just the current originally
flowing through the solenoid, and the maximum voltage is just the initial voltage:
Vmax = I 0 R
and the required resistance is
V
80V
R = max =
= 0.01Ω
I0
8000 A
Page 1 of 4 The peak power dissipated is
P = I 0Vmax = 8000 A × 80V = 0.64 MW
To be safe we should round up the size, so what we need is a 0.01Ω, 1MW resistor. This
is not something you find at your local Radio Shack1.
(c) The resistor and the solenoid form an LR circuit with an exponential time constant,
τ=L/R. The inductance per unit length for an infinite solenoid is
L
= µ0 n 2 × Area
l so the inductance of the solenoid is approximately
L = µ0 N2
l2 () πr 2 l = πµ0 N 2r2
l and the time constant is
τ= L
N 2r2
= πµ0
R
Rl = 4 π 2 × 10−7 kg ⋅ m 1200 2 (1.2 m)
C 2 (0.01Ω)5.3m 2 = 154 s The current decay is exponential, so to drop to one tenth of the original voltage will take
τ 1 = τ ln10 = (154 s)2.3 = 356 s
10 i.e. about 6 minutes.
3) Cracked!
The broken wire is equivalent to two resistances and the crack is a dielectric filled parallel
plate capacitor, all in series. Since they are part of the same wire, their resistivity, ρ, and
crosssectional area, A, are the same. The resistances of the two pieces of wire are L
R1 = ρ 1
A L
R2 = ρ 2
A And the total resistance is L + L2
L
R = R1 + R2 = ρ 1
=ρ
A
A
where L=10m is the total length of the wire.
The capacitance of the crack is C = κε 0 A
d where d is the width of the crack and κ is the dielectric constant.
The RC time constant is 1 http://www.radioshack.ca/en/ Page 2 of 4 L
A
L
τ = RC = ρ κε 0 = κε 0 ρ
A
d
d
So the crack width is d = κε 0 ρ L
τ
10 m = 7 ⋅ 8.854 × 108 F / m ⋅ 1.7 × 108 Ω ⋅ m 2 × 109 s
= 5 × 109 m
= 5nm
4) Gravity isn’t Bohring
(a) The energy levels of an atom are determined by the quantization condition that orbital
angular momentum is quantized in units of Planck’s constant over 2π, i.e.
h
Ln = mev n rn = n
= nh
n = 1, 2, 3,…
2π
Using a simple Bohr atomic model for our gravitational hydrogen atom, the attractive
gravitational force must balance the centripetal force, i.e.
2
mem p mev n
GN
=
2
rn
rn
Using our quantization condition and solving this gives
rn = n 2 h2
2
GN me m p The total (potential+kinetic) energy of the electron is thus
mem p
2
E n = 1 mev n − GN
2
rn
mem p
mem p
= 1 GN
− GN
2
rn
rn
2
= − 1 GN
2 = 3
m 2 me
p n 2 h2 −0.26 × 10−77 eV n2
The energy difference between the ground (n=1) and first excited state (n=2) is
1
1
∆E = −2.6 × 10−79 eV 2 − 2 = 2 × 10−78 eV
2
1
This is 79 orders of magnitude smaller than the corresponding energy in our universe
(10.2eV). Page 3 of 4 (b) The smallest radius of a gravitationally bound hydrogen atom is that of its n=1 ground
state
r1 = h2
2
GN me m p = 1.2 × 10 29 m = 1.2 × 1013 light years This is 3 orders of magnitude large than the universe, so a gravitationally bound hydrogen
atom cannot exist in the imaginary universe.
5) Resisting the light
(a) I used a 15W bulb, the measured resistance was about 85±1 ohms.
(b) The resistance of the bulb when operating should be R=V2/Power=(120V)2/W=960 ohms.
(c) The operating resistance is an order of magnitude higher than the measured room
temperature resistance, so the the resistance of the filament must increase with
temperature. Page 4 of 4 20022003 Physics Olympiad Preparation Program
– University of Toronto – Problem Set 1: General
Due 28 October 2002
Here are a few comments before starting your first problem set.
• If you want to know what kind of physics you might need to solve these problems, look at
the Physics Olympiad syllabus at http://www.jyu.fi/ipho/syllabus.html. You do not have
to know the syllabus by heart to do the problems, but you should recognize what you
need to know so you can look it up.
• Don’t forget to look at the information given in the POPBits™ section at the end of the
problem set. It sometimes has information helpful or necessary for particular problems.
• Pay attention to words like “estimate” or “about”. They indicate that the expected answer
is not exact because either the input data is not precisely known or approximations or
simplifying assumptions are necessary. Much of real physics is learning how to turn
insoluble exact problems into soluble approximations.
• “Nothing ventured, nothing gained.” Whether you finish a problem or not, please make
sure your reasoning and analysis are clear. If you write down nothing, we can only give
you zero. Your basic ideas may be right even if you make a mistake or get stuck.
1. Two Alberts
AE must throw a ball so that it reaches AM after hitting the ground one time. AM remains fixed
at his position. The distances shown in the diagram are in metres.
(a) What should AE do to achieve the above? Ignore air resistance.
(b) Is there more than one way for AE to throw the ball so that it hits the ground only once before
arriving at AM?
AE 10 AM
5
5 5 5 Page 1 of 2 2. Two violins
How much louder are two violins compared to one violin? Suppose that each violin produces a
single pitch (frequency) wave f(t)=Asin(2πνt+φ), where A is the constant amplitude, ν is the
frequency, and φ is a phase factor that might vary randomly in time for each instrument. You will
need to consider several cases.
3. Temperature of the planet Venus
All objects emit radiation. The energy flux density J of this radiation is the energy emitted per
unit time and per unit area. The StefanBoltzmann law is J=eσΤ4, where e=1 for a perfect
radiator, σ is a constant, and T is the temperature of the object. What is the equilibrium
temperature of the planet Venus? Assume that the only radiation incident on Venus is from the
Sun. Assume that Venus and the Sun are perfect radiators. If your calculated T is different from
the measured temperature of Venus of about 700 Kelvin, speculate on why this is the case.
4. Can the can roll?
Rest an empty aluminum pop can on its side and on a table. Bring a charged object close to the
can, but do not touch the can. Can you get the can to roll? If it does start to roll, why does it do
so? Can you halt the can’s motion by using the charged object, but without hitting the can?
Describe the charged object you used in this experiment.
5. Circuits
When the following circuit is connected, how bright are the bulbs A, B, C, D, and E relative to
one another? The bulbs are identical. If bulb A is removed from the circuit, what happens to the
brightness of B, C, D, and E? If bulb A is replaced by two new bulbs, what happens to the
brightness of the remaining bulbs? You can try this question experimentally and/or theoretically. A battery D B E C POPBits™ – Possibly useful bits of information
Constants and units1,2
StefanBoltzmann radiation constant
Mean SunVenus distance
Temperature of the Sun
Radius of the Sun 1
2 σ
RSV
TS
RS (5.670400±0.000040)×108 W/m2/K4
1.08×108 km
5800 K
6.96×105 km http://physics.nist.gov/cuu/Constants/index.html
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 Spring '12
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 Physics, Tess of the d'Urbervilles, Speak

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