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Unformatted text preview: FACULTY OF SCIENCE
FINAL EXAMINATION
PHYS 230 DYNAMICS OF SIMPLE SYSTEMS Examiner: Prof. R. Harris Friday December 8th, 2006 Associate Examiner: Prof. N. de Takacsy 2:00  5:00 pm INSTRUCTIONS: Answer all questions in the booklet provided. There are 5 questions at 10 points
each. Textbooks and/ or notes are not permitted. However, calculators are permitted,
and formulae are provided on the last page. Dictionaries, if required, are also
permitted. You may keep the question paper. This exam comprises 5 pages, With questions on pages numbered 1 through 3. PHYS 230 1 Remember that all questions carry points for writing down, in point form, the
steps of your “process”. 1. Process 5; solution 5. TWO boats are playing on a lake e or, rather, their drivers are playing. One boat, “Albert”,
is traveling due East (left to right), and the second, “Brigitte”, starts 500 metres to the
North, and tries to meet up with the ﬁrst. Brigitte Albert
.,_,_.— There are three scenarios. In all three, Albert always travels due East. At the beginning
of each scenario, Albert is always travelling at 4 m/s, and Brigitte at 5 m/s. Take the
origin of coordinates to be the original position of Albert. In each scenario, Brigitte always attempts to travel directly towards Albert, so
as to meet in the shortest possible time.. You are advised to solve all three scenarios by working in Brigitte’s frame of
reference: this should make the soluti0n(s) easier. (a) Albert has constant velocity 4 m/s due East, Brigitte has constant velocity 5 m/s. Show that, in order that they meet, Brigitte must travel at an angle 0 : arccos (4/ 5)
South of East. What is the time and place of the meeting? (b) Albert now accelerates at 0.01 m/s2, starting at 4 m/s due East, and in response, starting at 5 m/s, in the direction 0 : arccos (4/5) South of East, Brigitte accelerates
at 0.0125 m/s2. What is the time and place of the meeting?
(c) Brigitte maintains a constant speed (not constant velocity) of 5 m/s, starting in the direction 6 : arccos (4/5) South of East, But Albert accelerates at 0.01 m/sg,
starting at 4 m/s. Show that the meeting is impossible. What is the time of the closest distance of
approach? PHYS 230 {O 2. Process 5; solution 5. Two equal masses collide, along a straight line, in such a way that a small massless spring
is “trapped” between them. The spring at ﬁrst is attached to the right—hand mass, as
shown. The initial velocities are 31) and v, in the same direction, as shown. .m, 31) Win), 1) H .
spring
There are again three scenarios. (a) The spring at ﬁrst has its natural, unstretched, length.
What are the velocities of the two masses after the collision? (b) The spring at ﬁrst is compressed, and held at its compressed length by a string, so
that it has elastic energy 3mv2. The collision releases this energy by breaking the
string. What are the velocities of the two masses after the collision? (c) The spring is defective. When it is compressed during the collision, it does not expand
again. It gains energy from the masses, but does not give this energy back.
How much energy is stored in the spring? What are the velocities of the two masses after the collision? 3. Process 5; solution 5. A “simple” pendulum consists of a (point) mass m hanging on a (massless) string of length
6. Its angular displacement 6 is deﬁned with respect to the vertical equilibrium position,
as shown. When 0 is small, the energy of the pendulum has the form 1  1
E 2 57716202 + Emg€62 + constant
and this leads to the usual expression for the angular frequency of oscillation, to, not the same as (9, which is (4)2 = g/é. A more complicated pendulum is constructed from a (massless) T—shaped frame carrying
three equal masses at its corners. It hangs from the centre of its shortest side, as shown.  ,4
...——~"‘" 111
111
E I?
0 , ,9 ’
n n (a) Show that the energy of this arrangement, as a function of the same small angle 6, is 1 . 1
E 2 —2—m[€2 + 2h2]62 + Emg€02 + constant PHYS 230 3 and that the angular frequency of oscillation is given by w2 : gé/[ﬂ2 + 2h2] (b) The top right mass falls off. What is the new equilibrium position of the frame and
the remaining two masses? (c) What is the angular frequency of oscillation of the frame and the remaining two
masses? 4. Process 5; solution 5. Two men, each of mass 100 kg, stand at opposite ends of a diameter of a rotating turntable
of mass 200 kg and radius 3m. Initially, the turntable makes one revolution every 2 seconds.
The two men make their way to the middle of the turntable at equal rates. (a) Calculate the ﬁnal rate of revolution and the factor by which the kinetic energy of
rotation has been increased. (b) What force is responsible for this increase in energy? Do not attempt a direct calcu—
lation of the work done by this force! (c) At what radial distance from the axis of rotation do the men experience the greatest
centrifugal force as they make their way to the centre? Note: The moment of inertia of a solid disc, for a perpendicular axis through its centre, is alle For a point mass rotating around any axis, from which it has distance 7", the moment of inertia is mrz. 5. Process 7: solution 3. Notice: here, process is everything! A space—craft of proper length L0 is traveling at constant velocity V, relative to frame 3.
The nose of the space—craft is at the origin (90 : 0) at time t = 0. At this time, the origin
of frame coincides with that of 8. Also at this time, a laser beam is sent from the nose
of the spacecraft towards the back of the space—craft. (a) When, by space—craft time, does the laser beam reach the back of the spacecraft?
(b) At what time t1, measured in S, does the signal reach the back of the space—craft? (c) At what time, t2, measured in S, does the back of the space—craft pass the origin of
frame 5'? ((1) At what time, measured in , does the back of the space—craft pass the origin of
frame S? PHYS 230 Kinematics for constant
acceleration vf:vi+at CC — £170 : ’Uit—l— 20152 22% = "01.2 + 2a(:c — x0) Calculus of motion dc a
ﬁrzv
dc a
ﬁvza
dﬂ__.
Ep—
_dv$_ dvm
am—E_U$_(E Polar coordinates . A
A 17=7ir+r66 a: [f—r92]f+[ré+2f9]é
dA_ .A
a _00
dA $0 I —6T Rotational motion Useful Stuff . . . around the Centre of Mass [0 Z th" — R012 L} 2 low
I 1
K : §I0w2
Conservation laws
d —b —»
‘—P : Fem
dt t Z 0
KE 2 KC + K d ~ _,
aLC : Text Special Relativity l ’y : [1 — ,82]_2 with ,8 : v/c Lorentz transform:
:8 2 ﬁx — vt]; {2 'y[t — 053/02] 1: :7[:1’c+vt’] t=7[7§+vx’/c2] Lorentz contraction: Ax = 7A1)?
Time dilation: At = A7577 Calculus and other math ‘ dv
/ = 1n v — vol + constant
’0 — ’Uo ‘ d 1
/ ﬂ : — 1n [A122 :1: B+constant 2A
(1—a)p21—ap ifa<<1
‘1
e‘azl—a+§a2...a<<1 If b:e“ then azlnb ...
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 Winter '07
 Harris

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