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Unformatted text preview: AP® Physics C: Mechanics
2005 FreeResponse Questions The College Board: Connecting Students to College Success
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AP Central is the official online home for the AP Program and PreAP: apcentral.collegeboard.com. TABLE OF INFORMATION FOR 2005
CONSTANTS AND CONVERSION FACTORS
27 = 1.66 = 931 MeV/c ¥ 1 unified atomic mass unit, 10  1u UNITS
Name Symbol 10 9 giga G kilogram kg 10 6 mega M 10 3 kilo k centi c milli m micro µ nano n pico p mn = 1.67 × 10 −27 kg second s Electron mass, me = 9.11 × 10 −31 kg ampere A e = 1.60 × 10 C kelvin N0 = 6.02 × 10 23 mol −1
R = Universal gas constant, K mole mol hertz 8.31 J / ( mol K )
◊ Avogadro’s number, Hz Boltzmann’s constant, k B = 1.38 × 10 −23 J / K Speed of light, c = 3.00 × 10 8 m / s newton N Planck’s constant, h = 6.63 × 10 −34 J ⋅ s pascal Pa = 4.14 × 10 −15 eV ⋅ s hc = 1.99 × 10 −25 k = 1 / 4π θ sin θ cos θ tan θ C 0 0 1 0 V
Ω 30 1/2 3 /2 3 /3 H 37 3/5 4/5 3/4 farad = G 6.67 g = 9.8 m / s 10 11 F tesla T 45 2 /2 2 /2 1 degree
Celsius C 53 4/5 3/5 4/3 electronvolt eV 60 3 /2 1/2 3 90 3 m / kg s 2 1 atm = 1.0 × 10 5 N / m 2
= 1.0 × 10 5 Pa 1 electron volt, J
W henry k = µ 0 / 4π = 10 −7 (T ⋅ m) / A
¥ 1 atmosphere pressure, VALUES OF TRIGONOMETRIC
FUNCTIONS FOR COMMON ANGLES volt 2 = 9.0 × 10 9 N ⋅ m 2 / C 2  Acceleration due to gravity
at the Earth’s surface, 10 −12 1 0 ∞ ' Universal gravitational constant, 10 −9 ohm C / N⋅m
2 µ 0 = 4π × 10 −7 (T ⋅ m) / A Vacuum permeability,
Magnetic constant, 0 = 8.85 × 10 10 −6 coulomb ◊ Coulomb’s law constant, 0 10 −3 watt J⋅m −12 10 −2 joule = 1.24 × 10 3 eV ⋅ nm Vacuum permittivity, Symbol m Neutron mass, Magnitude of the electron charge, Prefix meter 2 −19 Factor kg m p = 1.67 × 10 −27 kg Proton mass, PREFIXES 1 eV = 1.60 × 10 −19 J 2 The following conventions are used in this examination.
I. Unless otherwise stated, the frame of reference of any problem is assumed to be inertial.
II. The direction of any electric current is the direction of flow of positive charge (conventional current).
III. For any isolated electric charge, the electric potential is defined as zero at an infinite distance from the charge. 2 ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2004 and 2005 t a= w Rs = t
12
t
2
a + + w = FM = qv × B
B•d = 1
f Bs = = ˆ
r e = p r2
Gm1m2
r 0 nI = B • dA dm
dt
dI
= −L
dt
1
U L = LI 2
2 e =
= p m g
Gm1m2 0I F= Id ×B m
k f = w
p = q FG 1
Ri z = 2 i 3 =− f q + w Tp 1
=
Rp Ri P = IV 12
kx
2 2 i z = w kx 0 2 = UG a = Ts A
V = IR w= u Us R= I 0t dQ
dt
1
1
Uc = QV = CV 2
2
2
I= m ¥=
w=
Â=
Ú=
0 1
1
=
Cs
i Ci m t = tÂ
=t ¥ Fs T rp
12
I
2 Ci i ∑ =
r m Cp = d ∑ Â
Â= mr mr 2 a
w u = w r 2 dm K q I net rcm
L m D = F 0A r = I r Q
V ∑ u r 2 0 q1 q 2
r Q
q
R
r
t
U
V =
=
=
=
=
=
=
=
=
m=
= k = r C= 4 1 f = u 2 C= ∑ Ú= ∑ ac UE = qV = r
u £ m F dr
12
K
m
2
dW
P
dt
P Fv
Ug
mgh k N p p acceleration
force
frequency
height
rotational inertia
impulse
kinetic energy
spring constant
length
angular momentum
mass
normal force
power
momentum
radius or distance
position vector
period
time
potential energy
velocity or speed
work done on a system
position
coefficient of friction
angle
torque
angular speed
angular acceleration p =
Ú= W D= = F fric = dp
dt
F dt
mv ma =
=
=
=
=
=
=
=
=
L=
m=
N=
P=
p=
r=
r=
T=
t=
U=
=
W=
x=
=
=
=
=
= ∑ =Â
=u u J
p Fnet a
F
f
h
I
J
K
k e + F ( F  = 2 0 ) + u u
2 12
at
2
2 a x x0 0t + = u x0 at + x 0 ELECTRICITY AND MAGNETISM
1 q1 q 2
A = area
F=
2
B = magnetic field
40r
C = capacitance
F
d = distance
E=
q
E = electric field
= emf
Q
E • dA =
F = force
0
I = current
dV
L = inductance
E=−
= length
dr
n = number of loops of wire
qi
1
V=
per unit length
4 0 i ri
P = power
p MECHANICS charge
point charge
resistance
distance
time
potential or stored energy
electric potential
velocity or speed
resistivity
magnetic flux
dielectric constant ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2004 and 2005
GEOMETRY AND TRIGONOMETRY
area
circumference
volume
surface area
base
height
length
width
radius d f d f du
dx du dx
dn
x
nxn 1
dx
dx
e
ex
dx
d
1
1n x
dx
x
d
sin x cos x
dx
d
cos x
sin x
dx
1 n1
xn dx
x ,n
n1
ex dx ex
dx
ln x
x
cos xdx sin x
(
(
( =) ( =) = Ú = Ú Ú
Ú
4 sin xdx =
= p q a
b = p
p p Ú
b + p
p p q + a
90° cos x 1 π ( =) c b
c = =) q
q tan = =) cos  A=
C=
V=
S=
b=
h=
=
w=
r= = Rectangle
A = bh
Triangle
1
A = bh
2
Circle
A = r2
C=2r
Parallelepiped
V = wh
Cylinder
V = r2
S = 2 r + 2 r2
Sphere
43
V=
r
3
S = 4 r2
Right Triangle
a 2 + b2 = c2
a
sin =
c CALCULUS 2005 AP® PHYSICS C: MECHANICS FREERESPONSE QUESTIONS
PHYSICS C
Section II, MECHANICS
Time—45 minutes
3 Questions
Directions: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions,
which are worth 15 points each. The parts within a question may not have equal weight. Show all your work in the
pink booklet in the spaces provided after each part, NOT in this green insert.
Mech. 1.
A ball of mass M is thrown vertically upward with an initial speed of 0 . It experiences a force of air resistance
k v , where k is a positive constant. The positive direction for all vector quantities is upward.
given by F
Express all algebraic answers in terms of M, k, 0 , and fundamental constants. u = u (a) Does the magnitude of the acceleration of the ball increase, decrease, or remain the same as the ball moves
upward?
____ increases ____ decreases ____ remains the same Justify your answer. u (b) Write, but do NOT solve, a differential equation for the instantaneous speed
as the ball moves upward. of the ball in terms of time t (c) Determine the terminal speed of the ball as it moves downward.
(d) Does it take longer for the ball to rise to its maximum height or to fall from its maximum height back to the
height from which it was thrown?
____longer to rise ____longer to fall Justify your answer.
(e) On the axes below, sketch a graph of velocity versus time for the upward and downward parts of the ball’s
flight, where t f is the time at which the ball returns to the height from which it was thrown. Copyright © 2005 by College Entrance Examination Board. All rights reserved.
Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). GO ON TO THE NEXT PAGE.
5 2005 AP® PHYSICS C: MECHANICS FREERESPONSE QUESTIONS
Mech. 2.
A student is given the set of orbital data for some of the moons of Saturn shown below and is asked to use the
data to determine the mass M S of Saturn. Assume the orbits of these moons are circular. 1.18
1.63
2.37 ¥
¥
¥
¥ 8.14 Orbital Radius, R
(meters) 10 4 1.85 10 5 2.38 105 2.95 10 5 3.77 ¥
¥
¥
¥ Orbital Period, T
(seconds) 108
108
108
108 (a) Write an algebraic expression for the gravitational force between Saturn and one of its moons.
(b) Use your expression from part (a) and the assumption of circular orbits to derive an equation for the orbital
period T of a moon as a function of its orbital radius R.
(c) Which quantities should be graphed to yield a straight line whose slope could be used to determine Saturn’s
mass?
(d) Complete the data table by calculating the two quantities to be graphed. Label the top of each column,
including units.
(e) Plot the graph on the axes below. Label the axes with the variables used and appropriate numbers to indicate
the scale. (f) Using the graph, calculate a value for the mass of Saturn.
Copyright © 2005 by College Entrance Examination Board. All rights reserved.
Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). GO ON TO THE NEXT PAGE.
6 2005 AP® PHYSICS C: MECHANICS FREERESPONSE QUESTIONS TOP VIEWS
Mech. 3.
A system consists of a ball of mass M 2 and a uniform rod of mass M 1 and length d. The rod is attached to a
horizontal frictionless table by a pivot at point P and initially rotates at an angular speed , as shown above
1
left. The rotational inertia of the rod about point P is M 1d 2 . The rod strikes the ball, which is initially at rest.
3
As a result of this collision, the rod is stopped and the ball moves in the direction shown above right. Express all w w answers in terms of M 1 , M 2 , , d, and fundamental constants. (a) Derive an expression for the angular momentum of the rod about point P before the collision. u (b) Derive an expression for the speed of the ball after the collision. (c) Assuming that this collision is elastic, calculate the numerical value of the ratio M 1 M 2 . (d) A new ball with the same mass M 1 as the rod is now placed a distance x from the pivot, as shown above.
Again assuming the collision is elastic, for what value of x will the rod stop moving after hitting the ball? END OF SECTION II, MECHANICS Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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This note was uploaded on 02/25/2010 for the course PHY 15166 taught by Professor Mr.donald during the Spring '10 term at 4.1.
 Spring '10
 mr.donald
 Physics, mechanics

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