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Don't forget to review the old stuff.
RIII-1. Don't forget to review the old stuff.
y
The free-body diagram for a car rounding a
banked curve is shown. The car is going so
fast that it is about to slip. The coordinate
system has been chosen.
What is the correct Y-equation?
N

x
a
A: N sin + sN cos – mg = 0
B: N – sN sin – mg cos = 0
C: N – mg cos = 0
D: N cos – sN sin – mg = 0
E: None of these

sN
mg
Answer: N cos – sN sin – mg = 0
RIII-2. A projectile is fired straight up and then it comes back down to its original height. There is air
resistance as the projectile is moving. During the entire flight of the projectile, the total work done by
the force of gravity is..
A: zero.
B: positive
C: negative
During the flight of the projectile, the total work done by the force of air resistance is...
A: zero.
B: positive
C: negative
During the flight of the projectile, the total work done by the net force is...
A: zero.
B: positive
C: negative
Answers:
Q1: Zero. The work done by gravity is negative on the way up, positive on the way down. The total
work done by gravity is zero.
Q2: Negative. The force of friction always opposes the motion, so the work done by friction is
negative both on the way up and on the way down.
Q3: Negative. Work done by net force = Work done both by gravity and by friction, so Wnet =
Wgrav+Wfric. Wgrav=0 and Wfric<0 (from above questions) so Wnet<0.
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RIII-3. A pendulum is launched in two different ways. During both
launches, the bob is given an initial speed of 3.0 m/s and the same
initial angle from the vertical. On launch 1, the speed is upwards,
on launch 2, the speed is downwards.
Which launch will cause the pendulum to swing the largest angle
from the equilibrium position to the left side?
A: Launch 1 B: Launch 2
maximum displacement.
1
2
C: Both launches give the same
Answer: Both launches give the same maximum displacement. Apply conservation of energy, Etot =
KEi + PEi = KEmax = PEmax. For both launch 1 and 2, the initial KE and the initial PE are the same. So
the Etot is the same for both launches.
RIII-4. A block of mass m with initial speed v slides up a frictionless ramp of height h inclined at an
angle  as shown. Assume no friction.
m
m

v
h

True A or False B: Whether the block makes it to the top of the ramp depends on the mass of block
and/or on the angle .
Answer: false. The block will make it to the top if (1/2)mv2 > or = mgh. The m’s cancel. Whether the
block makes it to the top depends only on v and h.
RIII-5. A block with an initial speed vo slides up a rough incline. A student wishes to know the speed
v when the block is at a height h. The friction between the block and the incline has coefficient of
kinetic friction K.
v=?
m K v
True(A) or False (B).
vo
h
1
1
mvo2  mv2  mgh
2
2
Answer: False. There is friction and thermal energy generated in this situation. So
KEi + PEi = KEf + PEf + Ethermal (Ethermal = heat generated = |Wfric | )
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RIII-6. A mass m is placed into a spring-loaded gun, with the spring compressed by an amount x.
The gun is fired so the mass is pushed forward and slides down a frictionless surface of height h1, and
then up to a final height h2 as shown. What is the final kinetic energy of the mass m when it is at its
final height h2? (NOTE! This question is asking for the KE, not the speed v).
x
m
h1
1
A) 2
C)
v
h2
2
kx
 2mg(h 2  h1 )
B)
kx 2  mg(h1  h 2 )
k 2
x  2g(h 2  h1 )
m
D)
k 2
x  2g(h 2  h1 )
m
E) None of these is the correct expression for the KE.
KEi  PEi  KE f  PE f
Answer:(A) 0  mgh1 
1
2
mv 2 
1
2
1
2
kx 2

1
2
mv 2  mgh 2
kx 2  mgh1  mgh 2
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RIII-7. A pendulum consists of a mass m at the end of a string of length L. When the string is
vertical, the mass has a speed vo, as shown.
What is the maximum height h, above the lowest point, to which the
mass swings?

2g
A) 2
vo
m
h
vo
B)
v 2o
D)
2g
v2
Answer: o Using conservation of energy:
2g
vo
2g
C)
mv 2o
2g
E) None of these
KEi  PEi  KE f  PE f
1
2
mv0 2  0
1
2
v0 2
 0  mgh
 gh
(m's cancel)
RIII-8. Ball 1 of mass m moving right with speed v bounces off ball 2 with mass M (M > m), and then
moves left with speed 2v.
What is the magnitude of the impulse I = p of ball 1?
M
m
m
v
M
2v
A) mv
B) 2mv
E) None of these
C) 3mv
By how much did the momentum of ball 2 decrease?
A) Mv
B) 2Mv
C) 3mv
D) (1/2)mv
D) (1/2)mv
E) zero
Answers: Let's choose right as the +x direction. p = pf – pi = –2mv – mv = –3mv. p  3mv . Or



graphically, p f  pi  p
pf
pi
p
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p  3mv is the same for both ball 1 and ball 2. The momentum of ball 2 decreased by 3mv. By
conservation of momentum, the change in momentum of ball 2 must be equal in magnitude (and
opposite in sign) to the change in momentum of ball 1.
RIII-9. A bullet of mass m traveling horizontally with initial speed v strikes a wooden block of mass
M resting on a frictionless table. The bullet buries itself in the block, and the block+bullet have a final
speed vf.
M+m
v
m
M
vf
Before
After
Fill in the blank: The total kinetic energy of the bullet+block after the collision is ___________ the
total KE before the collision.
A) greater than
B) less than
C) equal to
True(A) or False(B): In the problem above, the collision was elastic.
Answer: KE was lost in the collision. The total KE after the collision is less than before the collision.
There was lots of friction when the bullet hit the block and so lots of thermal energy was generated.
The collision was NOT elastic.
RIII-10. Two balls, labeled 1 and 2, collide. Their initial momenta and the final momentum of ball 1
is shown in the figure.
What is the x-component of the final momentum of ball 2?
A: –2 B: –1
C: 0
D: +1 E: +2
What is the y-component of the final momentum of ball 2?
A: –4 B: –3
C: –1
D: 0
E: +4
p1f
p1i
p2i
Answers: p2f,x = +2, p2f,y = 0 Total momentum must be conserved.
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RIII-11.
What is the torque about the origin?
A) r F sin
B) r F cos
Answer: = r F = r F cos (Whether you use sin or cos
depends on which angle is used!)

F
r
axis
Try again. What is the torque about the origin?
A) r F sin
B) r F cos
F

r

Answer: = r F = r F sin(Notice that this angle is different than the one
in the previous problem.)
axis
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RIII-12. A very light rod of length L has a mass M at the top end. The bottom end is placed on the
ground and the rod is released from rest at an angle q. It falls over, rotating about the bottom end.
Immediately after the rod is released, the upper end…. Hint: net = I 
M
L

has angular velocity 
A) zero
B) not zero.
has angular acceleration 
A) zero
B) not zero.
has tangential acceleration atan
A) zero
B) not zero.
has radial acceleration ar
A)zero
B) not zero.
As the rod falls over (as  decreases to zero), the angular acceleration 
A) increases
B) decreases
C) remains constant.
Answers: Initially (t = 0+)  = 0,   0, atan = r  0, aradial = v2/r = 0 (since v = r = 0, initially).
The instant the rod is released, there is a net torque on the rod due to gravity, so there is an  =  / I,
but  is initially zero. It takes time for the  to build up.
 is increasing since the torque  = rF = L mg cos is increasing. The component of the (straightdownward) weight mg that is perpendicular to the rod is increasing.
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RIII-13.
Consider a light rotating rod with 4 point masses attached
to it at distances 2 and 4, as shown. The axis of rotation is
thru the center of the rod, perpendicular to its length. The
4 masses are moved to new positions so that all masses are
now 3 units from the axis. Did the moment of inertia I
A: increase
B: decrease
2
4
C: remain the same
3
Answer: Let's say the units of length are cm. For the first
rod, I   mi ri 2  2m(22  42 )  2m(20cm)
i
For the second rod, I   mi ri 2  2m(32  32 )  2m(18cm) . So the first rod has larger moment of
i
inertia.
RIII-14. A frictionless sliding hockey puck and a rolling hoop each have the same mass m and the
same initial speed v. Each has more than enough initial KE to travel up an inclined plane a height h as
shown.
m
h
v
m
v
Which object has more total KE at the top of the hill?
A) Hoop
B) puck
C) Same KE at the top.
Answer: The one with the largest initial KE, which is the hoop.
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RIII-15. An irregularly-shaped object of total mass M is free to
rotate about a frictionless axis. The moment-of-inertia about the axis
is I. The object, initially at rest, is acted on by a torque of constant
magnitude  for a time t. There are no other torques acting. After the
time t, what is the final angular velocity  of the bar?
A)
t
I
B)
It

D) 0
E) None of these.
Answer:
t
I
Use
C)
  I  and  =
axis
M, I
t
M
M, I

t
RIII-16.
A disk of mass M and area A and radius R is reshaped into a uniform square of the same area A (and
the same mass M). The edge length L of the square is related to the radius R of the disk by L2 = R2
(same areas).
Consider an axis of rotation through the center of the disk/square perpendicular
to the plane of the disk/square. The moment of inertia I of the square is ..
A) larger
B) smaller
C) the same as
that of disk.
Answer: The square has larger moment-of-inertia than the disk. This is pretty tricky: to make the
square, starting with the disk, we have to remove some mass from the disk and move it outward to fill
out the corners of the square. So we are moving mass outward to make the square, and so the square
has more mass further f