The next three questions pertain to the situation described below.
A harmonic wave propagates horizontally along a taut
string of length 𝐿 = 8.0 m and mass 𝑀 = 0.23 kg. The
vertical displacement of the string along its length is given
by 𝑦(𝑥, 𝑡) = 0.1 m cos(1.5 𝑡 + 0.8 𝑥 ), where 𝑥 is
measured in meters and 𝑡 in seconds.
y
x
L
11. (7 points)
What is the tension, 𝑇, in the string ?
a. 𝑇 = 0.10 N
b. 𝑇 = 0.21 N
c. 𝑇 = 0.29 N
vprop
!
= =
k
d. 𝑇 = 0.049 N
e. 𝑇 = 0.078 N
s
T
!2
M !2
=) T = µ 2 =
µ
k
L k2
✓
◆2
0.23 1.5
=
= 0.10 N
8
0.8
12. (7 points)
What is the maximum acceleration in the 𝑦-direction of any point along the string?
a. 𝑎<=> = 1.50 m/s2
y(x, t) = A cos(!t + kx)
b. 𝑎<=> = 0.375 m/s2
c. 𝑎<=> = 0.150 m/s2
d. 𝑎<=> = 0.015 m/s2
e. 𝑎<=> = 0.225 m/s2
d2 y
= ay (x, t) = ! 2 A cos(!t + kx)
2
dt
2
=) amax,y = ! 2 A = (1.5)2 ⇥ 0.1 = 0.225 m/s
13. (5 points)
In which direction is the wave propagating?
a. +𝑥 direction
b. – 𝑥 direction
c. +𝑦 direction
y(x, t) = A cos (!t + kx)
When time (𝑡) increases, then the position
(𝑥) must decrease in order for the value of
the cosine to remain constant.
The next three questions pertain to the situation described below.
A block (which you should treat like a point particle) has mass 𝑚 = 1.515 kg
and slides with initial speed 𝑣B in the +𝑥 direction. It collides with a rod of
length 𝐿 = 1.3 m and mass 𝑀 = 30.3 kg, which is initially at rest and
oriented perpendicular to the path of the block. The block hits the rod a
distance 𝐷 = 0.39 m from the center of the rod. After the collision, the
block is at rest and the rod spins with angular velocity of 𝜔E = 28 rad/s.
Everything is on top of a horizontal frictionless table, and the rod has a fixed
frictionless pivot through its center that allows it to rotate freely but keeps its
center from moving.
14. (7 points)
Which of the following statements best describes the collision?
a. Neither the angular momentum about the pivot nor the linear (translational) momentum are conserved.
b. The angular momentum about the pivot is conserved, but the linear (translational) momentum is not conserved.
c. The angular momentum about the pivot and the linear (translational) momentum are both conserved.
d. The linear (translational) momentum is conserved, but the angular momentum about the pivot is not conserved.
e. There is not enough information provided to answer this question.
The pivot exerts an external force on the block/rod system to keep the center of the rod from
sliding to the right upon impact of the block.
15. (7 points)
What was the initial speed of the block?
a. 𝑣B = 808.89 m/s
b. 𝑣B = 46.99 m/s
c. 𝑣B = 93.98 m/s
d. 𝑣B = 1213.33 m/s
e. 𝑣B = 202.22 m/s
Li = Lf
mvi D = Irod !f =
=) vi =
1
2
12 M L
✓
◆
1
M L 2 !f
12
!f
mD
= 202.22 m/s
=
1
2
12 (30.3)(1.3)
28
(1.515)(0.39)
16. (5 points)
Suppose the experiment is repeated using new block that has the same mass and initial speed as in the original
situation, but is made of a different material so that it bounces back and ends up moving in the −𝑥 direction after
the collision. How would the final angular velocity of the rod in this new case, 𝜔HIJ , compare to the final angular
velocity of the rod in the original case 𝜔E ?
a. 𝜔HIJ = 𝜔E
b. 𝜔HIJ > 𝜔E
c. 𝜔HIJ < 𝜔E
In the second case, the block will have a negative value of angular momentum (if it
had positive before the collision), so to conserve angular momentum, the rod must
spin faster in the second case.
The next three questions pertain to the situation described below.
A block of mass 𝑀 slides on a frictionless surface with a velocity 𝑉. It
strikes an identical block of mass 𝑀 that is at rest and is attached to an
ideal spring of spring constant 𝑘. Before the collision, the spring is
not compressed and the collision takes place very quickly before the
spring has time to compress appreciably. After the collision, the
blocks stick together.
𝑉 𝑀 𝑘 𝑀 17. (7 points)
Immediately after the collision, the blocks slide to the right and the spring is compressed. By how much is the
spring shortened, 𝛥𝐿, before the blocks are pushed to the left?
a. 𝛥𝐿 =
2𝑀𝑉 P /𝑘
b. 𝛥𝐿 =
𝑀𝑉 P /(2𝑘)
c. 𝛥𝐿 =
𝑀𝑉 P /𝑘
d. 𝛥𝐿 =
4𝑀𝑉 P /𝑘
e. 𝛥𝐿 =
𝑀𝑉 P /(4𝑘)
Pi = Pf =) M V = (2M )Vf
=) Vf = V /2
✓ ◆2
1
V
1
=) Etot = (2M )
= MV 2
2
2
4
1
1
=) k( L)2 = M V 2
2
4
r
M
=) L =
V
2k
18. (7 points)
After the collision, the blocks undergo simple harmonic motion. How much time does it take after the collision for
the blocks to return to their starting position (spring at equilibrium length) for the first time ? (Hint: the blocks will
oscillate through this position repeatedly. The problem refers to the first occurrence.)
a. 𝛥𝑡 = 4𝜋
2𝑀
𝑘
b. 𝛥𝑡 = 2𝜋
𝑘
2𝑀
c. 𝛥𝑡 = 2𝜋
2𝑀
𝑘
d. 𝛥𝑡 =
1
𝑘
2𝜋 2𝑀
e. 𝛥𝑡 = 𝜋
2𝑀
𝑘
r
2⇡
k
! = 2⇡f =
=
P
Mtot
r
Mtot
=) P = 2⇡
k
r
2M
t = P/2 = ⇡
k
19. (5 points)
After the collision, the blocks undergo simple harmonic motion. If the initial velocity 𝑉 is increased, the frequency
of the oscillation
a. decreases.
b. stays the same.
c. increases.
The initial speed of the block will affect the oscillation
amplitude. However, for simple harmonic motion, oscillation
frequency does NOT depend on oscillation amplitude.
The next three questions pertain to the situation described below.
A T-shaped object is made by combining two identical,
uniform rods of equal mass 𝑀 and length 𝐿. The
thickness of each rod is negligible. The object can be
rotated around the four different axes shown by the
dashed lines.
~g
20. (5 points)
Rank in increasing order the moments of inertia, 𝐼T to 𝐼U for rotation about the dashed lines.
a. 𝐼P < 𝐼T < 𝐼U < 𝐼V
b. 𝐼T < 𝐼P < 𝐼U < 𝐼V
c. 𝐼T < 𝐼P < 𝐼V < 𝐼U
1
M L2
12
1
I2 = M L2
3
I1 =
1
4
M L2 + M L2 = M L2
3
3
1
7
I4 = M L2 + M (L/2)2 =
M L2
3
12
I3 =
21. (7 points)
Now consider the situation depicted above in diagram #2 and imagine that the object is oscillating as a pendulum
around the axis shown. Gravity acts downward as indicated. What is the correct expression for the oscillation
frequency, 𝜔 ?
Rod at top of “T” does not contribute to oscillation frequency
3𝑔
a. 𝜔 =
because it has no moment of inertia about the given axis
2𝐿
(width of each rod is neglected).
b. 𝜔 =
2𝑔
3𝐿
c. 𝜔 =
2𝑔
𝐿
d. 𝜔 = 2
e. 𝜔 =
𝑔
𝐿
1 𝑔
2 𝐿
22. (7 points)
Now consider the situation depicted above in diagram #3. If the mass of each rod is 𝑀 = 2.1 kg and each has a
length of 𝐿 = 1.7 m, what is the moment of inertia of the object around the axis shown?
a. 𝐼V = 5.1 kg m2
b. 𝐼V = 6.1 kg m2
c. 𝐼V = 8.1 kg m2
d. 𝐼V = 7.4 kg m2
e. 𝐼V = 4.0 kg m2
1
I3 = Itop,3 + Ivert,3 = 0 + M D2 + M L2
3
1
4
D = L =) I3 = M L2 + M L2 = M L2
3
3
4
= (2.1 kg)(1.7 m)2 = 8.1 kg m2
3
The next three questions pertain to the situation described below.
A uniform rod of length 𝑑 and mass 𝑀 is inclined at 45° with respect to the
horizontal. On one end of the rod is an ideal pivot. A massless string is
attached to the other end of the rod. The string runs vertically upward, then
over a frictionless pulley, and a block of mass 𝑚 is hung from the other end of
the string. In this configuration, the system is in equilibrium.
23. (7 points)
𝑚 𝑑 What is the mass of the block, 𝑚 ?
𝑀 a. 𝑚 = 3𝑀
b. 𝑚 = 2𝑀
c. 𝑚 = 2𝑀/3
d. 𝑚 = 𝑀/2
e. 𝑚 = 𝑀/4
𝑔 45˚ ✓
◆
d
Mg
cos 45 = T (d cos 45 )
2
✓
◆
d
T = mg =) M g
cos 45 = mg (d cos 45 )
2
=) m = M/2
24. (5 points)
What is the horizontal force on the pivot?
a. 0
b. 𝑀𝑔/ 2
c. 𝑀𝑔
The tension in the rope acts vertically, so the rope exerts no horizontal force on the
rod. Thus, the pivot can exert no horizontal force or the rod’s CM would
accelerate horizontally.
25. (7 points)
Consider the case where the mass of the rod is 𝑀 = 30 kg, and its length is 𝑑 = 3 m. At a particular moment, the
string is cut and the rod is free to rotate about the pivot. What is the angular acceleration, 𝛼, of the beam
immediately after the string is cut?
a. 𝛼 = 13.9 rad/s2
b. 𝛼 = 3.47 rad/s2
c. 𝛼 = 4.91 rad/s2
d. 𝛼 = 6.94 rad/s2
e. 𝛼 = 2.45 rad/s2
↵=
⌧
Mg
d
2 cos 45
1
2
3Md
=
Irod,end
3g
3g
3 · 9.81
=p = p
= p
2 2·d
8d
3 8
2
= 3.47 rad/s
The next three questions pertain to the situation described below.
Mass 𝑚T initially has a speed of 𝑉\ in the +𝑥 direction along a
frictionless horizontal floor. It collides with mass 𝑚P , which is
initially at rest. Immediately after the collision 𝑚T is moving in
the – 𝑥 direction with speed 𝑉TE = 𝑉\ /4 and 𝑚P is moving in
the +𝑥 direction with speed 𝑉PE = 𝑉\ /2.
26. (5 points)
Is this collision elastic?
a. There is not enough information provided to determine
whether the collision is elastic.
b. No, the collision is not elastic.
Vapproach = V0
c. Yes, the collision is elastic.
Vseparate = V0 /4 + V0 /2 = 3V0 /4
=)Vapproach > Vseparate
27. (7 points)
Which of the following correctly expresses the relationship between the masses?
a. 𝑚T = 2𝑚P
b. 𝑚T = 𝑚P /2
c. 𝑚T = (4/3)𝑚P d. 𝑚T = (3/4)𝑚P
e. 𝑚T = (2/5)𝑚P
m1 V0 = m2 V0 /2
5
1
m1 = m2
4
2
2
=) m1 = m2
5
m1 V0 /4 =)
28. (7 points)
Mass 𝑚P continues to the right and slides up a frictionless ramp. It momentarily comes to rest at a maximum
height ℎ above its starting height on the floor before sliding back down. Which of the following correctly expresses
the maximum height ℎ reached by 𝑚P ?
a. ℎ = 𝑉\P /(2𝑔)
b. ℎ = 4𝑉\P /(3𝑔)
c. ℎ = 𝑉\P /(8𝑔)
d. ℎ = 3𝑉\P /(4𝑔)
e. ℎ = 𝑉\P /(4𝑔)
Em2 ,after collision = Em2 ,top of ramp
✓ ◆2
1
V0
m2
= m2 gh
2
2
V02
=) h =
8g
The next three questions pertain to the situation described below.
Block A has mass 𝑀^ = 4 kg and slides on a frictionless inclined
plane. It is connected to a hanging block B of mass 𝑀_ by a massless
string that runs over a frictionless pulley. The incline makes an angle of
𝜃 = 30° with horizontal. The acceleration of block A up the incline
is 𝑎 = 2.1 m/s2 .
29. (7 points)
As block A moves a distance 1.3 m along the incline, what is the work done on it by the tension in the string, 𝑊^,b ?
a. 𝑊^,b = 36.43 J
b. 𝑊^,b = −14.59 J
T
MA g sin ✓ = MA a =) T = MA (a + g sin ✓)
c. 𝑊^,b = 55.1 J
= (4 kg)(2.1 + 9.81 sin 30 ) m/s
d. 𝑊^,b = 10.92 J
e. 𝑊^,b = 0 J
= 28.02 kg m/s
2
2
WA,T = T · d = 28.02 · 1.3 = 36.43 J
30. (5 points)
As block A moves a distance 1.3 m up the incline, how does the magnitude of the work done on it by the tension in
the string |𝑊^,b |, compare to magnitude of the work done on it by gravity, |𝑊^,d | ?
a. |𝑊^,b | = |𝑊^,d |
b. 𝑊^,b > |𝑊^,d |
c. 𝑊^,b < |𝑊^,d |
Wnet = WA,T
WA,g =
K>0
=) WA,T > WA,g
31. (7 points)
Now assume that there is friction between block A and the plane and that both blocks have the same mass 𝑀^ =
𝑀_ = 𝑀. For the geometry shown, what is the minimum coefficient of static friction, 𝜇f , needed to hold block A
stationary if it is initially released from rest?
a. 𝜇g = 0.74
b. c. d. e. 𝜇g
𝜇g
𝜇g
𝜇g
= 0.67
= 0.82
= 0.58
= 0.49
Block B: T = M g
Block A: T = M g sin ✓ + fs
=) fs = M g(1
sin ✓)
µs N = µs M g cos ✓ = M g(1 sin ✓)
1 sin ✓
1 0.5
=) µs =
=
= 0.58
cos ✓
0.866
The next three questions pertain to the situation described below.
A child is playing with a new toy that consists of a ball of
mass 𝑚 = 0.87 kg tied to the end of a string of length ℓ𝓁 = 0.3
m. The child swings the toy above her head so that it moves
with uniform circular motion in a horizontal plane. The string
makes an angle of 63° with respect to the vertical arm of the
child. The ball should be treated as a point particle.
32. (7 points)
What is the tension in the string?
a. 𝑇 = 18.8 N
b. 𝑇 = 55.83 N
c. 𝑇 = 8.53 N
d. 𝑇 = 4.35 N
e. 𝑇 = 9.58 N
⌃Fy = 0 =) T cos ✓ = mg
mg
0.87 · 9.81
T =
=
cos ✓
cos 63
= 18.8 N
33. (5 points)
Say the speed of the ball in the original problem is 𝑉\ . Suppose we increase mass of the ball but keep the length of
the string the same. If we want the angle that the string makes with the vertical to be the same as in the original
problem, what would the new speed of the ball have to be?
a. 𝑉HIJ > 𝑉\
b. 𝑉HIJ < 𝑉\
c. 𝑉HIJ = 𝑉\
V2
R
V2
=) mg tan ✓ = m
` sin ✓
2
V = `g sin ✓ tan ✓ (independent of m)
⌃Fx = macent =) T sin ✓ = m
34. (7 points)
Now consider once again the geometry shown in the figure above. Assuming the ball travels with tangential speed
𝑉\ as it travels along the circular path shown, what is the magnitude of the angular momentum of the ball as it
orbits the central vertical axis?
a. 𝐿 = 𝑚 ℓ𝓁 𝑉\ sin 𝜃
b. 𝐿 = 𝑚 ℓ𝓁 𝑉\ cos 𝜃
c. 𝐿 =
d. 𝐿 =
<
𝑉P
ℓ𝓁 \
< P
𝑉
ℓ𝓁 \
sin 𝜃
tan 𝜃
e. 𝐿 = 𝑚 ℓ𝓁 𝑉\ tan 𝜃
~ = p~ ⇥ R
~ =) L = mV0 R = mV0 ` sin ✓
L
The next three questions pertain to the situation described below.
Two houses are located on opposite sides of a river as shown. The
width of the river is 𝑊 = 247 m, and the river flows in the +𝑦
direction with speed 𝑉m,n = 1.3 m/s relative to the houses. A boat
sets off from the left house, moving with velocity 𝑉_,m relative to the
river. The 𝑥-component of 𝑉_,m is 2.1 m/s and the 𝑦-component
of 𝑉_,m is 3.1 m/s.
35. (7 points)
What is the speed of the boat as measured in the reference frame of
the houses?
a. |𝑉_,n | = 3.74 m/s
b. |𝑉_,n | = 2.77 m/s
c. |𝑉_,n | = 4.88 m/s
d. |𝑉_,n | = 4.4 m/s
e. |𝑉_,n | = 2.47 m/s
~B,H = V
~B,R + V
~R,H
V
x : (VB,H )x = (VB,R )x + (VR,H )x = 2.1 m/s + 0
y : (VB,H )y = (VB,R )y + (VR,H )y = 3.1 m/s + 1.3 m/s = 4.4 m/s
p
|VB,H | = 2.12 + 4.42 = 4.88 m/s
36. (7 points)
When the boat reaches the other side of the river, how far is it from the nearest house?
a. 517.52 m
b. 152.9 m
W
247
tcross =
=
= 117.62 s
c. 211.71 m
(VB,H )x
2.1
d. 167.32 m
Ddownstream = (VB,H )y tcross = 4.4 · 117.62 = 517.52 m
e. 364.62 m
37. (5 points)
Say the time it takes the boat to get to the other side in the above scenario is 𝑇\ . Now suppose the driver of the
boat steers in such a way that the boat moves along the 𝑥-axis, directly from the house on the left to the house on
the right. If the speed of the boat relative to the water is the same as in the original problem, compare the time it
takes the boat to get across the river in the new case, 𝑇HIJ , to 𝑇\ .
a. 𝑇HIJ > 𝑇\
b. 𝑇HIJ < 𝑇\
c. 𝑇HIJ = 𝑇\
The time it takes for the boat to cross the river is dictated by the 𝑥-component of its
velocity relative to the houses (or river, since the water flows in the 𝑦-direction). In the
first scenario, the 𝑥-component of the boat’s velocity relative to the river is 2.1 m/s and
the 𝑦-component is 3.1 m/s. In the second scenario, the boat travels directly across the
river, which means the 𝑦-component of the boat’s velocity relative to the water is only
1.3 m/s. Since the 𝑦-component of the boat’s velocity relative to the water is smaller
than in the first case, but its total speed relative to the water is the same as before, then
its 𝑥-component of velocity must be larger and it crosses the river more quickly.
The next three questions pertain to the situation described below.
A football is kicked across a level field. The ball spends
𝛥𝑡 = 4.8 seconds in the air and lands a distance 𝑋 = 30
meters from the point where it was kicked. The initial
speed of the ball is 𝑉\ . You should ignore air resistance.
38. (7 points)
What is the speed of the ball at the top of its trajectory?
a. 𝑣 = 23.54 m/s
b. 𝑣 = 3.06 m/s
Vx =
c. 𝑣 = 47.09 m/s
X
= 30/4.8 = 6.25 m/s
t
d. 𝑣 = 6.25 m/s
e. There is not enough information given to determine this.
39. (7 points)
What is the maximum height reached by the ball?
a. ℎ = 28.25 m
Vy
b. ℎ = 7.5 m
c. ℎ = 56.51 m
d. ℎ = 15 m
e. ℎ = 10 m
tpeak
(V0 )y =
gtpeak =) (V0 )y = gtpeak
1 2
h = (V0 )y tpeak
gt
2 peak
1 2
1
= gt2peak
gtpeak = gt2peak
2
2
2
t
t
=
=) h = g
= 28.25 m
2
8
40. (5 points)
After the kick, but before the ball hits the ground, which of the following statements best describes the acceleration
vector of the ball?
a. It always points toward the right.
b. It points upward before the ball reaches its maximum height, then downward.
c. It always points downward.