# practice test

```Momentum Test v3
PSI AP 1 Physics
Name____________________________________
Multiple Choice– Choose the correct answer for each question. No partial credit will be given.
1. A 2 kg toy car is traveling at a speed of 0.5 m/s. For how long must a 0.2 N force be applied
to raise the car’s momentum by 1.2 kg-m/s?
A)
B)
C)
D)
7s
2s
6s
9s
2. As shown, a baseball of mass m rebounds from a vertical wall with the same
speed as it had initially. What is the change in momentum of the baseball in the
vertical direction?
A)
B)
C)
D)
0
2mvcosθ
2mvcosθ
2mvsinθ
3. The velocity of a moving bicycle is tripled. Which of the following quantities must
also be tripled?
A)
B)
C)
D)
Acceleration
Kinetic Energy
Momentum
Potential Energy
4. A steel ball (1) moving at a constant speed, v on a horizontal frictionless surface, collides
obliquely with an identical ball (2) initially at rest. The velocity of the ball (1) before and after
the collision is shown in the diagram below. What is the approximate direction of the velocity
of ball (2) after the collision?
2
1
A)
C)
B)
D)
1
Questions 5-6:
Use the given graph below to answer questions 5-6. The graph represents the force applied
over time to an object of mass 2.0 kg that was traveling at a velocity of 0.50 m/s at t = 0 s.
5. What is the change in momentum of the object after 4.0 s?
A)
B)
C)
D)
2.0 N-m
3.0 N-m
4.0 N-m
0 N-m
6. What is the speed of the object at t = 2.0 s?
A)
B)
C)
D)
1.0 m/s
1.5 m/s
2.0 m/s
Not enough information is given.
7. Ball 1 of mass 3m is moving at a velocity of 4v m/s north and strikes Ball 2 of mass 4m which
is traveling at a velocity of v north. If the final velocity of ball 1 is 2v north, what is the final
velocity of ball 2?
A)
B)
C)
D)
3v/2 north
5v/2 north
2v north
v/2 north
8. Object 1 of mass 3.0 kg travels east with a speed of 5.0 m/s. Object 2 of mass 4.0 kg travels
south with a speed of 3.0 m/s. The two objects collide and stick together. What is the
momentum of the two objects (magnitude and the angle they make with the east-west axis)
after the collision
A)
B)
C)
D)
19 kg-m/s; -40º
19 kg-m/s; -39º
21 kg-m/s; -40º
21 kg-m/s; -39º
9. A physics student is standing on a boat in the middle of a river without a paddle. The
student has a very heavy chemistry book on deck. She wants to get to the north shore of the
river, so she decides to throw the book off the boat. Which direction should she throw it?
Which physics law applies to this situation?
A)
B)
C)
D)
North; conservation of energy
South; conservation of energy
North; conservation of momentum
South; conservation of momentum
10. A 5 kg block moves with a constant speed of 3 m/s on a horizontal frictionless surface and
collides elastically with an identical block initially at rest. The second block then moves and
collides with a third, identical block, also initially at rest. This time, it’s a perfect inelastic
collision and the two blocks move off together. What is the initial speed of the second block
after its collision? What is the initial speed of the second and third block system after their
collision?
A)
B)
C)
D)
3 m/s; 3/2 m/s
3 m/s; 5/2 m/s
3/2 m/s; 3 m/s
3/2 m/s; 2 m/s
Multi-Correct: For each of the questions or incomplete statements below, two of the suggested
answers will be correct. For each of these questions, you must select both correct choices to
earn credit. No partial credit will be earned if only one correct choice is selected. Select the two
that are best in each case and then enter both of the appropriate letters in the corresponding
11. A fisherman is standing in the back of his small fishing boat (the mass of the fisherman is the
same as the mass of the boat) and he is a few meters from shore. His is tired of fishing so
he starts walking on his boat towards the shore so he can get off the boat. What happens to
the boat and the fisherman? Assume there is no friction between the boat and the water.
A)
B)
C)
D)
The fisherman gets closer to the shore.
The fisherman gets further away from the shore.
The boat gets closer to the shore.
The boat gets further away from the shore.
12. A spring is compressed between two blocks with unequal masses, m 1 and m2, held together
by a spring. The objects are initially at rest on a horizontal surface with no friction. The
spring is then cut. What is true of the two object system after the spring is cut?
A)
B)
C)
D)
The net final momentum of the two objects is zero.
Both objects will travel with different constant speeds unless they hit another object.
The energy of the system increases as the blocks separate.
The net final momentum of the two objects is positive.
Free Response - Solve this problem, showing all work. Partial credit may be given.
A 5.00 kg piece of clay moves with a constant v = 30.0 m/s. The piece of clay collides and sticks
to a massive ball of mass 20.0 kg suspended at the end of a 2.00 m string. The combined
masses swing up in an arc, reach a maximum height, and return to the original position.
a. Calculate the initial momentum of the piece of clay.
b. A student predicts that the clay will stick to the ball in an inelastic collision.
i. Assuming his prediction is correct, find the net momentum of the clayball system directly after the collision.
ii. Find the velocity of the clay-ball system directly after the collision.
iii. Find the maximum height that the clay-ball system reaches on its
upswing.
c. The student conducted an experiment and found that the collision was inelastic, but
the clay did not stick to the ball – in fact, it rebounded from the ball with a velocity of
-10.0 m/s.
i.
Find the final velocity of the ball.
ii. Find the kinetic energy transferred into other forms of energy during the
inelastic collision.
Momentum Test v3
PSI AP 1 Physics
Weight 50% Multiple Choice, 50% Free Response
Multiple Choice1 point each, 12 points total
1. C
2. A
3. C
4. D
5. C
6. A
7. B
8. B
Free Response – 15 points total
a. 2 points
p=mv=(5kg)(30m/s)
p=150 kg-m.s
b. 6 points
i. momentum is conserved
150 kg-m/s
ii. po=pf
150kgm/s=(m1+m2)vf
150kgm/s=(20+5kg)(vf)
vf = 6.00 m/s
iii. ½(m1+m2)vf2=(m1+m2)gh
h=l(1-cosø)
½(m1+m2)vf2=(m1+m2)gl(1-cosø)
vf2/2=gl(1-cosø)
vf2/(2gl)=1-cosø
cosø=1-vf2/(2gl)=1-((6m/s)2/(2x9.8m/s2x2m))
ø=85.3º
h=2(1-cos26)
h = 1.84 m
c. 7 points
i. po=pf
Negative is opposite direction of initial path
150=m1v1f+m2v2f
150=-50+m2v2f
200=20v2f
v2f =10.0 m/s
ii. Ko=1/2mv2=1/2(5)(30)2=2250 J
Kf1=1/2mv2=1/2(5)(-20)2=1000 J
Kf2=1/2mv2=1/2(10)(12.5)2=781 J
∆K=(Kf1+Kf2)-K0=(1000J+781J)-2250J=-469J
KE transferred is 469 J
9. D
10. A
11. A, D
12. A, B
```