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Chapter 6 Test REVIEW: Momentum and Collisions
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1. A person sitting in a chair with wheels stands up, causing the chair to roll backward across the floor. The
momentum of the chair
a. was zero while stationary and increased when the person stood.
b. was greatest while the person sat in the chair.
c. remained the same.
d. was zero when the person got out of the chair and increased while the person sat.
2. A 0.2 kg baseball is pitched with a velocity of 40 m/s and is then batted to the pitcher with a velocity of 60
m/s. What is the magnitude of change in the ball’s momentum?
a. 2 kg•m/s
c. 8 kg•m/s
b. 4 kg•m/s
d. 20 kg•m/s
3. Alex throws a 0.15-kg rubber ball down onto the floor. The ball's speed just before impact is 6.5 m/s, and just
after is 3.5 m/s. What is the change in the magnitude of the ball's momentum?
a. 0.09 kg⋅m/s
b. 1.5 kg⋅m/s
c. 4.3 kg⋅m/s
d. 5.7 kg⋅m/s
e. 126 kg⋅m/s
4. A cannon of mass 1 500 kg fires a 10-kg shell with a velocity of 200 m/s at an angle of 45° above the
horizontal. Find the recoil velocity of the cannon across the level ground.
a. 1.33 m/s
b. 0.94 m/s
c. 2.41 m/s
d. 1.94 m/s
e. 0.87 m/s
5. A high-diver of mass 70 kg jumps off a board 10 m above the water. If, 1.0 s after entering the water his
downward motion is stopped, what average upward force did the water exert?
a. 100 N
b. 686 N
c. 980 N
d. 751 N
e. No answer is correct.
6. During a snowball fight two balls with masses of 0.4 and 0.6 kg, respectively, are thrown in such a manner
that they meet head-on and combine to form a single mass. The magnitude of initial velocity for each is 15
m/s. What is the speed of the 1.0-kg mass immediately after collision?
a. zero
b. 3 m/s
c. 6 m/s
d. 9 m/s
e. 10 m/s
1
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Short Answer
7. A 0.16 kg cue ball is traveling at 10.0 m/s toward a full rack of 15 billiard balls. What is the magnitude of the
total momentum of the system of 16 balls after the cue ball strikes the rack?
Problem
8. Compare the momentum of a 6160 kg truck moving at 3.00 m/s to the momentum of a 1540 kg car moving at
12.0 m/s.
9. A cricket ball with a mass of 0.11 kg moves at a speed of 12 m/s. Then the ball is hit by a bat and rebounds in
the opposite direction at a speed of 15 m/s. What is the change in momentum of the ball?
10. A pool cue strikes a 0.16 kg billiard ball with a force of 15 N. The cue remains in contact with the ball for
0.085 s. The ball was initially at rest. What is the final speed of the ball?
11. A player at first base catches a throw traveling 22 m/s. The baseball, which has a mass of 0.21 kg, comes to a
complete stop in the glove after 0.15 s. Assuming the force of the glove was uniform, what force did the glove
exert on the ball?
A 68.0 kg diver jumps off a diving platform, rises about 1 m above the platform, then falls to the pool.
12. The diver strikes the water at a speed of 14.7 m/s, then slows to a stop underwater in 0.35 s. What force does
the water exert on the diver?
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ID: A
Chapter 6 Test REVIEW: Momentum and Collisions
Answer Section
MULTIPLE CHOICE
1. ANS: A
2. ANS: D
Given
m = 2.0 kg
PTS: 1
DIF: I
OBJ: 6-1.2
v i = 40 m/s
v f = −60 m/s
Solution
Δp = mÊÁË v f − v i ˆ˜¯ = ÁÊË 0.2 kg ˜ˆ¯ (−60 m/s − 40 m/s) = −20 kgm/s
3.
4.
5.
6.
PTS:
ANS:
ANS:
ANS:
ANS:
TOP:
1
DIF: IIIA
OBJ:
B
PTS: 1
DIF:
B
PTS: 1
DIF:
E
PTS: 1
DIF:
B
PTS: 1
DIF:
6.3 Collisions | 6.4 Glancing Collisions
6-1.3
2
2
3
2
SHORT ANSWER
7. ANS:
1.6 kg•m/s
Given
m 1 = 0.16 kg
v 1 = 10.0 m/s
Solution
p f = p i = m 1 v 1 = (0.16 kg)(10.0 m/s) = 1.6 kg•m/s
PTS: 1
DIF: II
OBJ: 6-2.2
1
TOP: 6.1 Momentum and Impulse
TOP: 6.2 Conservation of Momentum
TOP: 6.2 Conservation of Momentum
ID: A
PROBLEM
8. ANS:
They have the same momentum. (1.85 × 10 4 kg•m/s)
Given
m1 = 6160 kg
v 1 = 3.00 m/s
m2 = 1540 kg
v 2 = 12.0 m/s
Solution
p 1 = m1 v 1 = ÊÁË 6160 kg ˆ˜¯ (3.00 m/s) = 1.85 × 10 4 kgm/s
p 2 = m2 v 2 = ÊÁË 1540 kg ˆ˜¯ (12.0 m/s) = 1.85 × 10 4 kgm/s
p1 = p2
PTS: 1
9. ANS:
–3.0 kg•m/s
DIF: IIIA
OBJ: 6-1.1
Given
m = 0.11 kg
v i = 12 m/s
v f = −15 m/s
Solution
Δp = mÊÁË v f − v i ˆ˜¯ = ÁÊË 0.11 kg ˜ˆ¯ (−15 m/s − 12 m/s) = −3.0 kgm/s
PTS: 1
DIF: IIIA
OBJ: 6-1.3
2
ID: A
10. ANS:
8.0 m/s
Given
m = 0.16 kg
F = 15 N
Δt = 0.085 s
v i = 0 m/s
Solution
FΔt = Δp = mv f − mv i
ˆ
Ê
FΔt + mv i (15 N) (0.085 s) − ÁË 0.16 kg ˜¯ (0 m/s)
vf =
=
= 8.0 m/s
0.16 kg
m
PTS: 1
DIF: IIIB
11. ANS:
31 N in the direction opposite the throw
OBJ: 6-1.4
Given
m = 0.21 kg
v i = 22 m/s
v f = 0 m/s
Δt = 0.15 s
Solution
FΔt = Δp
Ê
ˆ Ê
ˆ
Δp mÁË v f − v i ˜¯ ÁË 0.21 kg˜¯ (0 m/s − 22 m/s)
F=
=
=
= −31 N
Δt
Δt
0.15 s
PTS: 1
DIF: IIIB
OBJ: 6-1.4
3
ID: A
12. ANS:
2.9 × 10 3 N upward
Given
m = 68.0 kg
v i = 14.7 m/s downward; v i = −14.7 m/s
v f = 0 m/s
Δt = 0.35 s
Solution
Ê
ˆ Ê
ˆ ÍÈ
˙˘
Δp mÁË v f − v i ˜¯ ÁË 68.0 kg ˜¯ ÍÍÎ (0 m/s) − (−14.7 m/s) ˙˙˚
F=
=
=
= 2.9 × 103 N
Δt
Δt
0.35 s
F = 2.9 × 10 3 N upward
PTS: 1
DIF: IIIB
OBJ: 6-1.4
4