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DATE _______________ CLASS ____________________
Holt Physics
Problem 6C
STOPPING DISTANCE
PROBLEM
A high-speed train with a total mass of 9.25 105 kg travels north at a
speed of 220 km/h. Suppose it takes 16.0 s of constant acceleration for the
train to come to rest at a station platform. Calculate the force acting on
the train during this time. What is the train’s stopping distance?
SOLUTION
Given:
m = 9.25 × 105 kg
vi = 220 km/h to the north
vf = 0 km/h
∆t = 16.0 s
Unknown:
F=?
∆x = ?
Use the impulse-momentum theorem to solve for F. Use the kinematic equation
for ∆x in terms of initial velocity, final velocity, and time to solve for ∆x.
F∆t = ∆p
∆p mvf − mvi

F =  = 
∆t
∆t
(9.25 × 105 kg)(0 km/h) − (9.25 × 105 kg)(220 km/h)(103 m/km)(1 h/3600 s)
F = 
16.0 s
−(9.25 × 105 kg)(220 km/h)(103 m/km)(1 h/3600 s)
F = 
16.0 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
F = −3.5 × 106 N = 3.5 × 106 N to the south
1
∆x = 2(vi + vf)∆t
1
∆x = 2(220 km/h + 0 km/h)(103 m/km)(1 h/3600 s)(16.0 s)
∆x = 490 m to the north
ADDITIONAL PRACTICE
1. A race car has a velocity of 382 km/h to the right. If the car’s mass is
705 kg and the driver’s mass is 65 kg, what force is needed to bring the
car and driver to a stop in 12.0 s? What is the car’s stopping distance?
2. The danger that space debris poses to spacecraft can be understood in
terms of momentum. At 160 km above Earth’s surface, any object will
have a speed of about 7.82 × 103 m/s. Consider a meteoroid (a small
orbiting rock) with a mass of 42 g. Suppose this meteoroid collides
with a space shuttle and is brought to a full stop in 1.0 × 10–6 s. How
large is the force that stops the meteoroid? By how much would the
meteoroid dent the side of the shuttle?
Problem 6C
Ch. 6-5
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DATE _______________ CLASS ____________________
3. A 63 kg astronaut drifting with 7.0 m/s to the right with respect to a
spacecraft uses a jet pack to slow down. If it takes 14.0 s to come to a
stop with respect to the spacecraft, what is the force exerted by the jet
pack? How far does the astronaut travel before stopping?
4. A polar bear with a mass of 455 kg slides for 12.2 s across the surface of
a frozen lake. If the coefficient of kinetic friction between the bear and
the ice is 0.071, what is the change in the bear’s momentum as it comes
to a stop? How far does the polar bear slide?
5. A skyrocket that has consumed all of its fuel continues to move upward, slowed mostly by the force of gravity. If the rocket’s mass is
75.0 g and it takes 1.2 s for the rocket to stop, what is the change in the
rocket’s momentum? What is the rocket’s stopping distance?
6. A 4400 kg sailboat drifts in calm waters. The force of resistance that the
water exerts on the ship is 2200 N to the left. What is the change in the
ship’s momentum after 8.0 s? If the ship’s initial speed is 6.5 m/s, how
far has the ship drifted?
7. A jet of water exerts a 25.0 N force on a type of sail that is attached to a
small wagon. What is the magnitude of the change in the wagon’s momentum after 7.00 s? If the wagon’s mass is 14.0 kg and there are no
other forces acting on it, how far will the wagon travel during the
7.00 s? Assume the wagon is initially at rest.
9. Consider a Hornet jet landing on the flight deck of an aircraft carrier.
The jet’s mass is 1.35 × 104 kg and its initial velocity is 66.1 m/s to the
west. If the force required to bring the jet to a stop is 4.00 × 105 N to
the east, how long does it take the jet to come to rest? How far does the
jet travel during its deceleration?
10. The surface of Jupiter’s moon Europa is covered with ice. Suppose an
ice boat moving with a speed of 14.5 m/s drifts to a stop. The boat’s
mass is 1.50 × 103 kg and the coefficient of kinetic friction between the
boat’s runners and the ice is 0.065. However, free-fall acceleration on
Europa is only 1.305 m/s2. How long will it take the boat to stop? How
far does the ice boat glide before stopping?
Ch. 6-6
Holt Physics Problem Bank
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. How long will it take a 2.30 × 103 kg truck to go from 22.2 m/s to a
complete stop if acted on by a force of –1.26 × 104 N? What would be
its stopping distance?
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Givens
6. m = 1.1 × 103 kg
vf = 9.7 m/s to the east
vi = 0 m/s
∆t = 19 s
Solutions
∆p mvf − mvi

F =  = 
∆t
∆t
(1.1 × 103 kg)(9.7 m/s) − (1.1 × 103 kg)(0 m/s)
F =  = 560 N
19 s
F = 560 N to the east
7. m = 3.00 × 103 kg
vi = 0 m/s
vf = 8.9 m/s to the right
∆t = 5.5 s
8. m = 0.17 kg
∆v = −9.0 m/s
g = 9.81 m/s2
mk = 0.050
∆p mvf − mvi

F =  = 
∆t
∆t
(3.00 × 103 kg)(8.9 m/s) − (3.00 × 103 kg)(0 m/s)
(3.00 × 103 kg)(8.9 m/s)
F =  = 
5.5 s
5.5 s
F = 4.9 × 103 N to the right
F∆t = ∆p = m∆v
F = Fk = −mgmk
m∆v
∆v
−9.0 m/s
∆t =  =  = 
−mgmk −gmk
−(9.81 m/s2)(0.050)
∆t = 18 s
9. m = 12.0 kg
Fapplied = 15.0 N
q = 20.0°
Ffriction = 11.0 N
mvf − mvi
(12.0 kg)(4.50 m/s) − (12.0 kg)(0 m/s)
∆p
∆t =  =  = 
Fapplied(cos q) − Ffriction
(15.0 N)(cos 20.0°) − 11.0 N
F
vi = 0 m/s
5.40 kg•m/s
5.40 kg•m/s − 0 kg•m/s
∆t =  = 
3.1 N
14.1 N − 11.0 N
vf = 4.50 m/s
∆t = 1.7 s
10. vf = 15.8 km/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
F = Fapplied(cos q) − Ffriction
vi = 0 km/s
F = 12.0 N
m = 0.20 g
∆p mvf − mvi

∆t =  = 
F
F
(0.20 × 10−3 kg)(15.8 × 103 m/s) − (0.20 × 10−3 kg)(0 m/s)
∆t = 
12.0 N
(0.20 × 10−3 kg)(15.8 × 103 m/s)
∆t = 
12.0 N
∆t = 0.26 s
Additional Practice 6C
1. vi = 382 km/h to the right
vf = 0 km/h
mc = 705 kg
md = 65 kg
∆t = 12.0 s
∆p (mc + md)vf − (mc + md)vi
F =  = 
∆t
∆t
[(705 kg + 65 kg)(0 km/h) − (705 kg + 65 kg)(382 km/h)](103 m//km)(1 h/3600 s)
F = 
12.0 s
−(7.70 × 102 kg)(382 km/h)(103 m/km)(1 h/3600 s)
F =  = − 6.81 × 103 N
12.0 s
F = 6.81 × 103 N to the left
1
V
1
∆x = 2(vi + vf)∆t = 2(382 km/h + 0 km/h)(103 m/km)(1 h/3600 s)(12.0 s)
∆x = 637 m to the right
Section Five—Problem Bank
V Ch. 6–3
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Givens
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Solutions
2. vi = 7.82 × 103 m/s
vf = 0 m/s
m = 42 g
∆t = 1.0 × 10−6 s
∆p mvf − mv
F =  = i
∆t
∆t
(42 × 10−3 kg)(0 m/s) − (42 × 10−3 kg)(7.82 × 103 m/s)
F = 
1.0 × 10−6 s
−(42 × 10−3 kg)(7.82 × 103 m/s)
F = 
1.0 × 10−6 s
F = −3.3 × 108 N
∆x = 2(vi + vf)∆t = 2(7.82 × 103 m/s + 0 m/s)(1.0 × 10−6 s)
1
1
∆x = 3.9 × 10−3 m = 3.9 mm
3. m = 63 kg
vi = 7.0 m/s to the right
vf = 0 m/s
∆p mvf − mvi

F =  = 
∆t
∆t
(63 kg)(0 m/s) − (63 kg)(7.0 m/s)
−(63 kg)(7.0 m/s)
F =  =  = −32 N
14.0 s
14.0 s
∆t = 14.0 s
F = 32 N to the left
1
1
∆x = 2(vi + vf) ∆t = 2(7.0 m/s + 0 m/s)(14.0 s)
∆x = 49 m to the right
∆p = F∆t
4. m = 455 kg
∆t = 12.2 s
F = Fk = −mgmk
2
g = 9.81 m/s
vf = 0 m/s
∆p = −mgmk∆t = −(455 kg)(9.81 m/s2)(0.071)(12.2 s) = −3.9 × 103 kg•ms
∆p = 3.9 × 103 kg•ms opposite the polar bear’s motion
mvf − ∆p
(455 kg)(0 m/s) − (−3.9 × 103 kg•ms)
vi =  = 
m
455 kg
3.9 × 103 kg•ms
vi =  = 8.6 m/s
455 kg
1
1
∆x = 2(vi + vf)∆t = 2(8.6 m/s + 0 m/s)(12.2 s)
∆x = 52 m
5. m = 75.0 g
∆p = F∆t
∆t = 1.2 s
F = −mg
g = 9.81 m/s2
∆p = −mg∆t = −(75.0 × 10−3 kg)(9.81 m/s2)(1.2 s)
vf = 0 m/s
∆p = −0.88 kg•m/s
∆p = 0.88 kg•m/s downward
mvf − ∆p (75.0 × 10−3 kg)(0 m/s) − (−0.88 kg•m/s)
 = 
vi = 
m
75.0 × 10−3 kg
0.88 kg•m/s
vi = 
= 12 m/s upward
75.0 × 10−3 kg
1
V
V Ch. 6–4
1
∆x = 2(vi + vf)∆t = 2(12 m/s + 0 m/s)(1.2 s)
∆x = 7.2 m upward
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
mk = 0.071
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Givens
Solutions
∆p = F∆t = (−2200 N)(8.0 s) = −1.8 × 104 kg•m/s
6. m = 4400 kg
F = 2200 N to the left
= −2200 N
∆t = 8.0 s
vi = 6.5 m/s to the right
= +6.5 m/s
∆p = 1.8 × 104 kg•m/s to the left
∆p + mv
−1.8 × 104 kg•m/s + (4400 kg)(6.5 m/s)
vf = i = 
m
4400 kg
−1.8 × 104 kg•m/s + 2.9 × 104 kg•m/s
1.1 × 104 kg•m/s
vf =  =  = 2.5 m/s to the right
4400 kg
4400 kg
1
1
1
∆x = 2(vi + vf)∆t = 2(6.5 m/s + 2.5 m/s)(8.0 s) = 2(9.0 m/s)(8.0 s)
∆x = 36 m to the right
7. F = 25.0 N
∆p = F∆t = (25.0 N)(7.00 s) = 175 kg•m/s
∆t = 7.00 s
∆p + mv
175 kg•m/s + (14.0 kg)(0 m/s)
vf = i =  = 12.5 m/s
m
14.0 kg
m = 14.0 kg
vi = 0 m/s
1
1
∆x = 2(vi + vf)∆t = 2(0 m/s + 12.5 m/s)(7.00 s)
∆x = 43.8 m
8. m = 2.30 × 103 kg
vi = 22.2 m/s
vf = 0 m/s
F = −1.26 × 104 N
∆p mvf − mv
∆t =  = i
F
F
−5.11 × 104 kg•m/s
(2.30 × 103 kg)(0 m/s) − (2.30 × 103 kg)(22.2 m/s)
∆t = 
= 
4
−1.26 × 10 N
−1.26 × 104 N
∆t = 4.06 s
1
1
∆x = 2(vi + vf)∆t = 2(22.2 m/s + 0 m/s)(4.06 s)
∆x = 45.1 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. m = 1.35 × 104 kg
vi = 66.1 m/s to the west
= −66.1 m/s
vf = 0 m/s
F = 4.00 × 105 N to the east
= +4.00 × 105 N
∆p mvf − mvi

∆t =  = 
F
F
(1.35 × 104 kg)(0 m/s) − (1.35 × 104 kg)(−66.1 m/s)
8.92 × 105 kg•m/s
∆t = 
= 
5
4.00 × 10 N
4.00 × 105 N
∆t = 2.23 s
1
1
∆x = 2(vi + vf)∆t = 2(−66.1 m/s + 0 m/s)(2.23 s) = −73.7 m
∆x = 73.7 m to the west
vf = 0 m/s
∆p mvf − mv
∆t =  = i
F
F
m = 1.50 × 103 kg
F = Fk = mamk
mk = 0.065
vf − v
∆t = i
amk
10. vi = 14.5 m/s
2
a = −1.305 m/s
0 m/s − 14.5 m/s
∆t = 
(− 1.305 m/s2)(0.065)
∆t = 1.7 × 102 s
1
1
∆x = 2(vi + vf)∆t = 2(14.5 m/s + 0 m/s)(1.7 × 102 s)
V
∆x = 1.2 × 103 m = 1.2 km
Section Five—Problem Bank
V Ch. 6–5