Openstax College Physics
Instructor Solutions Manual
Chapter 8
CHAPTER 8: LINEAR MOMENTUM AND
COLLISIONS
8.1 LINEAR MOMENTUM AND FORCE
1.
(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of
7.50 m/s . (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg
tranquilizer dart fired at a speed of 600 m/s . (c) What is the momentum of the 90.0-kg
hunter running at 7.40 m/s after missing the elephant?
Solution
(a) pe  me ve  2000 kg  7.50 m/s  1.50 10 4 kg  m/s
(b) p b  mb v b  0.0400 kg  600 m/s  24.0 kg  m/s, so
pc 1.50  10 4 kg.m/s

 625
pb
24.0 kg.m/s
The momentum of the elephant is much larger because the mass of the elephant
is much larger.
(c) pb  mh vh  90.0 kg  7.40 m/s  6.66 10 2 kg  m/s . Again, the momentum is
smaller than that of the elephant because the mass of the hunter is much smaller.
2.
Solution
(a) What is the mass of a large ship that has a momentum of 1.60  109 kg  m/s , when
the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the
momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s .
(a) ps  ms vs  ms 
ps 1.60  10 9 kg m/s
1 km 3600 s



 1.20  10 8 kg
vs
48.0 km/h
1000 m 1 hr
6
(b) pa  ma va  (1100 kg) (1200 m/s)  1.32  10 kg m/s , so
ps 1.60  109 kg m/s

 1.21  103
pa 1.32  10 6 kg m/s
Openstax College Physics
3.
Solution
Instructor Solutions Manual
Chapter 8
(a) At what speed would a 2.00  104 - kg airplane have to fly to have a momentum of
1.60  109 kg  m/s (the same as the ship’s momentum in the problem above)? (b) What
is the plane’s momentum when it is taking off at a speed of 60.0 m/s ? (c) If the ship is
an aircraft carrier that launches these airplanes with a catapult, discuss the
implications of your answer to (b) as it relates to recoil effects of the catapult on the
ship.
(a) pa  ma va  va 
pa 1.60  10 9 kg m/s

 8.00  10 4 m/s
4
ma
2.00  10 kg
(b) pa  ma va  2.00 10 4 kg 60.0 m/s   1.20 106 kg. m/s
(c) Since the momentum of the airplane is 3 orders of magnitude smaller than the
ship, the ship will not recoil very much. The recoil would be
m v 1.20  10 9 kg m/s
vs  a a 
  0.0100 m/s , which is probably not noticeable.
ms
1.20  10 8 kg
4.
Solution
(a) What is the momentum of a garbage truck that is 1.20 10 4 kg and is moving at
10.0 m/s ? (b) At what speed would an 8.00-kg trash can have the same momentum as
the truck?
(a) p  mv  1.20 10 4 kg 30.0 m/s   3.60 105 kg  m/s
(b) p  mv  v 
p 3.60  10 5 kg  m/s

 4.50  10 4 m/s
m
8.00 kg
5.
A runaway train car that has a mass of 15,000 kg travels at a speed of 5.4 m/s down a
track. Compute the time required for a force of 1500 N to bring the car to rest.
Solution
p  net Ft so t 
6.
The mass of Earth is 5.972  10 24 kg and its orbital radius is an average of
1.496  1011 m . Calculate its linear momentum.
Solution
m  5.972 1024 kg, R  1.496 1011 m
T  365 days 
v
p
15,000 kg(5.4 m/s)

 54 s to stop the car.
net F
(1500 N)
24 h 3600 s

 3.15 107 s
1 day
1h
1.496 1011 m 5.972 1024 kg   1.78 1029 kg  m/s
2 R
2 Rm
; p  mv 
 2
T
T
3.15 107 s
8.2 IMPULSE
Openstax College Physics
Instructor Solutions Manual
Chapter 8
7.
A bullet is accelerated down the barrel of a gun by hot gases produced in the
combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to
accelerate it to a speed of 600 m/s in a time of 2.00 ms (milliseconds)?
Solution
net F 
8.
Professional Application A car moving at 10 m/s crashes into a tree and stops in 0.26
s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a
halt. The mass of the passenger is 70 kg.
Solution
net F 
9.
A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds
from an initial speed of 4.00 m/s. (a) What is the average force exerted on the leg,
taking the effective mass of the hand and forearm to be 1.50 kg? (b) Would the force
be any different if the woman clapped her hands together at the same speed and
brought them to rest in the same time? Explain why or why not.
Solution
(a) net F 
10.
Professional Application A professional boxer hits his opponent with a 1000-N
horizontal blow that lasts for 0.150 s. (a) Calculate the impulse imparted by this blow.
(b) What is the opponent’s final velocity, if his mass is 105 kg and he is motionless in
midair when struck near his center of mass? (c) Calculate the recoil velocity of the
opponent’s 10.0-kg head if hit in this manner, assuming the head does not initially
transfer significant momentum to the boxer’s body. (d) Discuss the implications of your
answers for parts (b) and (c).
p mv 0.0300 kg  600 m/s


 9.00  10 3 N
3
t
t
2.00  10 s
p
(10 m/s)(70 kg)

 2690 N
t
0.26 s
p mv 1.50kg0 m/s  4.00 m/s 


  2.40  10 3 N
3
t
t
2.50  10 s
(taking movement toward the leg as positive). Therefore, by Newton’s third law,
the net force exerted on the leg is 2.40  10 3 N , toward the leg.
(b) The force on each hand would have the same magnitude as that found in part (a)
(but in opposite directions by Newton’s third law) because the changes in
momentum and time interval are the same.
Openstax College Physics
Instructor Solutions Manual
Chapter 8
Solution (a) p  net F  t  1000 N  0.150 s  150 kg.m/s
p
p
150 kg.m/s
(b) p  mv  v 
 vf  vi , so that v f 
 vi 
 0 m/s  1.43 m/s
m
m
105 kg
(c) Assuming the mass of the head is 10.0 kg:
p
150 kg.m/s
vf 
 vi 
 0 m/s  15.0 m/s
m
10.0 kg
(d) The boxer’s head recoils much faster than the body, since its mass is smaller. To
knock someone out it is much more effective to hit them in the head than in the
torso.
11.
Professional Application Suppose a child drives a bumper car head on into the side
rail, which exerts a force of 4000 N on the car for 0.200 s. (a) What impulse is imparted
by this force? (b) Find the final velocity of the bumper car if its initial velocity was 2.80
m/s and the car plus driver have a mass of 200 kg. You may neglect friction between
the car and floor.
Solution
(a) net F 
12.
Professional Application One hazard of space travel is debris left by previous missions.
There are several thousand objects orbiting Earth that are large enough to be detected
by radar, but there are far greater numbers of very small objects, such as flakes of
paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft
window at a relative speed of 4.00  103 m/s , given the collision lasts 6.00  10 8 s .
Solution
p mv 1.00  10 -4 kg 4.00  103 m/s


 6.67  106 N
t
t
6.00  10 8 s
(assuming the chip sticks to the spacecraft)
13.
Professional Application A 75.0-kg person is riding in a car moving at 20.0 m/s when
the car runs into a bridge abutment. (a) Calculate the average force on the person if he
is stopped by a padded dashboard that compresses an average of 1.00 cm. (b)
Calculate the average force on the person if he is stopped by an air bag that
compresses an average of 15.0 cm.
p
 p  net F  t   4000 N  0.200 s   800kg.m/s
t
(taking the original direction of the car as positive)
(b) p  mvf  mvi
p  mvi  800 kg.m/s  200 kg  2.80 m/s
vf 

  1.20 m/s
m
200 kg
net F 



Openstax College Physics
Solution
Instructor Solutions Manual
Chapter 8
v 2  v02 0  20.0 m/s 

  2.00  10 4 m/s 2 .
(a) net F=ma , and v  v  2ax, so a 
2x
20.0100 m 
4
2
Thus, net F  ma  75.0 kg  2.00  10 m/s    1.50  10 6 N
2
2
(b)
2
0
v 2  v02 0  20.0 m/s 2
a

  1.33  10 3 m/s 2 . Or
2x
20.0150 m 


net F  75.0 kg   1.33  10 3 m/s 2   1.00  10 5 N
14.
Professional Application Military rifles have a mechanism for reducing the recoil
forces of the gun on the person firing it. An internal part recoils over a relatively large
distance and is stopped by damping mechanisms in the gun. The larger distance
reduces the average force needed to stop the internal part. (a) Calculate the recoil
velocity of a 1.00-kg plunger that directly interacts with a 0.0200-kg bullet fired at 600
m/s from the gun. (b) If this part is stopped over a distance of 20.0 cm, what average
force is exerted upon it by the gun? (c) Compare this to the force exerted on the gun if
the bullet is accelerated to its velocity in 10.0 ms (milliseconds).
Solution
(a) pi  0  pf  m1v1 '  m2 v2 '  v1 '  
(b)
m2 v 2 '
0.0200 kg 600 m/s    12.0 m/s

m1
1.00 kg
v 2  v02 0  12.0 m/s 2
v  v  2ax, so a 

  360 m/s 2 , so
2x
20.200 m 
2
2
0


net F  ma  1.00 kg   360 m/s 2   360 N
p mv'b  vb  0.0200 kg 600 m/s  0 
(c) net F 


  2.40  103 N.
3
t
t
10.0  10 s
Fplunger
 360 N

 0.300.
The force on the gun is then  1.20 103 N, so
Fw/o
 1.20 103 N
The plunger reduces the recoil force by 70%.
15.
Solution
A cruise ship with a mass of 1.00 107 kg strikes a pier at a speed of 0.750 m/s. It
comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain’s
finances. Calculate the average force exerted on the pier using the concept of impulse.
(Hint: First calculate the time it took to bring the ship to rest.)
x  vt 
1
v  v0 t so that t  2x  26.00 m   16.0 s.
2
v  v0 0  0.750  m/s


p mv  v0  1.00  10 7 kg 0  0750  m/s
net F 


  4.69  10 5 N.
t
t
16.0 s
So, by Newton’s third law, the net force on the pier is 4.69  10 5 N , in the direction
the ship was originally traveling.
Openstax College Physics
Instructor Solutions Manual
Chapter 8
16.
Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s
but collides head-on with a padded goalpost and experiences a backward force of
1.76 10 4 N for 5.50  10 -2 s .
Solution
p  net F  t  (1.76 x 10 4 N)(0.055s)  968 kgm/s
p  m(vf  vi )  110 kg(v f  (8.00 m/s))  968 kg  m/s,
 vf 
968 kg  m/s
 8.00 m/s  0.80 m/s.
110 kg
Note the change in direction of the rugby player's velocity.
17.
Water from a fire hose is directed horizontally against a wall at a rate of 50.0 kg/s and
a speed of 42.0 m/s. Calculate the force exerted on the wall, assuming the water’s
horizontal momentum is reduced to zero.
Solution
p
  50.0 kg/s 42.0 m/s    2.10  10 3 N.
t
The calculated value is the force on the water. By Newton’s third law, an equal but
opposite force is exerted on the wall (with a positive magnitude). Thus, the force is
2.10  10 3 N .
18.
A 0.450-kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes
to rest after driving the nail 1.00 cm into a board. (a) Calculate the duration of the
impact. (b) What was the average force exerted on the nail?
Solution
(a) x 
19.
Starting with the definitions of momentum and kinetic energy, derive an equation for
the kinetic energy of a particle expressed as a function of its momentum.
Solution
p  mv  p 2  m 2 v 2 
net F 
1
v  v0 t  Since v  0 m/s, t  2x  2  0.0100 m  2.86  10 3 s.
2
v
7.00 m/s
2
v
mv
0.450 kg 7.00 m/s 2  1.10  103 N.
p
(b) net F 
 mv0  0  0 
t
2x 2x
2  0.0100 m
KE 
20.
p2
p2 1
 mv 2 
 mv 2  KE
m
2m 2
p2
2m
A ball with an initial velocity of 10 m/s moves at an angle 60 above the  x -direction.
The ball hits a vertical wall and bounces off so that it is moving 60 above the  x direction with the same speed. What is the impulse delivered by the wall?
Openstax College Physics
Instructor Solutions Manual
Chapter 8
Solution
Assume the ball has a mass of m . The ball's vertical momentum does not change. The
ball's horizontal velocity reverses direction, so the impulse delivered to the ball by the
wall is: p  2mv x  2m10 m/s  cos 60  (10 m/s) m (in kg)
21.
When serving a tennis ball, a player hits the ball when its velocity is zero (at the
highest point of a vertical toss). The racquet exerts a force of 540 N on the ball for 5.00
ms, giving it a final velocity of 45.0 m/s. Using these data, find the mass of the ball.
Solution
net F 
p mv

, therefore ,
t
t
net F  t 540 N 5.00 10 -3 s
m

 6.00 10 -2 kg  60.0 g
v
45.0 m/s

22.

A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves
the foot with a speed of 18 m/s at an angle 55 above the horizontal. What is the
impulse delivered by the foot (magnitude and direction)?
Solution
The ball falls from rest, so its speed as it reaches the foot is


vi  2 gx  2 9.80 m/s 2 1.00 m   4.43 m/s .
If the ball has mass m , then the initial momentum is m  4.43 m/s  straight down.
The final velocity is given by v x  v cos   18 m/s cos55  10.3 m/s , and
v y  v sin   18 m/s sin 55  14.74 m/s . The change in velocity is then
vx  10.3 m/s; v y  14.74 m/s   4.43 m/s   19.17 m/s , or
v 
v x 2  v y 2
 21.8 m/s
 v y 
  61.8 above horizontal . The impulse equals the change in
at an angle   tan 1 
 v x 
momentum of the ball, which is mv , with v as above.
8.3 CONSERVATION OF MOMENTUM
23.
Professional Application Train cars are coupled together by being bumped into one
another. Suppose two loaded train cars are moving toward one another, the first
having a mass of 150,000 kg and a velocity of 0.300 m/s, and the second having a
mass of 110,000 kg and a velocity of  0.120 m/s . (The minus indicates direction of
motion.) What is their final velocity?
Solution
m1v1  m2v2  (m1  m2 )v' ,
Openstax College Physics
v' 
Instructor Solutions Manual
Chapter 8
m1v1  m2 v2 150,000 kg 0.300 m/s   110,000 kg  0.120 m/s 

 0.122 m/s
m1  m2
150,000 kg  110,000 kg
24.
Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a
speed of 0.750 m/s. It runs into another clay model, which is initially motionless and
has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is
their final velocity?
Solution
This is a perfectly inelastic collision, therefore:
m1v1  m2v2  (0.200 kg)(0.750 m/s)  0  (0.550 kg) vf so vf  0.272 m/s
25.
Professional Application Consider the following question: A car moving at 10 m/s
crashes into a tree and stops in 0.26 s. Calculate the force the seatbelt exerts on a
passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. Would
the answer to this question be different if the car with the 70-kg passenger had
collided with a car that has a mass equal to and is traveling in the opposite direction
and at the same speed? Explain your answer.
Solution
(10 m/s)(70 kg)
 2690N
0.26 s
In a collision with an identical car, momentum is conserved. Afterwards vf  0 for
both cars. The change in momentum will be the same as in the crash with the tree.
However, the force on the body is not determined since the time is not known. A
padded stop will reduce injurious force on body.
26.
What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150kg deer initially running at 12.0 m/s in the same direction? Assume the deer remains
on the car.
Solution
m1v1  m2 v 2  ( m1  m2 )v' ,
net F  p/t 
v' 
m1v1  m2 v 2 900 kg 30.0 m/s   150 kg 12.0 m/s 

 27.4 m/s
m1  m2
900 kg  150 kg
27.
A 1.80-kg falcon catches a 0.650-kg dove from behind in midair. What is their velocity
after impact if the falcon’s velocity is initially 28.0 m/s and the dove’s velocity is 7.00
m/s in the same direction?
Solution
m1v1  m2 v 2  (m1  m2 )v' ,
v' 
m1v1  m2 v 2 1.80 kg 28.0 m/s   0.650 kg 7.00 m/s 

 22.4 m/s
m1  m2
1.80 kg  0.650 kg
8.4 ELASTIC COLLISIONS IN ONE DIMENSION
Openstax College Physics
Instructor Solutions Manual
Chapter 8
28.
Two identical objects (such as billiard balls) have a one-dimensional collision in which
one is initially motionless. After the collision, the moving object is stationary and the
other moves with the same speed as the other originally had. Show that both
momentum and kinetic energy are conserved.
Solution
1
1
KE i  m1v12 , KE f  m2 v22 . Given m2  m1 and v2  v1 , so
2
2
1
KE f  m1v12  KE i
2
pi  m1v1 , pf  m2 v2  m1v1  pi
29.
Professional Application Two manned satellites approach one another at a relative
speed of 0.250 m/s, intending to dock. The first has a mass of 4.00  103 kg , and the
second a mass of 7.50 103 kg . If the two satellites collide elastically rather than dock,
what is their final relative velocity?
Solution In a one-dimensional elastic collision of two objects, the momentum and kinetic
energies before and after the collision are conserved:
2
2
2
2
'
'
m1 (v1  v1' )  m2 (v2'  v2 ) and m1 (v1  v1 )  m2 ( v 2  v1 )
Solving the two equations gives the result: v1  v2  (v1'  v'2 )
So, the relative velocity of approach equals the relative velocity of recession =
0.25 m/s
30.
A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped
at him at a velocity of 35.0 m/s. Suppose the goalie and the ice puck have an elastic
collision and the puck is reflected back in the direction from which it came. What
would their final velocities be in this case?
Openstax College Physics
Solution
Instructor Solutions Manual
Chapter 8
m1v1  m1v1 'm2v2 '
m1 v1  v1 '  m2v2 '
(i)
1
1
1
m1v12  m1v'1  m2 v'22
2
2
2


m1 v12  v'22  m2v'22
m1 v1  v1 'v1  v1 '  m2v'22
(ii)
Dividing Equation (ii) by Equation (i) gives v1  v'1  v' 2 , which can be used in Equation
(i) to eliminate either v'1 or v' 2 , yielding:
 m  m2
v'1   1
 m1  m2

v1

 2m1 
v1
v' 2  
 m1  m2 
0.150 kg  70.0 kg
35.0 m/s   34.850 m/s   34.9 m/s
v' puck 
0.150 kg  70.0 kg
2  0.150 kg
v' goalie 
(35.0 m/s)  0.150 m/s
0.150 kg  70.0 kg
8.5 INELASTIC COLLISIONS IN ONE DIMENSION
31.
A 0.240-kg billiard ball that is moving at 3.00 m/s strikes the bumper of a pool table
and bounces straight back at 2.40 m/s (80% of its original speed). The collision lasts
0.0150 s. (a) Calculate the average force exerted on the ball by the bumper. (b) How
much kinetic energy in joules is lost during the collision? (c) What percent of the
original energy is left?
Solution
(a) net F 
p mv mv'f v i  2.40 kg  2.40 m/s  3.00 m/s 



  86.4 N.
t
t
t
0.015 s
1
1 2 m 2
2
2
(b) KE  mv'  mv  v  v
2
2
2


2
2


0.240 kg 
 2.40 m/s   3.00 m/s  

  0.389 J
2
1
0.240 kg 3.00 m/s 2  1.08 J
2
1.08 J  0.389 J
 % KE left 
100  64.0%
1.08 J
(c) KE i 
Openstax College Physics
Instructor Solutions Manual
Chapter 8
32.
During an ice show, a 60.0-kg skater leaps into the air and is caught by an initially
stationary 75.0-kg skater. (a) What is their final velocity assuming negligible friction
and that the 60.0-kg skater’s original horizontal velocity is 4.00 m/s? (b) How much
kinetic energy is lost?
Solution
(a) m1v1  (m1  m2 )v' , v' 
33.
Professional Application Using mass and speed data from Example 8.1 and assuming
that the football player catches the ball with his feet off the ground with both of them
moving horizontally, calculate: (a) the final velocity if the ball and player are going in
the same direction and (b) the loss of kinetic energy in this case. (c) Repeat parts (a)
and (b) for the situation in which the ball and the player are going in opposite
directions. Might the loss of kinetic energy be related to how much it hurts to catch
the pass?
Solution
(a) Use conservation of momentum for the player and the ball:
m1v1  m2 v2  m1  m2 v' so that
m v  m2 v2 110 kg 8.00 m/s   0.410 kg 25.0 m/s 
v'  1 1

 8.063 m/s  8.06 m/s
m1  m2
110 kg  0.410 kg
(b) KE  KE'  KE1  KE 2 
1
1
1
1
1
1
 m1v'12  m2 v'22  m1v12  m2 v22  m1  m2 v'2  m1v'12  m2 v'22 
2
2
2
2
2
2
1
1
2
2
2
 110.41 kg 8.063 m/s   110 kg 8.00 m/s   0.400 kg 25.0 m/s 
2
2
  59.0 J
110 kg 8.00 m/s   0.410 kg  25.0 m/s   7.88 m/s
(c) (i) v' 
110.41 kg
(ii) KE  1 110.41 kg 7.877 m/s 2 
2
1
110 kg 8.00 m/s 2  1 0.410 kg  25.0 m/s 2    223 J

2
2

m1v1
60.0 kg 4.00 m/s   1.778 m/s  1.78 m/s

m1  m2
60.0 kg  75.0 kg
1
2 1
2
(b) KE  (m1  m2 )v'  m1v1
2
2
2
2
 0.560.0 kg  75.0 kg 1.778 m/s   0.560.0 kg 4.00 m/s 
 266.613 J   267 J


Openstax College Physics
34.
Solution
35.
Solution
Instructor Solutions Manual
Chapter 8
A battleship that is 6.00 107 kg and is originally at rest fires a 1100-kg artillery shell
horizontally with a velocity of 575 m/s. (a) If the shell is fired straight aft (toward the
rear of the ship), there will be negligible friction opposing the ship’s recoil. Calculate its
recoil velocity. (b) Calculate the increase in internal kinetic energy (that is, for the ship
and the shell). This energy is less than the energy released by the gun powder—
significant heat transfer occurs.
(a) m1v1  m2 v 2  0  m1v'1  m2 v' 2 
 m1v'1  1100 kg  575 m/s
v' 2 

 1.054  10  2 m/s   1.05  10  2 m/s
7
m2
6.00  10 kg
(b) KE  KE'  KE  KE'  0  KE'
1
1
KE  m1v'12  m2 v' 22
2
2
2
2
 0.51100 kg 575 m/s   0.5 6.00 10 7 kg  1.054  10 2 m/s 


8
3
8
ΔKE  1.818  10 J  3.333 10 J  1.818  10 J
Professional Application Two manned satellites approaching one another, at a
relative speed of 0.250 m/s, intending to dock. The first has a mass of 4.00 103 kg ,
and the second a mass of 7.50  103 kg . (a) Calculate the final velocity (after docking)
by using the frame of reference in which the first satellite was originally at rest. (b)
What is the loss of kinetic energy in this inelastic collision? (c) Repeat both parts by
using the frame of reference in which the second satellite was originally at rest.
Explain why the change in velocity is different in the two frames, whereas the change
in kinetic energy is the same in both.
(a) m1v1  m2 v2  m1  m2 v' ( v1  0 in the frame)


m2 v 2
7.50  10 3 kg 0.250 m/s 

 0.163 m/s
m1  m2 4.00  10 3 kg  7.50  10 3 kg
1
1
2
2
(b) KE  KE'  KE  m1  m2 v'  m2 v2
2
2
1
1
2
2
KE  4.00  10 3 kg  7.50  10 3 kg 0.163 m/s   7.50  10 3 kg 0.250 m/s 
2
2
  81.6 J
v' 

(c)
v' 





m1v2
4.00  10 3 kg  0.250 m/s 

  0.0870 m/s
m1  m2
11.5  103 kg
1
m1  m2 v' 1 m1v12
2
2
2
2
3
KE  0.5 11.5  10 kg  0.0870 m/s   0.5 4.00  10 3 kg  0.250 m/s 
KE 

  81.5 J



Openstax College Physics
Instructor Solutions Manual
Chapter 8
The velocity of the center of mass of the two-satellite system is unchanged by the
collision because there are no external forces. The two velocities calculated above
are the velocity of the center of mass in each of the two different individual
reference frames. The loss in KE is the same for both reference frames because
the KE lost to internal forces (heat, friction, etc.) is the same regardless of the
chosen coordinate system.
36.
Professional Application A 30,000-kg freight car is coasting at 0.850 m/s with
negligible friction under a hopper that dumps 110,000 kg of scrap metal into it. (a)
What is the final velocity of the loaded freight car? (b) How much kinetic energy is
lost?
Solution
(a) m1v1  m'1 v'1 ; v'1 
 m1v1 30,000 kg 0.850 m/s 

 0.182 m/s
m'1
30,000 kg  110,000 kg
1
1
m'1 v'12  m1v'12
2
2
2
2
 0.5140,000 kg 0.182 m/s   0.530,000 kg 0.850 m/s   8.52  10 3 J
(b) KE 
37.
Professional Application Space probes may be separated from their launchers by
exploding bolts. (They bolt away from one another.) Suppose a 4800-kg satellite uses
this method to separate from the 1500-kg remains of its launcher, and that 5000 J of
kinetic energy is supplied to the two parts. What are their subsequent velocities using
the frame of reference in which they were at rest before separation?
Solution
By conservation of momentum: m1v1  m2 v2  0  v2 
By conservation of energy:
 m1v1
m2
1
1
m1v12  m2 v22  KE , so that m1v12  m2v22  2KE or
2
2
m12v12
m v  m2 2  2KE
m2
2
1 1

 2 

2KE
25000 J 
  
v1  

2
2
 4800 kg  (4800 kg) /1500 kg 
 m1  m1 m2 
(assuming three significant figure accuracy)
 m1v1  4800 kg 0.7043 m/s 
v2 

  2.25 m/s
m2
1500 kg
1




1
2
 0.7043 m/s  0.704 m/s
Openstax College Physics
Instructor Solutions Manual
Chapter 8
38.
A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The
pain of the rifle’s kick is much worse if you hold the gun loosely a few centimeters from
your shoulder rather than holding it tightly against your shoulder. (a) Calculate the
recoil velocity of the rifle if it is held loosely away from the shoulder. (b) How much
kinetic energy does the rifle gain? (c) What is the recoil velocity if the rifle is held
tightly against the shoulder, making the effective mass 28.0 kg? (d) How much kinetic
energy is transferred to the rifle-shoulder combination? The pain is related to the
amount of kinetic energy, which is significantly less in this latter situation. (e) See
Example 8.1 and discuss its relationship to this problem.
Solution
(a) m1v1  m2 v 2  0  m1v'1  m2 v' 2 so that
 m2 v' 2  0.0250 kg 550 m/s 
v'1 

  4.583 m/s  4.58 m/s
m1
3.00 kg
1
1
2
(b) KE  m1v'12  3.00 kg  4.58 m/s   31.5 J
2
2
 m2 v2  0.0250 kg 550 m/s 
(c) v'1 

  0.491 m/s
m1
28.0 kg
1
1
2
(d) KE  m1v'12  28.0 kg  0.491 m/s   3.376 J  3.38 J
2
2
(e) Example 8.1 makes the observation that if two objects have the same momentum,
the heavier object will have a smaller kinetic energy. Keeping the rifle close to the
body increases the effective mass of the rifle, hence reducing the kinetic energy of
the recoiling rifle. Pain is related to the amount of kinetic energy, so a rifle hurts
less if it is held against the body.
39.
Professional Application One of the waste products of a nuclear reactor is plutonium239  239 Pu  . This nucleus is radioactive and decays by splitting into a helium-4 nucleus
and a uranium-235 nucleus  4 He  235 U , the latter of which is also radioactive and
will itself decay some time later. The energy emitted in the plutonium decay is
8.40 10 13 J and is entirely converted to kinetic energy of the helium and uranium
nuclei. The mass of the helium nucleus is 6.6810 27 kg , while that of the uranium is
3.92 10 25 kg (note that the ratio of the masses is 4 to 235). (a) Calculate the
velocities of the two nuclei, assuming the plutonium nucleus is originally at rest. (b)
How much kinetic energy does each nucleus carry away? Note that the data given
here are accurate to three digits only.
Openstax College Physics
Solution
Instructor Solutions Manual
(a) m U v U  m He v He  0, v U 
Chapter 8
 m He v He
mU
1
1
2
m U v U2  m He v He
, so that
2
2
m2 v2
2
2KE  m U He 2He  m He v He
mU
KE 

1
v He
v He


 2 

2KE
2 8.40  10 -13 J
 
  2

- 27
2
- 25
- 27

 (6.68  10 kg) /3.92  10 kg  6.68  10 kg 
 mHe mU  mHe 
 1.573  10 7 m/s  1.57  10 7 m/s





1
2


 6.68  10 -27 kg 1.573  10 7 m/s
vU 
 2.68  10 5 m/s
- 25
3.92  10 kg
(b) KE  1 m v 2  0.56.68  10 27 kg 1.573  10 7 m/s 2
He
He He
2
 8.264  10 13 J  8.26  10 13 J
2
1
KE U  mU vU2  0.53.92  10 27 kg 2.68  10 5 m/s   1.41  10 14 J
2
40.
Professional Application The Moon’s craters are remnants of meteorite collisions.
Suppose a fairly large asteroid that has a mass of 5.00  1012 kg (about a kilometer
across) strikes the Moon at a speed of 15.0 km/s. (a) At what speed does the Moon
recoil after the perfectly inelastic collision (the mass of the Moon is 7.36  10 22 kg ) ?
(b) How much kinetic energy is lost in the collision? Such an event may have been
observed by medieval English monks who reported observing a red glow and
subsequent haze about the Moon. (c) In October 2009, NASA crashed a rocket into the
Moon, and analyzed the plume produced by the impact. (Significant amounts of water
were detected.) Answer part (a) and (b) for this real-life experiment. The mass of the
rocket was 2000 kg and its speed upon impact was 9000 km/h. How does the plume
produced alter these results?
Solution
(a) mm vm  ma va  (mm  ma )v'm , given vm  0 m/s



ma va
5.00  1012 kg 1.50  10 4 m/s

 1.019  10 -6 m/s  1.02  10 -6 m/s
mm  ma 7.36  10 22 kg  5.00  1012 kg
2
1
1
(b) KE i  ma v a2  5.00  1012 kg 1.50  10 4 m/s   5.625  10 20 J,
2
2
1
2
KEf  ma  mm v'm 
2
v' m 
 0.55.00  1012 kg  7.36  10 22 kg 1.019  10 6 m/s   3.821  1010 J
2
KE  KEf  KEi  3.821  1010 J  5.625  10 20 J   5.625  10 20 J   5.63  10 20 J
Openstax College Physics
Instructor Solutions Manual
Chapter 8
In other words, almost all of the KE is lost in this collision.
(c) Yes, this is sufficient to cause such a large-scale observable event, since it is
greater than the annual energy use of the entire United States!
41.
Professional Application Two football players collide head-on in midair while trying to
catch a thrown football. The first player is 95.0 kg and has an initial velocity of 6.00
m/s, while the second player is 115 kg and has an initial velocity of –3.50 m/s. What is
their velocity just after impact if they cling together?
Solution
m1v1  m2v2  (m1  m2 )v' ,
m v  m2 v2 95.0 kg 6.00 m/s   115 kg  3.50 m/s 
v'  1 1

 0.798 m/s
m1  m2
95.0 kg  115 kg
42.
What is the speed of a garbage truck that is 1.20 10 4 kg and is initially moving at
25.0 m/s just after it hits and adheres to a trash can that is 80.0 kg and is initially at
rest?
Solution
m1v1  m2v2  (m1  m2 )v' , and v2  0 , so that



m1v1
1.20  10 4 kg 25.0 m/s 
v' 

 24.8 m/s
m1  m2
1.20  10 4 kg  80.0 kg

43.
During a circus act, an elderly performer thrills the crowd by catching a cannon ball
shot at him. The cannon ball has a mass of 10.0 kg and the horizontal component of
its velocity is 8.00 m/s when the 65.0-kg performer catches it. If the performer is on
nearly frictionless roller skates, what is his recoil velocity?
Solution
m1v1  m2v2  (m1  m2 )v' , so that v' 
44.
(a) During an ice skating performance, an initially motionless 80.0-kg clown throws a
fake barbell away. The clown’s ice skates allow her to recoil frictionlessly. If the clown
recoils with a velocity of 0.500 m/s and the barbell is thrown with a velocity of 10.0
m/s, what is the mass of the barbell? (b) How much kinetic energy is gained by this
maneuver? (c) Where does the kinetic energy come from?
Solution
(a) m1v1  m2 v2  m1v1 'm2 v2 ' gives:
 m1v1  80.0 kg  0.500 m/s 
0  m1v1  m2 v 2  m2 

 4.00kg
v2
10.0 m/s
m1v1
10.0 kg 8.00 m/s   1.07 m/s

m1  m2
10.0 kg  65.0 kg
Openstax College Physics
Instructor Solutions Manual
Chapter 8


1
1
1
1
1
m1v'12  m2 v'22  m1v12  m2 v22  m1v'12  m2 v'22
2
2
2
2
2
1
1
2
2
 80.0 kg  0.500 m/s   4.00 kg 10.0 m/s    210 J
2
2

(c) The clown does work to throw the barbell, so the kinetic energy comes from the
muscles of the clown. The muscles convert the chemical potential energy of ATP
into kinetic energy.
(b) KE 
8.6 COLLISIONS OF POINT MASSES IN TWO DIMENSIONS
45.
Solution
Two identical pucks collide on an air hockey table. One puck was originally at rest. (a)
If the incoming puck has a speed of 6.00 m/s and scatters to an angle of 30.0 ,what is
the velocity (magnitude and direction) of the second puck? (You may use the result
that 1   2  90 for elastic collisions of objects that have identical masses.) (b)
Confirm that the collision is elastic.
y
(a)
m
m
1
x
2
m
1  30 and 1   2  90 so that  2  60
By conservation of momentum in x-direction:
m1v1  m1v'1 cos 1  m2v'2 cos 2
(i)
v1  v'1 cos 1  v'2 cos 2
By conservation of momentum in the y-direction:
v ' sin  2
m1v1 ' sin 1  m1v2 ' sin  2 , or v1 '  2
sin 1
Substituting this expression into (i) gives:
v ' sin  2 cos 1
v1 '  2
 v2 ' cos  2 so that
sin 1
v1
6.00 m/s
v2 ' 

cos  2  sin  2 cos 1 / sin 1  cos 60  sin 60cos 30 / sin 30
v2 '  3.00 m/s in a direction 90 opposite that of the incoming puck
( 60 south of east if the incoming puck scatters 30 north of east)
Openstax College Physics
(b)
v1 ' 
Instructor Solutions Manual
Chapter 8
3.00 m/s  sin 60  5.20 m/s, KE  KE' ?
sin 30
1 2
mv1
2
2
v12

KE
6.00 m/s 

 2 2 
 1.00
KE' 1
v'1 v' 2 3.00 m/s 2  5.20 m/s 2
2
2
m v'1 v' 2
2


46.
Confirm that the results of Example 8.7 do conserve momentum in both the x  and
y  directions.
Solution
(i) x-direction
m1v1  m1v'1 cos 1  m2 v' 2 cos  2
0.250 kg 2.00 m/s   0.250 kg 1.50 m/s cos45  0.400 kg 0.886 m/s cos 48.5
0.50 kg  m/s  0.265 kg  m/s  0.265 kg  m/s  0.50 kg  m
(ii) y-direction
0  m1v1 ' sin 1  m1v2 ' sin  2
0  0.250 kg 1.50 m/s sin45   0.400 kg 0.886 m/s sin  48.5
0  0.265 kg.m/s   0.265 kg.m/s   0
47.
A 3000-kg cannon is mounted so that it can recoil only in the horizontal direction. (a)
Calculate its recoil velocity when it fires a 15.0-kg shell at 480 m/s at an angle of 20.0
above the horizontal. (b) What is the kinetic energy of the cannon? This energy is
dissipated as heat transfer in shock absorbers that stop its recoil. (c) What happens to
the vertical component of momentum that is imparted to the cannon when it is fired?
Solution
(a) Conversation of momentum applies in the horizontal (x) direction only, so since
the initial momentum is zero,
m v cos  2 15.0480 m/s cos20.0
m1v1   m2 v2 cos  2 or v1   2 2

  2.26 m/s
m1
3000 kg
1
1
2
(b) KE  m1v12  3000 kg  2.255 m/s   7.63  10 3 J
2
2
(c) The ground will exert a normal force to oppose recoil of the cannon in the vertical
direction. The momentum in the vertical direction is transferred to the earth. The
energy is transferred into the ground, making a dent where the cannon is. After
long barrages, cannons have erratic aim because the ground is full of divots.
Openstax College Physics
Instructor Solutions Manual
Chapter 8
48.
Professional Application A 5.50-kg bowling ball moving at 9.00 m/s collides with a
0.850-kg bowling pin, which is scattered at an angle of 85.0 to the initial direction of
the bowling ball and with a speed of 15.0 m/s. (a) Calculate the final velocity
(magnitude and direction) of the bowling ball. (b) Is the collision elastic? (c) Linear
kinetic energy is greater after the collision. Discuss how spin on the ball might be
converted to linear kinetic energy in the collision.
Solution
(a) m1v1  m1v'1 cos 1  m2 v' 2 cos  2 or m1v'1 cos 1  m1v1  m2 v' 2 cos  2
0  m1v'1 cos 1  m2 v' 2 cos  2 or m1v'1 cos 1   m2 v' 2 cos  2 . Therefore,
tan  2 

 m2 v1 ' sin  2
m1v1  m2 v 2 ' cos  2
 0.850 kg 15.0 m/s sin85 
  0.2625
5.50 kg 9.00 m/s   0.850 kg 15.0 m/s cos85
θ1  tan 1  0.2625   14.71   14.7
Using the second equation:
 m2 v2 ' sin  2  0.850 kg 15.0 m/s sin85 
v1 ' 

 9.10 m/s
5.50 kg sin  14.71
m1sin 1
1 / 2m1v12
5.50 kg 9.00 m/s 
KE


 0.689.
2
2
KE' 1 / 2 m1v'1  m2 v'2 5.50 kg 9.096 m/s 2  0.850 kg 15.0 m/s 2
Since KE is not constant, it is not an elastic collision.
(c) Right after the collision, the ball spins faster than it is rolling. A frictional force
opposes the spinning of the ball. This force has two effects: it creates a torque to
slow the spinning and it produces a linear acceleration. The net effect is to convert
rotational kinetic energy into linear kinetic energy. This process continues until the
ball rolls without slipping (when v  r ).
2
(b)


49.
Professional Application Ernest Rutherford (the first New Zealander to be awarded
the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by
scattering helium-4 nuclei  4 He  from gold-197 nuclei 197 Au  . The energy of the
incoming helium nucleus was 8.00  1013 J , and the masses of the helium and gold
nuclei were 6.68 1027 kg and 3.29  1025 kg , respectively (note that their mass ratio
is 4 to 197). (a) If a helium nucleus scatters to an angle of 120 during an elastic
collision with a gold nucleus, calculate the helium nucleus’s final speed and the final
velocity (magnitude and direction) of the gold nucleus. (b) What is the final kinetic
energy of the helium nucleus?
Solution
 2KEi 
 2 8.00  10 13 J 
1
2


(a) m1v1  KEi  vi  
   6.68  10 27 kg 
2


 m1 
Conservation of internal kinetic energy gives:
1/ 2


1/ 2
 1.548  10 7 m/s
Openstax College Physics
Instructor Solutions Manual
(i)
1
1
1
2
m1v1  m1v'12  m1v' 22
2
2
2

Chapter 8

(i’)
m1 2 2
v'1 v' 21 '  v' 22
m2
Conservation of momentum along the x-axis gives:
or
(ii)
m1v1  m1v'1 cos 1  m2 v' 2 cos  2
Conservation of momentum along the y-axis gives:
0  m1v'1 sin 1  m2 v' 2 sin  2
Rearranging Equations (ii) and (iii) gives:
(iii)
m1v1  m1v'1 cos 1  m2 v' 2 cos  2
(ii’)
(iii’)
 m1v'1 sin 1  m2 v' 2 sin  2
Squaring Equation (ii’) and (iii’) and adding gives:
2
m22 v' 22 cos 2  2  m22 v' 22 sin 2  2  (m1v1  m1v'1 cos 1 ) 2   m1v'1 cos 1 
or m22 v' 22  m12 v'12 2m12 v1v'1 cos 1  m12 v'12
Solving for v'22 and substituting into (i’):






m1 2
m12 2
2
v1  v'1  2 v1  v'12 2v1v'1 cos 1 so that
m2
m2
m1 2
v1  v'12 2v1v'1 cos 1
m2
Using v1  1.548 107 m/s ;1  120; m1  6.68 1027 kg; m2  3.29 1025 kg
v12  v'12 
 m1  2  m1

 m 
1 
v'1  2
v1 cos 1 v'1 1  1 v12  0
 m2 
 m2

 m2 
a  1
m1
2m
 1.0203, b   1 v1 cos 1  3.143 105 m/s,
m2
m2
 m 
c  1  1 v12  2.348  1014 m 2 /s 2 so that
 m2 
v'1 

 b  b 2  4ac
2a
 3.143  10 5 m/s 
3.143  10
v'1  1.50  10 7 m/s and v' 2 




m/s  41.0203  2.348  1014 m 2 /s 2
or
21.0203
5
2

m1 2
v1  v'1  5.36  10 5 m/s
m2
Openstax College Physics
Instructor Solutions Manual
tan  2 

Chapter 8

 v'1 sin 1
 1.50  107 m/s sin120 

 0.56529
v1  v'1 cos 1 1.58  107 m/s  1.50  107 m/s cos120
or  2  tan  0.56529   29.5
(b) The final kinetic energy is then:


1


KEf  0.5m1v'12  0.5 6.68 10 27 kg 1.50 107 m/s
50.
Solution

2
 7.52 10 13 J
Professional Application Two cars collide at an icy intersection and stick together
afterward. The first car has a mass of 1200 kg and is approaching at 8.00 m/s due
south. The second car has a mass of 850 kg and is approaching at 17.0 m/s due west.
(a) Calculate the final velocity (magnitude and direction) of the cars. (b) How much
kinetic energy is lost in the collision? (This energy goes into deformation of the cars.)
Note that because both cars have an initial velocity, you cannot use the equations for
conservation of momentum along the x -axis and y -axis; instead, you must look for
other simplifying aspects.
(a) m1v1  m1  m2 v' sin 
m2 v 2  m1  m2 v' cos 
m1
8.00 m/s
p1
p2
p3
tan  

m2
17.0 m/s
v'
( m 1+ m )2
mv 
 1200 kg  8.00 m/s 
 m1v1
  33.6
   tan 1  1 1   tan 1 
m2 v 2
m
v
850
kg

17.0
m/s


 2 2
m1v1
1200 kg  8.00 m/s

 8.462 m/s  8.46 m/s
m1  m2 sin  1200 kg  850 kg sin33.6 
(b) Initial energy is given by
v' 
1
1
KE1  KE 2  m1v12  m2 v22
2
2
2
2
 0.51200 kg 8.00 m/s   850 kg 17.0 m/s   161,225 J


The final kinetic energy is
1
2
KE'  m1  m2 v'2  0.52050 kg 8.462 m/s   73,396 J. Thus,
2
KE  73,396 J  161,225 J   8.78 10 4 J
Openstax College Physics
Instructor Solutions Manual
Chapter 8
The negative answer indicates a loss of kinetic energy.
51.
Solution
Starting with equations m1v1  m1v1cos1  m2v2 cos2 and 0  m1v1sin 1  m2v2 sin 2
for conservation of momentum in the x - and y -directions and assuming that one
object is originally stationary, prove that for an elastic collision of two objects of equal
1
1
1
masses, mv12  mv1 2  mv2 2  mv1v2 cos 1   2  as discussed in the text.
2
2
2
m1v1  m1v'1 cos 1  m2 v' 2 cos  2
(i)
(ii)
0  m1v'1 sin 1  m2 v' 2 sin  2
Squaring Equation (i) and (ii) and adding gives:
m 2 v12  (mv'1 cos1  mv'2 cos 2 ) 2  (mv'1 sin 1  mv'2 sin  2 ) 2



 m 2 v'12 cos 2 1  sin 2 1  m 2 v'22 cos 2  2  sin 2  2
 2m 2 v1v2 cos1 cos 2  sin 1 sin  2 
Dividing by 2m and using trigonometric identities:

1
1
1
2
mv1  mv'12  mv' 22  mv'1 v' 2 cos1   2 
2
2
2
52.
Solution
Integrated Concepts A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck
a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far
does the player recoil in the time it takes the puck to reach the goal 15.0 m away?
 0.150 kg 45.0 m/s 
  7.50  10 2 m/s
mh
90.0 kg
dp
15.0 m

 0.333 s, and the hockey player
The puck travels the 15.0 m in t p 
v p 45.0 m/s
The recoil velocity is: v h  
mp v p

travels d h  vh t  7.50  10 2 m/s 0.333 s   0.0250 m  2.50 cm in the opposite
direction.
Openstax College Physics
Instructor Solutions Manual
Chapter 8
8.7 INTRODUCTION TO ROCKET PROPULSION
53.
Solution
Professional Application Antiballistic missiles (ABMs) are designed to have very large
accelerations so that they may intercept fast-moving incoming missiles in the short
time available. What is the takeoff acceleration of a 10,000-kg ABM that expels 196
kg of gas per second at an exhaust velocity of 2.50  103 m/s?
a
v e m
2.50  10 3 m/s
196 kg/s   9.80 m/s 2
g 
m t
10,000 kg
 4.90 m/s 2t  9.80 m/s 2  39.2 m/s 2
54.
Solution
Professional Application What is the acceleration of a 5000-kg rocket taking off from
the Moon, where the acceleration due to gravity is only 1.6 m/s 2 , if the rocket expels
8.00 kg of gas per second at an exhaust velocity of 2.20  103 m/s?
a
v e m
2.20  10 3 m/s
8.00 kg/s   1.60 m/s 2
g 
m t
5000 kg
 3.52 m/s 2  1.60 m/s 2  1.92 m/s 2
55.
Solution
Professional Application Calculate the increase in velocity of a 4000-kg space probe
that expels 3500 kg of its mass at an exhaust velocity of 2.00  103 m/s . You may
assume the gravitational force is negligible at the probe’s location.
m 
v  v 0  v e ln  0 ,
 m 
 4000 kg 
  4.159  10 3 m/s  4.16  10 3 m/s
v  v 0  2.00  10 3 m/s ln 
 500 kg 

56.

Professional Application Ion-propulsion rockets have been proposed for use in space.
They employ atomic ionization techniques and nuclear energy sources to produce
extremely high exhaust velocities, perhaps as great as 8.00  106 m/s . These
techniques allow a much more favorable payload-to-fuel ratio. To illustrate this fact:
(a) Calculate the increase in velocity of a 20,000-kg space probe that expels only 40.0kg of its mass at the given exhaust velocity. (b) These engines are usually designed to
produce a very small thrust for a very long time—the type of engine that might be
useful on a trip to the outer planets, for example. Calculate the acceleration of such an
engine if it expels 4.50  10 6 kg/s at the given velocity, assuming the acceleration due
to gravity is negligible.
Openstax College Physics
Solution
Instructor Solutions Manual
Chapter 8
 m0 
, v 0  0
(a) v  v 0  ve ln 
 m 


20,000 kg
  1.602  10 4 m/s  1.60  10 4 m/s
v  8.00  10 6 m/s ln 
20,000
kg

40.0
kg


(b) a 
v e m 8.00  10 6 m/s

4.50  10 -6 kg/s  1.80  10 -3 m/s 2
m t
20,000 kg


57.
Derive the equation for the vertical acceleration of a rocket.
Solution
The force needed to give a small mass m an acceleration a m is F  mam . To
accelerate this mass in the small time interval t at a speed ve requires ve  am t ,
m
so F  ve
. By Newton’s third law, this force is equal in magnitude to the thrust
t
m
force acting on the rocket, so Fthrust  ve
, where all quantities are positive.
t
v m
Applying Newton’s second law to the rocket gives Fthrust  mg  ma  a  e
g,
m t
where m is the mass of the rocket and unburnt fuel.
58.
Professional Application (a) Calculate the maximum rate at which a rocket can expel
gases if its acceleration cannot exceed seven times that of gravity. The mass of the
rocket just as it runs out of fuel is 75,000-kg, and its exhaust velocity is 2.40  103 m/s .
Assume that the acceleration of gravity is the same as on Earth’s surface 9.80 m/s 2  .
(b) Why might it be necessary to limit the acceleration of a rocket?
Solution
v e m
 g  7 g , so that
m t
m 8 gm 8  9.80 m/s 2  75,000 kg


 2.45  10 3 kg/s
3
t
ve
2.40  10 m/s
(b) It might be necessary to limit the acceleration of a rocket because of the impact of
such large forces on the human body. It might not be possible to survive
acceleration greater than 7 g . Also, equipment might get crushed under such high
forces.
(a)
a
59.
Given the following data for a fire extinguisher-toy wagon rocket experiment,
calculate the average exhaust velocity of the gases expelled from the extinguisher.
Starting from rest, the final velocity is 10.0 m/s. The total mass is initially 75.0 kg and
is 70.0 kg after the extinguisher is fired.
Solution
v
10.0 m/s
m 
v  ve ln  0 , so that ve 

 145 m/s
ln m0 / m  ln(75.0 kg/70.0 kg)
 m 
Openstax College Physics
60.
Solution
Instructor Solutions Manual
Chapter 8
How much of a single-stage rocket that is 100,000 kg can be anything but fuel if the
rocket is to have a final speed of 8.00 km/s , given that it expels gases at an exhaust
velocity of 2.20  103 m/s?
m 
m  v
v  ve ln  0 . Assume v0  0 so that ln  0  
 m
 m  ve
m  m0 e
v ve
 8.00103 m/s   2.20103 m/s 
 

 100,000 kg e 
 2634.8 kg  2.63  103 kg
61.
Professional Application (a) A 5.00-kg squid initially at rest ejects 0.250-kg of fluid
with a velocity of 10.0 m/s. What is the recoil velocity of the squid if the ejection is
done in 0.100 s and there is a 5.00-N frictional force opposing the squid’s movement?
(b) How much energy is lost to work done against friction?
Solution
(a) First, find v'1 , the velocity after ejecting the fluid:
m1  m2 v  0  m1v'1  m2 v'2 , so that
 m2 v'2  0.250 kg 10.0 m/s 

  0.526 m/s
m1
4.75 kg
Now, the frictional force slows the squid over the 0.100 s
p  ft  m1v1' ,f  m2 v2' , gives :
v'1 
ft  m2 v2' 5.00 N 0.100 s   0.250 kg 10.0 m/s 

  0.421 m/s
m1
4.75 kg
1
1
1
(b) KE  m1v'12,f  m1v'12  m1 v1,2f  v'12
2
2
2
1
2
2
 4.75 kg 0.421 m/s   0.526 m/s    0.236 J
2
v1' ,f 



62.
Solution

Unreasonable Results Squids have been reported to jump from the ocean and travel
30.0 m (measured horizontally) before re-entering the water. (a) Calculate the initial
speed of the squid if it leaves the water at an angle of 20.0 , assuming negligible lift
from the air and negligible air resistance. (b) The squid propels itself by squirting
water. What fraction of its mass would it have to eject in order to achieve the speed
found in the previous part? The water is ejected at 12.0 m/s ; gravitational force and
friction are neglected. (c) What is unreasonable about the results? (d) Which premise
is unreasonable, or which premises are inconsistent?
(a) R 
v 02
 sin 2 so that
g
 gR 
v0  

 sin 2 
1/ 2
 9.80 m/s 2  30.0 m 

 
sin40 


1/ 2
 21.39 m/s  21.4 m/s
Openstax College Physics
Instructor Solutions Manual
Chapter 8
(b) v  v e ln  m0 , so that m  e v / ve  e 21.39 m/s  12.0 m/s   0.168  fraction retained
m0
 m
m
1
 0.832  fraction ejected
m0
(c) It is unreasonable to think that the squid could eject 83% of its mass.
(d) It is unreasonable that the squid can travel 30.0 m horizontally through the air.
This file is copyright 2016, Rice University. All Rights Reserved.