Download Chapter 7 LINEAR MOMENTUM

Chapter 7
LINEAR MOMENTUM
Conceptual Questions
1. The likelihood of injury resulting from jumping from a second floor window is primarily determined by the
average force acting to decelerate the body.
(a) The deceleration time interval for a person landing stiff legged on pavement is very short. The impulsemomentum theorem tells us that the average force acting on the person’s feet must therefore be very large—
such a person is likely to incur injuries.
(b) Jumping into a privet hedge increases the time interval over which the body decelerates. This decreases the
average force on the person’s limbs and therefore decreases the likelihood of injury.
(c) Jumping into a firefighter’s net is the best option of the three. The net stretches downward, gradually
bringing the person to rest. Additionally, the firefighters lower the net with their hands as the person lands to
further lengthen the time interval during which the person is brought to rest.
2. (a) A body’s momentum change is equal to the impulse that has acted on it. Impulse is defined as the product of
the average force acting on a body and the time interval over which it acts—the bodies therefore experience
the same impulse and so have equal momentum changes.
(b) The change in a body’s velocity is defined as the ratio of the change in its momentum to its mass—the less
massive body therefore incurs a larger velocity change.
(c) The acceleration of a body is defined as the ratio of the force acting on it to its mass—the less massive body
therefore has the larger acceleration.
3. The muzzle speed is determined by the change in the bullet’s momentum. The impulse-momentum theorem tells
us that this momentum change is determined by the impulse acting on the bullet. The force acting on the bullet
due to the expanding hot gases is roughly constant throughout the muzzle. A shorter muzzle produces a shorter
time interval over which the bullet is accelerated by the firing force. This results in a smaller impulse and
therefore a smaller momentum change—thus producing lower bullet velocities.
4. After the explosion, each piece of the firecracker has a momentum vector associated with it that points in the
direction of its motion. The law of conservation of linear momentum tells us that the vector sum of the
momentum of all the pieces of the firecracker must equal the initial momentum of the whole firecracker—in this
case, both the initial and the final net momentum vectors equal zero.
5. The law of the conservation of linear momentum states that in the absence of external interactions, the linear
momentum of a closed system is constant. Floating in free space, the astronaut and the wrench form a closed
system free from interactions with other bodies. If the astronaut throws the wrench in the direction opposite the
ship, conservation of momentum dictates that he must in turn move toward the ship.
6. The horizontal component of the golf ball’s momentum is conserved since no external force acts on the ball in the
horizontal direction. The vertical component of the ball’s momentum is not conserved however because the
Moon’s gravitational force interacts with it and changes its momentum.
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Chapter 7: Linear Momentum
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7. In an elastic collision between the hammer and nail, the kinetic energy of the system is conserved while in a
perfectly inelastic collision, the greatest percentage of the kinetic energy is lost. The energy lost by the system in
a perfectly inelastic collision is used to do the work required to bring the hammer and nail together. In an elastic
collision, this work is available to drive the nail into the wood—the total work available to drive the nail is
therefore greater for an elastic collision. Thus, for equal applied forces, the hammer will drive the nail further into
the wood if the collision is elastic.
8. The momentum of the squid (including the water inside its cavity) comprises a system for which momentum is
conserved. The means the momentum of the squid (plus water) must be the same before and after some of the
water has been ejected. When the squid expels some water, the water gains momentum in the direction it is being
expelled. To conserve momentum, the squid must gain an equal amount of momentum in the opposite direction,
propelling it forward. Similarly, a rocket engine expels exhaust from burning fuel to propel itself forward.
9. First law: The momentum of an object is constant unless acted upon by an external force. Second law: The net
force acting on an object is equal to the rate of change of the object’s momentum. Third law: When two objects
interact, the changes in momentum that each imparts to the other are equal in magnitude and opposite in direction.
10. Noting that the (translational) kinetic energy can be written as p 2 /(2m), and that both objects have the same
kinetic energy, it is evident that the object with the greater mass has the larger magnitude of momentum.
11. The woman’s center of mass is not necessarily 0.80 m above the floor, because her mass is not necessarily
distributed uniformly with height. Normally, the upper body of a person is more massive than the lower body and
thus we would expect the woman’s center of mass to be slightly higher than 0.80 m.
12. The frictional force of the road on the tires supplies the external force to change the bicycle’s momentum.
Changes in the bicycle’s kinetic energy do not require an external force. For example, the rider could throw her
helmet away hard, increasing both her and the helmet’s speed. The kinetic energy of the system (bicycle, rider,
and helmet) would increase, while the momentum would remain the same. Note that the work-energy theorem
(total work done equals change in kinetic energy) cannot be used here, because the internal structure of the
system cannot be ignored.
13. An impulse must be supplied to the egg to change its momentum and bring it to rest. A good strategy is to make
the time interval over which the stopping force is applied as large as possible. This will reduce the magnitude of
the force required to stop the egg. One should therefore attempt to catch the egg with a swinging motion, moving
the hand backwards as it is being caught, to bring it to rest as slowly and gently as possible.
14. The collisions of the balls in the “executive toy” are nearly perfectly elastic. The kinetic energy of the system just
before and after a collision must therefore be the same. This is the reason we never see three balls moving away
after a collision in which two balls were initially pulled back and released—such an event would not conserve
kinetic energy.
15. According to the impulse-momentum theorem, the change in momentum of the baseball is equal to the impulse it
receives from the bat. Impulse is equal to the average force times the time interval over which the force is applied.
To give the ball the greatest possible momentum, one should attempt to maximize the amount of time during
which the force is being applied.
16. Jeremy has it right. By momentum conservation, Micah needs to throw the balls forward if he wants to propel
himself backward, but the balls need not strike any surface. You can also consider Newton’s third law and see
that it is the force by the balls on Micah’s hand that pushes Micah backward.
17. Daryl has done his homework. If he falls when rock climbing, his rope will stretch and stop him more gradually
than the rope Mary wants to buy. In a fall, the climber’s momentum must go from some initial value to zero. If
the time over which the momentum is decreased to zero is longer, the average force delivered by the rope is
smaller.
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Physics
Chapter 7: Linear Momentum
Problems
1. Strategy Use the definition of linear momentum.
Solution Find the magnitude of the total momentum of the system.

 


 


p total  p1  p 2  mv1  mv 2  m( v1  v 2 )  m[ v1  ( v1 )]  0, so the magnitude is 0 .
2. Strategy Use the definition of linear momentum.
Solution Find the momentum of the automobile.

9800 N
 W 
p  mv  v 
(35 m s south)  3.5  104 kg  m s south
g
9.80 m s 2
3. Strategy and Solution Impulse  F t , so the SI unit is N  s  kg  m s 2  s  kg  m s. p  mv, so the SI unit is
kg  m s. Therefore, the SI unit of impulse is the same as the SI unit of momentum.
4. Strategy Use the impulse-momentum theorem.
Solution Find the final speed of the cue ball.
F t (24 N)(0.028 s)
p  pf  pi  mvf  m(0)  Fav t , so vf  av 
 4.2 m s .
m
0.16 kg
5. Strategy Add the momenta of the three particles.
Solution Find the total momentum of the system.

 




p tot  p1  p 2  p3  m1v1  m2 v 2  m3 v3  m1v1 north  m2 v2 south  m3v3 north
 (m1v1  m2 v2  m3v3 ) north   (3.0 kg)(3.0 m s)  (4.0 kg)(5.0 m s)  (7.0 kg)(2.0 m s)  north
 3 kg  m s north
6. (a) Strategy Form a ratio of the magnitudes of the final and initial momenta.
Solution Compute the ratio.
pf mvf vf 60.0 mi h



 3.00
pi mvi vi 20.0 mi h
(b) Strategy Form a ratio of the final and initial kinetic energies.
Solution Compute the ratio.
Kf
Ki

1 mv 2
f
2
1 mv 2
i
2
v
  f
 vi
2

2
  3.00  9.00

7. Strategy The initial momentum is toward the wall and the final momentum is away from the wall.
Solution Find the change in momentum.
p  pf  pi  mvf  mvi  m(vf  vi )  (5.0 kg)(2.0 m s  2.0 m s)  20 kg  m s , so

p  20 kg  m s in the x-direction .
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Chapter 7: Linear Momentum
Physics
8. Strategy The final and initial velocities are the same, since air resistance is ignored. Use the definition of the
1
linear momentum. Use Eq. y  viy t  g (t )2 to find the initial speed. Let up be the positive direction.
2
Solution Find the initial speed.
1
1
y  0  viy t  g (t ) 2 , so viy  g t.
2
2
Find p.
1

p  pfy  piy  m(vfy  viy )  m(viy  viy )  2m  g t   mg t  (3.0 kg)(9.80 m s 2 )(3.4 s)
2


 1.0  102 kg  m s

So, p  1.0 102 kg  m s downward .
9. Strategy Use the definition of linear momentum. Let up be the positive direction.
Solution vf  vfy  viy  g t   g t , since the object starts from rest.
Find p.
p  pf  pi  m(vf  vi )  m( g t  0)   mg t  (3.0 kg)(9.80 m s 2 )(3.4 s)  1.0  102 kg  m s, so

p  1.0  102 kg  m s downward .
10. Strategy Use the impulse-momentum theorem.
Solution Find the average force.
p mv (50.0 kg)(3.0 m s  0)


 7.5 N
Fav 
t
t
20.0 s
The force necessary is 7.5 N in the direction of the sled’s velocity.
11. Strategy Use the impulse-momentum theorem. Let the forward direction be positive.
Solution Find the time interval for which the engine must be fired.
p mv (3800 kg)(1.1 104 m s  2.6  104 m s)
t 


 320 s
Fav
Fav
1.8  105 N
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Physics
Chapter 7: Linear Momentum
12. (a) Strategy Use the component method of subtracting vectors.
y
Solution Compute the magnitude of the change in momentum.
p x  mvx and p y  mv y .

p  m (vx )2  (v y ) 2  (0.15 kg) [0  (20 m s)]2  (15 m s  0)2  3.8 kg  m s
x
pf
pi
Find the angle of the change in momentum.
p y
v y
15
  tan 1
 tan 1
 tan 1
 37
p x
vx
20
The change in momentum of the baseball is

3.8 kg  m s at 37 above the horizontal direction opposite vi .
(b) Strategy Use the impulse-momentum theorem.
Solution Find the average force of the bat on the ball.


p 3.75 kg  m s
Fav 

 75 N, so F  75 N in the same direction as p .
0.050 s
t
13. Strategy Use the impulse-momentum theorem. Let the positive direction be in the direction of motion.
Solution Find the average horizontal force exerted on the automobile during breaking.
p m(vf  vi ) (1.0  103 kg)(0  30.0 m s)


 6.0  103 N
Fav 
t
t
5.0 s

So, Fav  6.0  103 N opposite the car’s direction of motion .
14. Strategy Use the impulse-momentum theorem.
Solution
(a) Compute the changes in momenta for each direction.
pnorth  0 and peast  Fav t  mveast  mveast .

Find the magnitude and direction of v f .
15 m/s
N
15 N
2
2
 (15 N)(4.0 s) 
 F t 
vf  vnorth 2  veast 2  vnorth 2   av   (15 m s)2  
  25 m s and
m


 3.0 kg 
v
15 m s

  tan 1 north  tan 1
 37 north of east, so v f  25 m s at 37 north of east .
veast
20 m s
(b) Let +y be north and +x be east. Compute the change in momentum.
p  Fav t  (15 N)(4.0 s)  60 kg  m s.

The entire change in momentum is due to the force, so p  60 kg  m s east .
375
Δp
pi
pf
y
x
Chapter 7: Linear Momentum
Physics
15. (a) Strategy Use the definition of linear momentum. Use vf2y  viy 2  2a y y to find the speed after the fall.
Solution Find the initial speed, which is the final speed after the fall.
vf2y  viy 2  vf2y  0  2 g y  2 gh, so vfy  2 gh .


If up is positive, v y  2 gh down   2 gh up  vi .


p  m(vf  vi )  m  0   2 gh   m 2 gh  (60.0 kg) 2(9.80 m s 2 )(8.0 m)  750 kg  m s , so



p  750 kg  m s upward .
(b) Strategy The impulse on the net is equal to the boy’s weight times t plus the change in momentum of the

boy due to the net, p.
Solution Find the impulse on the net.

mg t downward  p  (60.0 kg)(9.80 N kg)(0.40 s) downward  750 kg  m/s downward
 990 N  s downward
(c) Strategy Use Eq. (7-3).
Solution Find the average on the net due to the boy.


p 990 kg  m/s downward

 2500 N downward
Fav 
t
0.40 s
16. Strategy The impulse is equal to the area under the graph. Use the impulse-momentum theorem. Let the positive
direction be to the right.
Solution Each rectangle of the grid is equal to (100 N)(0.0010 s)  0.10 kg  m s. The area can be divided easily
into three right triangles and one rectangle. Thus, there are 12 (7)(4)  12 (6)(2)  12 (8)(6)  (6)(4)  68 rectangles
under the graph and the magnitude of the impulse is p  68(0.10 kg  m s)  6.8 kg  m s  mv. The impulse is
opposite the direction of motion of the initial velocity. Compute the final speed.
p
p
6.8 kg  m s
v 
 vf  vi , so vf  vi 
 30 m s 
 29 m s .
m
m
0.115 kg
17. (a) Strategy Use conservation of energy to find the speed with which the pole-vaulter lands on the padding.
Solution
1
K  mv 2  U  mgh, so v  2 gh  2(9.80 m s 2 )(6.0 m)  10.84 m/s  11 m s .
2
(b) Strategy The padding exerts an upward force on the pole-vaulter while gravity continues to exert a
downward force. Use the impulse-momentum theorem and let Fav represent the average force exerted by the
padding.
Solution
p  Fnet t , and Fnet  Fav  mg , so
m(vf  vi )
m[0  ( 2 gh )]
p
Fav  Fnet  mg 
 mg 
 mg 
 mg
t
t
t
 2 gh

 10.84 m/s

Fav  m 
 g   60.0 kg  
 9.80 m s 2   1900 N .
 t

0.50
s




376
Physics
Chapter 7: Linear Momentum
18. Strategy Use conservation of momentum.
Solution Find the recoil speed of the rifle.
m




0.0100 kg




p rf  p bf  mr v rf  mb v bf  p ri  p bi  0  0, so v rf  b  v bf 
(820 m s)  1.8 m s .
mr
4.5 kg
19. (a) Strategy Right after the collision, the bullet and baseball combination must have the same momentum as the
bullet had just before it stuck the baseball.
Solution Before the collision, the momentum of the bullet is pi  mbullet vbullet . After the collision, the
momentum of the bullet and baseball combination is pf  (mbullet  mbaseball )vf  pi .
Thus, the speed of the bullet and baseball combination right after the collision was
mbullet vi
pi
(0.030 kg)(200 m s)
vf 


 33 m s .
mbullet  mbaseball mbullet  mbaseball
0.030 kg  0.15 kg
(b) Strategy Use conservation of energy to determine the work done by air resistance on the bullet and baseball
combination.
Solution Determine the work done by air resistance. Let up be the positive direction.
Wtotal  Wc  Wnc  U  Wair  K , so
1
 1

Wair  K  U  0  mv 2  mg y  (0.18 kg)   (33.333 m s) 2  (9.80 m s 2 )(37 m)   34.73 J.
2
 2

W
34.73 J
Since Wair  Fair, av y, Fair, av  air 
 0.94 N. Therefore, the average force of air resistance
37 m
y
was 0.94 N down .
20. Strategy Use conservation of momentum.
Solution Find the recoil speed of the submarine.
m 




250 kg



(100.0 m s)  0.010 m s .
psf  p tf  ms vsf  mt v tf  psi  p ti  0  0, so vsf  t  v tf 
ms
2.5  106 kg
21. Strategy Use conservation of momentum.
Solution Find the recoil speed of the thorium nucleus.


pi  0  p f , so if n = nucleus and p = particle,
mp 
4.0 u 





0.050(2.998  108 m s)   2.6  105 m s .
p n  p p  mn v n  mp v p  0, so v n 
vp 

234 u 
mn
22. Strategy Use the law of conservation of linear momentum to determine the speed Dash must throw the balls.
Solution According to the law of conservation of linear momentum, Dash and his skateboard will move
backward with linear momentum equal in magnitude to the magnitude of the combined momentum of the balls.
Find the speed of the balls.
m v
(60 kg)(0.50 m s)
 100 m s (224 mph) .
pb  3mb vb  mD vD  pD , so vb  D D 
3mb
3(0.10 kg)
Since 224 mph is faster than any human can throw a ball, Dash will not succeed .
377
Chapter 7: Linear Momentum
Physics
23. Strategy Use the law of conservation of linear momentum to determine the astronaut’s speed.
Solution According to the law of conservation of linear momentum, the astronaut will move toward the ship with
linear momentum equal in magnitude to the magnitude of the combined momentum of the objects thrown. Find
the speed of the astronaut after he throws the mallet.
pobjects  pw  ps  pm  mw vw  ms vs  mm vm  pA  mA vA , so
vA 
mw vw  ms vs  mm vm
mA
(0.72 kg)(5.0 m s)  (0.80 kg)(8.0 m s)  (1.2 kg)(6.0 m s)

 0.30 m s .
58 kg
24. Strategy Use conservation of energy to determine the skier’s speed and momentum just before he grabs the
backpack. Then, use conservation of linear momentum to find his new speed after he grabs the backpack. Finally,
from Chapter 4, use the equations for motion with a changing velocity.
Solution Use conservation of energy to find the speed of the skier just before he grabs the backpack.
1
1
E  K  U  K f  Ki  U f  U i  ms vs 2  0  0  ms gh  ms vs 2  ms gh  0, so vs  2 gh .
2
2
Use conservation of linear momentum to find his new speed after he grabs the backpack.
pi  ms vs  pf  (ms  mb )vx , so vx 
ms vs
ms  mb

ms 2 gh
ms  mb
.
Now, find the time it takes for an object to fall 2.0 m from rest.
1
1
1
2 y
y  viy t  g (t )2  (0)t  g (t )2   g (t )2 , so t  
.
g
2
2
2
The skier will travel a horizontal distance of
m 2 gh
2y 2ms yh 2(65 kg) (2.0 m)(5.0 m)
x  vx t  s



 4.8 m .
ms  mb
g
ms  mb
65 kg  20 kg
25. Strategy Use conservation of momentum.
Solution Find the recoil speed of the railroad car.


pi  0  p f , and since we are only concerned with the horizontal direction, we have:
m
98 kg
(105 m s) cos 60.0  0.10 m s .
mc vcx  ms vsx , so vcx  s vsx 
mc
5.0  104 kg
26. Strategy Use conservation of momentum.
Solution Find the mass of the man and the car.


pi  0  p f , and since we are only concerned with the horizontal direction, we have:
mmc vmc  mb vb , so mmc 
vb
vmc
mb 
(173 m s) cos 30.0
1.0  103 m s
378
(0.010 kg)  1500 kg .
Physics
Chapter 7: Linear Momentum
39. Strategy Use conservation of momentum. Let the positive direction be to the right.
Solution Find the final velocity of the helium atom.
mHe vHef  mO vOf  mHe vHei  mO vOi , so
vHef 

mO (vOi  vOf )  mHe vHei
mHe
mO
mHe
(vOi  vOf )  vHei 
32.0 u
 412 m s  456 m s   618 m s  270 m s.
4.00 u
Thus, the velocity of the helium atom after the collision is 270 m s to the right .
40. Strategy Linear momentum is conserved, so pf  pi .
Solution Find the change in speed of the car.
pf  (mcar  mclay )vf  pi  mcar vi , so vf 
v  vf  vi 
mcar
mcar  mclay
mcar
mcar  mclay
vi , and


mcar


120 g
vi  vi  
 1 vi  
 1 (0.75 m s)  0.15 m s .
 mcar  mclay

120
g
30.0
g





41. Strategy Use conservation of momentum.
Solution
(a) The collision is perfectly inelastic, so v1f  v2f  vf . Find the speed of the two cars after the collision.
v
1.0 m s
m1v1i  m2 v2i  mv1i  4.0m(0)  m1v1f  m2 v2f  mvf  4.0mvf , so vf  1i 
 0.20 m s .
5.0
5.0
(b) The cars are at rest after the collision, so v1f  v2f  0.
mv1i  4.0mv2i  0, so v2i  
v1i
4.0

1.0 m s
 0.25 m s. The initial speed was 0.25 m s .
4.0
42. Strategy Use conservation of momentum. The block is initially at rest, so v2i  0. Let east be in the +xdirection.
Solution Find the final velocity of the block.
m1v1f  m2 v2f  m1v1i  m2 v2i  m1v1i  m2 (0), so
m (v  v ) 0.020 kg
v2f  1 1i 1f 
 200.0 m s  (100.0 m s)  3.0 m s.
m2
2.0 kg

Thus, v block  3.0 m s east .
43. Strategy Use conservation of momentum. The collision is perfectly inelastic, so v1f  v2f  vf . Also, the block
is initially at rest, so v2i  0.
379
Chapter 7: Linear Momentum
Physics
Solution Find the speed of the block of wood and the bullet just after the collision.
m1v1f  m2 v2f  (m1  m2 )vf  m1v1i  m2 v2i  m1v1i  m2 (0), so
m1
0.050 kg
vf 
v 
(100.0 m s)  5.0 m s .
m1  m2 1i 0.050 kg  0.95 kg
44. Strategy The collision is perfectly inelastic since the bullet embeds in the wood. Friction does negative work on
the block and bullet combination. Use Newton’s second law and conservation of momentum.
Solution Let the +x-direction be in the direction of motion. Find the acceleration of the
block and bullet due to friction.
Fy  N  (mbullet  mblock ) g  0, so N  (mbullet  mblock ) g.
Fx   f k   k N   k (mbullet  mblock ) g  (mbullet  mblock )ax , so a x   k g .
Find the initial speed of the block and bullet (just after the collision).
N
x
fk
(mbullet + mblock)g
vfx  vix  0  vix  2a x x  2 k g x, so vix  2k g x  v.
Use conservation of momentum to find the speed of the bullet just before its collision with the block.
mbullet vbullet  (mbullet  mblock )v  (mbullet  mblock ) 2k g x , so
2
2
vbullet 
2
(mbullet  mblock ) 2 k g x
mbullet

(2.02 kg) 2(0.400)(9.80 m s 2 )(1.50 m)
0.020 kg
 350 m s .
45. Strategy Use conservation of momentum.
Solution Find the total momentum of the two blocks after the collision.
p2  p1
p2f  p2i  p1i  p1f
p1f  p2f  p1i  p2i
(m1  m2 )vf  m1v1i  m2 v2i
pf  (2.0 kg) 1.0 m s   (1.0 kg)(0)  2.0 kg  m s  p1i
Since p1i was directed to the right, and pf  p1i , the total momentum of the two blocks after the collision is
2.0 kg  m s to the right .
46. Strategy The collision is perfectly inelastic, so v1f  v2f  vf . Use conservation of momentum. Let the positive
direction be the initial direction of motion.
Solution Find the speed of the man (1) just after he catches the ball (2).
m1v1f  m2 v2f  m1vf  m2 vf  m1v1i  m2 v2i  m1 (0)  m2 v2i , so
m2
0.20 kg
vf 
v 
(25 m s)  0.066 m s .
m1  m2 2i 75 kg  0.20 kg
47. Strategy Use conservation of momentum. Let the positive direction be in the initial direction of motion.
Solution Find the speed of the Volkswagen after the collision.
pV  mV vVf  mV vVi  pB  mB vB , so
vVf 
mV vVi  mB vB
mV

(1.0  103 kg)  25 m s   (2.0  103 kg)(33 m s  42 m s)
1.0  103 kg
380
 43 m s .
Physics
Chapter 7: Linear Momentum
48. Strategy Use conservation of linear momentum. Since the collision is perfectly elastic, kinetic energy is
conserved.
Solution The 100-g ball is (1) and the 300-g ball is (2). Note that m2  3m1.
m1v1i  m2 v2i  m1v1i  m2 (0)  m1v1i  m1v1f  m2 v2f , so v1i  v1f 
m2
m1
v2f  v1f  3v2f .
1
1
1
1
1
1
1
m1v1i 2  m2 (0)2  m1v1i 2  m1v1f 2  m2 v2f 2  m1v1f 2  (3m1 )v2f 2 , so v1i 2  v1f 2  3v2f 2 .
2
2
2
2
2
2
2
Substitute for v1i .
 v1f  3v2f 2  v1f 2  6v1f v2f  9v2f 2  v1f 2  3v2f 2 , so 6v1f v2f
 6v2f 2 , or v1f  v2f .
Find the final velocities of each ball.
v1i  v1f  3v2f  v2f  3v2f  2v2f , so v2f 
1
1
v  (5.00 m s)  2.50 m s.
2 1i 2
Since v1f  v2f , v1f  2.50 m s. So, the 300-g ball moves at 2.50 m s in the +x -direction
and the 100-g
ball moves at 2.50 m s in the  x-direction .
49. Strategy Use conservation of momentum. Let the positive direction be the initial direction of motion.
Solution Find the speed of the 5.0-kg body after the collision.
m1v1f  m2 v2f  m1v1i  m2 v2i , so
m (v  v )  m2 v2i (1.0 kg) 10.0 m s  ( 5.0 m s)   (5.0 kg)(0)

 3.0 m s .
v2f  1 1i 1f
5.0 kg
m2
50. Strategy The collision is perfectly inelastic, so v1f  v2f  vf . Use conservation of momentum. Let the positive
direction be the initial direction of motion.
Solution Find the speed of the combination.
m (0)  m2 v2i
3.0 kg
m1vf  m2 vf  m1v1i  m2 v2i , so vf  1
(8.0 m s)  4.8 m s .

m1  m2
2.0 kg  3.0 kg
51. Strategy The spring imparts the same (in magnitude) impulse to each block. (The same magnitude force is
exerted on each block by the ends of the spring for the same amount of time.) So, each block has the same final
magnitude of momentum. (The initial momentum is zero.)
Solution Find the mass of block B.
mB vB  mA vA , so
v
d t
d
1.0 m
mB  A mA  A
mA  A mA 
(0.60 kg)  0.20 kg .
3.0 m
vB
d B t
dB
381
−p
p
A
B
Chapter 7: Linear Momentum
Physics
52. (a) Strategy Use conservation of momentum. Let the +x-direction be to the right.
Solution Find the final velocity of the other glider.
mv1f  mv2f  mvi  mvi , so v2f  vi  vi  v1f  0.50 m s  0.50 m s  1.30 m s  0.30 m s.
So, the velocity of the other glider is 0.30 m s to the left .
(b) Strategy Form a ratio of the final to the initial kinetic energies.
Solution Compute the ratio.
1 mv 2  1 mv 2
Kf
v 2  v2f 2 (1.30 m s)2  (0.30 m s) 2
1f
2f
2
 2
 1f

 3.6
Ki
2vi 2
2(0.50 m s)2
2 1 mvi 2
2

The final kinetic energy is greater than the initial kinetic energy. The extra kinetic energy
comes from the elastic potential energy stored in the spring.
53. Strategy The collision is perfectly inelastic, so v1f  v2f  v. The block is initially at rest, so v2i  0 and
v1i  vi . Use conservation of momentum.
Solution Find the speed of the bullet and block system.
mbul
(mbul  mblk )v  mbul vi  mblk (0), so v 
v.
mbul  mblk i
Determine the time it takes the system to hit the floor.
1
1
2h
y  h  viy t  g (t )2  0  g (t ) 2 , so t 
.
2
2
g
h = 1.2 m
Δx
Find the horizontal distance traveled.
x  vix t  vt 
mbul
2h
0.010 kg
2(1.2 m)

 0.49 m
vi
 400.0 m s 
0.010 kg  4.0 kg
mbul  mblk
g
9.80 m s 2
54. Strategy Use conservation of momentum. K i  Kf , since the collision is elastic.
Solution Show that the final speed of each object is the same as the initial speed.
m1v1f  m2 v2f  m1v1i  m2 v2i and p1i   p2i. So, m1v1i  m2 v2i and m1v1f  m2 v2f  0, or m1v1f   m2 v2f .
1
1
1
1
m v 2  m2 v2i 2  m1v1f 2  m2 v2f 2 , so m1v1i 2  m2 v2i 2  m1v1f 2  m2 v2f 2 .
2 1 1i
2
2
2
Eliminate v1i and v1f .
2
2
 m

 m

m1   2 v2i   m2 v2i 2  m1   2 v2f   m2 v2f 2
 m1

 m1

 m22
 2  m22
 2
 m2  v2i  
 m2  v2f

 m

 m

 1

 1

2
2
v2i  v2f
382
Physics
Chapter 7: Linear Momentum
Therefore, the initial and final speeds of object 2 are the same. Eliminate v2i and v2f .
2
2
 m

 m

m1v1i  m2   1 v1i   m1v1f 2  m2   1 v1f 
 m2

 m2

2
2


m 
m 
 m1  1  v1i 2   m1  1  v1f 2



m2 
m2 


v1i 2  v1f 2
Therefore, the initial and final speeds of object 1 are the same.
2
55. Strategy Use conservation of linear momentum. Since the collision is perfectly elastic, kinetic energy is
conserved as well.
Solution Let the +x-direction be in the original direction of motion of the 2.0-kg object. The 2.0-kg object is (1)
and the 6.0-kg object is (2). Note that m2  3m1.
m
m1v1i  m2 v2i  m1v1i  m2 (0)  m1v1i  m1v1f  m2 v2f , so v1i  v1f  2 v2f  v1f  3v2f .
m1
1
1
1
1
1
1
1
m1v1i 2  m2 (0)2  m1v1i 2  m1v1f 2  m2 v2f 2  m1v1f 2  (3m1 )v2f 2 , so v1i 2  v1f 2  3v2f 2 .
2
2
2
2
2
2
2
Substitute for v1i .
 v1f  3v2f 2  v1f 2  6v1f v2f  9v2f 2  v1f 2  3v2f 2 , so 6v1f v2f
 6v2f 2 , or v1f  v2f .
Find the final speed of the 6.0-kg object.
1
1
v1i  v1f  3v2f  v2f  3v2f  2v2f , so v2f  v1i  (10 m s)  5.0 m s .
2
2
56. Strategy Look at the collision in its center of mass frame. Assume a one-dimensional collision. The initial
velocities are v1ix and v2ix . The masses are m1 and m2.
Solution Transform the initial velocities to the CM frame by subtracting vCMx from each.
v1ix  v1ix  vCMx
v2ix  v2ix  vCMx
According to the result from Problem 54, the final speeds of the objects must be the same as the initial speeds, but
the final and initial velocities are oppositely directed since the objects rebound after colliding. Therefore,
v1fx  v1ix  v1ix  vCMx
v2fx  v2ix  v2ix  vCMx
Transform back to the original frame of reference.
v1fx  v1fx  vCMx  v1ix  2vCMx
v2fx  v2fx  vCMx  v2ix  2vCMx
The relative speed after the collision is v1fx  v2fx  v1ix  2vCMx  v2ix  2vCMx  v1ix  v2ix , which is the
relative speed before the collision.
57. Strategy Use conservation of momentum. Let each of the first two pieces be 45 from the positive x-axis (one
CW, one CCW).
Solution Find the speed of the third piece.
Find v3 x .
383
Chapter 7: Linear Momentum
Physics
p1x  p2 x  p3 x  mv1x  mv2 x  mv3 x  0, so v3 x  v1x  v2 x  v cos 45  v cos(45)  
v
2

v
2
 v 2.
Similarly,
v3 y  v1 y  v2 y  v sin 45  v sin( 45)  
v
2
v

2
 0, so v3  v3 x  v 2  120 m s  2  170 m s .
58. Strategy Use conservation of momentum.
Solution Find vBfx .
pix  MvAix  pfx  MvAfx  MvBfx , so vBfx  vAix  vAfx .
Find vBfy .
piy  0  pfy  MvAfy  MvBfy , so vBfy  vAfy .
Calculate vBf .
vBf  (vAix  vAfx ) 2  (vAfy )2 
 6.0
m s  1.0 m s    2.0 m s   5.4 m s
2
2
59. Strategy Use conservation of momentum. Refer to Practice Problem 7.11.
Solution
(a) Find the momentum change of the ball of mass m1.
1

p1x  p2 x  m2 v2ix  m2 v2fx  m2 (0  v2fx )   m2 v2fx  5m1  vi cos( 36.9)   1.00m1vi
4


1

p1 y  p2 y  m2 v2iy  m2 v2fy  5m1 (0  v2fy )  5m1v2fy  5m1  vi sin(36.9)   0.751m1vi
4

(b) Find the momentum change of the ball of mass m2 .
p2 x  p1x  m1 (v1ix  v1fx )  m1 (vi  0)  m1vi
p2 y  p1 y  m1 (v1iy  v1fy )  m1 (0  v1 )   m1v1  m1 (0.751vi )  0.751m1vi
The momentum changes for each mass are equal and opposite.
60. Strategy Use conservation of momentum. Let right be +x and +y be in the initial direction of the puck.
y
Solution Find v2fx .
mv1fx  mv2fx  mv1ix  mv2ix  0  0, so v2fx  v1fx .
Find v2fy .
mv1fy  mv2fy  mv1iy  mv2iy  v1iy  0, so v2fy  v1iy  v1fy .
384
v1i
v1f
θ1 = 37°
x
Physics
Chapter 7: Linear Momentum
Calculate v2f .
v2f  v2fx 2  v2fy 2  (v1fx ) 2  (v1iy  v1fy ) 2  (v1f sin 1 ) 2  (v1i  v1f cos 1 ) 2
    0.36 m s  sin 37  0.45 m s   0.36 m s  cos 37  0.27 m s
Calculate the direction of the second puck.
  0.36 m s  sin 37
v1fx
v
 tan 1
 53 to the left
  tan 1 2fx  tan 1
v2fy
v1iy  v1fy
0.45 m s   0.36 m s  cos 37

Thus, v 2f  0.27 m s at 53 to the left .
2
2
61. Strategy Use conservation of momentum.
Solution Find v2f in terms of v1f .
mv1fy  mv2fy  mv1f sin 1  mv2f sin  2  mv1iy  mv2iy  0  0, so
 sin 1
 sin 60.0
v2f 
v 
v  1.73v1f .
sin  2 1f sin(30.0) 1f
y
v1i
v1f
60.0°
30.0° x
v2f
62. Strategy The collision is perfectly inelastic, so the final velocities of the blocks are identical. Use conservation
of momentum.
Solution Find the initial speed of block B.
pix  mA vAix  mB vBix  0  mB vBix  mB vBix  pfx  (mA  mB )vfx ,
(m  mB )vfx (mA  mB )vf cos 
so vBix  A

 vBi . Compute vBi .
mB
mB
vBi 
y
vf
vBi
42.5°
(220 g  300 g)(3.13 m s) cos(180  42.5)
 4.0 m s
300 g
N
x
vAi
Thus the initial speed of block B was 4.0 m s .
63. Strategy Use conservation of momentum. Let +x be along the initial direction of the projectile.
Solution Find the magnitude of the momentum of the target body after the collision.
Find p2fx .
p2 x  p2fx  p2ix  p2fx  0  p1x  p1ix  p1fx  mvi  mvfx  m(vi  vf cos  ), so p2fx  m(vi  vf cos  ).
Find p2fy .
p2 y  p2fy  p2iy  p2fy  0  p1 y  p1iy  p1fy  0  mvf sin  , so p2fy  mvf sin  .
Calculate p2f .
p2f 
p2fx 2  p2fy 2  m (vi  vf cos  )2  vf 2 sin 2 
 (2.0 kg) 5.0 m s   3.0 m s  cos 60.0   3.0 m s  sin 2 60.0  8.7 kg  m s
2
2
385
Chapter 7: Linear Momentum
Physics
64. (a) Strategy The collision is perfectly inelastic, so the final velocities of the cars are identical. Use conservation
of momentum.
y
Solution Let the 1500-kg car be (1) and the 1800-kg car be (2).
m1
pix  m1v1ix  m2 v2ix  m1v1ix  0  pfx  (m1  m2 )vfx , so vfx 
v .
m1  m2 1ix
m2
piy  m1v1iy  m2 v2iy  0  m2 v2iy  pfy  (m1  m2 )vfy , so vfy 
v .
m1  m2 2iy
v2i
v1i
N
x
Compute the final speed.
2
2
 m1v1ix   m2 v2iy 
[(1500 kg)(17 m s)]2  [(1800 kg)(15 m s)]2
vf  vfx  vfy  
 
  
1500 kg  1800 kg
 m1  m2   m1  m2 
 11 m s
Compute the direction.
vfy
(1800 kg)(15 m s)
  tan 1
 tan 1
 47
(1500 kg)(17 m s)
vfx
2
2
Thus, the final velocity of the cars is 11 m s at 47 S of E .
(b) Strategy Find the change in kinetic energy.
Solution
1
1
1
K  K f  Ki  (m1  m2 )vf 2  m1v1i 2  m2 v2i 2
2
2
2
1
1
2 1
 (1500 kg  1800 kg)(11.254 m s)  (1500 kg)(17 m s) 2  (1800 kg)(15 m s) 2  210 kJ
2
2
2
Thus, 210 kJ of the initial kinetic energy was converted to another form of energy during the collision.
65. Strategy The collision is perfectly inelastic, so the final velocities of the cars are identical. Use conservation of
momentum.
386
Physics
Chapter 7: Linear Momentum
y
Solution Let the 1700-kg car be (1) and the 1300-kg car be (2).
m1
pix  m1v1ix  m2 v2ix  m1v1ix  0  pfx  (m1  m2 )vfx , so vfx 
v .
m1  m2 1ix
piy  m1v1iy  m2 v2iy  pfy  (m1  m2 )vfy , so vfy 
N
m1v1iy  m2 v2iy
.
m1  m2
v1i
Compute the final speed and the direction.
 mv
vf  vfx 2  vfy 2   1 1ix
 m1  m2

2
  m1v1iy  m2 v2iy
  
m1  m2
 



2
[(1700 kg)(14 m s) cos 45]2  [(1700 kg)(14 m s)sin 45  (1300 kg)(18 m s)]2
  tan 1
1700 kg  1300 kg
vfy
vfx
 tan 1
 6.0 m s
(1700 kg)(14 m s) sin 45  (1300 kg)(18 m s)
 21
(1700 kg)(14 m s) cos 45
Thus, the final velocity of the cars is 6.0 m s at 21 S of E .
66. Strategy Use conservation of momentum.
y
Solution Find the components of the deuteron’s velocity after the collision.
Find vdfx .
mn vnix  md vdix  mn vi  0  mn vnfx  md vdfx  0  md vdfx , so vdfx
vi / 3
m
 n vi .
md
vi
x
Find vdfy .
mn vniy  md vdiy  0  0  mn vnfy  md vdfy  mn
vi
3
 md vdfy , so vdfy  
1
3
vi
mn
.
md
Find the components, vdfx and vdfy .
m
m   m
m
1
1
(vdfx , vdfy )   n vi , 
vi n    n vi , 
vi n
3 md   2mn
3 2mn
 md

vi 
 vi
   , 

2 3
2

67. Strategy Use conservation of linear momentum.
y
Solution Find vBfx .
pix  mvAix  pfx  mvAfx  mvBfx , so vBfx  vAix  vAfx .
Find vBfy .
piy  mvAiy  0  pfy  mvAfy  mvBfy , so vBfy  vAfy .
Compute the final speed of puck B.
vBf  (vBfx )2  (vBfy )2  (vAix  vAfx ) 2  (vAfy )2
 [2.0 m s  (1.0 m s) cos 60]2  [(1.0 m s) sin 60]2  1.7 m s
Compute the direction of puck B.
vBfy
(1.0 m s)sin 60
 tan 1
 30
  tan 1
2.0 m s  (1.0 m s) cos 60
vBfx
387
vAf
vAi
60°
x
v2i
x
Chapter 7: Linear Momentum
Physics
Thus, the speed and direction of puck B after the collision is 1.7 m s at 30 below the x-axis .
68. Strategy The collision is perfectly inelastic, so the final velocities of the acrobats are identical. Use conservation
of momentum.
Solution Let the first acrobat be (1) and the second acrobat be (2).
pix  m1v1ix  m2 v2ix  pfx  (m1  m2 )vfx , so vfx  (m1v1ix  m2 v2ix ) (m1  m2 ).
piy  m1v1iy  m2 v2iy  pfy  (m1  m2 )vfy , so vfy  (m1v1iy  m2 v2iy ) (m1  m2 ).
y
2.0 m/s
20°
3.0 m/s
10°
Compute the final speed.
vf  vfx 2  vfy 2 

(m1v1ix  m2 v2ix )2  (m1v1iy  m2 v2iy )2
m1  m2
[(60)(3.0 m s) cos10  (80)(2.0 m s) cos160]2  [(60)(3.0 m s) sin10  (80)(2.0 m s) sin160]2
60  80
 0.64 m s
Compute the direction.
vfy
(60)(3.0 m s) sin10  (80)(2.0 m s) sin160
 tan 1
 73
  tan 1
vfx
(60)(3.0 m s) cos10  (80)(2.0 m s) cos160
Thus, the final velocity of the acrobats is 0.64 m s at 73 above the +x-axis .
69. Strategy Use conservation of momentum.
388
x
Physics
Chapter 7: Linear Momentum
Solution Let swallow 1 and its coconut be (1) and swallow 2 and its coconut be (2)
(before the collision). After the collision, let swallow 1’s coconut be (3), swallow 2’s
coconut be (4), and the tangled-up swallows be (5).
m v  m4 v4 x
pix  m1v1x  m2 v2 x  0  0  pfx  m3v3 x  m4 v4 x  m5 v5 x , so v5 x   3 3 x
.
m5
piy  m1v1 y  m2 v2 y  m1v1  m2 v2  pfy  m3v3 y  m4 v4 y  m5v5 y , so
m1v1  m2 v2  m3v3 y  m4 v4 y
v5 y 
.
m5
y
v2 30°
N
x
10°
v3
v4
v1
Compute the final speed of the tangled swallows, v5 .
v5  v5x 2  v5y 2 
1
[ (m3v3 x  m4 v4 x )]2  (m1v1  m2 v2  m3v3 y  m4 v4 y )2
m5
[(0.80 kg)(13 m s) cos 260  (0.70 kg)(14 m s) cos 60]2
[(1.07 kg)(20 m s)  (0.92 kg)( 15 m s)  (0.80 kg)(13 m s) sin 260  (0.70 kg)(14 m s) sin 60]2
0.270 kg  0.220 kg
 20 m s
Compute the direction.
v5 y
  tan 1
v5 x
(1.07
kg)(20 m s)  (0.92 kg)(15 m s)  (0.80 kg)(13 m s) sin 260  (0.70 kg)(14 m s) sin 60
 tan 1
(0.80 kg)(13 m s) cos 260  (0.70 kg)(14 m s) cos 60
 72
Since v5 x  0 and v5 y  0, the velocity vector is located in the second quadrant, so the angle is

180  72  108 from the positive x-axis or 18° west of north. Thus, the velocity of the birds immediately
after the collision is 20 m s at 18 W of N .
70. Strategy Use conservation of momentum. The collision is perfectly inelastic, so v1f  v2f  vf .
Solution Find the speed of the sled once the book is on it.
m1v1f  m2 v2f  (m1  m2 )vf  m1v1i  m2 v2i  m1vi  0, so
m1
5.0 kg
vf 
vi 
(1.0 m s)  0.83 m s .
m1  m2
5.0 kg  1.0 kg
71. Strategy Use conservation of momentum. The collision is perfectly inelastic, so v1f  v2f  vf .
Solution Find the speed of the cars just after the collision.
m1v1f  m2 v2f  (m1  m2 )vf  m1v1i  m2 v2i  m1vi  0, so
m1 g
13.6 kN
vf 
v 
17.0 m s   10.2 m s .
(m1  m2 ) g i 13.6 kN  9.0 kN
72. Strategy Use Eqs. (7-10) and (7-11).
Solution Find the velocity of the center of mass of the system.





p CM  Mv CM  m1v1  m2 v 2  m3 v3 , so
389
Chapter 7: Linear Momentum
Physics
(3.0 kg)(290 m s)  (5.0 kg)(120 m s)  (2.0 kg)(52 m s)
 37 m s
3.0 kg  5.0 kg  2.0 kg

 0, so v CM  37 m s in the +x-direction .
vCMx 
vCMy
73. Strategy Use the definition of linear momentum.
Solution Find the magnitude of the total momentum of the ship and the crew.
ptot  mtot v  (2.0  103 kg  4.8  104 kg)(1.0  105 m s)  5.0  109 kg  m s
74. Strategy Use the definition of linear momentum and the impulse-momentum theorem.
Solution
(a) Compute the magnitude of the change in momentum of the ball.
p  pf  pi  mvf  mvi  m(vf  vi )  (0.145 kg) 37 m s   41 m s    11 kg  m s
(b) Compute the impulse delivered to the ball by the bat.
Impulse  p  11 kg  m s
(c) Compute the magnitude of the average force exerted on the ball by the bat.
p 11.31 kg  m s
Fav 

 3.8 kN
t
3.0  103 s
75. Strategy Use the impulse-momentum theorem.
y
Solution Find the average force exerted by the ground on the ball.
p
x
Fav 
t
54 m/s
53 m/s
m (vx )2  (v y ) 2
18°
22°

t
0.060 kg
2
2

 53 m s  cos18   54 m s  cos(22)    53 m s  sin18   54 m s  sin(22)   34 N
0.065 s 
76. Strategy The center of each length is its center of mass. Use the component form of the definition of center of
mass.
Solution Find the location of the center of mass of the rod.
mx mx mx
1
1 3 2 3 3
xCM  3
 (0  5.0 cm  10.0 cm)  5.00 cm
3
m
390
Physics
Chapter 7: Linear Momentum
my my my
1
1 3 2
3 3
yCM  3
 (5.0 cm  10.0 cm  5.0 cm)  6.67 cm
3
m
Thus, ( xCM , yCM )  (5.00 cm, 6.67 cm) .
77. Strategy The center of mass of each block is its center. Add up the individual center of mass components to find
the components of the center of mass of the block structure.
Solution
mx  2mx2  5mx3  4mx4 x1  2 x2  5 x3  4 x4 0  2(1.0 in)  5(2.0 in)  4(3.0 in)


 2.0 in
xCM  1
12m
12
12
6my1  4my2  my3  my4 6 y1  4 y2  y3  y4 6(0)  4(1.0 in)  2.0 in  3.0 in


 0.75 in
yCM 
12m
12
12
9mz1  3mz2 9 z1  3 z2 9(0)  3(1.0 in)


 0.25 in
zCM 
12m
12
12
The center of mass of the block structure is located at (2.0 in, 0.75 in, 0.25 in).
78. Strategy Use the impulse-momentum theorem.
Solution Compute the force exerted by the stream on a person in the crowd.
p mv m

 v  (24 kg s)(17 m s)  410 N
Fav 
t
t
t
79. Strategy Use the impulse-momentum theorem.
Solution Compute the average forces imparted to the two gloved hands during the catches.
Inexperienced: Fav

130 km h  10kmm
p mv


 (0.14 kg)
t
t
103 s
Experienced: Fav  (0.14 kg)
130
km h 

103 m
km
3
10  10

1h
3600 s

3

1h
3600 s

 5000 N
 500 N
s
80. Strategy The fly splatters on the windshield, so the collision is perfectly inelastic (vfly, final  vcar, final  vf ).
Use conservation of momentum. Let the positive direction be along the velocity of the automobile.
Solution
(a) Compute the change in momentum.
pcar  pfly  mfly (vfly, f  vfly, i )   mfly (vcar, i  0)  (0.1103 kg)(100 km h)  0.01 kg  km h
So, the change in the car’s momentum due to the fly is 0.01 kg  km/h opposite the car’s motion.
(b) Compute the change in momentum.
pfly  pcar  0.01 kg  km h , or 0.01 kg  km h along the car’s velocity.
(c) Compute the number of flies N required to slow the car.
(1000 kg)  1 km h 
m v
 105 flies .
N pfly  mcar vcar , so N   car car  
pfly
0.01 kg  km h
391
Chapter 7: Linear Momentum
Physics
81. (a) Strategy The initial momentum of the baseball is pi  mvi . The final momentum is zero.
Solution Compute the change in momentum.
p  pf  pi  0  mvi   mvi  (0.15 kg)(35 m s)  5.3 kg  m s
Thus, the change in momentum was 5.3 kg  m s opposite the ball’s direction of motion .
(b) Strategy and Solution According to the impulse-momentum theorem, the impulse applied to the ball is
equal to the change in the momentum of the ball, or 5.3 kg  m s opposite the ball’s direction of motion .
(c) Strategy Use the impulse momentum theorem.
Solution Since the acceleration is assumed constant, the time it takes for the ball to come to a complete stop
is t  x vav . Compute the average force applied to the ball by the catcher’s glove.



p
p  35 m s  5.25 kg  m s opposite the direction of motion
Fav 
 vav


t
x  2 
0.050 m
 1.8 kN opposite the ball’s direction of motion
82. Strategy Use conservation of momentum. The collision is perfectly elastic, so Ki  K f . Also, v1ix  v1i and
v1fy  v1f .
Solution Find the speed of the target body after the collision.
x-direction:
m1v1fx  m2 v2fx  0  m2 v2fx  m1v1ix  m2 v2ix  m1v1ix  0
y-direction:
m1v1fy  m2 v2fy  m1v1iy  m2 v2iy  0  0, so m2 v2fy  m1v1fy .
y
v1f = 6.0 m/s
v1i = 8.0 m/s
x
Square the results and add.
m22 v2fx 2  m22 v2fy 2  m2 2 v2f 2  m12 v1ix 2  m12 v1fy 2  m12 v1i 2  m12 v1f 2 , so m2 v2f 2 
Calculate the kinetic energies.
1
1
1
1
1
m v 2  m2 v2f 2  m1v1i 2  m2 v2i 2  m1v1i 2  0, so m1v1f 2  m2 v2f 2  m1v1i 2 .
2 1 1f
2
2
2
2
392
m12
m2
(v1i 2  v1f 2 ).
Physics
Chapter 7: Linear Momentum
Thus, m2 v2f 2  m1 (v1i 2  v1f 2 ) 
m12
m2
(v1i 2  v1f 2 ) and m2  m1
v1i 2  v1f 2
v1i 2  v1f 2
.
From the kinetic energies,
v2f 
m1
m2
(v1i 2  v1f 2 ) 
m1 (v1i 2  v1f 2 )
 v 2 v 2 
m1  1i 2 1f 2 
 v1i v1f 

(v1i 2  v1f 2 )2
v1i 2  v1f 2

[(8.0 m s) 2  (6.0 m s) 2 ]2
(8.0 m s)2  (6.0 m s) 2
 2.8 m s .
83. Strategy Use conservation of momentum. Let e = electron,   neutrino, and n = nucleus.
Solution
(a) Find the direction of motion of the recoiling daughter nucleus.



pe  p  p n  0, so
pnx   pex  p x   pex  0   pe and pny   pey  p y  0  p y   p ( p  0).
Find the angle with respect to the electron’s direction.
p
5.00  1019
  tan 1   tan 1
 31.4  180  148.6 CCW from the electron’s direction
 pe
8.20  1019
(b) Find the momentum of the recoiling daughter nucleus.
pn 
pnx 2  pny 2  ( pe ) 2  ( p ) 2  (8.20  1019 kg  m s)2  (5.00 1019 kg  m s)2
 9.60  1019 kg  m s

Thus, p n  9.60  1019 kg  m/s in the direction found in (a) .
84. Strategy Use conservation of momentum and the definition of center of mass. Let the pier be to the left of the
raft and woman at x  0.
Solution

(a) Since p CM  0, as the woman walks toward the pier, the raft moves away from the pier, and the center of
mass does not change. So, xCM 
mw xwi  mr xri
mw  mr

mw xwf  mr xrf
.
mw  mr
Initially, xCM is to the right of xri . When the woman has walked to the other end of the raft, xCM is to the
left of xrf . By symmetry, the distance xCM  xri equals the distance xrf  xCM , thus
xrf  xCM  xCM  xri , so xrf  2 xCM  xri .
The final distance of the raft from the dock, df , is equal to the difference between xrf and half its length,
3.0 m.
393
Chapter 7: Linear Momentum
Physics
df  xrf  3.0 m  2 xCM  xri  3.0 m  2 xCM  (3.0 m  0.50 m)  3.0 m  2 xCM  6.5 m
Calculate xCM .
xCM 
(60.0 kg)(6.5 m)  (120 kg)(3.5 m)
 4.5 m, so df  2(4.5 m)  6.5 m  2.5 m .
60.0 kg  120 kg
(b) Find the distance the woman walked relative to the pier.
xw  xwf  xwi  df  xwi  2.5 m  6.5 m  4.0 m
85. Strategy We must determine the initial speeds of the two cars. The collision is perfectly inelastic, so the final
velocities of the cars are identical. Use conservation of momentum and the work-kinetic energy theorem.
y
Solution Let the 1100-kg car be (1) and the 1300-kg car be (2). Use the work-kinetic
energy theorem to determine the kinetic energy and, thus, the initial speed of the wrecked
vf
cars, which is the final speed of the collision.
v1i
30°
1
W  F r  f k r   k mg r  K  0  mvi 2 , so vi  2k g r .
v2i
2
Thus, the final speed of the collision is vf  2k g r .
Find the initial speeds.
pix  m1v1ix  m2 v2ix  m1v1i  0  pfx  (m1  m2 )vfx , so
m  m2
m  m2
 1 km h
2400 kg
v1i  1
vfx  1
2k g r cos150 
2(0.80)(9.80 m s 2 )(17 m) cos150 
m1
m1
1100 kg
 0.2778 m
 110 km h .
N
x


s
piy  m1v1iy  m2 v2iy  0  m2 v2i  pfy  (m1  m2 )vfy , so
m1  m2
m  m2
vfy  1
m2
m2
 54 km h .
v2i 
2 k g r sin150 
 1 km h
2400 kg
2(0.80)(9.80 m s 2 )(17 m) sin150 
1300 kg
 0.2778 m
Since 110 > 70, the lighter car was speeding .
86. Strategy Use the impulse-momentum theorem.
Solution Find the speed of the expelled gas relative to the ground.
Fav
p (m)vgas m
6.0  104 N
Fav 


vgas , so vgas 

 740 m s .
t
t
t
m t
81 kg s
87. Strategy Use the definition of linear momentum and the impulse-momentum theorem.
Solution Find the force the kinesin molecule needs to deliver in order to accelerate the organelle.
p mv (0.01 1015 kg)(1 106 m s  0)
Fav 


 1018 N
t
t
10  106 s
88. Strategy Use conservation of momentum. The collision is perfectly inelastic, so vAf  vBf  vf .
394


s
Physics
Chapter 7: Linear Momentum
Solution Find the final speed in terms of the initial speed.
1
1
1

mA vAf  mB vBf  (mA  mB )vf   m  m  vf  mA vAi  mB vBi  mA vi  0  mvi , so vf  vi .
2
3
2

Calculate the ratio of the final kinetic energy to the initial kinetic energy.
Kf
Ki

1 ( m  m )v 2
A
B f
2
1m v2
2 A i
 1 m  m   13 vi 
 2
1 mv 2
i
2
2


3m 1v2
2
9 i
1 mv 2
i
2

1
3
89. Strategy Use conservation of energy and momentum. Let 2m  mB  2mA .
Solution Find the maximum kinetic energy of A alone and, thus, its speed just before it strikes B.
1
K  mv12  0  U  mgh  0, so v1  2 gh .
2
Use conservation of momentum to find the speed of the combined bobs just after impact. The collision is
perfectly inelastic, so vAf  vBf  v2 .
1
mA vAf  mB vBf  (m  2m)v2  mA vAi  mB vBi  mv1  0, so v2  v1.
3
Find the maximum height.
2
1
1 1
1

K  0  mv2 2   m 
h .
2 gh   U  0  mgh2 , so h2 
2
2 3
9

90. Strategy The center of mass of the disk prior to drilling is ( xCM , yCM )  (0, 0). Let S stand for the small circle
removed and L stand for the large circle that remains.
Solution Find the center of mass of the metal disk after the hole has been drilled.
m x  mS xS
 0, so
xCM  L L
mL  mS
xL  
mS
mL
xS  
AS
AL
xS  
 rS2
 rL 2   rS2
xS  
(1.5 cm)2
(3.0 cm) 2  (1.5 cm)2
(1.5 cm)  0.50 cm.
By symmetry, yL  yS  yCM  0, so ( xL , yL )  (0.50 cm, 0) .
91. Strategy Use conservation of momentum and energy. The collision is elastic, so kinetic energy is conserved.
Solution Find the speed of bob B immediately after the collision.
Momentum conservation:
mvAf  mvBf  mvAi  mvBi  mvAi  0, so vBf  vAi  vAf .
Perfectly elastic collision ( Ki  K f ):
1
1
1
1
1
mvAf 2  mvBf 2  mvAi 2  mvBi 2  mvAi 2  0, so vAf 2  vBf 2  vAi 2 .
2
2
2
2
2
Energy conservation:
1
mv 2  mgh, so vAi  2 gh .
2 Ai
Find vAf in terms of vAi .
395
Chapter 7: Linear Momentum
Physics
vAi 2  vAf 2  vBf 2  vAf 2  (vAi  vAf ) 2  vAf 2  vAi 2  2vAi vAf  vAf 2 , so vAf (vAf  vAi )  0.
Thus, vAf  0 or vAi . The only way vAf could equal vAi is if bob B didn’t exist, so vAf  0.
Calculate vBf .
vBf  vAi  vAf  2 gh  0  2(9.80 m s 2 )(5.1 m)  10 m s
92. Strategy Use conservation of momentum and energy. The collision is elastic, so kinetic energy is conserved. Let
the positive direction be to the right.
Solution Find the velocities of the gliders after the collision.
Momentum conservation:
mv1f  mv2f  mv1i  mv2i  mv1i  0, so v2f  v1i  v1f .
Perfectly elastic collision due to bumpers ( K i  K f ):
1
1
1
1
1
mv 2  mv2f 2  mv1i 2  mv2i 2  mv1i 2  0, so v1f 2  v2f 2  v1i 2 .
2 1f
2
2
2
2
Find v1f in terms of v1i .
v1f 2  (v1i  v1f ) 2  v1f 2  v1i 2  2v1i v1f  v1f 2  v1i 2 , so v1f (v1f  v1i )  0.
So, v1f  0 or v1i . The only way v1f could equal v1i is if glider 2 didn’t exist, so v1f  0.
Calculate v2f .
v2f  v1i  v1f  0.20 m s  0  0.20 m s
After the collision, glider 1 is stationary and glider 2 has a velocity of 0.20 m s in the direction of
glider 1’s initial velocity.
93. Strategy Use conservation of momentum and Eq. (6-6) for the kinetic energies. Since the radium nucleus is at


rest, pi  p Ra  0.
Solution
(a) Find the ratio of the speed of the alpha particle to the speed of the radon nucleus.
pf  mRn vRn  m v  pi  0, so m v  mRn vRn . Therefore,
v
vRn

mRn
m

222 u 222
111
, where the negative was dropped because speed is nonnegative.


4u
4
2
396
Physics
Chapter 7: Linear Momentum


(b) Since the initial momentum is zero, p Rn  p ; therefore,
(c) Find the ratio of the kinetic energies.
1m v 2
m
 
 2
 
K Rn 1 mRn vRn 2 mRn
2
K
 v

 vRn
2
2

4 u  111 
111



 
222
u
2
2



397

p
p
   1 .

pRn
p Rn