Download Homework #5: Momentum

Homework #5: Momentum
1.
(I) What is the magnitude of the momentum of a 28-g sparrow flying with a speed of 8.4 m s ?
2.
(I) A constant friction force of 25 N acts on a 65-kg skier for 20 s. What is the skier’s change in
velocity?
3.
(II) A 0.145-kg baseball pitched at 39 .0 m s is hit on a horizontal line drive straight back toward the
4.
5.
pitcher at 52 .0 m s . If the contact time between bat and ball is 3.00  10 3 s, calculate the average
force between the ball and bat during contact.
(II) A child in a boat throws a 6.40-kg package out horizontally with a speed of 10.0 m s . Calculate
the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is
26.0 kg, and that of the boat is 45.0 kg. Ignore water resistance.
(II) A 12,600-kg railroad car travels alone on a level frictionless track with a constant speed of
18 .0 m s . A 5350-kg load, initially at rest, is dropped onto the car. What will be the car’s new speed?
6.
(II) A 9300-kg boxcar traveling at 15 .0 m s strikes a second boxcar at rest. The two stick together
and move off with a speed of 6.0 m s . What is the mass of the second car?
7.
(II) An atomic nucleus initially moving at 420 m s emits an alpha particle in the direction of its
velocity, and the remaining nucleus slows to 350 m s . If the alpha particle has a mass of 4.0 u and
the original nucleus has a mass of 222 u, what speed does the alpha particle have when it is emitted?
(II) A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m s . The golf club was in contact
8.
with the ball for 3.5 10 3 s. Find (a) the impulse imparted to the golf ball, and (b) the average force
exerted on the ball by the golf club.
9.
A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to
reach a height of 0.960 m. What impulse was given to the ball by the floor?
10.
A pitcher claims he can throw a 0.145-kg baseball with as much momentum as a 3.00-g bullet
moving with a speed of 1.50 × 103 m/s. (a) What must the baseball’s speed be if the pitcher’s claim is
valid? (b) Which has greater kinetic energy, the ball or the bullet?
11.
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its
speed from 0 to 5.20 m/s in 0.832 s. What are the magnitudes of the linear impulse and the average total
force experienced by a 70.0-kg passenger in the car during the time the car accelerates?
12.
The front 1.20 m of a 1 400-kg car is designed as a “crumple zone” that collapses to absorb the
shock of a collision. If a car traveling 25.0 m/s stops uniformly in 1.20 m, (a) how long does the collision
last, (b) what is the magnitude of the average force on the car, and (c) what is the acceleration of the car?
Express the acceleration as a multiple of the acceleration of gravity.
13.
A 730-N man stands in the middle of a frozen pond of radius 5.0 m. He is unable to get to the
other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he
throws his 1.2-kg physics textbook horizontally toward the north shore at a speed of 5.0 m/s. How long
does it take him to reach the south shore?
14.
High-speed stroboscopic photographs show that the head of a 200-g golf club is traveling at 55
m/s just before it strikes a 46-g golf ball at rest on a tee. After the collision, the club head travels (in the
same direction) at 40 m/s. Find the speed of the golf ball just after impact.
15.
A railroad car of mass 2.00 × 104 kg moving at 3.00 m/s collides and couples with two coupled
railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s. (a)
What is the speed of the three coupled cars after the collision? (b) How much kinetic energy is lost in the
collision?
16.
A 7.0-g bullet is fired into a 1.5-kg ballistic pendulum. The bullet emerges from the block with a
speed of 200 m/s, and the block rises to a maximum height of 12 cm. Find the initial speed of the bullet.
17.
A 0.030-kg bullet is fired vertically at 200 m/s into a 0.15-kg baseball that is initially at rest. How
high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the
ball?
Homework #6: Conservation of Momentum: Elastic Collisions
18.
19.
20.
(II) A ball of mass 0.440 kg moving east (  x direction) with a speed of 3.30 m s collides head-on
with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of
each ball after the collision?
(II) Two billiard balls of equal mass undergo a perfectly elastic head-on collision. If one ball’s initial
speed was 2.00 m s , and the other’s was 3.00 m s in the opposite direction, what will be their
speeds after the collision?
(II) A 0.060-kg tennis ball, moving with a speed of 2.50 m s , collides head-on with a 0.090-kg ball
initially moving away from it at a speed of 1.15 m s . Assuming a perfectly elastic collision, what are
the speed and direction of each ball after the collision?
21. (II) A softball of mass 0.220 kg that is moving with a speed of 8.5 m s collides head-on and
elastically with another ball initially at rest. Afterward the incoming softball bounces backward with
a speed of 3.7 m s . Calculate (a) the velocity of the target ball after the collision, and (b) the mass of
the target ball.
22. (II) A 28-g rifle bullet traveling 230 m s buries itself in a 3.6-kg pendulum hanging on a 2.8-m-long
string, which makes the pendulum swing upward in an arc. Determine the vertical and horizontal
components of the pendulum’s displacement.
23. (II) A 920-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The
bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police
officer, knowing that the coefficient of kinetic friction between tires and road is 0.80, calculates the
speed of the sports car at impact. What was that speed?
24.
A 5.00-g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0-g
object that is initially at rest. Find (a) the velocity of each object after the collision and (b) the fraction of
the initial kinetic energy transferred to the 10.0-g object.
25.
A 10.0-g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 15.0-g
object moving in the opposite direction at 30.0 cm/s. Find the velocity of each object after the collision.
26.
A 25.0-g object moving to the right at 20.0 cm/s overtakes and collides elastically with a 10.0-g
object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision.
27. (II) An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5
times the mass of the other. If 7500 J were released in the explosion, how much kinetic energy did
each piece acquire?
28. (II) A wooden block is cut into two pieces, one with three times the mass of the other. A depression is
made in both faces of the cut, so that a firecracker can be placed in it with the block reassembled. The
reassembled block is set on a rough-surfaced table, and the fuse is lit. When the firecracker explodes,
the two blocks separate and slide apart. What is the ratio of distances each block travels?
Solutions #5 Momentum:
1.
2.
p  mv   0.028 kg  8.4 m s   0.24 kg m s
From Newton’s second law, p  Ft . For a constant mass object, p  mv . Equate the two
expressions for p .
Ft  mv  v 
Ft
.
m
If the skier moves to the right, then the speed will decrease, because the friction force is to the left.
v  
F t

 25 N  20 s 
m
65 kg
The skier loses 7.7 m s of speed.
3.
 7.7 m s
Choose the direction from the batter to the pitcher to be the positive direction. Calculate the average
force from the change in momentum of the ball.
p  F t  mv 
F m
4.
v
t
 52.0 m s  39.0 m s 
3
  4.40 10 N, towards the pitcher
3
3.00  10 s


  0.145 kg  
The throwing of the package is a momentum-conserving action, if the water resistance is ignored. Let
“A” represent the boat and child together, and let “B” represent the package. Choose the direction
that the package is thrown as the positive direction. Apply conservation of momentum, with the
initial velocity of both objects being 0.
 mA  mB  v  mAvA  mB vB 
 6.40 kg 10.0 m s 

 0.901m s
 26.0 kg  45.0 kg 
pinitial  pfinal 
vA  
5.
mB vB
mA
The boat and child move in the opposite direction as the thrown package.
Consider the horizontal motion of the objects. The momentum in the horizontal direction will be
conserved. Let “A” represent the car, and “B” represent the load. The positive direction is the
direction of the original motion of the car.
pinitial  pfinal  mAv A  mB vB   mA  mB  v 
v 
6.
mAv A  mB vB
mA  mB

12, 600 kg 18.0 m s   0
 12.6 m
12, 600 kg    5350 kg 
s
Consider the motion in one dimension, with the positive direction being the direction of motion of the
first car. Let “A” represent the first car, and “B” represent the second car. Momentum will be
conserved in the collision. Note that vB  0 .
pinitial  pfinal  mAv A  mB vB   mA  mB  v 
mB 
7.
mA  vA  v   9300 kg 15.0 m s  6.0 m s 

 1.4  104 kg

v
6.0 m s
Consider the motion in one dimension, with the positive direction being the direction of motion of the
original nucleus. Let “A” represent the alpha particle, with a mass of 4 u, and “B” represent the new
nucleus, with a mass of 218 u. Momentum conservation gives the following.
 mA  mB  v  mAvA  mB vB 
 mA  mB  v  mB vB  222 u  420 m s    218 u  350 m s 
pinitial  pfinal 
vA 
mA

4.0 u
 4.2  103 m s
Note that the masses do not have to be converted to kg, since all masses are in the same units, and a
ratio of masses is what is significant.
8.
(a)
The impulse is the change in momentum. The direction of travel of the struck ball is the
positive direction.


p  mv  4.5  102 kg  45 m s  0   2.0 kg m s
(b)
The average force is the impulse divided by the interaction time.
F
9.
p
t

2.0 kg m s
 5.8  10 2 N
3
3.5  10 s
2
2
The velocity of the ball just before impact is found from vy  v0y  2ay y as


v1   v02y  2ayy   0  2 9.80 m s2  1.25 m    4.95 m s
and the rebound velocity with which it leaves the floor is


v2   v2f  2ayy   0  2 9.80 m s2  0.960 m    4.34 m s
The impulse given the ball by the floor is then
 

I Ft  m v  m v2  v1

  0.150 kg   4.34 m s  4.95 m s    1.39 N  s  1.39 N s upw ard
10.
(a)
If pball  pbullet ,



3.00  103 kg 1.50  103 m s
m bulletvbullet
then vball 

 31.0 m s
m ball
0.145 kg
(b)
The kinetic energy of the bullet is



2
3.00  10-3 kg 1.50  103 m s
1
2
KEbullet  m bulletvbul

 3.38  103 J
let
2
2
2
0.145 kg 31.0 m s

1
2
 69.7 J
while that of the baseball is KEball  m ballvball 
2
2
The bullethas the largerkinetic energy by a factor of 48.4.
11.
I Fav  t  p  m  v



Thus, I  m v  70.0 kg 5.20 m s 0  364 kg  m s , and
Fav 
or
364 kg  m s
I

 438 kg  m s2
t
0.832 s
Fav  438 N directed forw ard
12.
13.
(a)
t
21.20 m 
x 2 x


 9.60  102 s
vav vf  vi 0  25.0 m s
(b)
Fav 
p m  v 1400 kg 25.0 m s


 3.65  105 N
t
t
9.60  102 s
(c)
aav 


25.0 m s
1g
v

 260 m s2  260 m s2 
 26.6 g
2
2
t 9.60  10 s
 9.80 m s 


We shall choose southward as the positive direction.
The mass of the man is m 
w
730 N

 74.5 kg . Then, from conservation of
g 9.80 m s2
momentum, we find
 m m anvm an  m bookvbook  f   m m anvm an  m bookvbook i or
74.5 kg v  1.2 kg 5.0 m s  0  0 and vm an  8.1 102
m an
m s
Therefore, the time required to travel the 5.0 m to shore is
t
14.
x
5.0 m

 62 s
vm an 8.1 102 m s
Requiring that total momentum be conserved gives
 m clubvclub  m ballvball f   m clubvclub  m ballvballi
15.
or
 200 g 40 m s   46 g v
and
vball  65 m s
(a)
ball
  200 g 55 m s  0
If M is the mass of a single car, conservation of momentum gives
 3M  vf  M  3.00 m s   2M  1.20 m s , or vf  1.80 m s
(b)
The kinetic energy lost is KElost  KEi  KE f , or
KElost 
2
1
1
M  3.00 m s   2M
2
2
 1.20 m s
2

1
 3M
2
4
With M  2.00  104 kg , this yields KElost  2.16  10 J
 1.80 m s
2
16.
Let us apply conservation of energy to the block from the time just after the bullet has passed
through until it reaches maximum height in order to find its speed V just after the collision.
1 2
1
1
m vi  m gyi  m v2f  m gyf becomes m V 2  0  0  m gyf
2
2
2
or
V
2gyf 


2 9.80 m s2  0.120 m   1.53 m s
Now use conservation of momentum from before until just after the collision in order to find the
initial speed of the bullet, v.
7.0  10
3


v  5.3  102 m s
from which
17.

kg v  0  1.5 kg1.53 m s  7.0  103 kg  200 m s
Let M = mass of ball, m = mass of bullet, v = velocity of bullet, and V = the initial velocity of the
ball-bullet combination. Then, using conservation of momentum from just before to just after
collision gives
M
 m 
 m  V  m v  0 or V  
v
 M  m 
Now, we use conservation of mechanical energy from just after the collision until the ball reaches
maximum height to find
0   M  m  ghm ax 
2
1
V2
1
m  2
 M  m  V 2  0 or hm ax   
 v
2
2g 2g M  m 
With the data values provided, this becomes


0.030 kg
1
hm ax 
2 
2 9.80 m s  0.15 kg  0.030 kg 


2
 200 m s
2
 57 m
Homework #6 Solutions
18. Let A represent the 0.440-kg ball, and B represent the 0.220-kg ball. We have vA  3.30 m s and
vB  0 . Use Eq. 7-7 to obtain a relationship between the velocities.
vA  vB    vA  vB   vB  vA  vA
Substitute this relationship into the momentum conservation equation for the collision.
mA vA  mB vB  mA vA  mB vB  mA vA  mA vA  mB  vA  vA  
vA 
0.220 kg
 mA  mB 
vA 
 3.30 m s   1.10 m s  east 
0.660 kg
 mA  mB 
vB  vA  vA  3.30 m s  1.10 m s  4.40 m s  east 
19. Let A represent the ball moving at 2.00 m/s, and call that direction the positive direction. Let B
represent the ball moving at 3.00 m/s in the opposite direction. So vA  2.00 m s and
vB  3.00 m s . Use Eq. 7-7 to obtain a relationship between the velocities.
vA  vB    vA  vB   vB  5.00 m s  vA
Substitute this relationship into the momentum conservation equation for the collision, noting that
mA  mB .
mA vA  mB vB  mA vA  mBvB  vA  vB  vA  vB 
1.00 m s  vA   vA  5.00 m s   2vA  6.00 m s  vA  3.00 m s
vB  5.00 m s  vA  2.00 m s
The two balls have exchanged velocities. This will always be true for 1-D elastic collisions of objects
of equal mass.
20. Let A represent the 0.060-kg tennis ball, and let B represent the 0.090-kg ball. The initial direction of
the balls is the positive direction. We have vA  2.50 m s and vB  1.15 m s . Use Eq. 7-7 to
obtain a relationship between the velocities.
vA  vB    vA  vB   vB  1.35 m s  vA
Substitute this relationship into the momentum conservation equation for the collision.
mA vA  mB vB  mA vA  mB vB  mA vA  mB vB  mA vA  mB 1.35 m s  vA  
vA 
mA vA  mB  vB  1.35 m s 
mA  mB

 0.060 kg  2.50 m s    0.090 kg 1.15 m s  1.35 m s 
0.150 kg
 0.88 m s
vB  1.35 m s  vA  2.23 m s
Both balls move in the direction of the tennis ball’s initial motion.
21. Let A represent the moving softball, and let B represent the ball initially at rest. The initial direction
of the softball is the positive direction. We have vA  8.5 m s , vB  0 , and vA  3.7 m s .
(a) Use Eq. 7-7 to obtain a relationship between the velocities.
vA  vB    vA  vB   vB  vA  vB  vA  8.5 m s  0  3.7 m s  4.8 m s
(b)
Use momentum conservation to solve for the mass of the target ball.
mA vA  mBvB  mA vA  mBvB 
mB  mA
 vA  vA 
 8.5 m s  3.7 m s 
  0.220 kg 

4.8 m s
 vB  vB 
0.56 kg
22.
From the analysis in the Example 7-10, we know that
mM
v
2 gh 
m
1  mv 
1
  0.028 kg  230 m s  
h

 

2 
2g  m  M 
2 9.8 m s  0.028 kg  3.6kg 
2

2

 0.1607 m  0.16 m
From the diagram we see that
L2   L  h   x 2
2
x
L2   L  h  
2
 2.8 m    2.8 m  0.1607 m 
2
2
 0.94 m
Use conservation of momentum in one dimension. Call the direction of the sports car’s velocity the
positive x direction. Let A represent the sports car, and B represent the SUV. We have vB  0 and
23
vA  vB . Solve for vA .
pinitial  pfinal  mA vA  0   mA  mB  vA  vA 
mA  mB
mA
vA
The kinetic energy that the cars have immediately after the collision is lost due to negative work done
by friction. The work done by friction can also be calculated using the definition of work. We
assume the cars are on a level surface, so that the normal force is equal to the weight. The distance
the cars slide forward is x . Equate the two expressions for the work done by friction, solve for
vA ,
and use that to find vA .
Wfr   KEfinal  KEinitial after
collision
 0  12  mA  mB  vA2
Wfr  Ffr x cos180o    k  mA  mB  g x
 12  mA  mB  vA2    k  mA  mB  g x  vA 
vA 
mA  mB
mA
vA 
mA  mB
mA
2  k g x 
920 kg  2300 kg
920 kg
2  k g x

2  0.80  9.8 m s 2
  2.8 m 
 23.191m s  23 m s
24
(a)
From conservation of momentum,
 5.00 g v
1f
 10.0 g v2 f   5.00 g 20.0 cm s  0
(1)
Also for an elastic, head-on, collision, we have v1i  v1 f  v2i  v2 f , which becomes
20.0 cm s v1 f  v2 f .
Solving (1) and (2) simultaneously yields
v1 f   6.67 cm s , and v2 f  13.3 cm s
(2)
(b)
KEi  KE1i  KE2i 
KE2 f 




2
1
1
m 2v22 f  10.0  103 kg 13.3  10-2 m s  8.89  105 J, so
2
2
KE2 f
KEi
25.

2
1
5.00  103 kg  0.200 m s  0  1.00  104 J
2

8.89  105 J
 0.889
1.00  104 J
Using conservation of momentum gives
10.0 g v
1f
 15.0 g v2 f  10.0 g 20.0 cm s  15.0 g 30.0 cm s
(1)
For elastic, head on collisions, v1i  v1 f  v2i  v2 f which becomes
20.0 cm s v1f   30.0 cm s+v2 f
(2)
Solving (1) and (2) simultaneously gives v1 f   40.0 cm s ,
and v2 f  10.0 cm s
26.
Conservation of momentum gives
 25.0 g v
1f
 10.0 g v2 f   25.0 g 20.0 cm s  10.0 g15.0 cm s
(1)
For head-on, elastic collisions, we know that v1i  v1 f  v2i  v2 f .
Thus,
20.0 cm s v1f  15.0 cm s+v2 f
(2)
Solving (1) and (2) simultaneously yields
v1 f  17.1 cm s , and v2 f  22.1 cm s
27. Use conservation of momentum in one dimension, since the particles will separate and travel in
opposite directions. Call the direction of the heavier particle’s motion the positive direction. Let A
represent the heavier particle, and B represent the lighter particle. We have mA  1.5mB , and
vA  vB  0 .
pinitial  pfinal  0  mA vA  mB vB  vA  
The negative sign indicates direction.
mB vB
mA
  23 vB
Since there was no mechanical energy before the explosion, the kinetic energy of the particles after
the explosion must equal the energy added.
Eadded  KE A  KEB  12 mAvA2  12 mB vB2 
KEB  53 Eadded 
3
5
 7500 J   4500 J
1
2
1.5mB   23 vB 
2
 12 mB vB2 
5
3

1
2

mB vB2  53 KEB
KE A  Eadded  KEB  7500 J  4500 J  3000 J
Thus KE A  3.0  10 J KEB  4.5  10 J
3
3
28. Let A represent the more massive piece, and B the less massive piece. Thus mA  3mB . In the
explosion, momentum is conserved. We have vA  vB  0 .
pinitial  pfinal  0  mA vA  mBvB  3mBvA  mBvB  vA   13 vB
For each block, the kinetic energy gained during the explosion is lost to negative work done by
friction on the block.
Wfr  KE f  KEi   12 mv 2
But work is also calculated in terms of the force doing the work and the distance traveled.
Wfr  Ffr x cos180o    k FN x    k mg x
Equate the two work expressions, solve for the distance traveled, and find the ratio of distances.
 mv    k mg x  x 
1
2
And so
2
 x heavy  x light  1 9
v2
g k
 x A
 x B
vA2
g
v2   13 vB 
1
 2k  A2 

2
vB
vB
vB
9
g k
2