Ch7. Impulse and Momentum
There are situations in which force acting on an
object is not constant, but varies with time.
Two new ideas: Impulse of
the force and Linear
momentum of an object.
1
Impulse-Momentum Theorem
2
Definition of Impulse
The impulse J of a force is the product of the
average force F and the time interval t
during which the force acts:
J=
F t
Impulse is a vector quantity and has the same
direction as the average force.
SI Unit of Impulse: newton.second (N.s)
3
Definition of Linear Momentum:
The linear momentum p of an object is the
product of the object’s mass m and velocity v:
p=mv
Linear momentum is a vector quantity that
points in the same direction as the velocity.
SI Unit of Linear Momentum:
kilogram.meter/second(kg.m/s)
4
vt  v 0
a
t
 F  ma
vt  v 0
mvt  mv0
 F  m( t )  t
5
Impulse-Momentum Theorem
When a net force acts on an object,
the impulse of this force is equal to
the change in momentum of the
object:
( F )t  mv f  mv0
Impulse
Final
Initial
momentum momentum
Impulse=Change in momentum
6
Example 1. A Well-Hit Ball
A baseball (m=0.14kg) has initial velocity of v0=-38m/s
as it approaches a bat. The bat applies an average
force F that is much larger than the weight of the
ball, and the ball departs from the bat with a final
velocity of vf=+58m.
(a) Determine the impulse applied to the ball by the bat.
(b) Assuming time of contact is t =1.6*10-3s, find the
average force exerted on the ball by the bat.
7
(a)
J  mv f  mv0
 (0.14kg)(58m / s)  (0.14kg)( 38m / s)
= +13.4 kg.m/s
(b)
J 13.4kg.m / s
F

 8400 N
3
t 1.6  10 s
8
Example 2. A Rain Storm
Rain comes straight down with velocity of v0=-15m/s
and hits the roof of a car perpendicularly. Mass of
rain per second that strikes the car roof is 0.06kg/s.
Assuming the rain comes
to rest upon striking the
car (vf=0m/s), find the
average force exerted by
the raindrop.
9
F
mv f  mv0
t
m
 ( ) v 0
t
F = -(0.06kg/s)(-15m/s)=0.9 N
According to action-reaction law, the force
exerted on the roof also has a magnitude of
0.9 N points downward: -0.9N
10
Conceptual Example 3.
Hailstones Versus Raindrops
Suppose hail is falling. The hail comes straight
down at a mass rate of m/ t =0.06kg/s and an
initial velocity of v0=15m/s and strikes the roof
perpendicularly. Hailstones bounces off the roof.
Would the force on the
roof be smaller than,
equal to, or greater than
that in example 2?
Greater.
11
Check your understanding 1
Suppose you are standing on the edge of a dock and jump
straight down. If you land on sand your stopping time is much
shorter than if you land on water. Using the impulse-momentum
theorem as a guide, determine which one is correct.
A In bringing you to a halt, the sand exerts a greater impulse on
you than does the water.
B In bringing you to a halt, the sand and the water exert the
same impulse on you, but the sand exerts a greater average
force.
C In bringing you to a halt, the sand and the water exert the
same impulse on you, but the sand exerts a smaller average
force.
12
B
The Principle of Conservation of
Linear Momentum
13
Two types of forces act on the system:
1. Internal forces: Forces that the objects within
the system exert on each other.
2. External forces: Forces exerted on the objects by
agents external to the system.
(W1  F 12 )t  m1v f 1  m1v01
External Internal
force
force
(W2  F 21 )t  m2 v f 2  m2 v02
External Internal
force
force
14
(W1  W2  F 12  F 21 )t  (m1v f 1  m2 v f 2 )  (m1v01  m2 v02 )
(
) t
sum of average sum of average
external forces + internal forces
= pf - p0
internal forces cancel F12 = -F21
(Sum of average external forces)
If sum of external forces is zero
Then 0 = pf - p0
or
t
= pf - p0
(an isolated system)
p f = p0
m1vf1+m2vf2 = m1v01+m2v02
pf
p0
15
Principle of Conservation of Linear Momentum:
The total linear momentum of an isolated system remains
constant(is conserved). An isolated system is one for which
the vector sum of the average external forces acting on the
system is zero.
16
Conceptual Example 4.
Is the Total Momentum Conserved?
Two balls collide on the billiard table
that is free of friction.
(a) Is the total momentum of the two
ball system the same before and after
the collision?
(b) Answer (a) for a system that contains
only one ball.
(a) The total momentum is conserved.
(b) The total momentum of one ball
system is not conserved.
17
Example 5.
Assembling a Freight Train
Car 1 has a mass of m1=65*103kg and moves at a
velocity of v01=+0.8m/s. Car 2 has a mass of
m2=92*103kg and a velocity of v02=+1.3m/s.
Neglecting friction, find the common velocity vf
of the cars after they become coupled.
18
(m1+m2) vf = m1v01 + m2v02
After collision
Before collision
m1v01  m2 v02
vf 
m1  m2
(65  10 3 kg)(0.8m / s)  (92 10 3 kg)(1.3m / s)

3
3
(65  10 kg  92 10 kg)
=+1.1 m/s
19
Example 6. Ice Skaters
Starting from rest, two skaters
push off against each other on
smooth level ice (friction is
negligible). One is a woman
(m1=54kg), and one is a
man(m2=88kg). The woman
moves away with a velocity of
vf1=2.5m/s. Find the recoil
velocity vf2 of the man.
20
For the two skater system, what are the forces?
In horizontal direction
internal forces
F12 F21
system taken together
No external forces.
isolated system
0
conservation of momentum
21
m1vf1 + m2vf2 = 0
after pushing
vf 2 
 m1v f 1
m2
before pushing
 (54kg)( 2.5m / s)

 1.5m / s
88kg
It is important to realize that the total linear
momentum may be conserved even when the kinetic
energies of the individual parts of a system change.
22
Check your understanding 2
A canoe with two people aboard is coasting with an initial
momentum of 110kg.m/s. Then person 1 dives off the back of
the canoe. During this time, the net average external force
acting on the system is zero. The table lists four possibilities
for the final momentum of person 1 and final momentum of
person 2 plus the canoe, immediately after person 1 leaves the
canoe. Which possibility is correct?
a
b
c
d
Person 1
-60kg.m/s
-30kg.m/s
-40kg.m/s
+80kg.m/s
Person 2 & canoe
+170kg.m/s
+110kg.m/s
-70kg.m/s
-30kg.m/s
a
23
Collisions in One Dimension
Elastic collision: One in which the total kinetic
energy of the system after the collision is equal to
the total kinetic energy before the collision.
Inelastic collision: One in which the total kinetic
energy of the system is not the same before and
after the collision; if the objects stick together
after colliding, the collision is said to be
completely inelastic.
24
Example 7.
A Collision in One Dimension
A ball of mass m1=0.25kg
and velocity v01=5m/s
collides head-on with a ball
of mass m2=0.8kg that is
initially at rest(v02=0m/s).
No external forces act on
the balls. If the collision in
elastic, what are the
velocities of the balls after
the collision?
25
m1v f 1  m2 v f 2  m1v01  0
Total momentum
after collision
(2)
Total momentum
before collision
1
1
1
2
2
2
m1v f 1  m2 v f 2  m1v01  0
2
2
2
Total kinetic energy
before collision
Total kinetic energy
after collision
v f1 
(1)
m1v01  m2 v f 2
m1
(3)
26
Substitute in (2)
1  m1v01  m2 v f 2
m1 
2 
m1
2
 1
1
2
2
  m2 v f 2  m1v01
2
 2
1 m v  m v  2m1m2 v01 v f 2 1 m2 m v

2
m1
2 m1
2 2
1 01
2 2
2 f2
2
1 f2
1
2
 m1v01
2
 m22  m1m2  2 2m1m2v01 v f 2
v f 2 

 
0
2m1
2
2m1
 2m1

m12v021
m1v021
27
m2
2m1m2
2
m1  m2 v f 2 
v01 v f 2
2m1
2m1
m2  m1 v f
2
2
 2m1v01 v f 2
2m1
v f2 
v01
m1  m2
28
Substitute in (1)
m1v f1  m1v01  m2 v f 2
2m1
 m1v01  m2
v01
m1  m2

2m2 
v01
 m1 1 
 m1  m2 
29
 m1  m2  2m2 
m1v f1  m1 
v01
 m1  m2 
 m1  m2 
v f1  
 v01
 m1  m2 
30
m1=0.25, m2=0.8
v01 =5 m/s, v02= 0
 0.25  0.8 
v f1  
5  2.62m / s
 0.25  0.8 
2  0.25
v f2 
 5  2.38m / s
0.25  0.8
31
Types of Collisions
32
Example 8.
A Ballistic Pendulum
The ballistic pendulum consists
of a block of wood(mass
m2=2.5kg)suspended by a wire
of negligible mass. A bullet(mass
m1=0.01kg)is fired with a speed
v01. After collision, the block has
a speed vf and then swings to a
maximum height of 0.65m
above the initial position. Find
the speed v01 of the bullet,
assuming air resistance is
negligible.
33
Just before
collision
Just after
collision
Is conservation of energy valid?
No (completely inelastic)
Is conservation of momentum valid?
m1+m2
Yes (no external forces
)
vf
34
(m1  m2 )v f  m1v01
Total momentum
after collision
Total momentum
before collision
m1  m2
v01 
vf
m1
35
hf=0.65 m
vf
Applying conservation of energy
1
2
(m1  m2 ) gh f  (m1  m2 )v f
2
Total mechanical energy at top
of swing, all potential
Total mechanical energy at
bottom of swing, all kinetic
36
 896m/ s
37
Check your understanding 3
Two balls collide in a one-dimensional, elastic
collision. The two balls constitute a system,
and the net external force acting on them is
zero. The table shows four possible sets of
values for the initial and final momenta of the
two balls as well as their initial and final
kinetic energies. Which one is correct?
38
a
b
c
d
Ball 1
Ball 2
Ball 1
Ball 2
Ball 1
Ball 2
Ball 1
Ball 2
Initial
Final
Momentum Kinetic
Energy
Momentum Kinetic
Energy
+4 kg.m/s
-3 kg.m/s
+7 kg.m/s
+2 kg.m/s
-5 kg.m/s
-8 kg.m/s
+9 kg.m/s
+4 kg.m/s
-5 kg.m/s
-1 kg.m/s
+5 kg.m/s
+4 kg.m/s
-6 kg.m/s
-9 kg.m/s
+6 kg.m/s
+7 kg.m/s
12 J
5J
22 J
8J
12 J
31 J
25 J
15 J
10 J
7J
18 J
15 J
15 J
25 J
18 J
22 J
d
39
Example 9.
A Collision in Two Dimensions
40
Collisions in Two Dimensions
vf2=0.7
m/s
Ball 1
m1=0.15 kg
Ball 2
m2=0.26 kg
before
after
before
after
v01sin50=
vf1cos 0
v02
vf2cos35
y component -v01cos50=
vf1sin 0
0
-vf2sin35
x component
41
x component
m1v f 1x  m2 v f 2 x  m1v01x  m2 v02 x
Pfx
P0x
y component
m1v f 2 y  m2 v f 2 y  m1v01y  m2 v02 y
Pfy
P0y
42
Use momentum conservation to determine the
magnitude and direction of the final velocity of ball 1
after the collision.
x component
(0.15kg)(v f 1x )  (0.260kg)(0.700m / s)(cos 35.0 )
0
Ball 1 after
Ball 2 after
 (0.150kg)(0.900m / s)(sin 50.0 0 )  (0.260kg)(0.540m / s)
Ball 1 before
Ball 2 before
43
y component
(0.150kg)(v f 1 y )  (0.260kg)[(0.700m / s)(sin 35.00 )]
Ball 1 after
Ball 2 after
 (0.150kg)[(0.900m / s)(cos 50.0 )]  0
0
Ball 1 before
Ball 2 before
44
v f 1  (0.63m / s)  (0.12m / s)  0.64m / s
2
2
0.12m / s
0
  tan (
)  11
0.63m / s
1
45
Center of Mass
xcm
m1 x1  m2 x2

m1  m2
46
Suppose
xcm
m1=5kg,
m2=12kg
x1=2m,
x2=6m
(5.0kg)( 2.0m)  (12kg)(6.0m)

 4.8m
5.0kg  12kg
47
xcm
vcm
m1 x1  m2 x2

m1  m2
m1v1  m2 v2

m1  m2
During a time T displacements of the
particles, x1 x2 displacement of cm xcm
48
m1=0.25 kg, m2=0.8 kg
v01=5 m/s,
v02=0 m/s
Before collision
vcm
(0.250kg)( 5.00m / s)  (0.800kg)(0m / s)

 1.19m / s
0.250kg  0.800kg
After collision
vcm
(0.250kg)( 2.62m / s)  (0.800kg)( 2.38m / s)

 1.19m / s
0.250kg  0.800kg
49
Check your understanding 4
Water, dripping at a constant rate from a faucet,
falls to the ground. At any instant, there are many
drops in the air between the faucet and the ground.
Where does the center of mass of the drops lie
relative to the halfway point between the faucet
and the ground: above it, below it, or exactly at
(consider gravity)
the halfway point?
Above the halfway point
50
Concepts & Calculations Example 10.
A Scalar and a Vector
51
Jim and Tom are both running at a speed of
4m/s. Jim has a mass of 90kg, and Tom has
a mass of 55kg. Find the kinetic energy
and momentum of the two-jogger system
when
(a) Jim and Tom are both running due north.
(b)Jim is running due north and Tom is
running due south.
52
(a)
KE system
1
1
2
2
 m Jim v Jim  mTom vTom
2
2
1
1
2
 (90.0kg)( 4.00m / s )  (55.0kg)( 4.00m / s) 2  1160 J
2
2
Psystem  m Jim v Jim  mTom vTom
 (90.0kg)( 4.00m / s)  (55.0kg)( 4.00m / s)  580kg  m / s
53
(b)
KE system
1
1
2
2
 m Jim v Jim  mTom vTom
2
2
1
1
2
 (90.0kg)( 4.00m / s )  (55.0kg)( 4.00m / s) 2  1160 J
2
2
Psystem  m Jim v Jim  mTom vTom
 (90.0kg)( 4.00m / s)  (55.0kg)( 4.00m / s)  140kg  m / s
54
Concepts & Calculations Example 11.
Momentum and Kinetic Energy
Mass
Speed
Object A 2.0 kg
6.0 m/s
Object B 6.0 kg
2.0 m/s
Find the magnitude of the momentum and
the kinetic energy for each object.
55
PA  mAv A  (2.0kg)(6.0m / s)  12kg  m / s
PB  mB vB  (6.0kg)(2.0m / s)  12kg  m / s
1
1
2
2
KE A  m A v A  (2.0kg)(6.0m / s)  36 J
2
2
1
1
2
2
KE B  m B v B  (6.0kg)( 2.0m / s )  12 J
2
2
56
Problem 6
REASONING The impulse that the roof of the car
applies to the hailstones can be found from the impulsemomentum theorem, Equation 7.4. Two forces act on the
hailstones, the average force F exerted by the roof, and
the weight of the hailstones. Since it is assumed that F is
much greater than the weight of the hailstones, the net
average force  F  is equal to F .
SOLUTION From Equation 7.4, the impulse that the
roof applies to the hailstones is:
F t 
Impulse
mvf
Final
momentum

mv 0
Initial
momentum
 m  vf  v 0 
57
Solving for
F
(with up taken to be the positive direction) gives
m
F    ( vf  v 0 )  (0.060 kg/s)  (15 m/s)  (15 m/s) = +1.8 N
 t 
This is the average force exerted on the hailstones by the roof
of the car. The positive sign indicates that this force points
upward. From Newton's third law, the average force exerted
by the hailstones on the roof is equal in magnitude and
opposite in direction to this force. Therefore,
Force on roof = -1.8 N
The negative sign indicates that this force points downward.
58
Problem 20
REASONING During the time that the skaters are pushing
against each other, the sum of the external forces acting on
the two-skater system is zero, because the weight of each
skater is balanced by a corresponding normal force and
friction is negligible. The skaters constitute an isolated
system, so the principle of conservation of linear momentum
applies. We will use this principle to find an expression for
the ratio of the skater’s masses in terms of their recoil
velocities. We will then obtain expressions for the recoil
velocities by noting that each skater, after pushing off, comes
to rest in a certain distance. The recoil velocity, acceleration,
and distance are related by Equation 2.9 of the equations of
kinematics.
59
Skater 1 glides twice as far as skater 2
Conservation of momentum
Initial momentum = final momentum
0  m1v f 1  m2 v f 2
vf2
m1

m2
vf1
Ignore kinetic friction
60
vf1 and vf2 are just after pushing each other. That will be the
initial velocity for the motion of the skaters and will come to
rest.
x2
stop
Vend=0
For skater 1
vf2
x1
vf1
a1
a2
Vend=0
v f  vi  2ax
2
2
vend  v f 1  2a1 x1
2
2
v f 1  vend  2a1 x1  0  2a1 x1
2
2
61
For skater 2
v f  vi  2ax
2
2
vend  v f 2  2a 2 x2
2
2
v f 2  2a 2 x2
2
Magnitude of the acceleration is the same, i.e.,
a1  a2
62
If x1 is positive, x2 is negative,
a2  a1 (opposite direction)
x1  ()2x2
( 1 glides twice as far as 2 )
v f 1  2a1 x1
2
vf 2
2
x1
 2(a1 )(  )  a1 x1
2
m1

m2
 a1 x1
1

 0.707
 2a1 x1
2
63
Problem 34
vf
REASONING AND SOLUTION
Momentum is conserved in the
horizontal direction during the
"collision." Let the coal be
object 1 and the car be object 2.
Then
(m1  m2 )v f  m1v1 cos 25.0  m2 v 2
m1v1 cos 25  m2 v2
vf 
m1  m2
(150kg)(0.8m / s) cos 25  (440kg)(0.5m / s)

 0.56m / s
150kg  440kg
The direction of the final velocity is to the right.
64
Problem 39
REASONING The two balls
constitute the system. The tension in
the wire is the only non-conservative
force that acts on the ball. The tension
does no work since it is perpendicular
to the displacement of the ball.
Since Wnc=0 J, the principle of conservation of mechanical
energy holds and can be used to find the speed of the 1.50-kg ball
just before the collision. Momentum is conserved during the
collision, so the principle of conservation of momentum can be
used to find the velocities of both balls just after the collision.
Once the collision has occurred, energy conservation can be used
to determine how high each ball rises.
65
SOLUTION
a. Applying the principle of energy conservation to the
1.50-kg ball, we have
mv f2  mghf  21 mv 02  mgh0


 


Ef
E0
1
2
m1=
h0=
hf=0, vf=?
v0=
m2=
If we measure the heights from the lowest point in the
swing, hf=0 m, and the expression above simplifies to
1
2
mv  mv  mgh0
Solving for vf, we have
2
f
1
2
2
0
v f  v 02  2 gh0  ( 5.00 m / s) 2  2(9.80 m / s 2 )(0.300 m)  5.56 m / s
66
Just before
v01
Just after
v02
m1
vf1
m2
vf2
Conservation of momentum
m1v01  m2 v02  m1v f 1  m2 v f 2
(v02  0)
67
Conservation of energy
1
1
1
1
2
2
2
m1 (v01 )  m2 (v02 )  m1 (v f 1 )  m2 (v f 2 ) 2
2
2
2
2
1
1
1
2
2
m1 (v01 )  m1 (v f 1 )  m2 (v f 2 ) 2
2
2
2
vf1
m1  m2
(
)v01
m1  m2
vf 2
2m1
(
)v01
m1  m2
68
b. If we assume that the collision is elastic, then the
velocities of both balls just after the collision can be obtained
from Equations 7.8a and 7.8b:
 m1  m2 
v01 and
v f 1  
 m1  m2 
vf 2
 2m1 
v01
 
 m1  m2 
Since v01 corresponds to the speed of the 1.50-kg ball just
before the collision, it is equal to the quantity vf calculated
in part (a). With the given values of m1  1.50 kg
and m2  4.60 kg ,and the value of v 01 = 5.56 m / s obtained
in part (a), Equations 7.8a and 7.8b yield the following
values:
v f1 v
=f1–2.83
m / sm/s
and
v f2 = +2.73 m / s
= -2.83
The minus sign in vf1 indicates that the first ball reverses its
direction as a result of the collision.
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c. If we apply the conservation of mechanical energy to
either ball after the collision we have
mv f2  mghf  21 mv 02  mgh0


 


Ef
E0
1
2
where v0 is the speed of the ball just after the collision, and hf
is the final height to which the ball rises. For either ball, h0 = 0
m, and when either ball has reached its maximum height,
vf = 0 m/s. Therefore, the expression of energy conservation
reduces to
ghf  v
1
2
2
0
or
hf 
v
2
0
2g
70
Thus, the heights to which each ball rises after the collision are
1.50 - kg ball
4.60 - kg ball
v 02
(2.83 m / s) 2
hf 

 0.409 m
2
2 g 2 (9.80 m / s )
v 02
(2.73 m / s) 2
hf 

 0.380 m
2
2 g 2 (9.80 m / s )
71
Problem 41
REASONING AND SOLUTION The location of the center
of mass for a two-body system is given by Equation 7.10:
xcm
m1x1  m2 x2

m1  m2
72
where the subscripts "1" and "2" refer to the earth and the moon,
respectively. For convenience, we will let the center of the earth be
coincident with the origin so that x1  0 and x 2  d , the center-tocenter distance between the earth and the moon. Direct calculation
then gives
xcm
(7.35 1022 kg)(3.85 108 m)
6



4.67

10
m
24
22
m1  m2 5.98 10 kg  7.35 10 kg
m2d
73
Problem 50
REASONING AND SOLUTION
a. According to Equation 7.4, the impulse-momentum
theorem,  F  t  mvf  mv 0 . Since the only horizontal force
exerted on the puck is the force F exerted by the
goalie F  F . Since the goalie catches the puck, vf  0 m/s .
Solving for the average force exerted on the puck, we have
F
m( vf  v0 )
t

 0.17 kg   0 m/s    +65 m/s 
5.0 103 s
3
3N
-2.2*10
 –2.2
10
N
74
By Newton’s third law, the force exerted on the goalie by the
puck is equal in magnitude and opposite in direction to the
force exerted on the puck by the goalie. Thus, the average
force exerted on the goalie is 2.2*103N
b. If, instead of catching the puck, the goalie slaps it with his
stick and returns the puck straight back to the player with a
velocity of –65 m/s, then the average force exerted on the puck
by the goalie is
-3
= -4.4*103 N
75
The average force exerted on the goalie by the puck
is thus +4.4*103N.
The answer in part (b) is twice that in part (a). This
is consistent with the conclusion of Conceptual
Example 3. The change in the momentum of the puck
is greater when the puck rebounds from the stick.
Thus, the puck exerts a greater impulse, and hence a
greater force, on the goalie.
76