Ch7. Impulse and Momentum

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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.
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Impulse-Momentum Theorem
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Definition of Impulse The impulse J of a force is
the product of the average force and the
time interval during which the force acts
J
Impulse is a vector quantity and has the same
direction as the average force. SI Unit of
Impulse newton.second (N.s)
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Definition of Linear Momentum The linear
momentum p of an object is the product of the
objects mass m and velocity v
pmv 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)

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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
Final momentum
Initial momentum
Impulse
ImpulseChange in momentum
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Example 1. A Well-Hit Ball

• A baseball (m0.14kg) has initial velocity of
  v0-38m/s as it approaches a bat. The bat applies
  an average force that is much larger than
  the weight of the ball, and the ball departs from
  the bat with a final velocity of vf58m.
• Determine the impulse applied to the ball by the
  bat.
• Assuming time of contact is 1.610-3s,
  find the average force exerted on the ball by the
  bat.

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(a)
13.4 kg.m/s
(b)
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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 (vf0m/s), find the average force exerted by
the raindrop.
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-(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
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Conceptual Example 3. Hailstones Versus Raindrops
Suppose hail is falling. The hail comes straight
down at a mass rate of m/ 0.06kg/s and an
initial velocity of v015m/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.
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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.
B
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The Principle of Conservation of Linear Momentum
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• Two types of forces act on the system
• Internal forces Forces that the objects within
  the system exert on each other.
• External forces Forces exerted on the objects by
  agents external to the system.

External force
Internal force
External force
Internal force
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(
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sum of average external forces
sum of average internal forces
pf - p0

internal forces cancel F12 -F21
(Sum of average external forces)
pf - p0
If sum of external forces is zero (an
isolated system)
Then 0 pf - p0 or pf p0
m1vf1m2vf2 m1v01m2v02
pf
p0
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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.
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Conceptual Example 4. Is the Total Momentum
Conserved?

• Two balls collide on the billiard table that is
  free of friction.
• Is the total momentum of the two ball system the
  same before and after the collision?
• Answer (a) for a system that contains only one
  ball.

1. The total momentum is conserved.

(b) The total momentum of one ball system is not
conserved.
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Example 5. Assembling a Freight Train
Car 1 has a mass of m165103kg and moves at a
velocity of v010.8m/s. Car 2 has a mass of
m292103kg and a velocity of v021.3m/s.
Neglecting friction, find the common velocity vf
of the cars after they become coupled.
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(m1m2) vf m1v01 m2v02
After collision
Before collision
1.1 m/s
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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 (m154kg), and one is
a man(m288kg). The woman moves away with a
velocity of vf12.5m/s. Find the recoil velocity
vf2 of the man.
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For the two skater system, what are the
forces? In horizontal direction
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m1vf1 m2vf2 0
after pushing
before pushing
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.
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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?
Person 1 Person 2 canoe
a -60kg.m/s 170kg.m/s
b -30kg.m/s 110kg.m/s
c -40kg.m/s -70kg.m/s
d 80kg.m/s -30kg.m/s
a
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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.
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Example 7. A Collision in One Dimension
A ball of mass m10.25kg and velocity v015m/s
collides head-on with a ball of mass m20.8kg
that is initially at rest(v020m/s). No external
forces act on the balls. If the collision in
elastic, what are the velocities of the balls
after the collision?
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(1)
(2)
(3)
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Substitute in (2)
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Substitute in (1)
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m10.25, m20.8
v01 5 m/s, v02 0
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Types of Collisions
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Example 8. A Ballistic Pendulum
The ballistic pendulum consists of a block of
wood(mass m22.5kg)suspended by a wire of
negligible mass. A bullet(mass m10.01kg)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.
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Is conservation of energy valid?
No (completely inelastic)
Is conservation of momentum valid?
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Applying conservation of energy
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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?
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Initial Final
Momentum Kinetic Energy Momentum Kinetic Energy
a Ball 1 4 kg.m/s 12 J -5 kg.m/s 10 J
Ball 2 -3 kg.m/s 5 J -1 kg.m/s 7 J
b Ball 1 7 kg.m/s 22 J 5 kg.m/s 18 J
Ball 2 2 kg.m/s 8 J 4 kg.m/s 15 J
c Ball 1 -5 kg.m/s 12 J -6 kg.m/s 15 J
Ball 2 -8 kg.m/s 31 J -9 kg.m/s 25 J
d Ball 1 9 kg.m/s 25 J 6 kg.m/s 18 J
Ball 2 4 kg.m/s 15 J 7 kg.m/s 22 J
d
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Example 9. A Collision in Two Dimensions
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Collisions in Two Dimensions
Ball 1 m10.15 kg Ball 1 m10.15 kg Ball 2 m20.26 kg Ball 2 m20.26 kg
before after before after
v01sin50 vf1cos 0 v02 vf2cos35
-v01cos50 vf1sin 0 0 -vf2sin35
x component
y component
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x component
P0x
Pfx
y component
P0y
Pfy
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Use momentum conservation to determine the
magnitude and direction of the final velocity of
ball 1 after the collision.
x component
Ball 1 after
Ball 2 after
Ball 2 before
Ball 1 before
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y component
Ball 1 after
Ball 2 after
Ball 1 before
Ball 2 before
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Center of Mass
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Suppose m15kg, m212kg
x12m, x26m
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During a time displacements of the
particles, displacement of cm
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m10.25 kg, m20.8 kg v015 m/s, v020
m/s
Before collision
After collision
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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 the halfway point?
(consider gravity)
Above the halfway point
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Concepts Calculations Example 10. A Scalar
and a Vector
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• 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
• Jim and Tom are both running due north.
• Jim is running due north and Tom is running due
  south.

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(a)
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(b)
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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.
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Problem 6
REASONING The impulse that the roof of the
car applies to the hailstones can be found from
the impulse-momentum theorem, Equation 7.4. Two
forces act on the hailstones, the average force
exerted by the roof, and the weight of the
hailstones. Since it is assumed that is much
greater than the weight of the hailstones, the
net average force is equal to .
SOLUTION   From Equation 7.4, the impulse that
the roof applies to the hailstones is
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Solving for (with up taken to be the
positive direction) gives
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.
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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 skaters 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.
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Skater 1 glides twice as far as skater
2 Conservation of momentum Initial momentum
final momentum
Ignore kinetic friction
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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.
x1
x2
vf1
vf2
stop
a1
a2
Vend0
Vend0
For skater 1
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For skater 2
Magnitude of the acceleration is the same, i.e.,
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If x1 is positive, x2 is negative,
(opposite direction)
( 1 glides twice as far as 2 )
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Problem 34
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
vf
The direction of the final velocity is to the
right.
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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 Wnc0 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.
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SOLUTION
a. Applying the principle of energy conservation
to the 1.50-kg ball, we have
m1
h0
hf0, vf?
v0
m2
If we measure the heights from the lowest
point in the swing, hf0 m, and the expression
above simplifies to
Solving for vf, we have
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Just before
Just after
v01
v02
vf1
vf2
m1
m2
Conservation of momentum
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Conservation of energy
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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
and
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 and
,and the value of
obtained in part (a), Equations 7.8a and 7.8b
yield the following values
vf1 -2.83 m/s
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
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
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Thus, the heights to which each ball rises after
the collision are
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Problem 41
REASONING AND SOLUTION The location of the
center of mass for a two-body system is given by
Equation 7.10
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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
and , the center-to-center distance
between the earth and the moon. Direct
calculation then gives
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Problem 50
REASONING AND SOLUTION
a. According to Equation 7.4, the
impulse-momentum theorem,
. Since the only horizontal force exerted
on the puck is the force exerted by the
goalie . Since the goalie catches
the puck, . Solving for the
average force exerted on the puck, we have
-2.2103 N
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By Newtons 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 2.2103N
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
-4.4103 N
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The average force exerted on the goalie by the
puck is thus 4.4103N.
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.