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Chapter 7 - Giancoli
Momentum and
Impulse
Chapter 7
Linear Momentum
Units of Chapter 7
•Momentum and Its Relation to Force
•Conservation of Momentum
•Collisions and Impulse
•Conservation of Energy and Momentum in
Collisions
•Elastic Collisions in One Dimension
Units of Chapter 7
•Inelastic Collisions
•Collisions in Two or Three Dimensions
•Center of Mass (CM)
•CM for the Human Body
•Center of Mass and Translational Motion
7-1 Momentum and Its Relation to Force
Momentum is a vector symbolized by the
symbol p, and is defined as
(7-1)
The rate of change of momentum is equal to the
net force:
(7-2)
This can be shown using Newton’s second law.
7-2 Conservation of Momentum
During a collision, measurements show that the
total momentum does not change:
(7-3)
7-2 Conservation of Momentum
More formally, the law of conservation of
momentum states:
The total momentum of an isolated system of
objects remains constant.
7-2 Conservation of Momentum
Momentum conservation works for a rocket as
long as we consider the rocket and its fuel to
be one system, and account for the mass loss
of the rocket.
6.
(II) A 95-kg halfback moving at
4.1 m s
on an apparent breakaway for a touchdown is tackled from
behind. When he was tackled by an 85-kg cornerback
running at 5.5 m s
in the same direction, what was their mutual speed
immediately after the tackle?
6.
The tackle will be analyzed as a one-dimensional
momentum conserving situation. Let “A” represent the
halfback, and “B” represent the tackling cornerback.
pinitial  pfinal  mAvA  mB vB   mA  mB  v 
v 
mAvA  mB vB
mA  mB
 95 kg  4.1m s   85 kg  5.5m s 

 4.8m s
 95 kg   85 kg 
Problem 7/3
• 3. (II) A 0.145-kg baseball pitched at
39.0 m/s is hit on a horizontal line
drive straight back toward the pitcher
at 52.0 m/s. If the contact time
between bat and ball is 3.00 x 10-3 s,
calculate the average force between
the ball and bat during contact
Answer to: 7/3
• 3. 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 
v
 52.0 m s  39.0 m
F m
  0.145 kg  
t
3.00 103 s

s
3

4.40

10
N, towards the pitcher


7-3 Collisions and Impulse
During a collision, objects
are deformed due to the
large forces involved.
Since
write
, we can
(7-5)
The definition of impulse:
7-3 Collisions and Impulse
Since the time of the collision is very short, we
need not worry about the exact time dependence
of the force, and can use the average force.
7-3 Collisions and Impulse
The impulse tells us that we can get the same
change in momentum with a large force acting for a
short time, or a small force acting for a longer time.
This is why you should bend
your knees when you land;
why airbags work; and why
landing on a pillow hurts less
than landing on concrete.
Problem: 7/15
• 15.(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 with the ball for 3.5 x 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.
15. (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 kg  45 m s  0   2.0 kg m s
2
7/15
• (b) The average force is the impulse
divided by the interaction time.
F
p
t

2.0 kg m s
3
3.5  10 s
 5.8  10 N
2
7-4 Conservation of Energy and Momentum
in Collisions
Momentum is conserved
in all collisions.
Collisions in which
kinetic energy is
conserved as well are
called elastic collisions,
and those in which it is
not are called inelastic.
7-5 Elastic Collisions in One Dimension
Here we have two objects
colliding elastically. We
know the masses and the
initial speeds.
Since both momentum
and kinetic energy are
conserved, we can write
two equations. This
allows us to solve for the
two unknown final
speeds.
Problem: 7/22
• 22.(II) A ball of mass 0.440 kg moving east
( direction) with a speed of collides headon 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?
22. Let A represent the 0.440-kg ball, and
B represent the 0.220-kg ball. We have
and . 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
7/22
mA vA  mB vB  mA vA  mB vB  mA vA  mA vA  mB  vA  vA  
mA  mB 
0.220 kg

vA 
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 
7-6 Inelastic Collisions
With inelastic collisions, some of
the initial kinetic energy is lost to
thermal or potential energy. It
may also be gained during
explosions, as there is the
addition of chemical or nuclear
energy.
A completely inelastic collision is
one where the objects stick
together afterwards, so there is
only one final velocity.
Problem: 7/31
• 31.(I) In a ballistic pendulum experiment,
projectile 1 results in a maximum height h
of the pendulum equal to 2.6 cm. A
second projectile causes the the pendulum
to swing twice as high, h2 = 5.2 cm. The
second projectile was how many times
faster than the first?
See page 179 for Ex. 7-10.
• 31.From the analysis in Example 7-10, the
initial projectile speed is given by .
v
mM
m
2 gh
• Compare the two speeds with the same
masses.
7/31
mM
v2
 m
v1 m  M
m
2 gh2
2 gh1

h2
h1

h2
h1

5.2
2.6
 2 
v2  2v1
Problem: 7/32
• 32.(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.
32. From the analysis
in the Example 7-10,
we know that
L
q
L-h
x
h
v
mM
m
2 gh 
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

 0.1607 m  0.16 m

2
7/32
L
q
L-h
x
From the diagram we
see that
h
L   L  h  x2
2
2
x
L   L  h 
2
2
 2.8 m    2.8 m  0.1607 m 
2
2
 0.94 m
End of Chapter 7
Only Sections 1-6 are covered in
APPB
7-7 Collisions in Two or Three Dimensions
Conservation of energy and momentum can also
be used to analyze collisions in two or three
dimensions, but unless the situation is very
simple, the math quickly becomes unwieldy.
Here, a moving object
collides with an object
initially at rest. Knowing
the masses and initial
velocities is not enough;
we need to know the
angles as well in order to
find the final velocities.
7-7 Collisions in Two or Three Dimensions
Problem solving:
1. Choose the system. If it is complex,
subsystems may be chosen where one or
more conservation laws apply.
2. Is there an external force? If so, is the
collision time short enough that you can
ignore it?
3. Draw diagrams of the initial and final
situations, with momentum vectors labeled.
4. Choose a coordinate system.
7-7 Collisions in Two or Three Dimensions
5. Apply momentum conservation; there will be
one equation for each dimension.
6. If the collision is elastic, apply conservation
of kinetic energy as well.
7. Solve.
8. Check units and magnitudes of result.
7-8 Center of Mass
In (a), the diver’s motion is pure translation; in (b)
it is translation plus rotation.
There is one point that moves in the same path a
particle would
take if subjected
to the same force
as the diver. This
point is called the
center of mass
(CM).
7-8 Center of Mass
The general motion of an object can be
considered as the sum of the translational
motion of the CM, plus rotational, vibrational, or
other forms of motion about the CM.
7-8 Center of Mass
For two particles, the center of mass lies closer
to the one with the most mass:
where M is the total mass.
7-8 Center of Mass
The center of gravity is the point where the
gravitational force can be considered to act. It is
the same as the center of mass as long as the
gravitational force does not vary among different
parts of the object.
7-8 Center of Mass
The center of gravity can be found experimentally
by suspending an object from different points.
The CM need not be within the actual object – a
doughnut’s CM is in the center of the hole.
7-9 CM for the Human Body
The x’s in the small diagram mark the CM of
the listed body segments.
7-9 CM for the Human Body
The location of the center of
mass of the leg (circled) will
depend on the position of
the leg.
7-9 CM for the Human Body
High jumpers have
developed a technique
where their CM actually
passes under the bar as
they go over it. This allows
them to clear higher bars.
7-10 Center of Mass and Translational Motion
The total momentum of a system of particles is
equal to the product of the total mass and the
velocity of the center of mass.
The sum of all the forces acting on a system is
equal to the total mass of the system multiplied
by the acceleration of the center of mass:
(7-11)
7-10 Center of Mass and Translational Motion
This is particularly useful in the analysis of
separations and explosions; the center of
mass (which may not correspond to the
position of any particle) continues to move
according to the net force.
Summary of Chapter 7
• Momentum of an object:
• Newton’s second law:
•Total momentum of an isolated system of objects is
conserved.
• During a collision, the colliding objects can be
considered to be an isolated system even if external
forces exist, as long as they are not too large.
• Momentum will therefore be conserved during
collisions.
Summary of Chapter 7, cont.
•
• In an elastic collision, total kinetic energy is
also conserved.
• In an inelastic collision, some kinetic energy
is lost.
• In a completely inelastic collision, the two
objects stick together after the collision.
• The center of mass of a system is the point at
which external forces can be considered to
act.