Chapter 12: Momentum
 12.1 Momentum
 12.2 Force is the Rate of Change of
Momentum
 12.3 Angular Momentum
Chapter 12 Objectives

Calculate the linear momentum of a moving object given the mass
and velocity.

Describe the relationship between linear momentum and force.

Solve a one-dimensional elastic collision problem using
momentum conservation.

Describe the properties of angular momentum in a system—for
instance, a bicycle.

Calculate the angular momentum of a rotating object with a
simple shape.
Chapter Vocabulary
 angular momentum
 collision
 law of conservation of
 momentum
 elastic collision
 gyroscope
 impulse
 inelastic collision
 linear momentum
 momentum
Inv 12.1 Momentum
Investigation Key Question:
What are some useful properties of
momentum?
12.1 Momentum
 Momentum is a property of moving matter.
 Momentum describes the tendency of objects
to keep going in the same direction with the
same speed.
 Changes in momentum result from forces or
create forces.
12.1 Momentum
 The momentum of a ball depends on its mass
and velocity.
 Ball B has more momentum than ball A.
12.1 Momentum and Inertia
 Inertia is another property of mass that resists
changes in velocity; however, inertia depends
only on mass.
 Inertia is a scalar quantity.
 Momentum is a property of moving mass that
resists changes in a moving object’s velocity.
 Momentum is a vector quantity.
12.1 Momentum




Ball A is 1 kg moving 1m/sec, ball B is 1kg at 3 m/sec.
A 1 N force is applied to deflect the motion of each ball.
What happens?
Does the force deflect both balls equally?
 Ball B deflects much
less than ball A
when the same force
is applied because
ball B had a greater
initial momentum.
12.1 Kinetic Energy and Momentum
 Kinetic energy and momentum are different quantities,
even though both depend on mass and speed.
 Kinetic energy is a scalar quantity.
 Momentum is a vector, so it always depends on
direction.
Two balls with the same mass and speed have the same kinetic energy
but opposite momentum.
12.1 Calculating Momentum
 The momentum of a moving object is its
mass multiplied by its velocity.
 That means momentum increases with both
mass and velocity.
Momentum
(kg m/sec)
Mass (kg)
p=mv
Velocity (m/sec)
Comparing momentum
A car is traveling at a velocity of 13.5 m/sec (30
mph) north on a straight road. The mass of the car
is 1,300 kg. A motorcycle passes the car at a speed
of 30 m/sec (67 mph). The motorcycle (with rider)
has a mass of 350 kg. Calculate and compare the
momentum of the car and motorcycle.
1.
2.
3.
You are asked for momentum.
You are given masses and velocities.
Use: p = m v
4.
5.
Solve for car: p = (1,300 kg) (13.5 m/s) = 17,550 kg m/s
Solve for cycle: p = (350 kg) (30 m/s) = 10,500 kg m/s

The car has more momentum even though it is going much slower.
12.1 Conservation of Momentum
 The law of conservation of momentum states
when a system of interacting objects is not
influenced by outside forces (like friction), the
total momentum of the system cannot change.
If you throw a rock forward from a
skateboard, you will move
backward in response.
12.1 Conservation of Momentum
12.1 Collisions in One Dimension
 A collision occurs when two or more objects hit
each other.
 During a collision, momentum is transferred
from one object to another.
 Collisions can be elastic or inelastic.
12.1 Collisions
Elastic collisions
Two 0.165 kg billiard balls roll toward
each other and collide head-on.
Initially, the 5-ball has a velocity of 0.5
m/s.
The 10-ball has an initial velocity of -0.7
m/s.
The collision is elastic and the 10-ball
rebounds with a velocity of 0.4 m/s,
reversing its direction.
What is the velocity of the 5-ball after
the collision?
Elastic
collisions
1.
You are asked for 10-ball’s velocity after collision.
2.
You are given mass, initial velocities, 5-ball’s final velocity.
3.
Diagram the motion, use m1v1 + m2v2 = m1v3 + m2v4
4.
Solve for V3 : (0.165 kg)(0.5 m/s) + (0.165 kg) (-0.7 kg)=
(0.165 kg) v3 + (0.165 kg) (0.4 m/s)
5.
V3 = -0.6 m/s
Inelastic collisions
A train car moving to the right at 10 m/s
collides with a parked train car.
They stick together and roll along the
track.
If the moving car has a mass of 8,000 kg
and the parked car has a mass of 2,000
kg, what is their combined velocity after
the collision?
1.
You are asked for the final velocity.
2.
You are given masses, and initial velocity of moving train
car.
Inelastic collisions
3.
Diagram the problem, use m1v1 + m2v2 = (m1v1 +m2v2) v3
4.
Solve for v3= (8,000 kg)(10 m/s) + (2,000 kg)(0 m/s)
(8,000 + 2,000 kg)
v3= 8 m/s
The train cars moving together to right at 8 m/s.
12.1 Collisions in 2 and 3 Dimensions
 Most real-life collisions do not occur in one
dimension.
 In a two or three-dimensional collision, objects
move at angles to each other before or after
they collide.
 In order to analyze two-dimensional collisions
you need to look at each dimension separately.
 Momentum is conserved separately in the x and
y directions.
12.1 Collisions in 2 and 3 Dimensions
Chapter 12: Momentum
 12.1 Momentum
 12.2 Force is the Rate of Change of
Momentum
 12.3 Angular Momentum
12.2 Force is the Rate of Change of
Momentum
Investigation Key Question:
How are force and momentum
related?
12.2 Force is the Rate of Change of
Momentum
 Momentum changes when
a net force is applied.
 The inverse is also true:
 If momentum changes,
forces are created.
 If momentum changes
quickly, large forces are
involved.
12.2 Force and Momentum Change
The relationship between force and motion
follows directly from Newton's second law.
Force (N)
Change in time (sec)
F=Dp
Dt
Change in momentum
(kg m/sec)
Calculating force
Starting at rest, an 1,800 kg rocket takes off, ejecting
100 kg of fuel per second out of its nozzle at a speed of
2,500 m/sec. Calculate the force on the rocket from the
change in momentum of the fuel.
1.
You are asked for force exerted on rocket.
2.
You are given rate of fuel ejection and speed of rocket
3.
Use F = Δ ÷Δt
4.
Solve: Δ = (100 kg) (-25,000 kg m/s) ÷ (1s) = - 25,000 N

The fuel exerts and equal and opposite force on rocket of +25,000 N.
12.2 Impulse
 The product of a force and
the time the force acts is
called the impulse.
 Impulse is a way to
measure a change in
momentum because it is
not always possible to
calculate force and time
individually since
collisions happen so fast.
12.2 Force and Momentum Change
To find the impulse, you rearrange the
momentum form of the second law.
Impulse (N•sec)
FD t=Dp
Change in
momentum
(kg•m/sec)
Impulse can be expressed in kg•m/sec
(momentum units) or in N•sec.
Chapter 12: Momentum
 12.1 Momentum
 12.2 Force is the Rate of Change of
Momentum
 12.3 Angular Momentum
Inv 12.3 Angular Momentum
Investigation Key Question:
How does the first law apply to
rotational motion?
12.3 Angular Momentum
 Momentum resulting
from an object moving in
linear motion is called
linear momentum.
 Momentum resulting
from the rotation (or
spin) of an object is
called angular
momentum.
12.3 Conservation of Angular Momentum
 Angular momentum is
important because it
obeys a conservation
law, as does linear
momentum.
 The total angular
momentum of a closed
system stays the same.
12.3 Calculating angular momentum
Angular momentum is calculated in a similar way to
linear momentum, except the mass and velocity are
replaced by the moment of inertia and angular
velocity.
Angular
momentum
(kg m/sec2)
L=Iw
Moment of
inertia
(kg m2)
Angular
velocity
(rad/sec)
12.3 Calculating angular momentum
 The moment of inertia of an
object is the average of
mass times radius squared
for the whole object.
 Since the radius is
measured from the axis of
rotation, the moment of
inertia depends on the axis
of rotation.
Calculating angular momentum
An artist is making a moving metal sculpture. She takes two
identical 1 kg metal bars and bends one into a hoop with a
radius of 0.16 m. The hoop spins like a wheel. The other bar is
left straight with a length of 1 meter. The straight bar spins
around its center. Both have an angular velocity of 1 rad/sec.
Calculate the angular momentum of each and decide which
would be harder to stop.
1.
You are asked for angular momentum.
2.
You are given mass, shape, and angular velocity.
 Hint: both rotate about y axis.
3.
Use L= Iw, Ihoop = mr2, Ibar = 1/12 ml2
Calculating angular momentum
3.
Solve hoop: Ihoop= (1 kg) (0.16 m)2 = 0.026 kg m2
 Lhoop= (1 rad/s) (0.026 kg m2) = 0.026 kg m2/s
4.
Solve bar: Ibar = (1/12)(1 kg) (1 m)2 = 0.083 kg m2
 Lbar = (1 rad/s) (0.083 kg m2) = 0.083 kg m2/s
5.
The bar has more than 3x the angular momentum
of the hoop, so it is harder to stop.
12.3 Gyroscopes angular momentum
 A gyroscope is a device that contains a spinning object
with a lot of angular momentum.
 Gyroscopes can do amazing tricks because they
conserve angular momentum.
 For example, a spinning gyroscope can easily balance
on a pencil point.
12.3 Gyroscopes angular momentum
 A gyroscope on the space shuttle is mounted at the
center of mass, allowing a computer to measure
rotation of the spacecraft in three dimensions.
 An on-board computer is able to accurately measure
the rotation of the shuttle and maintain its orientation
in space.
Jet Engines
 Nearly all modern airplanes use jet propulsion to fly. Jet engines
and rockets work because of conservation of linear momentum.
 A rocket engine uses the same principles as a jet, except that in
space, there is no oxygen.
 Most rockets have to carry so much oxygen and fuel that the
payload of people or satellites is usually less than 5 percent of
the total mass of the rocket at launch.