Momentun

3. Chapter12: Momentum
 12.1 Momentum
 12.2 Force is the Rate of Change of
Momentum
 12.3 AngularMomentum

4. Chapter12 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.

5. Chapter Vocabulary
 angular momentum
 collision
 law of conservation of
 momentum
 elastic collision
 gyroscope
 impulse
 inelastic collision
 linear momentum
 momentum

6. Inv 12.1 Momentum
Investigation Key Question:
What are so m e use fulpro pe rtie s o f
m o m e ntum ?

7. 12.1 Momentum
 Momentumis a property of moving matter.
 Momentumdescribes the tendency of objects to
keep going in the same direction with the same
speed.
 Changes in momentumresult fromforces or
create forces.

8. 12.1 Momentum
 The momentumof a ball depends on its mass and
velocity.
 Ball Bhas more momentumthan ball A.

9. 12.1 Momentum and Inertia
 Inertia is anotherproperty of mass that resists
changes in velocity; however, inertia depends only
on mass.
 Inertia is a scalarquantity.
 Momentumis a property of moving mass that
resists changes in a moving object’s velocity.
 Momentumis a vectorquantity.

10. 12.1 Momentum
 Ball A is 1 kg moving 1m/sec, ball Bis 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 Bdeflects much
less than ball A when
the same force is
applied because ball B
had a greaterinitial
momentum.

11. 12.1 Kinetic Energy and Momentum
 Kinetic energy and momentumare different quantities,
even though both depend on mass and speed.
 Kinetic energy is a scalarquantity.
 Momentumis 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. 12.1 Calculating Momentum
 The momentumof a moving object is its mass
multiplied by its velocity.
 That means momentumincreases with both
mass and velocity.
Velocity (m/sec)
Mass (kg)
Momentum
(kg m/sec)
p = m v

13. 1. You are asked for momentum.
2. You are given masses and velocities.
3. Use: p = m v
4. Solve for car: p = (1 , 30 0 kg ) (1 3. 5 m /s) = 1 7 , 550 kg m /s
5. Solve for cycle: p = (350 kg ) (30 m /s) = 1 0 , 50 0 kg m /s
 The car has more momentum even though it is going much
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.

14. 12.1 Conservation of Momentum
 The law of conservationof momentumstates when
a systemof interacting objects is not influenced
by outside forces (like friction), the total
momentumof the systemcannot change.
If you throw a rock forward from a
skateboard, you will move
backward in response.

15. 12.1 Conservation of Momentum

16. 12.1 Collisions in One Dimension
 A collisionoccurs when two ormore objects hit
each other.
 During a collision, momentumis transferred from
one object to another.
 Collisions can be elastic orinelastic.

17. 12.1 Collisions

18. 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?

19. 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)
1. V3 = -0.6 m/s
Elastic
collisions

20. 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.

21. 3. Diagram the problem, use m 1 v1 + m 2v2 = (m 1 v1 + m 2v2) v3
4. Solve for v3= (8 , 0 0 0 kg )(1 0 m /s) + (2, 0 0 0 kg )(0 m /s)
(8 , 0 0 0 + 2, 0 0 0 kg )
v3= 8 m /s
The train cars moving together to right at 8 m/s.
Inelastic collisions

22. 12.1 Collisions in 2 and 3 Dimensions
 Most real-life collisions do not occurin one
dimension.
 In a two orthree-dimensional collision, objects
move at angles to each otherbefore orafterthey
collide.
 In orderto analyze two-dimensional collisions you
need to lookat each dimension separately.
 Momentumis conserved separately in the xand y
directions.

23. 12.1 Collisions in 2 and 3 Dimensions

24. Chapter12: Momentum
 12.1 Momentum
 12.2 Force is the Rate of Change of
Momentum
 12.3 AngularMomentum

25. 12.2 Force is the Rate of Change of
Momentum
Investigation Key Question:
How are force and momentum
related?

26. 12.2 Force is the Rate of Change of
Momentum
 Momentumchanges when a
net force is applied.
 The inverse is also true:
 If momentumchanges,
forces are created.
 If momentumchanges
quickly, large forces are
involved.

27. 12.2 Force and MomentumChange
The relationship between force and motion
follows directly fromNewton's second law.
Change in momentum
(kg m/sec)
Change in time (sec)
Force (N) F = ∆ p
∆ t

28. 1. You are asked for force exerted on rocket.
2. You are given rate of fuel ejection and speed of rocket
3. Use F = Δρ ÷Δt
4. Solve: Δρ = (1 0 0 kg ) (-25, 0 0 0 kg m /s) ÷ (1 s) = - 25, 0 0 0 N
 The fuel exerts and equal and opposite force on rocket of +25,000
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.

29. 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 changeinmomentum
because it is not always
possible to calculate force
and time individually since
collisions happen so fast.

30. 12.2 Force and MomentumChange
To find the impulse, you rearrange the
momentumformof the second law.
Change in
momentum
(kg•m/sec)
Impulse (N•sec)
F ∆ t = ∆ p
Impulse can be expressed in kg•m/sec
(momentumunits) orin N•sec.

31. Chapter12: Momentum
 12.1 Momentum
 12.2 Force is the Rate of Change of
Momentum
 12.3 AngularMomentum

32. Inv 12.3 AngularMomentum
Investigation Key Question:
How does the first law apply to
rotational motion?

33. 12.3 AngularMomentum
 Momentumresulting from
an object moving in linear
motion is called linear
momentum.
 Momentumresulting from
the rotation (orspin) of an
object is called angular
momentum.

34. 12.3 Conservation of AngularMomentum
 Angularmomentumis
important because it obeys
a conservation law, as does
linearmomentum.
 The total angular
momentumof a closed
systemstays the same.

35. 12.3 Calculating angularmomentum
Angularmomentumis calculated in a similarway to linear
momentum, except the mass and velocity are replaced
by the moment of inertia and angularvelocity.
Angular
velocity
(rad/sec)
Angular
momentum
(kg m/sec2
)
L = I ω
Moment of inertia
(kg m2
)

36. 12.3 Calculating angularmomentum
 The moment of inertia of an
object is the average of mass
times radius squared forthe
whole object.
 Since the radius is measured
fromthe axis of rotation, the
moment of inertia depends on
the axis of rotation.

37. 1. You are asked for angular momentum.
2. You are given mass, shape, and angular velocity.
 Hint: both rotate about y axis.
1. Use L= Iω, Ihoop = m r2
, Ibar = 1
/1 2 m l2
Calculating angularmomentum
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.

38. 3. Solve hoop: Ihoop= (1 kg ) (0 . 1 6 m )2
= 0 . 0 26 kg m 2
 Lhoop= (1 rad/s) (0 . 0 26 kg m 2
) = 0 . 0 26 kg m 2
/s
3. Solve bar: Ibar = (1
/1 2)(1 kg ) (1 m )2
= 0 . 0 8 3 kg m 2
 Lbar= (1 rad/s) (0 . 0 8 3 kg m 2
) = 0 . 0 8 3 kg m 2
/s
3. The bar has more than 3x the angular momentum
of the hoop, so it is harder to stop.
Calculating angularmomentum

39. 12.3 Gyroscopes angularmomentum
 A gyroscopeis a device that contains a spinning object with
a lot of angularmomentum.
 Gyroscopes can do amazing tricks because they conserve
angularmomentum.
 Forexample, a spinninggyroscope can easily balance on a
pencil point.

40. 12.3 Gyroscopes angularmomentum
 A gyroscope on the space shuttle is mounted at the center
of mass, allowing a computerto measure rotation of the
spacecraft in three dimensions.
 An on-board computeris able to accurately measure the
rotation of the shuttle and maintain its orientation in
space.

41.  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.
Jet Engines