Chapter 9

Linear Momentum and Collisions

1

Review of Newton’s Third Law

If two objects interact, the force F12 exerted by object 1 on object is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1:

F12 = -F21

2

m1 m2

a1

a2

Linear Momentum

doesn’t change w.r.t. time t.

Call each individual mivi linear momentum.3

m1 m2

a1

a2

F12 = - F21

F12 = - F21 m1a1+m2a2=0

02

2

1

1 dt

vdm

dt

vdm

0)( 2211 vmvmdt

d

2211 vmvm

Linear Momentum

The linear momentum of a particle, or an object that can be modeled as a particle, of mass m moving with a velocity is defined to be the product of the mass and velocity:

The terms momentum and linear momentum will be used interchangeably in the text

v

mp v

4

Linear Momentum, cont

Linear momentum is a vector quantity Its direction is the same as the direction of the

velocity The dimensions of momentum are ML/T The SI units of momentum are kg · m / s Momentum can be expressed in component

form: px = m vx py = m vy pz = m vz

5

Newton and Momentum

Newton’s Second Law can be used to relate the momentum of a particle to the resultant force acting on it

with constant mass

d md dm m

dt dt dt

vv pF a

6

Newton’s Second Law

The time rate of change of the linear momentum of a particle is equal to the net force acting on the particle This is the form in which Newton presented the

Second Law It is a more general form than the one we used

previously This form also allows for mass changes

Applications to systems of particles are particularly powerful

7

Conservation of Linear Momentum

Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant The momentum of the system is conserved, not

necessarily the momentum of an individual particle

This also tells us that the total momentum of an isolated system equals its initial momentum

8

Conservation of Momentum, 2

Conservation of momentum can be expressed mathematically in various ways

In component form, the total momenta in each direction are independently conserved pix = pfx piy = pfy piz = pfz

Conservation of momentum can be applied to systems with any number of particles

This law is the mathematical representation of the momentum version of the isolated system model

total 1 2p = p + p = constant

1i 2i 1f 2fp + p = p + p

9

Conservation of Momentum, Archer Example

A 60-kg archer stands at rest on a frictionless surface (ice) and fire a 0.5-kg arrow horizontally at 50m/s.

With what velocity does the archer move across the ice after firing the arrow?

10

Conservation of Momentum, Archer Example

Approaches: Newton’s Second Law – no

information about F or a Energy approach – no

information about work or energy

Momentum – yes

11

Archer Example, 2

Conceptualize The arrow is fired one way and the archer recoils in the

opposite direction Categorize

Momentum Let the system be the archer with bow (particle 1) and the

arrow (particle 2) There are no external forces in the x-direction, so it is

isolated in terms of momentum in the x-direction Analyze

Total momentum before releasing the arrow is 0

12

Archer Example, 3 Analyze, cont.

The total momentum after releasing the arrow is

Solve v1.

Finalize The final velocity of the archer is

negative Indicates he moves in a direction

opposite the arrow Archer has much higher mass than

arrow, so velocity is much lower

1 2 0f f p p

13

02211 vmvm

Impulse and Momentum

From Newton’s Second Law, Solving for gives Integrating to find the change in momentum

over some time interval

The integral is called the impulse, , of the force acting on an object over t

f

i

t

f i tdt p p p F I

d

dt

pF

dp d dtp F

I

14

Impulse-Momentum Theorem

This equation expresses the impulse-momentum theorem: The impulse of the force acting on a particle equals the change in the momentum of the particle This is equivalent to Newton’s Second Law

p I

15

More About Impulse Impulse is a vector quantity The magnitude of the

impulse is equal to the area under the force-time curve The force may vary with

time Dimensions of impulse are

M L / T Impulse is not a property of

the particle, but a measure of the change in momentum of the particle

16

Impulse, Final

The impulse can also be found by using the time averaged force

This would give the

same impulse as the time-varying force does

t I F

17

Impulse Approximation

In many cases, one force acting on a particle acts for a short time, but is much greater than any other force present

When using the Impulse Approximation, we will assume this is true Especially useful in analyzing collisions

The force will be called the impulsive force The particle is assumed to move very little during

the collision represent the momenta immediately before

and after the collisioni fandp p

18

Impulse-Momentum: Crash Test Example

A car of mass 1500kg collides with a wall.

Vi=-15 m/s and Vf=2.6 m/s. The collision lasts 0.15 seconds.

Find the impulse caused by the collision and the average force exerted on the car.

19

Impulse-Momentum: Crash Test Example

Categorize Assume force exerted by

wall is large compared with other forces

Gravitational and normal forces are perpendicular and so do not effect the horizontal momentum

Can apply impulse approximation

20

Crash Test Example, 2

Analyze The momenta before and after the collision between the car

and the wall can be determined Find

Initial momentum Final momentum Impulse Average force

Finalize Check signs on velocities to be sure they are reasonable

21

f

i

t

f i tdt p p p F I

ii vmP ff vmP

if vmvmPI

t

PF average

Collisions – Characteristics

We use the term collision to represent an event during which two particles come close to each other and interact by means of forces May involve physical contact, but must be generalized to

include cases with interaction without physical contact The time interval during which the velocity changes

from its initial to final values is assumed to be short The interaction forces are assumed to be much

greater than any external forces present This means the impulse approximation can be used

22

Collisions – Example 1

Collisions may be the result of direct contact

The impulsive forces may vary in time in complicated ways This force is internal to

the system Observe the variations in

the active figure Momentum is

conserved

23

Collisions – Example 2

The collision need not include physical contact between the objects

There are still forces between the particles

This type of collision can be analyzed in the same way as those that include physical contact

24

Types of Collisions

In an elastic collision, momentum and kinetic energy are conserved Perfectly elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elastic

collisions actually occur Generally some energy is lost to deformation, sound, etc.

In an inelastic collision, kinetic energy is not conserved, although momentum is still conserved If the objects stick together after the collision, it is a

perfectly inelastic collision

25

Collisions, cont

In an inelastic collision, some kinetic energy is lost, but the objects do not stick together

Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types

Momentum is conserved in all collisions

26

Perfectly Inelastic Collisions

Since the objects stick together, they share the same velocity after the collision

Compare Ki with Kf

1 1 2 2 1 2i i fm m m m v v v

27

Elastic Collisions

Both momentum and kinetic energy are conserved

1 1 2 2

1 1 2 2

2 21 1 2 2

2 21 1 2 2

1 1

2 21 1

2 2

i i

f f

i i

f f

m m

m m

m m

m m

v v

v v

v v

v v

28

Elastic Collisions, cont Typically, there are two unknowns to solve for and so you

need two equations The kinetic energy equation can be difficult to use With some algebraic manipulation, a different equation can be

used

v1i – v2i = - (v1f - v2f) why?

This equation, along with conservation of momentum, can be used to solve for the two unknowns It can only be used with a one-dimensional, elastic collision

between two objects

29

Elastic Collisions, cont

Momentum:

kinetic energy:

30

ffii vmvmvmvm 22112211

222

211

222

211 2

1

2

1

2

1

2

1ffii vmvmvmvm

)()( 22

222

21

211 fifi vvmvvm

)()( 222111 fifi vvmvvm

fifi vvvv 2211 )( 2121 ffii vvvv ….(B)

….(A)

Elastic Collisions, cont

By (A) and (B),

31

iif vmm

mv

mm

mmv 2

21

21

21

211

2

iif vmm

mmv

mm

mv 2

21

121

21

12

2

Elastic Collisions, final

Example of some special cases m1 = m2 – the particles exchange velocities

When a very heavy particle collides head-on with a very light one initially at rest, the heavy particle continues in motion unaltered and the light particle rebounds with a speed of about twice the initial speed of the heavy particle

When a very light particle collides head-on with a very heavy particle initially at rest, the light particle has its velocity reversed and the heavy particle remains approximately at rest

32

Problem-Solving Strategy: One-Dimensional Collisions

Conceptualize Image the collision occurring in your mind Draw simple diagrams of the particles before and

after the collision Include appropriate velocity vectors

Categorize Is the system of particles isolated? Is the collision elastic, inelastic or perfectly

inelastic?

33

Problem-Solving Strategy: One-Dimensional Collisions

Analyze Set up the mathematical representation of the

problem Solve for the unknown(s)

Finalize Check to see if the answers are consistent with

the mental and pictorial representations Check to be sure your results are realistic

34

Example: Stress Reliever

Conceptualize Imagine one ball coming in

from the left and two balls exiting from the right

Is this possible? Categorize

Due to shortness of time, the impulse approximation can be used

Isolated system Elastic collisions

35

Example: Stress Reliever, cont

Analyze Check to see if momentum is conserved

It is Check to see if kinetic energy is conserved

It is not Therefore, the collision couldn’t be elastic

Finalize Having two balls exit was not possible if only one

ball is released

36

Example: Stress Reliever, final

What collision is possible

Need to conserve both momentum and kinetic energy Only way to do so is with

equal numbers of balls released and exiting

37?

Example: Stress Reliever, final

What if balls 4 and 5 are glued together?

v45 = ?

Pi = Pf mv1i = mv1f + 2mv45

Ki = Kf

38

?

245

21

21 )2(

2

1

2

1

2

1vmmvmv fi

ifi vvvv 11145 3

1

3

2

Collision Example – Ballistic Pendulum

Conceptualize Observe diagram Determine the velocity of the bullet in

terms of h Categorize

Isolated system of projectile and block Perfectly inelastic collision – the bullet is

embedded in the block of wood Momentum equation will have two

unknowns Use conservation of energy from the

pendulum to find the velocity just after the collision

Then you can find the speed of the bullet

39

Collision Example – Ballistic Pendulum

Pi = Pf m1v1A = (m1+m2)vB

KB = (1/2)(m1+m2)vB 2

= (m12 v1A

2)/2(m1+m2)

KB = UC = (m1+m2)gh

v1A =(m1+m2)2gh/m1

40

Ballistic Pendulum, cont

A multi-flash photograph of a ballistic pendulum

Analyze Solve resulting system of

equations Finalize

Note different systems involved

Some energy was transferred during the perfectly inelastic collision

41

Two-Dimensional Collisions

The momentum is conserved in all directions Use subscripts for

Identifying the object Indicating initial or final values The velocity components

If the collision is elastic, use conservation of kinetic energy as a second equation Remember, the simpler equation can only be used for one-

dimensional situations

42

Two-Dimensional Collision, example

Particle 1 is moving at velocity and particle 2 is at rest

In the x-direction, the initial momentum is m1v1i

In the y-direction, the initial momentum is 0

1iv

43

Two-Dimensional Collision, example cont

After the collision, the momentum in the x-direction is m1v1f cos m2v2f cos

After the collision, the momentum in the y-direction is m1v1f sin m2v2f sin

If the collision is elastic, apply the kinetic energy equation

This is an example of a glancing collision

44

Problem-Solving Strategies – Two-Dimensional Collisions

Conceptualize Imagine the collision Predict approximate directions the particles will

move after the collision Set up a coordinate system and define your

velocities with respect to that system It is usually convenient to have the x-axis coincide

with one of the initial velocities In your sketch of the coordinate system, draw and

label all velocity vectors and include all the given information

45

Problem-Solving Strategies – Two-Dimensional Collisions, 2

Categorize Is the system isolated? If so, categorize the collision as elastic, inelastic or

perfectly inelastic Analyze

Write expressions for the x- and y-components of the momentum of each object before and after the collision Remember to include the appropriate signs for the

components of the velocity vectors Write expressions for the total momentum of the system in

the x-direction before and after the collision and equate the two. Repeat for the total momentum in the y-direction.

46

Problem-Solving Strategies – Two-Dimensional Collisions, 3

Analyze, cont If the collision is inelastic, kinetic energy of the system is

not conserved, and additional information is probably needed

If the collision is perfectly inelastic, the final velocities of the two objects are equal. Solve the momentum equations for the unknowns.

If the collision is elastic, the kinetic energy of the system is conserved Equate the total kinetic energy before the collision to the

total kinetic energy after the collision to obtain more information on the relationship between the velocities

47

Problem-Solving Strategies – Two-Dimensional Collisions, 4

Finalize Check to see if your answers are consistent with

the mental and pictorial representations Check to be sure your results are realistic

48

Two-Dimensional Collision Example

A 1500-kg car traveling east with a speed of 25 m/s collides at an intersection with a 2500-kg van traveling north at a speed of 20 m/s

Find the direction and magnitude of the velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.

49

Two-Dimensional Collision Example

Conceptualize See picture Choose East to be the

positive x-direction and North to be the positive y-direction

Categorize Ignore friction Model the cars as particles The collision is perfectly

inelastic The cars stick together

50

Two-Dimensional Collision Example, cont

Analyze Before the collision, the car has the total momentum in the

x-direction and the van has the total momentum in the y-direction

After the collision, both have x- and y-components Write expressions for initial and final momenta in both

directions Evaluate any expressions with no unknowns

Solve for unknowns Finalize

Check to be sure the results are reasonable

51

Two-Dimensional Collision Example

Pxi = (1500 kg)(25 m/s)

Pxf = (4000 kg)(vfcos m/s)

Pyi = (2500 kg)(20 m/s)

Pxf = (4000 kg)(vfsin m/s)

Solve vf and .

52

The Center of Mass

There is a special point in a system or object, called the center of mass, that moves as if all of the mass of the system is concentrated at that point

The system will move as if an external force were applied to a single particle of mass M located at the center of mass M is the total mass of the system

53

Center of Mass, Coordinates

The coordinates of the center of mass are

M is the total mass of the system Use the active figure to

observe effect of different masses and positions

CM

CM

CM

i ii

i ii

i ii

m xx

Mm y

yMm z

zM

54

Center of Mass, Extended Object

Similar analysis can be done for an extended object

Consider the extended object as a system containing a large number of particles

Since particle separation is very small, it can be considered to have a constant mass distribution

For x-axis, xCM (1/M)iximi

xCM = (1/M)xdm

rCM = (1/M) r dm 55

Center of Mass, position

The center of mass in three dimensions can be located by its position vector, For a system of particles,

is the position of the ith particle, defined by

For an extended object, CM

1dm

M r r

ˆ ˆ ˆi i i ix y z r i j k

CMr

1CM i i

i

mM

r r

ir

56

Center of Mass, Symmetric Object

The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry If the object has uniform density

57

Finding Center of Mass, Irregularly Shaped Object

Suspend the object from one point

The suspend from another point

The intersection of the resulting lines is the center of mass

58

Center of Gravity

Each small mass element of an extended object is acted upon by the gravitational force

The net effect of all these forces is equivalent to the effect of a single force acting through a point called the center of gravity If is constant over the mass distribution, the

center of gravity coincides with the center of mass

Mg

g

59

Center of Mass, Rod

Conceptualize Find the center of mass

of a rod of mass M and length L

The location is on the x-axis (or yCM = zCM = 0)

Categorize Analysis problem

Analyze Use equation for xcm

xCM = L / 2 why?

60

Center of Mass, Rod

61

22

2

0

L

M

Lxdx

xdmx

L

CM

L

M

Motion of a System of Particles

Assume the total mass, M, of the system remains constant

We can describe the motion of the system in terms of the velocity and acceleration of the center of mass of the system

We can also describe the momentum of the system and Newton’s Second Law for the system

62

Velocity and Momentum of a System of Particles

The velocity of the center of mass of a system of particles is

The momentum can be expressed as

The total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass

CMCM

1i i

i

dm

dt M r

v v

CM toti i ii i

M m v v p p

63

Acceleration of the Center of Mass

The acceleration of the center of mass can be found by differentiating the velocity with respect to time

CMCM

1i i

i

dm

dt M v

a a

64

Forces In a System of Particles

The acceleration can be related to a force

If we sum over all the internal forces, they cancel in pairs and the net force on the system is caused only by the external forces

CM ii

M a F

65

Newton’s Second Law for a System of Particles

Since the only forces are external, the net external force equals the total mass of the system multiplied by the acceleration of the center of mass:

The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the net external force on the system

ext CMMF a

66

Impulse and Momentum of a System of Particles

The impulse imparted to the system by external forces is

The total linear momentum of a system of particles is conserved if no net external force is acting on the system

ext CM totdt M d I F v p

0CM tot extM constant when v p F

67

Motion of the Center of Mass, Example A projectile is fired into the air

and suddenly explodes With no explosion, the

projectile would follow the dotted line

After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion

Use the active figure to observe a variety of explosions

68

Deformable Systems

To analyze the motion of a deformable system, use Conservation of Energy and the Impulse-Momentum Theorem

If the force is constant, the integral can be easily evaluated

0system

tot ext

E T K U

dt m

I p F v

69

Deformable System (Spring) Example

Conceptualize See figure Push on left block, it moves to

right, spring compresses At any given time t, the

blocks are generally moving with different velocities

The blocks oscillate back and forth with respect to the center of mass

Find the resulting speed vCM of the center of mass of the system

70

Spring Example, cont

Categorize Non isolated system

Work is being done on it by the applied force It is a deformable system The applied force is constant, so the acceleration

of the center of mass is constant Model as a particle under constant acceleration

Analyze Apply impulse-momentum Solve for vcm

71

Spring Example, final

Analyze, cont. Find energies

Finalize Answers do not depend on spring length, spring

constant, or time interval

72

Deformable System (Spring) Example

Impulse-momentum theorem

F t = Pf – Pi = 2mvCM – 0

t = (xCMf-xCMi)/vCM

= (L+x2+x1-L)/2vCM

= (x1+x2)/2vCM

F t = F(x1+x2)/2vCM = 2mvCM

vCM = F(x1+x2)/ 2m73

Rocket Propulsion

The operation of a rocket depends upon the law of conservation of linear momentum as applied to a system of particles, where the system is the rocket plus its ejected fuel

74

Rocket Propulsion, 2

The initial mass of the rocket plus all its fuel is M + m at time ti and speed v

The initial momentum of the system is

i = (M + m) p v

75

Rocket Propulsion, 3

At some time t + t, the rocket’s mass has been reduced to M and an amount of fuel, m has been ejected

The rocket’s speed has increased by v

76

Rocket Propulsion, 4

Because the gases are given some momentum when they are ejected out of the engine, the rocket receives a compensating momentum in the opposite direction

Therefore, the rocket is accelerated as a result of the “push” from the exhaust gases

In free space, the center of mass of the system (rocket plus expelled gases) moves uniformly, independent of the propulsion process

77

Rocket Propulsion, 5

(M+m)v = M(v+v) + m(v-ve)

Mv = vem

The increase in the exhaust mass dm corresponds to an equal decrease in the rocket mass, so dm = -dM.

Mdv = vedm = -vedM78

v-ve

f

i

f

i

M

M

e

v

v

dMM

vdv1

Rocket Propulsion, 6

The basic equation for rocket propulsion is

The increase in rocket speed is proportional to the speed of the escape gases (ve) So, the exhaust speed should be very high

The increase in rocket speed is also proportional to the natural log of the ratio Mi/Mf

So, the ratio should be as high as possible, meaning the mass of the rocket should be as small as possible and it should carry as much fuel as possible

ln if i e

f

Mv v v

M

79

Thrust

The thrust on the rocket is the force exerted on it by the ejected exhaust gases

The thrust increases as the exhaust speed increases

The thrust increases as the rate of change of mass increases The rate of change of the mass is called the burn rate

e

dv dMthrust M v

dt dt

80