Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 1

Chapter 7 Linear Momentum

From page 234

In Chapter 4 we learned how to determine the acceleration of an object by finding the net

force acting on it and applying Newton’s second law of motion. If the force happens to

be constant, then the resulting constant acceleration enables us to calculate changes in

velocity and position. Calculating velocity and position changes when the forces are not

constant is much more difficult. In many cases, the forces cannot even be easily

determined. Conservation of energy is one tool that enables us to draw conclusions about

motion without knowing all the details of the forces acting….

In this chapter we develop another conservation law. Conservation laws are

powerful tools. If a quantity is conserved, then no matter how complicated the situation,

we can set the value of the conserved quantity at one time equal to its value at a later

time….

The new conserved quantity, momentum, is a vector quantity, in contrast to

energy, which is a scalar. When momentum is conserved, both the magnitude and

direction of the momentum must be constant. Equivalently, the x- and y-components of

momentum are constant. When we find the total momentum of more than one object, we

must add the momentum vectors according to the procedure by which vectors are always

added.

Definition of linear momentum

An object with mass m and moving with velocity v

has linear momentum p

vp

m

Impulse-Momentum Theorem

Starting with Newton’s second law

aF

m

and using the definition of acceleration

t

va

we have

tm

vF

If the mass is constant, we can pull it inside the operator

t

m

)( vF

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 2

t

pF

ttotal FJp

This states that the change in linear momentum is caused by the impulse. The quantity

tFJ

is called the impulse. For situations where the force is not constant, we use the average force,

tavFJ

Restatement of Newton’s second law

There is a more general form of Newton’s second law

tt

pF

0lim

The net force is the rate of change of momentum.

Many Particles

If many particles m1, m2, etc. are moving with velocities ,, 21 vv

etc., the total linear momentum

of the system is the vector sum of the individual momenta,

i

ii

i

i m vpp

Linear momentum is a vector quantity. We have to use our familiar rules for vector addition

when dealing

with

momentum.

The

applications of

the impulse-

momentum

theorem are

unlimited. In

an automobile

we have

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 3

crumple zones, air bags, and bumpers. What do seat

belts do?

Here is another example:

The idea is to lengthen the time t during which the

force acts, so that the force is diminished while

changing the momentum a defined amount.

Problem 17. An automobile traveling at a speed of

30.0 m/s applies its brakes and comes to a stop in 5.0

s. If the automobile has a mass of 1.0×103 kg, what

is the average horizontal force exerted on it during

braking. Assume the road is level.

Solution: A FBD for the situation would be

We would use the impulse-momentum theorem

t Fp

To use a vector equation we need to take components

tFpxx

The change in momentum is

m/skg3000)m/s0.3)(kg1000(0 ixfxixfxx mvmvppp

N600m/skg600s5

m/skg3000 2

,

,

t

pF

tF

tFp

xxav

xav

xx

The average braking force acts to the left and is 600 N. Additionally, the work done on the car to

stop it is

mg

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 4

J1050.4)m/s30)(kg1000(00)( 52

2

1 ifnc KKUKW

and the corresponding power is

W000,90s0.5

J1050.4 5

t

WP

Conservation of Linear Momentum

Consider the collision between two pucks. When they collide, they exert forces on each other,

12F

and 21F

By Newton’s third law,

2112 FF

The total force acting on both pucks is then

02112 FFF

(Recall that since ay = 0 for both pucks, the y-component of the net force is zero.) By the

impulse-momentum theorem

0

tFp

Which leads to

fi pp

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 5

In a system composed of more than two objects, interactions between objects inside the system

do not change the total momentum of the system – they just transfer some momentum from one

part of the system to another. Only external interactions can change the total momentum of the

system.

The total momentum of a system is the vector sum of the momenta of each object in the

system

External interactions can change the total momentum of a system.

Internal interactions do not change the total momentum of a system.

The Law of Conservation of Linear Momentum

If the net external force acting on a system is zero, then the momentum of the system is

conserved.

fiext ppF

,0If

Linear momentum is always conserved for an isolated system. Of course, we deal with

components

fyiyfxixpppp and

I like to say that momentum is conserved in collisions and explosions.

Note: The total momentum of the system is conserved. The momentum of an individual particle

can change.

Problem 23. A rifle has a mass of 4.5 kg and it fires a bullet of mass 10.0 g at a muzzle speed of

820 m/s. What is the recoil speed of the rifle as the bullet leaves the gun barrel?

Solution: Draw a sketch of the initial and final situations.

Since only internal forces act, linear momentum is conserved.

fi pp

The motion is in the x-direction,

fxix pp

Initially, everything is at rest and pix = 0. The final momentum is the vector sum of the momenta

of the bullet and the rifle. From the diagram,

Initial Final

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 6

rrbbfx vmvmp

Using conservation of momentum

m/s82.1kg5.4

)m/s820)(kg010.0(

0

r

bbr

rrbb

fxix

m

vmv

vmvm

pp

The heavier the rifle, the smaller the recoil speed.

Center of Mass

From page 245:

We have seen that the momentum of an isolated system is conserved even though parts of

the system may interact with other parts; internal interactions transfer momentum

between parts of the system but do not change the total momentum of the system. We

can define a point called the center of mass (CM) that serves as an average location of the

system….

What if a system is not isolated, but has external interactions? Again imagine all

of the mass of the system concentrated into a single point particle located at the CM. The

motion of this fictitious point particle is determined by Newton’s second law, where the

net force is the sum of all the external forces acting on any part of the system. In the case

of a complex system composed of many

parts interacting with each other, the

motion of the CM is considerably

simpler than the motion of an arbitrary

particle of the system.

For the two particles pictured to the right, the

CM is

M

xmxm

mm

xmxmxCM

2211

21

2211

Notice that the CM is closer to the more

massive object.

For many particles the definition is generalized

to

M

m ii

CM

rr

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 7

The more useful component form is

M

xmx

ii

CM

M

ymy

ii

CM

M

zmz

ii

CM

Problem 38. Find the x-coordinate of the CM of the composite object shown in the figure. The

sphere, cylinder, and rectangular solid all have uniform composition. Their masses and

dimensions are: sphere: 200 g, diameter = 10 cm; cylinder: 450 g, length = 17 cm, radius = 5.0

cm; rectangular solid: 325 g, length in x-direction = 16 cm, height = 10 cm, depth = 12 cm.

Solution: Because the different pieces have uniform composition, the center of mass of each

piece is located at the geometrical center of that piece. (See the note at the top of page 240.) For

the sphere, ms = 200 g, xs = 5 cm. For the cylinder, mc = 450 g, xc = 10 cm + 8.5 cm = 18.5 cm.

For the rectangular solid, mr = 325 g, xr = 10 cm + 17 cm + 8 cm = 35 cm. The center of mass is

cm2.21

)g325()g450()g200(

)cm35)(g325()cm5.18)(g450()cm5)(g200(

rcs

rrccss

ii

CM

mmm

xmxmxm

M

xmx

Motion of the Center of Mass

How is the velocity of the CM related to the velocities of the various parts of the system of

particles?

During a short time interval t, each particle changes position by ir

. The CM changes

M

m ii

CM

rr

Divide each side by the time interval t

M

tm

t

ii

CM

r

r

In general, the velocity is defined as

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 8

tt

rv

0lim

Using the definition of velocity

iiCM

ii

CM

mM

M

m

vv

vv

The right hand side is the total linear momentum of the system of particles. We have the very

useful relation

CMtotal Mvp

For a complicated system consisting of many particles moving in different directions, the total

linear momentum of the system can be found from the total mass of the system and the velocity

of the center of mass.

(Page 247) … we have showed that, for an isolated system, the total linear momentum of

the system is conserved. In such a system, the equation above implies that the CM must

move with constant velocity regardless of the motions of the individual particles. On the

other hand, what if the system is not isolated? If a net force acts on a system, the CM

does not move with constant velocity. Instead, it moves as if all the mass were

concentrated into a fictitious point particle with all the external forces acting on that

point. The motion of the CM obeys the following statement of Newton’s second law:

CMext MaF

A complicated system is reduced to treating the system as a point particle located at its center of

mass reacting only to external forces!

Collisions in One Dimension

In general, conservation of momentum is not enough to predict what will happen after a

collision. Here is a collision between two objects with the same mass. Many other outcomes are

possible besides the two given,

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 9

Some vocabulary

A collision in which the total kinetic energy is the same before and after is called elastic.

See (a) above.

When the final kinetic energy is less than the initial kinetic energy, the collision is said to

be inelastic. Collisions between two macroscopic objects (for example, billiard balls) are

generally inelastic to some degree, but sometimes the change in kinetic energy is so small

that we treat the collision as elastic.

When a collision results in two objects sticking together, the collision is perfectly

inelastic. See (b) above. The decrease of kinetic energy is a perfectly inelastic collision

is as large as possible (consistent with the conservation of momentum).

Problem-Solving Strategy for Collisions Involving Two Objects (pages 251-2)

1. Draw before and after diagrams of the collision.

2. Collect and organize information on the masses and velocities of the two objects before

and after the collision. Express the velocities in component form (with correct algebraic

signs).

3. Set the sum of the momenta of the two before and after the collision equal to the sum of

the momenta after the collision. Write one equation for each direction:

fxfxixix vmvmvmvm 22112211

fyfyiyiy vmvmvmvm 22112211

4. If the collision is known to be perfectly inelastic, set the final velocities equal:

fyfyfxfx vvvv 2121 and

5. If the collision is known to be perfectly elastic, then set the final kinetic energy equal to

the initial kinetic energy:

2

22212

11212

22212

1121

ffii vmvmvmvm

or, for a one-dimensional collision, set the relative speeds equal:

)( 1212 fxfxixix vvvv

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 10

6. Solve for the unknown quantities.

Example: A one-dimensional elastic collision.

Conservation of momentum gives

ffii vmvmvmvm 22112211

Elastic collision means that kinetic energy is conserved

2

222

12

112

12

222

12

112

1ffii vmvmvmvm

Cancelling the ½ and grouping like masses on the same side of the equation,

)()(2

2

2

22

2

1

2

11 iffi vvmvvm

Doing the same with the linear momentum equation

)()( 222111 iffi vvmvvm

Divide the two equations

iffi

if

if

fi

fi

vvvv

vvm

vvm

vvm

vvm

2211

222

2

2

2

22

111

2

1

2

11

)(

)(

)(

)(

This can be rewritten as

ffii vvvv 1221

The equation states that m1 approaches m2 with the same speed as m2 moves away from m1 after

the collision. Important result. Solve the above for v2f

iiff vvvv 2112

and substitute into the momentum equation

v1i v2i m1 m2

v1f v2f m1 m2

initial final

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 11

iif

fii

iiff

iiff

ffii

vmm

mv

mm

mmv

vmmvmvmm

vmvmvmvm

vvvmvm

vmvmvmvm

2

21

21

21

211

12122121

22121211

211211

22112211

2

)(2)(

)(

Using very similar reasoning, or swapping the 1 and 2 subscripts, we can find an expression for

the final speed of m2

iif vmm

mmv

mm

mv 2

21

121

21

12

2

Be careful when you use these equations. A mass moving to the left would have a negative

speed.

If m2 is initially at rest, v2i = 0 and we have

if

if

vmm

mv

vmm

mmv

1

21

12

1

21

211

2

After the collision, m2 moves to the right.

If m1 > m2, m1 continues to move to the right.

If m1 = m2, m1 stops and m2 to move to the right with speed v1i.

If m1 < m2, m1 bounces back to the left.

Example: The masses are m1 = 2 kg and m2 = 3 kg and they collide elastically in one dimension.

Before the collision, m1 is moving at 4 m/s to the right and m2 is moving at 1 m/s to the left.

What are the velocities of m1 and m2 after the collision?

Solution: Use the equations derived above,

m/s2

)m/s1(kg3kg2

kg)3(2)m/s4(

kg3kg2

kg3kg2

22

21

21

21

211

iif v

mm

mv

mm

mmv

Lesson 9: Impulse, Momentum, Center of Mass, Collisions (Sections 7.1-7.7)

Lesson 9, page 12

m/s3

)m/s1(kg3kg2

kg2kg3)m/s4(

kg3kg2

)kg2(2

22

21

121

21

12

iif v

mm

mmv

mm

mv

To review relative velocity, the velocity of 2 relative to 1 is

1221 GG vvv

Taking components,

m/s5m/s)4(m/s11221 xGGxx vvv

As seen from m1, m2 is moving to the left at 5 m/s. After the collision,

m/s5m/s)2(m/s31221 xGGxx vvv

The masses approach at 5 m/s and separate at 5 m/s.

1 m/s 4 m/s m1 m2

3 m/s 2 m/s m1 m2