1

PHYS 218sec. 517-520

ReviewChap. 8

Momentum, Impulse, and Collisions

2

What you have to know

• Momentum and impulse• Momentum conservation• Various collision problems

– The typical examples where momentum conservation is used are collisions!

• Center of mass

• Section 8.6 is not in the curriculum.

3

Newton’s 2nd law of motion

: Newton's 2nd law when the mass is constant.F ma m=år r

This cannot be applied to the problems where mass m is changing.E.g.Rocket propulsion problem. (See Sec 9.6)

Original form of the 2nd law of motion written by Newton

( )dF mv

dt=å

r r( )when is constant, it becomes

d dvm F mv m ma

dt dt= = =å

rr r r

This defines the (linear) momentum ( . momenta)plr rp = mv Newton's 2nd law can be written as

dpF

dt=å

rrMomentum is a vector quantity.

4

The impulse-momentum theorem

Impulse ( )J F t= Dår r

Impulse is a vector quantity.

2 1

2 1

time interval

if the force is acting

between and

t t

F

t t

-

å2 1

2 1

If the net force is constant,

p pF

t t

-=

-år r

Note that Newton's 2nd law is dp

Fdt

=årr

( )( )2 1 2 1F t t p p- = -år r

2 1J p p= -r r r ( )since J F t= Då

r r

Impulse-momentum theoremThe change in momentum of a particle

during a time interval equals the impulse of the net force.

5

Impulse

2 2 2

1 1 12 1

If forces are changing with time,t t t

t t t

dpFdt dt dpJ p p

dtJ= - == = =åò ò ò

r rrr r rr

General definition Impulse-momentum theorem

Impulse: area under the curve of net force in F-t graph

Få

tSame area Same impulse

Large force for a short time

Smalle force for a longer time

6

Momentum vs. Kinetic Energy

2 1

Impulse-momentum theorem

change in a particle's momentum

is due to impulse

depends on

J p p= -

®

Þ time

2 1

Work-energy theorem

change in a particle's kinetic energy

is due to work

depends on

W K K= -

®

Þ distance

Integral form of Newton’s 2nd law

Differential form of Newton’s 2nd lawdp

Fdt

=årr

identical

7

Ex 8.2

x

A ball hits a wall

before

after

1v

2v

1 20.4 kg, 30 m/s, 20 m/s, 0.01 s

impulse? average force that the wall exerts on the ball?

m v v t= =- = D =

( )

( ) ( ) ( ){ }

1 1

2 2

2 1 2 1 2 1

Initial momentum:

Final momentum:

Impulse

0.4 kg 20 m/s 30 m/s

20 N s

p mv

p mv

J p p mv mv m v v

=

=

= = - = - = -

= ´ - -

= ×

average force:

20 N s2000 N

0.01 s

av

av

J F t

JF

t

= D

×Þ = = =

D

8

Ex 8.31 2kicking a soccer ball: 0.4 kg, 20 m/s, 30 m/s,

45 , 0.01 s

impulse of the net force? average net force?

m v v

tq

= = =

= ° D =

q

Before

After

1v

2v

x

y

This is a 2-dim. motion.First decompose the momentum into

x and y directions

( )

( )

1 1 1

2 22 2 2 2

22 1 2 1 1

22 1 2 1

20 m/s, 0

cos 45 , sin 452 2

The impulse is

16.5 kg m/s2

0 8.5 kg m/s2

x y

x y

x x x x x

y y y y y

v v v

v vv v v v

vJ p p m v v m v

vJ p p m v v m

=- =- =

= °= = °=

Þ

æ ö÷ç= - = - = + = ×÷ç ÷÷çè ø

æ ö÷ç= - = - = - = ×÷ç ÷÷çè ø

, ,

,2 2 3, ,

,

1650 N, 850 N

8501.9 10 N, tan 27

1650

yxav x av y

av yav av x av y

av x

JJF F

t tF

F F FF

q q

= = = =D D

Þ = + = ´ = = Þ = °

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Conservation of Momentum

Internal forces

External force

Isolated system: External force = 0

A B

B on AFr

A on BFr

B on A A on B,A Bdp dpF F

dt dt= =

r rr r

Newton’s 2nd law

Newton’s 3rd law

B on A A on B 0F F+ =rr r

( )B on A A on B 0A BA B

dp dp dF F p p

dt dt dt+ = + = + =

r r rr r r r

You can add these forces to obtain the net (internal) force for

the systemThe sum of these forces cannot be

applied to a single object.

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Conservation of Momentum

( )B on A A on B 0

If we define , 0

A B

A B

dF F p p

dt

dPP p p

dt

+ = + =

= + =

rr r r r

rrr r r

Total momentum

Total momentum of an isolated system is conserved.

When the system contains many particles,

the total momentum is

A B C A A B B C CP p p p m v m v m v= + + + = + + +r r r r r r r

L L

constantP =r

Momentum conservation gives a vector equation.

Therefore, each component of the total momentum is conserved!

constant

constant

constant

x Ax Bx Cx

y Ay By Cy

z Az Bz Cz

P p p p

P p p p

P p p p

= + + + =

= + + + =

= + + + =

L

L

L

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Ex 8.4 Recoil of a rifle

Before

After

x

x

Rifle + bullet

Rifle bullet

RvBv

Note that momentum conservation is valid for an

isolated system.So it is useful when the system

changes. Therefore, draw diagrams for “before” &

“after” the event

mass of the rifle 3 kg, mass of the bullet 5 g

300 m/s. Then (recoil speed of the rifle)?

Momentum and kinetic energy of the rifle? Of the bullet?

R B

B R

m m

v v

= =

=

( )

Use momentum conservation

Initial momentum: 0 ( the rifle+bullet is at rest)

Final momentum:

gives 0

0.005 kg300 m/s 0

3 kg

initial

final R R B B

initial final R R B B

BR B

R

P

P m v m v

P P m v m v

mv v

m

=

= +

= + =

æ ö÷çÞ =- =- ÷́ =-ç ÷ç ÷çè ø

Q

.5 m/s

change the unit so that the both masses have the same unit

The negative sign means that the rifle is moving back!

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Ex 8.4 (cont’d)

( ) ( )

( ) ( )

( ) ( )22

momentum of the bullet

0.005 kg 300 m/s 1.5 kg m/s

momentum of the rifle

3 kg 0.5 m/s 1.5 kg m/s

kinetic energy of the bullet

1 10.005 kg 300 m/s 225 J

2 2kinetic energy of the rif

B B B

R R R

B B B

p m v

p m v

E m v

= = ´ = ×

= = ´ - =- ×

= = ´ =

( ) ( )22

le

1 13 kg 0.5 m/s 0.375 J

2 2R R RE m v= = ´ - =

The sum of these two momenta vanishes as it should be.

The kinetic energy is NOT conserved.

The kinetic energy before the event was 0.

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Ex 8.5 Collision along a straight line

A B

A B

Before

After

1Av 1Bv

2Av 2Bv

1 1

2 2

0.5 kg, 0.3 kg

initial speeds: 2.0 m/s, 2.0 m/s

final speeds: 2.0 m/s, ?

A B

A B

B A

m m

v v

v v

= =

= =-

= =

If you have a collision problem, always use momentum conservation.

initial 1 1

final 2 2

initial final

1 1 2 2

1 12

total momentum before the collision:

total momentum after the collision:

momentum conservation gives

A A B B

A A B B

A A B B A A B B

A A B B B BA

P m v m v

P m v m v

P P

m v m v m v m v

m v m v m vv

= +

= +

=

+ = +

+ -Þ =

( ) ( ) ( ) ( ) ( ) ( )20.5 kg 2.0 m/s 0.3 kg 2.0 m/s 0.3 kg 2.0 m/s

0.5 kg

0.4 m/sAm

´ + ´ - - ´=

=-

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Ex 8.6 Collision in a horizontal plane

Before

After1Av 1 0Bv =

2Av

2Bv

x

x

y

a

b

1

1

2 2

20 kg, 12 kg

initial speeds: 2.0 m/s in the (+) -direction,

0 m/s

final speeds: 1.0 m/s with 30 , ?

A B

A

B

A A

m m

v x

v

v va

= =

=

=

= = ° =r

Again, we use momentum conservation.

1 1 2 2

1 1

by momentum conservation

A A B B A A B B

A A x B B x

m v m v m v m v

m v m v

+ = +

Þ +

r r r r

2 2

1

in -directionA A x B B x

A A y

m v m v x

m v

= +

1B B ym v+ 2 2

1 2 1 22

2 22

in -direction

cos1.89 m/s

sin0.83 m/s

A A y B B y

A A x A A x A A A AB x

B B

A A y A AB y

B B

m v m v y

m v m v m v m vv

m m

m v m vv

m m

a

a

= +

- -Þ = = =

=- =- =-

Also note that momentum is a vector quantity.

2 22 2 2

2

2

2.1 m/s,

arctan 24

B B x B y

B y

B x

v v v

v

vb

= + =

æ ö÷ç ÷= =- °ç ÷ç ÷çè ø

Use the numbers above

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Momentum conservation and collisions

The key to solve collision problems is Momentum conservation.

1. At the instant of collision, very complex and very large forces are acting on the bodies.

2. Then the other forces can be neglected and the system is almost an isolated system. momentum conservation can be applied

3. The forces are very complex and we do not know much about the forces. momentum conservation can give useful information for the motions of the colliding bodies

Elastic collision

If the forces between the colliding bodies are conservative, no mechanical energy is lost. The total kinetic energy of the system is conserved.

For elastic collisions, the total momentum and the total kinetic energy of the system are conserved.

initial final initial finaland K K P P= =r r

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Inelastic collision

In this collision, the total kinetic energy of the system after the collision is less than before the collision.

initial final initial finalbut K K P P> =r r

The forces during the collision are non-conservative.

Completely inelastic collision

This is a special case of inelastic collisions.

This is where the colliding bodies stick together and move as one body after the collision.

1,final 2,final

initial final

,

the mass of the combined body after the collision is

But we still have

A B

v v

m m

P P

=

+

=

r r

r r

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Completely inelastic collisions

What happens to momentum and kinetic energy in a completely inelastic collision of two bodies

A B

A B

Before

After

1Avr

1Bvr

2 2 2A Bv v v= =r r r

( ),1 ,1 2

momentum conservation gives

A A B B A Bm v m v m m v+ = +r r r

Consider a 1-dim collision where vB1 = 0

,1 ,1

momentum conservation:

A A B Bm v m v+ ( ) 2

2 ,1

A B

AA

A B

m m v

mv v

m m

= +

Þ =+

( )

( ) ( )

21 ,1 ,1

2

2 22 ,1 ,1

2

1

Kinetic energies

10

2

1 1

2 2

< 1

A A B

AA B A A B A

A B

A

A B

K m v v

mK m m v m m v

m m

K m

K m m

= =

æ ö÷ç ÷= + = + ç ÷ç ÷ç +è ø

Þ =+

Q

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Ex 8.8 The ballistic pendulum

This is a device which can measure the speed of a bullet.

y

Before collision

Immediatelyafter collision

Swing of the bodyafter collision

Analyze this event in two stages !

Collision processOne-body

motion

This gives a relation between v1 and v2.

This gives a relation between v2 and the height y.

(A) (B)1v

2v

mass of the bullet , mass of the wood blockB Wm m= =

0v =

2v

So, if you know y, you can know v2 and v1

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Ex 8.8 (cont’d)

The first stage: (A) collision

( )1 2

1 2

momentum conservation

completely inelastic scattering & is at rest initially

So we have

B B W

W

B W

B

m v m m v

m

m mv v

m

Þ = +

æ ö+ ÷ç ÷=ç ÷ç ÷çè ø

Q

The second stage: (B) one-body motion

1 1Energy conservation K U+ 2K=

( ) ( )

2

22 2

12

2 B W B W

U

m m v m m gy v gy

+

+ = + Þ =

Combining the two

1 2B W

B

m mv gy

m

+= By measuring the height y, you can know the

speed of the bullet

20

Ex 8.9 Automobile collision

x

y

Tvr

Cvr

Cm

Tm

Before

x

y

T Cm m+

vr

After

Completely inelastic collision

( ) ( )ˆ ˆ2000 kg, 10 m/s ; 1000 kg, 15 m/sT T C Cm v m v= = = =i jr r

initial

initial, ,

Total momentum:

-component:

C C T T

x C C x

P m v m v

x P m v

= +

=

r r r

, , initial,y , ,; -component: T T x T T x C C y T T ym v m v y P m v m v+ = = +

( ) ( )

( )

,

final initial , ,

22 4final , ,

final, ,

final, ,

final final

ˆ ˆ

2.5 10 kg m/s

tan 0.75 37

8.3 m/s

C C y

T T x C C y

T T x C C y

y C C y

x T T x

C T

m v

P P m v m v

P m v m v

P m v

P m v

V P m m

q q

=

= = +

Þ = + = ´ ×

Þ = = = Þ = °

= + =

i jr r

q

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Elastic collisions

initial final initial finalIn an elastic collision, and K K P P= =r r

Consider 1-dim elastic collision.

A B

A B

Before

After

1Av 1Bv

2Av 2Bv

1 1 2 2

2 2 2 21 1 2 2

Momentum conservation

Kinetic energy is conserved for elastic collision

1 1 1 1

2 2 2 2

A A B B A A B B

A A B B A A B B

m v m v m v m v

m v m v m v m v

+ = +

+ = +Am Bm

1 2 1 2

If the masses are known, the unknown quantities are

, , ,A A B Bv v v v

These give TWO equations.

1 1You can set up the initial condition, then you can know the initial velocities and A Bv v

You now have TWO unknowns (the final velocities).

Therefore, you can completely determine the unknowns.

22

Elastic collisions: one body initially at rest

1

1 2 2

2 2 21 2 2

0. Then we have

1 1 1

2 2 2

B

A A A A B B

A A A A B B

v

m v m v m v

m v m v m v

=

= +

= +

rearrange( )

( ) ( )( )2 1 2

2 2 22 1 2 1 2 1 2

B B A A A

B B A A A A A A A A

m v m v v

m v m v v m v v v v

= -

= - = - +rearrange

( ) ( )

2 1 2

1 2 1 2

2 1 2 1

Using the first equation, the second equation gives

This relation is used to rewrite the first equation as

2 and

B A A

B A A A A A

A B AA A B A

A B A B

v v v

m v v m v v

m m mv v v v

m m m m

= +

+ = -

-Þ = =

+ + A B

A B

Before

After

1Av 1 0Bv =

2 0Av =2 1B Av v=

2 2 1

If ,

0 and A B

A B A

m m

v v v

=

Þ = =

23

AB

A

Before

After

1Av 1 0Bv =

2 1A Av v» - 2 0Bv »

2 1 2

If ,

and 0A B

A A B

m m

v v vÞ » - »

=

B

AB

B

Before

After

1Av 1 0Bv =

2 1A Av v» 2 12B Av v»

2 1 2 1

If ,

and 2A B

A A B A

m m

v v v vÞ » »

?

A

1. The motion of the heavy object doesn’t change.

2. The light object moves to the opposite direction but with same speed.

1. The motion of the heavy object doesn’t change.

2. The light object moves in the direction of the heavy one but (two times) faster

than the heavy one.

24

Elastic collisions & relative velocity

/ / / /

/ / /

relative velocity: , where is the velocity of measured by .

Therefore, A P A B B P X Y

A B A P B P

v v v v X Y

v v v

= +

= -

2 1 2

1 2 2

From the slide (p.22), B A A

A B A

v v v

v v v

= +

Þ = - The velocity of B relative to A after the collision

( ) ( )1 1 1A B Av v v=- - =- - The negative of the velocity of B relative to A before the collision

1

1

we have used that 0,

but this is true for 0B

B

v

v

=

¹ ( )2 2 1 1

In general,

for elastic collisions.B A B Av v v v- =- -

r r r r

If you watch the motion of A sitting on B, then you will see:A is approaching to B with speed v before the collision.

After the collision, A is moving away from B with the same speed

25

Ex 8.12 Two-dim elastic collision (If this is a head-on collision, it is the same as the 1-dim collision.)

Before

After1Av 1 0Bv =

2Av

2Bv

x

x

y

a

b

FOUR unknowns after collision (two velocity vectors)

Momentum conservation gives two eqs.Kinetic energy conservation gives one eq.

So totally THREE equations.

To uniquely determine the final velocities, we need ONE more information for the

final velocities

26

Ex 8.12 (cont’d) ( )1 1

2

2

0.5 kg, 0.3 kg, 4.0 m/s , 0

2.0 m/s

Then what are , , and ?

A B A B

A

B

m m v v

v

v a b

= = = =

=

ir rr r

2 22 2 2 2 1 2

1 2 2 2 2

This is an elastic collision, so

1 1 14.47 m/s

2 2 2

i f

A A A AA A A A B B B B

B

K K

m v m vm v m v m v v v

m

=

-= + Þ = Þ =

1 2 2

2 2

2 2

momentum conservation

-direction: 2 cos 1.34cos

-direction: 0 0 sin 1.34sin

By solving the coupled equations (use sin cos 1)

36.9 , 26.6

A A A A x B B x

A A y B B y

x m v m v m v

y m v m v

a b

a b

q q

a b

= + Þ = +

= + Þ = -

+ =

= ° = °

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Center of mass

We now introduce the concept of center of mass (CM)

( ), ,i i i ir x y z=r

i-th particle

( )

1 1 2 2 3 3 4 4

1 2 3 4

1 1 2 2 3 3 4 4

1 2 3 4

The center of the mass of the system has

the position , , defined as

where is the t

cm cm cm

i i i icm

i

i i i icm

i

x y z

m x m xm x m x m x m xx

m m m m m M

m y m ym y m y m y m yy

m m m m m M

M

+ + + += = =

+ + + +

+ + + += = =

+ + + +

å ååå åå

LL

LL

otal mass of the system,

iM m=å

Then,

1 : the postion of the center of mass of the systemcm i ir m r

M= å

r r

28

Motion of the center of mass

( ) ( )

1Since ,cm i i

icm cm i i i i i i

r m rM

drd dM v r m r m m v p P

dt dt dt

=

= = = = = =

å

å å å å

r r

r rr r r r r

Total momentum of the systemMomentum of the CM

You can represent an extended object as a point-like particle; the total mass of the

object is at its Center of Mass (CM)

( )

If there is no external force, is constant, so is the velocity of the CM

const.

If there is net external force,

cm

cmcmext cm

P

Pv

M

d Mvdv dPF M a M

dt dt dt

= =

= = = =å

r

r

rrrr r

29

Motion of the CM

`

When a shell explodes

This is the trajectory of the shell if it doesn’t explode.