Example: An eagle of mass, mA = 6.0 kg moving with speed vA = 5.0 m/s is on collision course with a second eagle of mass mB = 4.0 kg moving at vB = 10.0 m/s in a direction perpendicular to the first. After they collide, they hold onto one another. At what speed they moving after the collision?

mA=6.0 kgvA=5.0 m/smB=4.0 kgvB=10.0 m/sv - ?

Av

Bv

Ap

Bp

smkgsmkgpmp

smkgsmkgpmp

mmm

BBB

AAA

AA

/0.40/0.100.4

/0.30/0.50.6

kg 10.0kg 4.0kg 6.0

AA ppp

22BA ppp

sm

kg

smkgsmkg

m

pp

m

pv BA /5

10

/0.40/0.30 2222

1. Increases2. Does not change3. Decreases

Example: Suppose rain falls vertically into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating water, the speed of the cart

Mass is increasing P = mv must be conserved (Fext = 0) Speed must decrease

Ballistic PendulumA simple device for measuring the speed of projectile

M

01 U

h

;2

21

1

mvK ghmMU )(3

mM

;03 K

)()( 3322 UKUK ghmMvMm

)(2

)( 22

ghm

Mmv 2

)(1

1,vm mM

p1=mv1

2. After collision:

p2=(m+M)v2

;2

)( 22

2

vMmK

02 U

1. Before collision: 3. At the highest point:

p3=0

p1=p2 mv1 = (m+M)v2 21

)(v

m

Mmv

2v

h1

m1

m2 h2

2

211

11

vmghm 11 2ghv

vmmvm )( 2111 121

11

21

1 2ghmm

mv

mm

mv

221

221

2ghmm

vmm

1

2

21

12

2 2h

mm

m

g

vh

Example:

p

p’ p+p’

rubber

p

putty

Example: You want to knock down a large bowling pin by throwing a ball at it.

You can choose between two balls of equal mass and size. One is made of

rubber and bounces back when it hits the pin. The other is made of putty and

sticks to the pin. Which ball do you choose?

A. The rubber ball.

B. The putty ball.

C. It makes no difference.

Elastic Collisions (1D)

The kinetic energy of the system is conserved: after the collision it is

the same as that before

A

A

A

AB B

BB

x

x

x

x

Before Before

AfterAfter

vv

AvBv Av

Bv

Elastic Collisions (1D)

BBAABBAA vmvmvmvm ''

2

'

2

'

22

2222BBAABBAA vmvmvmvm

BA

BBABAA mm

vmvmmv

2'

BA

AABABB mm

vmvmmv

2'

'' )( BBBAAA vvmvvm

2222 '' BBBAAA vvmvvm

(1)

(2)

'' )( BBAA vvvv '')( BABA vvvv (2a)

Example: A steel ball with mass m1 = 1 kg and initial speed v0 collides head-on with another ball of mass m2 = 2 kg that is initially at rest. What are the final speeds of the balls?

1)Conservation of momentum:

2) Conservation of energy:222

1212

1202

1 0 vvv

We have 2 equations and 2 unknown: v1 and v2.

It is more convenient to use eq. (2a) instead of eq.(2)

2a) Relative velocity:

Velocity of 1 relative to 2 before the collision

210 0 vvv

Velocity of 1 relative to 2 after the collision

210 20 vvv

Adding eq. 1and 2a: 20 32 vv 032

2 vv 031

1 vv

Example: Carefully place a small rubber ball (mass m) on top of a much bigger basketball (mass M>>m) and drop these from some height h. What is the velocity of the smaller ball after the basketball hits the ground, reverses direction, and then collides with small rubber ball?

Remember that relative velocityrelative velocity has to be equal before and after collision! Before the collision, the basketball bounces up with vv and the rubber ball is coming down with vv, so their relative velocity is –2v–2v. After the collision, it therefore has to be +2v+2v!!

This means that, after collision, the velocity of the smaller ball after is 3v.

v

v

m

M v

vv

3vV?

Relative speed is 2v Here too!

A. One ball rises on the right, but higher than h

pi = pf 2mvi = mvf vf = 2vi ,

but then kinetic energy would not be conserved: KEi = 2½mvi

2 = mvi2 KEf = ½mvf

2 = 2mvi2

B. Two balls rise on the right at height h

C. A ball will rise on each side to height h

pi = pf 2mvi = 2mvf vf = vi.

KEi = 2 ½mvi2 = mvi

2 KEf = 2 ½mvf2 = mvi

2 Ok!

If two balls on the left are pulled to a certain height h and released, what happens?

Newton’s craddle (a row of adjacent steel-ball )

Same height means |vi|= |vf| (from conservation of energy),but this violates conservation of momentum: pi = 2mvi ; pf = mvi – mvi = 0

In 2D and 3D the initial velocities of the two particles dose not determinethe final velocities. You also need to know the direction of one of the final particles.

''BBAABBAA vmvmvmvm

2

'

2

'

22

2222BBAABBAA vmvmvmvm

Elastic Collisions in 2D and 3D

2222

),,(

AzAyAxA

AzAyAxA

vvvv

vvvv

Kinetic energy: m

p

m

mvmvK

222

222

3 unknowns: M, Px, Py

m M (at rest)

pi

pf

P

A particle of unknown mass M is initially at rest. A particle of known mass m is “shot” against it with initial momentum pi. After the collision, the momentum of the particle of known mass is measured again, and it is pf. If the collision is elastic, that’s all we need to determine M and the final momentum of the target, P.

2D Elastic: Nuclear scattering

3 equations: conservation of momentum in the x directionconservation of momentum in the y directionconservation of kinetic energy

22 2

(2 equations)

2 2 2

i f

fi

p p P

pp Pm m M

222fi pp

m

MP

222ii ppPP

2

2 2

( )

( )i f

i f

p pM m

p p

i fP p p