Linear Momentum For an individual mass we define the linear momentum to be: From the 2 nd law we have: This is actually how Newton originally formulated.

Linear Momentum

For an individual mass we define the linear momentum to be:

vmp

From the 2nd law we have: am

dt

pd

Fnetdt

pd

This is actually how Newton originally formulated the 2nd law. The “ma” is a special case when m is not changing.

Linear Momentum of a System of Particles

For a system of many particles, we can define the “total linear momentum”:

...21 ppP

Then we can write:

...2211 vmvmP

CMCM VMdt

rdMP

The total linear momentum is a result of the total mass moving with the velocity of the CM

Let us differentiate the total momentum:

exttotalCM FaM

dt

Pd

If there are no external forces, then momentum is conserved:

fPPdt

Pd

00

This is a vector equation. If there are not external forces in say the x-direction, then Pxo = Pxf even if there are external forces in the y-direction.

Elastic Collisions in 1-D

1v 2v

1v 2v

22112211 vmvmvmvm fPP 0

Elastic means KE stays the same:

222

211

222

211 2

1

2

1

2

1

2

1vmvmvmvm

222111 vvmvvm

22222

21

211 vvmvvm

2222211111 vvvvmvvvvm

Dividing two equations gives (upon rearrangement):

1221 vvvv The relative velocity changes direction but keeps same magnitude.

Only true in 1-D, elastic collisions.

Example: Ping pong ball collides with stationary bowling ball.

211 vMvmmv

121 vvv Multiply by M and subtract

11 vMmvMm 1111 vvvv

Mm

Mm

Bounces back with same speed

Example: Bowling ball hits stationary ping pong ball.

211 vmvMMv

121 vvv Multiply by M and add

212 vMmMv

112 22

vvmM

Mv

Barry Bonds uses a light bat

A two dimensional collision

fPP

0

xfx PP 0

1v

1v

2v

coscos 221111 vmvmvm

yfy PP 0

sinsin0 2211 vmvm

Example: a 4 kg mass heading in the – y direction at 12 m/s collides and sticks to a 6 kg mass moving in the + x direction at 10 m/s. Find the magnitude and direction of the final velocity.

xfx PP 0 s

mvv xx 610106

yfy PP 0 s

mvv yy 8.410124

8.

6

8.4tan

22 8.46 v

Impulse

For a constant force, let is define a vector quantity called impulse as the product of the force times the time over which it acts:

tFJ

In one dimension, we need only worry about the sign. If the force is not constant during the time over which the force acts, we define through an integral:

2

1

t

t

dtFJ

Note the analogy to our definition of work. Of course a huge difference is work is a scalar and impulse is a vector. For any component, the impulse will be the area under the Fi vs t graph.

So why bother with impulse?Suppose we focus on the impulse delivered by the net force.

2

1

2

1

t

t

t

t

dtdt

pddtFnetJ

ppppdJ

p

p

12

2

1

curveunderareadtFJt

t

xx 2

1

Impulse – Momentum Theorem

p = pf – pi = F dttf

ti

The impulse of the force F acting on a particle equals the change in the

momentum of the particle.

J = p

Example: A 0.1 kg mass moving with a speed of 10 m/s along the x-axis collides head on with a stationary 0.1kg mass. The magnitude of the force between the two is shown below as a function of time. Find the final speed and direction of each mass.

areaJp

)]12)(10*20(2

11210*30)12)(10*20(

2

1[01. 333 fv

s

mv f 6

For 2nd block:

)]12)(10*20(2

11210*30)12)(10*20(

2

1[101. 333 fv

s

mv f 4

For 1st block:

Impulse is negative by Newton’s 3rd Law