1

Warm Up:

Which of the following has the greatest momentum?

a) A 100 g bullet moving at 300 m/sb) A 5000 kg train car moving at 1.5 m/sc) A 75 kg motorcyclist moving at 35 m/s

Linear Momentum

What is Linear Momentum?

“Trains are big. They’re hard to stop.”

p = Mv

TRAIN BULLET…Also hard to stop

p ≡ mv

p = mv

More on Momentum

• Units: kg·m/s• Momentum is a vector: px=mvx, py=mvy, etc.• Like KE affected by mass and velocity• UNLIKE KE increases linearly with both mass

and velocity• UNLIKE KE, can be negative or positive!

Why Study Momentum?Another powerful way to more

efficiently analyze systems of interacting objects:

ESPECIALLY COLLISIONS

and EXPLOSIONS

2

Force and Momentum: Derivation

dtpd

F�

�=�

Recall Newton’s Second Law: F=ma

dtvd

mFdtvd

a�

��

� =�=

If mass is constant, build it into the derivative.

So force and momentum are related…

dtpd

F�

�=� What if ΣF=0?

0=dtpd� constant=p

�

Conservation of Linear Momentum

Systems with no net external forces have constant momentum. (Impulse approximation helps here.)

pi = pf

2-Particle System

dtpd

F 212

��

=

p1

p2

F21F12m1

m2

system

dtpd

F 121

��

=

Internal Forces

Newton’s 3rd Law gives:F12 = −−−−F21 ΣFinternal = 0

Conservation of Linear Momentum

dtpd

dtpd

FFF 21intexttot

�����

+=+=� ��

dtppd )(

0 21�� +=

If there is no net external force, then

0 0

21 ppptot��� += is conserved.

3

In general

�� = exttot FF��

ΣFinternal = 0 Why?

0ext� =F�

If: Then:

Σpinit = Σpfinalsystem system

Note: this is 3 equations: px, py, pz

= constant

Conservation of Linear Momentum

Example two particle system:

A bullet with mass 35 g moving at 320 m/s embeds itself in a 2.40 kg wooden block sitting on a frictionless surface. How fast do the bullet and block move together after the collision?

mbulletvbullet,i= (mbullet+mblock)vf

Define system: bullet and block, so

pi = pf

.035(320)= (2.4+.035)vf vf = 4.6 m/s

demo

dp = F dt

t

F

∆p = pf −−−− pi = �f

i

dtF�

≡≡≡≡ Iimpulse

ΙΙΙΙ = ∆pImpulse-momentum theorem:

An Impulse is a change in momentum:- The effect of a force acting over a time- The agent of momentum transfer.- Handy for time-varying forces.

What if force �0? Impulse approximation

t

FConsider a very large force acting over a very short time.ex: hitting a hockey puck with a stick.

During the short time of the impulse from the large force, we can safely ignore other forces acting on the puck.

F

Useful to approximate the impulse as due to a constant force F acting over the same time ∆∆∆∆t.

Applications: collisions & explosions! demo

∆t

Ι = F ∆t

4

Impulse Example

A 60.0 g tennis ball is initially horizontally east at 50.0 m/s. A player hits the ball with a racket. The ball is in contact with the racket for 35 ms. After the hit, the ball is moving west at 40.0 m/s. What average force did the racket exert on the ball?

Impulse =�p=pf- pi - .060 (50) =-5.4 kg m/s

But, Impulse also =F�t => F(0.035 s)=-5.4 kg m/s=> F= -154 N

Negative => WestwardNote: Compare with force/impulse to stop ball…

=.060 (-40)

Collisions

Systems of particles interacting via impulsive forces (large, short time)

initialm1 m2

v1i v2i

finalm2

v1f v2f

m1

Choose system.

Impulse approximation:

Apply linear momentum conservation.

Two massesOnly consider large contact forces;Instants just before & after collision

1D collision:initial

m1 m2

v1i v2i

finalm2

v1f v2f

m1

Linear momentum conservation:

ΣPi = ΣPf

m1v1i – m2v2i = m1v1f + m2v2f

Given v1i and v2i we have one equation, but two unknowns.

Types of collisions:

�Elasticlinear momentum is conservedtotal kinetic energy is conserved

�Inelasticlinear momentum is conservedtotal kinetic energy is NOT conservedTotally inelastic collisions: particles stick together. Where does the energy go? demo

5

1D Perfectly Inelastic Collisionsinitial

m1 m2

v1i v2i

finalm2

vf

m1+m2

v1f = v2f = vf

m1v1i – m2v2i = (m1+m2)vf

m1v1i – m2v2i = m1v1f + m2v2f

vf = (m1v1i - m2v2i)/(m1+m2)

Caution with signs!

Momentum conservation:

One more inelastic example: Explosion!

pi=pf

2.0 kg, v1f =3.6 m/s

3.0 kg, v2f =?

5.0 kg vi=0

0 = 2(-3.6) +3(v2f) => v2f = 2.4 m/s

1D Elastic Collisionsinitial

m1 m2

v1i v2i

finalm2

v1f v2f

m1

m1v1i – m2v2i = m1v1f + m2v2f

Momentum conservation:

Kinetic Energy is conserved:

½m1v21i + ½m2v2

2i = ½ m1v21f + ½ m2v2

2f

Two equations ����can solve for two unknowns.

1D Elastic Collisions cont.Some algebra shows:

v1i – v2i = − (v1f − v2f)

iif

iif

mmmm

mmm

mmm

mmmm

221

121

21

12

221

21

21

211

vv2

v

v2

vv

���

����

+−+��

�

����

+=

���

����

++��

�

����

+−=

Use these: