Linear Momentum What is Linear Momentum? “Trains are big. They’re hard to stop.”

p ≡ mv TRAIN

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

BULLET…Also hard to stop

p = mv

Why Study Momentum? Another powerful way to more efficiently analyze systems of interacting objects:

• UNLIKE KE increases linearly with both mass

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

ESPECIALLY COLLISIONS

and EXPLOSIONS

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Force and Momentum: Derivation Recall Newton’s Second Law:

a=

dv dt

F =m

If mass is constant, build it into the derivative.

F=ma

dv dt

F =

dp dt

So force and momentum are related… F=

dp dt

What if ΣF=0?

dp =0 dt

p = constant

Conservation of Linear Momentum pi = pf

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

2-Particle System p1 m1

F21

Conservation of Linear Momentum dp1 dt dp2 F12 = dt

If there is no net external force, then

Internal Forces

0=

F21 =

F12

p2 m2

system

Newton’s 3rd Law gives: F12 = −F21

ΣFinternal = 0

Ftot =

Fext +

Fint = 0

0

dp1 dp2 + dt dt

d ( p1 + p2 ) dt

ptot = p1 + p2

is conserved.

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In general

Example two particle system:

ΣFinternal = 0 Ftot =

If:

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? demo

Why?

Fext

Fext =0

Then:

Define system: bullet and block, so

Σpinit = Σpfinal = constant

pi = pf mbulletvbullet,i= (mbullet+mblock)vf

system

system

vf = 4.6 m/s

.035(320)= (2.4+.035)vf

Conservation of Linear Momentum

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

Impulse approximation

What if force 0?

dp = F dt

F

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

f

F dt ≡ I

∆p = pf − pi = i

impulse

Impulse-momentum theorem:

Ι = ∆p 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.

t

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

F

F t

∆t Ι = F ∆t

Useful to approximate the impulse as due to a constant force F acting over the same time ∆t. Applications: collisions & explosions!

demo

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Impulse Example

Collisions

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?

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

Impulse = p=pf- pi =.060 (-40) - .060 (50) =-5.4 kg m/s But, Impulse also =F t => F(0.035 s)=-5.4 kg m/s => F= -154 N Note: Compare with force/impulse to stop ball… Negative => Westward

1D collision: final m2 v1i

v2i

final m1

m2 v1i

Choose system.

m2 v1f

v2i

v2f

Two masses

Impulse approximation:

Only consider large contact forces; Instants just before & after collision

Apply linear momentum conservation.

Types of collisions:

initial m1

initial m1

m1

m2 v1f

v2f

Linear momentum conservation:

ΣPi = ΣPf m1v1i – m2v2i = m1v1f + m2v2f Given v1i and v2i we have one equation, but two unknowns.

Elastic linear momentum is conserved total kinetic energy is conserved Inelastic linear momentum is conserved total kinetic energy is NOT conserved Totally inelastic collisions: particles stick together. Where does the energy go? demo

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1D Perfectly Inelastic Collisions initial m1

final m1+m2

m2 v1i

One more inelastic example: Explosion! 5.0 kg vi=0

m2 vf

v2i

v1f = v2f = vf

2.0 kg, v1f =3.6 m/s

Momentum conservation:

m1v1i – m2v2i = m1v1f + m2v2f m1v1i – m2v2i = (m1+m2)vf

3.0 kg, v2f =?

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

vf = (m1v1i - m2v2i)/(m1+m2) Caution with signs!

1D Elastic Collisions initial m1

final m1

m2 v1i

1D Elastic Collisions cont.

v1f

v2i

Some algebra shows: m2

v1i – v2i = − (v1f − v2f) v2f

Momentum conservation:

m1v1i – m2v2i = m1v1f + m2v2f

Kinetic Energy is conserved:

½m1v21i + ½m2v22i = ½ m1v21f + ½ m2v22f Two equations

can solve for two unknowns.

Use these:

v1 f =

m1 − m2 2m2 v1i + v 2i m1 + m2 m1 + m2

v2 f =

m − m1 2m1 v1i + 2 v 2i m1 + m2 m1 + m2

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