Momentum

Newton’s 2nd law:

∑~F =

d~P

dt

where ~P is the total momentum of all particles in a system:

~P =∑

i

~pi = ~p1 + ~p2 + ~p3 + · · ·

=∑

mi~vi

– Typeset by FoilTEX – 1

Momentum

~Ptotal =∑

i

~pi = ~p1 + ~p2 + ~p3 + · · ·

=∑

mi~vi =∑

mi

d~ri

dt

=d

dt

∑

i

mi~ri

=d

dtM~rcm

= M~vcm

– Typeset by FoilTEX – 2

Momentum and Impulse

Note that if the net force is constant,

~Fnet =∆~p

∆t

so that the change in momentum is given by

∆~p = p2 − p1 = ~Fnet∆t ≡ ~J

where ~J=impulse

– Typeset by FoilTEX – 3

Momentum and Impulse

impulse– the change in a particle’s momentum

For a constant force, ~J = ~Fnet∆t.

For a varying force,

~J =

∫

~Fnetdt

and the integral is taken over the time of the collision

– Typeset by FoilTEX – 4

Comparing Momentum & Kinetic Energy

Both momentum and kinetic energy depend on m and ~v, so what is thedifference?

~p = m~v K = 1

2mv2

momentum (kg m/s) energy (J=kg m2/s2)vector scalar

∆~p = ~p2 − ~p1 = ~J = ~Fnet∆t ∆K = K2 − K1 = Wtotal = ~Fnet · ∆~r

– Typeset by FoilTEX – 5

Clicker QuestionConsider two carts of masses m and 2m at rest on an air track. If you

push first one cart for 3 seconds then the other for the same length of time,exerting equal force on each, the momentum of the light cart is

A. four times

B. twice

C. equal to

D. one-half

E. one-quarter

the momentum of the heavy cart.

– Typeset by FoilTEX – 6

Clicker Question

Consider two carts of masses m and 2m at rest on an air track. If youpush first one cart for 3 seconds then the other for the same length of time,exerting equal force on each, the kinetic energy of the light cart is

A. larger than

B. equal to

C. smaller than

the momentum of the heavy cart.

– Typeset by FoilTEX – 7

Clicker Question

Suppose a ping-pong ball and a bowling ball are rolling toward you.Both have the same momentum, and you exert the same force to stop each.How do the time intervals to stop them compare?

A. It takes less time to stop the ping-pong ball.

B. Both take the same time.

C. It takes more time to stop the ping-pong ball.

– Typeset by FoilTEX – 8

Clicker Question

Suppose a ping-pong ball and a bowling ball are rolling toward you.Both have the same momentum, and you exert the same force to stop each.How do the distances needed to stop them compare?

A. It takes a shorter distance to stop the ping-pong ball.

B. Both take the same distance.

C. It takes a longer distance to stop the ping-pong ball.

– Typeset by FoilTEX – 9

Conservation of Momentum

2 astronauts are floating in space (neglect gravitational forces). Thepush each other, what happens?

BA

– Typeset by FoilTEX – 10

Conservation of Momentum

During the push, ~FBonA = −~FAonB (Newton’s 3rd law)

BA

so

~FBonA∆t = −~FAonB∆t (1)

~pA2 − ~pA1 = −(~pB2 − ~pB1) (2)

*Changes in momentum are equal and opposite*

– Typeset by FoilTEX – 11

Conservation of Momentum

Consider a system (of particles, objects, astronauts, etc....)

Two kinds of forces:

internal forces– forces that particles in the sytem exert on each other

external forces– forces exerted on a object in the system by an outsideobject

If the only forces acting on the system are internal, the system isisolated.

– Typeset by FoilTEX – 12

Total momentum in an isolated system

How does the total momentum in an isolated system change?

~ptotal = ~pA + ~pB = total momentum

d~ptotal

dt=?

– Typeset by FoilTEX – 13

Total momentum in an isolated system

How does the total momentum in an isolated system change?

~ptotal = ~pA + ~pB = total momentum

d~ptotal

dt=

d

dt(~pA + ~pB) =

d~pA

dt+

d~pB

dt= ~FBonA + ~FAonB = 0

Total momentum doesn’t change, so ~ptotal = constant and totalmomentum is conserved in an isolated system.

– Typeset by FoilTEX – 14

Total momentum in a non-isolated system

What if external forces exist?

∑~F =

∑~Finternal +

∑~Fexternal =

d~ptotal

dt

– Typeset by FoilTEX – 15

Total momentum in a non-isolated system

What if external forces exist?

∑~F =

∑~Finternal

︸ ︷︷ ︸=0

+∑

~Fexternal =d~ptotal

dt

If∑

~Fexternal = 0, then d~ptotaldt

= 0

⇒ momentum is still conserved!

– Typeset by FoilTEX – 16

Total momentum in a non-isolated system

2 ice-skaters on a frictionless pond push each other

BA

system: 2 ice-skaters

internal forces:

external forces:

– Typeset by FoilTEX – 17

Using conservation of momentum to solve problems

1. Make sure you can use conservation of momentum! (∑

~Fexternal = 0)

2. Set-up

• Define coordinate system (in an inertial reference frame!)• treat each body as a particle. Draw “before” and “after” sketches and

label all velocities and masses• identify target variables

3. Execute

• break into components (ptotalx = pxA + pxB + · · ·). Beware of signs!

• may need information from energy considerations

– Typeset by FoilTEX – 18

• solve problem

4. Assess –Does answer make sense?

– Typeset by FoilTEX – 19

Example: collision along a straight lineConsider two blocks sliding toward each other on a frictionless table as

shown below. After the collision, block B moves to the right with a speedof 2.0 m/s. What is the final velocity of A?

A B

m = 0.5 kg m = 0.3 kgA B

v = 2.0 m/s v = −2.0 m/sAxi Bxi

before

A B

m = 0.5 kg m = 0.3 kgA B

v = 2.0 m/sv = ?Axf Bxfafter

– Typeset by FoilTEX – 20

Collisions

If forces between bodies are much larger than any external force, we canneglect external forces and treat bodies as an isolated system (i.e., neglectfriction in a car crash).

– Typeset by FoilTEX – 21

Classification of collisions

elastic collision: If forces between bodies are conservative, then the totalkinetic energy is the same before and after the collision and the collisionis elastic

inelastic collision: Collision is inelastic if any kinetic energy is lost in thecollision

completely inelastic collision: If 2 bodies stick together after the collisionand move together as a single body, the collision is completely inelastic

– Typeset by FoilTEX – 22

Completely inelastic collision

vAxi Bxiv

mA mB

A B

before

Bxf

A B

beforevAxf v

– Typeset by FoilTEX – 23

Example: Ballistic PendulumThe bullet (mass m) is fired into a block of wood with mass M , suspendedlike a pendulum, and makes a completely inelastic collision with it. Afterthe impact of the bullet, the block swings to a height y. Given values m,M , y, what is the initial velocity, vx, of the bullet?

vx

y

before during after

M+m

M

m

– Typeset by FoilTEX – 24

System of particles

Each particle has it’s own velocity. Let ~vi be the velocity of the ithparticle. The velocity of the center of mass is

~vcm =d

dt~rcm

Let~vi = ~vcm + ~vi rel

with ~vi rel the velocity of the ith particle relative to the center of mass.

Consider the total momentum & total kinetic energy of the system...

– Typeset by FoilTEX – 25