1. MOMENTUM
 Momentum
= mass times velocity


p  mv
Units - kg m/s
2. IMPULSE

Collisions involve forces (there is a Dv).

Impulse = force times time.
 
I  FDt
Units - N s
3.
IMPULSE CHANGES MOMENTUM
Impulse = change in momentum


F  ma 

Dv
F m
Dt


FDt  mDv


FDt  mDv

 
FDt  m(v f  vi )



FDt  (mv f  mvi )



FDt  ( p f  pi )


I  Dp
Case 1
Increasing Momentum
Follow through
 
Dt  I  Dp



FDt  I  Dp
Examples:
Long Cannons
Driving a golf ball
Can you think of others?
Case 2
Decreasing Momentum over a Long Time

 
Dp  I  F
Dt

F
Dt
Warning – May be dangerous
Examples:
Bungee Jumping
Can you think of others?
Case 3
Decreasing Momentum over a Short Time

 
Dp  I  F
Dt
Examples:
Boxing (leaning into punch)
Head-on collisions
Can you think of others?
5. CONSERVATION OF MOMENTUM
Example:
Rifle and bullet
Demo - Rocket balloon
Demo - Clackers
Video - Cannon recoil
Video - Rocket scooter
Consider two objects, 1 and 2, and assume that
no external forces are acting on the system
composed of these two particles.



Impulse applied to object 1
F1Dt  m1v1 f  m1v1i



Impulse applied to object 2
F2 Dt  m2 v2 f  m2 v2 i


Apply Newton’s Third Law
F1   F2


or F1Dt   F2 Dt
Total impulse
applied
to system
or




0  m1v1 f  m1v1i  m2 v2 f  m2 v2 i




m1v1i  m2 v2 i  m1v1 f  m2 v2 f
 Internal
forces cannot cause a
change in momentum of the
system.
 For conservation of momentum, the
external forces must be zero.
6. COLLISIONS
 Collisions
involve forces internal to
colliding bodies.
 Elastic
collisions - conserve energy and
momentum
 Inelastic
 Totally
collisions - conserve momentum
inelastic collisions - conserve
momentum and objects stick together
Elastic and Inelastic Collisions

In an elastic collision the
total kinetic energy is
conserved


Momentum is conserved in
any collision
Example:
Let m2  2m1  2m and v1i   v 2i  v
P  mv  mv1 f  2mv 2 f
K  32 mv2  12 mv1 f  mv2 f
2

2
What are signs of final
velocities?
Elastic and Inelastic Collisions


Example (cont.):
Consider reference frame where CM is at rest
v CM
Lab

1
3m
(mv  2mv )   13 v
v1iCM  v1i -v CM
Lab
 43 v, and v 2iCM  v 2i -v CM
Lab
  23 v
P  0  mv1iCM  2mv 2iCM  v1iCM  2 v 2iCM
v * fCM   v *iCM
v1 fCM   43 v, v 2 fCM  23 v
v1 f  v1 fCM  v CM
Lab
  53 v, v 2 f  v 2 fCM  v CM
Lab
 13 v
Elastic and Inelastic Collisions

In an inelastic collision
the total kinetic energy
is not conserved


Momentum is conserved
in any collision
Example: case where
particles stick together
Let m2  2m1  2m and v1i   v 2i  v
P  mv  mv1 f  2mv 2 f  3mv f
v f   13 v
Elastic and Inelastic Collisions

Example: Ballistic
Pendulum
m1v1 A  (m1  m2 )vB
vB 
1
2
m1
m1  m2
v1 A
(m1  m2 )vB  (m1  m2 ) gh
v1 A 
2
m1  m2
m1
2 gh
Simple Examples of Head-On Collisions
(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
Collision between two objects. One at rest initially has twice the mass.
Collision between two objects. One not at rest initially has twice the mass.
Head-On Totally Inelastic
Collision Example
vtruck  60mph
vcar  60mph
Let the mass of the truck be 20 times the
mass of the car.
 Using conservation of momentum, we get

initial momentum of system = final momentum of system
20 m(60 mph)  m(60 mph)  (21 m)v
19(60 mph)  21v
19
v  (60 mph)
21
v  54.3 mph
Remember that the car and the truck exert
equal but oppositely directed forces upon
each other.
 What about the drivers?
 The truck driver undergoes the same
acceleration as the truck, that is

(54.3  60) mph  5.7 mph

Dt
Dt

The car driver undergoes the same
acceleration as the car, that is
54.3 mph  (60 mph) 114.3 mph

Dt
Dt
The ratio of the magnitudes of these two accelerations is
114.3
 20
5.7
Remember to use Newton’s Second Law to
see the forces involved.

For the truck driver his mass times his
acceleration gives
ma F
For the car driver his mass times his greater acceleration
gives
a F
m

Don’t mess with T
TRUCKS, big trucks that is.

Your danger is of the order of twenty times
greater than that of the truck driver.
Collisions in 2D

Use Conservation of
Momentum in each
direction
m1v1i  m1v1 f cos   m2v2 f cos 

Consider case where
m1v1 f sin   m2v2 f sin 
one particle is at rest

In CM frame particles
are back-to-back!
Rocket Propulsion

“Rockets can’t fly in
vacuum. What do they
have to push against?”

Nonsense. Rockets
don’t push; they
conserve momentum,
and send parts (fuel)
away from the body as
fast as possible
Rocket Propulsion

How fast do rockets
accelerate?


Start at rest, with mass
M+Dm
Some time Dt later,
have expelled Dm at
speed ve, to conserve
momentum, rest of
rocket (M) must have
velocity (in the other
direction) of
Dv = ve Dm/M
Rocket Propulsion
a

dv
dt
ve dm
m dt

t

f
e dmrockets
 vHow
 v fast vdo
dt  v
f
i
F  ve

ti
m dt
e
mf
 
m
dm
accelerate?

v
ln
e
m
m
mi
f
i
dm
dt
Thrust: (instantaneous) force on rocket