Chapter 6
Momentum
1. MOMENTUM
• Momentum - inertia in motion
• Momentum = mass times velocity


p  mv
Units - kg m/s or sl ft/s
2. IMPULSE
• Collisions involve forces (there is a Dv).
• Impulse = force times time.


I  F Δt
Units - N s or lb s
3.
IMPULSE CHANGES MOMENTUM
Impulse = change in momentum


F  ma

Dv
F m
Dt



F Dt  mDv


F Dt  mDv

 
F Dt  m( vf vi )



F Dt  ( mvf  mvi )

 
FDt  ( pf  pi )


I  Dp
Case 1
Increasing Momentum
Follow through
 
 DD
pp
 II 
FD
Dt
t


Examples:
Long Cannons
Driving a golf ball
Can you think of others?
Video Clip
Tennis racquet and ball
Case 2
Decreasing Momentum over a Long Time

 
Dp  I  F
Dt

F
Dt
Warning – May be dangerous
Examples:
Rolling with the Punch
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?
4. BOUNCING
There is a greater impulse
with bouncing.
Example:
Pelton Wheel
Demo – Impulse Pendulum
• Consider a hard ball and a clay ball
that have +10 units of momentum each
just before hitting a wall.
• The clay ball sticks to the wall and
the hard ball bounces off with -5
units of momentum.
• Which delivered the most “punch” to
the wall?
Initial momentum of the clay ball is 10.
Final momentum of clay ball is 0.
The change is 0 - 10 = - 10.
It received - 10 impulse so it
applied + 10 to the wall.
Initial momentum of the hard ball is 10.
Final momentum of hard ball is - 5.
The change is
- 5 - 10 = - 15.
It received - 15 impulse so it
applied + 15 to the wall.
5. CONSERVATION OF MOMENTUM
Example:
Rifle and bullet
Demo - Rocket balloons (several)
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
Impulse applied to object 2
Apply Newton’s Third Law
Total impulse
applied
to system
or




F1 Dt  m1v1 f  m1v1 i



F2 Dt  m2v2 f  m2v2 i


F1  F2


or F1 Dt  F2 Dt



0  m1v1 f  m1v1 i  m2v2 f  m2v2 i




m1v1 i  m2v2 i  m1v1 f  m2v2 f
• Internal forces cannot cause a
change in momentum of the
system.
• For conservation of momentum,
the external forces must be
zero.
Chapter 6 Review Questions
The product of mass times velocity is
most appropriately called
(a) impulse
(b) change in momentum
(c) momentum
(d) change in impulse
You jump off a table. When you land on
the floor you bend your knees during
the landing in order to
(a) make smaller the impulse applied to
you by the floor
(b) make smaller the force applied to
you by the floor
(c) both (a) and (b)
An egg dropped on carpet has a better
chance of surviving than an egg dropped
on concrete. The reason why is
because on carpet the time of impact is
longer than for concrete and thus the
force applied to the egg will be smaller.
(a) True
(b) False
6. COLLISIONS




Collisions involve forces internal to
colliding bodies.
Elastic collisions - conserve energy and
momentum
Inelastic collisions - conserve
momentum
Totally inelastic collisions - conserve
momentum and objects stick together
Demos and Videos
Terms in parentheses represent what is conserved.
Demo
– Air track collisions (momentum & energy)
Demo - Momentum balls (momentum & energy)
Demo - Hovering disks (momentum & energy)
Demo - Small ball/large ball drop
Demo - Funny Balls
Video - Two Colliding Autos (momentum)
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  60 mph
vcar  60 mph
• 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
a F
For the car driver his mass times his greater
acceleration gives
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.
7. More Complicated Collisions
Vector Addition of Momentum
Example of Non-Head-On Collisions
(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
If you vector add the total momentum after collision,
you get the total momentum before collision.
Examples:
Colliding cars
Exploding bombs
Video - Collisions in 2-D
Chapter 6 Review Questions
In which type of collision is energy
conserved?
(a) elastic
(b) inelastic
(c) totally inelastic
(d) All of the above
(e) None of the above
A Mack truck and a Volkswagen have a
collision head-on. Which driver
experiences the greater force?
(a) Mack truck driver
(b) Volkswagen driver
(c) both experience the same force