Chapter 8
Impulse and Momentum
THE LINEAR MOMENTUM
 Momentum
= mass times velocity
 “Think of it as inertia in motion”


p  mv
Units - kg m/s or sl ft/s
AN IMPULSE

Collisions involve forces (there is a Dv).

Impulse = force times time.
 
I  FDt
Units - N s or lb s
AN IMPULSE CAUSES A CHANGE
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



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:
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?
BOUNCING
There is a greater impulse with
bouncing.
Example:
Pelton Wheel
Pelton Wheel Water Sprinkler
 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.
CONSERVATION OF LINEAR
MOMENTUM
Example:
Rifle and bullet
Demo - Rocket balloon
Demo - Clackers
Video - Cannon Shoot
Video – Scooter Propulsion
IN COLISIONS AND EXPLOSIONS
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  m1u1



Impulse applied to object 2
F2 Dt  m2v2  m2u2


Apply Newton’s Third Law
F1   F2


or F1Dt   F2 Dt
Total impulse
applied
to system
or




0  m1v1  m1u1  m2v2  m2u2




m1u1  m2u2  m1v1  m2v2
In one dimension in component form,
m1u1x  m2u2 x  m1v1x  m2v2 x
 Internal
forces cannot cause a change
in momentum of the system.
 For conservation of momentum, the
external forces must be zero.
IN COLLISIONS AND
EXPLOSIONS

Collisions involve forces internal to colliding
bodies.

Inelastic collisions - conserve momentum

Totally inelastic collisions - conserve
momentum and objects stick together
A PERFECTLY ELASTIC
COLLISION

Perfectly elastic collisions - conserve energy
and momentum
1
2
mu  m u  mv  m v
2
1 1
1
2
2
2 2
1
2
2
1 1
1
2
2
2 2
Demos
Demo
- Momentum balls
Demo
- Small ball/large ball drop
Demo
- Funny Balls
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

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.

Your danger is of the order of twenty times
greater than that of the truck driver.
COEFFICIENT OF
RESTITUTION
For any collision between two bodies moving
along a single straight line, the coefficient of
restitution e is defined as
v2 x  v1x
e
u1x  u2 x
v2 x  v1x
e
u1x  u2 x
u’s are velocities before impact.
v’s are velocities after impact.
For perfectly elastic collisions e = 1.
For inelastic collisions e < 1.
For totally inelastic collisions e = 0.
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.
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.
THE CENTER OF MASS
The center of mass of an object of mass m is the single
point that moves in the same way as a point mass would
move when subjected to the same external forces that
act on the object.


F
acm 
m
The coordinates of the center of mass are
xi mi
xcm 
mi
yi mi
ycm 
mi
zi mi
zcm 
mi
CENTER OF MASS AND
CENTER OF GRAVITY
Center of mass - average position of mass
Center of gravity - average position of weight
..
Earth
Path of center of mass of a rotating object will
be a straight line if no external forces act on
the object.
Locating the Center of Gravity

Demo - Meterstick

Demo - Map of Texas

Demo - Balancing eagle

Demo - Curious George
Center can be outside of the object.
Examples: high jump and pole vaulting