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Transcript
```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 v).
Impulse = force times time
 
I  Ft
Units – Ns or lbs
3.
IMPULSE CHANGES MOMENTUM
Impulse = change in momentum



v
F  ma  m
t


Ft  mv


I  p
Case 1: Increasing Momentum
 
t  I  p

F
Examples: long cannons, driving a golf ball, can you think of others?
Video – Tennis racquet and ball

F  t
Case 2: Decreasing Momentum over a Long Time
 
p  I 
t

F
Examples: rolling with the punch, bungee jumping, can you think of others?
Case 3: Decreasing Momentum over a Short Time
 
p  I 

F
t
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
5.
CONSERVATION OF MOMENTUM
Example - Rifle and bullet
Demo – Rocket balloon
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.



F1 t  m1v1 f  m1v1i
Impulse applied to object 1
Impulse applied to object 2
Apply Newton’s Third Law
Total impulse
applied
to system
or



F2  t  m 2 v 2 f  m 2 v 2i


F1   F2




0  m1v1 f  m1v1i  m 2 v 2 f  m 2 v 2i




m1v1i  m 2 v 2i  m1v1 f  m 2 v 2 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 collisions - conserve momentum
Totally inelastic collisions - conserve momentum and objects stick together
Demo - Collisions on air track
Demo - Momentum balls
Demo - Small ball/large ball drop
Demo - Funny Balls
Video - Two Colliding Autos
20 m (60mph )  m ( 60 mph )  ( 21m )v
19(60mph)  21v
19
(60mph)
21
v  54.3mph
v
Remember that the car and the truck exert equal but oppositely directed forces upon each
other.
The truck driver undergoes the same acceleration as the truck, that is
(54.3  60)mph  5.7 mph

t
t
Remember to use Newton’s Second Law to
see the forces involved.

The car driver undergoes the same
acceleration as the car, that is
54.3mph  ( 60mph) 114.3mph

t
t
The ratio of these two accelerations is
114 .3
 20
5 .7
7
MORE COMPLICATED COLLISIONS