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Chapter 5: Momentum
Momentum: a measure of motion
Force: a cause of change in motion
What changes when a force is applied?
v
F  ma  m ; but mass does not change
t
mv
F
; mv changes wi th time
t
Linear Momentum: p = mv
the tendency of an object to pursue straight line motion
Kinetic Energy in terms of momentum:
p2
KE 
2m
p150c5:1
Impulse: the change in motion


I  p  mv
 
since v  at


and F  ma



I  Ft  p  mv
Example: The head of a golf club is in contact with a 46 g golf ball for 0.50 ms,
which results in the golf ball flying off at 70 m/s. Find the impulse and the average
force acting on the ball during the impact.
p150c5:2
Conservation of momentum
two bodies + action/reaction + no other forces
FAB = - FBA
=> equal but opposite impulses
=> pA + pB = 0
When the vector sum of the external forces acting on a system of
particles equals zero, the total linear momentum remains
constant.
p1 + p2 + p3 + ...
is constant
p150c5:3
An application of conservation of momentum: explosions
initially:
m
mv = 0
after explosion:
m1v1 + m2v2 + ...= 0
m
m
v
or (for two fragments)
m1v1 = -m2v2 same size p, opposite directions
1
1
2
v2
Example: An astronaut just outside of the space shuttle throws her 800 g camera away
when it jams. If she and her space suit have a combined mass of 100 kg and the speed
of the camera is 20 m/s, how far is she from the shuttle in 10 seconds?
p150c5:4
Rocket propulsion (note: no problems)
“continual explosion”
continual conservation of momentum:
 ( mv )
m
F
v
t
t
combustion chamber
ejected gases: m at speed v
All space propulsion systems depend upon conservation of
momentum.
most require some “reaction mass”
p150c5:5
Collisions
Elastic Collisions
conserve KE (total KE is same before and after collision)
Inelastic Collisions
some KE is lost during collision (heat, sound, etc.)
Completely Inelastic Collisions
objects stick together
maximum possible loss of KE
In all collisions, the total momentum is conserved!
p150c5:6
Example 5.4 A 5.0 kg lump of clay that is moving at 10 m/s to the left strikes a 6.0 kg
lump of clay moving 12 m/s to the right. The two lumps stick together after they
collide. Find the final speed of the composite lump of clay and the kinetic energy lost
during the collisions.
Example 5.5 (discussion only) A 60 kg man sliding east on a frictionless surface of a
frozen pond at a velocity of .50 m/s is struck by a 1 kg snowball whose velocity is 20
m/s towards the north. If the snowball sticks to the man, what is his final velocity?
p150c5:7
Elastic collisions
conservation of KE + conservation of momentum
=> initial relative velocity = -(final relative velocity)
v1 - v2 = -(v1' - v2 ')
For an object striking head-on a second object (initially at rest)
v2  0
v1 - v2  -(v1 '-v2 ' )
m1v1  m2 v2  m1v1 ' m2 v2 '
 algebra
m1 - m2
2m1
v1 ' 
v1 v2 ' 
v1
m1  m2
m1  m2
KE 2 '
4(m2 m1 )

KE1 (1  m2 m1 ) 2
p150c5:8
Example: A 5.0 Kg mass moving at 10 m/s collides elastically head-on with a
stationary 10 Kg mass. What are the final velocities of both masses? What is the initial
KE of the 5.0 kg mass? What is the final KE of the 10 kg mass?
Example: A 10 Kg mass moving at 10 m/s collides elastically head-on with a stationary
5.0 Kg mass. What are the final velocities of both masses? What is the initial KE of
the 10 kg mass? What is the final KE of the 5.0 kg mass?
p150c5:9
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