Conservation of Momentum
Sec. 9.2
The Conservation of Momentum
Three scenarios:
 Objectives
 Elastic Collisions
Recognize the conditions under which the
momentum of a system is conserved
Apply conservation of momentum to explain the
propulsion of rockets
Solve conservation of momentum problems in
two dimensions by using vector analysis
example:
pool balls colliding
 Inelastic Collisions
example:
train cars colliding and coupling
 Internal Explosions
example:
people spontaneously splitting into two pieces.
The Law of Conservation on Momentum
Collisions and the Conservation of Momentum
 The total momentum of two objects before
a collision or explosion is equal to the total
momentum of the two objects after the
collision or explosion.
 When 2 objects collide, they push off of each other with
equal and opposite impulses
So when they are done colliding, they have given
each other equal and opposite changes in momentum.
Collisions and the Conservation of Momentum
Collisions and the Conservation of Momentum
In a closed system,
This is called the:
(meaning no other forces are acting on either object)
The Law of the Conservation of Momentum
The total momentum of the system stays the same.
In a closed system, the total momentum
of the system stays the same.
15kgm/s
-12kgm/s
-12kgm/s
15kgm/s
Before
After
Total = 3kgm/s
Total = 3kgm/s
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Elastic vs. Inelastic Collisions
Scenario # 1:
Elastic Collision
 Elastic collisions: (particles “rebound”)
 Inelastic collisions: (particles stick together)
In both cases, momentum is conserved.
Before
After
PA + PB =
PA ´ + PB ´
mAvA + mBvB = mAvA´+ mBvB ´
Collisions and the Conservation of Momentum
A 200kg bumper car going 10m/s hits a 300kg bumper car
going -5m/s. If the first car bounces back at -7.5m/s what is
the velocity of the 2nd car after the collision?
ma = 200kg
p =p´
Va = 10m/s
mAvA
+ mBvB = mAvA´
+ mBvB ´
mb = 300kg
p
=
m
v
´
+ mBvB ´
A
A
Vb = -5m/s
v
´)
=
m
v
´
p
(m
A A
B B
V ´ = -7.5m/s
a
Vb´ = ?
These marks
are called
primes and
they tell you
these are the
velocities after
the collision
Scenario # 2:
Perfectly Inelastic Collision
Before
PA + PB
After
=
PAB´
mAvA + mBvB = (mA + mB)vAB´
vB ´ = 6.67m/s
Collisions and the Conservation of Momentum
Elastic or Inelastic Collision?
A 10kg chunk of putty moving at 10.0 m/s collides with and
sticks to a 20.0kg bowling ball that is initially at rest. The
bowling ball with its putty passenger will then be set in
motion with what velocity?
Ma = 10kg
mAvA + mBvB = (mA + mB) vAB´
Va = 10.0m/s
p
= (mA + mB) vAB´
Mb = 20.0kg
Vb = 0m/s
Vab´= ?
p / (mA + mB) = vAB´
3.33m/s = vAB´
2
Scenario # 3:
Internal Explosion
Before
After
PAB
=
PA´ + PB´
(mA + mB)vAB = mAvA´+ mBvB ´
Collisions and the Conservation of Momentum
Practice:
Two stationary ice skaters push off of each other during the
Olympics. If the first skater is 45kg and she moves off at
12.5m/s then what is the speed of the other 75kg skater
going the opposite direction?
1.
Practice:
More Problems:
A 2.0-kg chunk of putty moving at 5.0 m/s collides with and
sticks to a 8.0-kg bowling ball that is initially at rest.
The bowling ball with its putty passenger will then be
set in motion with what momentum?

Superman, with a mass of 150kg, goes into outer
space to push a 125000kg alien spaceship out of
stationary orbit. He uses his super strength and
pushes the ship away ay at speed of 150m/s. What is
the speed of Superman after his big push?
A 125kg bumper car going 12m/s hits a 235kg bumper
car going -13m/s. If the first car bounces back at 12.5m/s what is the velocity of the 2nd car after the
collision?
Elastic or Inelastic?
vB’ = 0.32 m/s
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More Problems:
More Problems:


Drew Brees has a mass of 85kg and is running at
8.5m/s to the goal line and a 125kg opponent is
running at -2.5m/s straight at Brees. If they bounce off
of each other and Brees still manages to go forward at
0.5m/s for the touchdown, then what must the velocity
of the opponent be as he is knocked back and
humiliated by Brees?
A 5.0-kg chunk of putty moving at 10.0 m/s collides
with and sticks to a 7.0-kg bowling ball that is initially at
rest. The bowling ball with its putty passenger will then
be set in motion with what momentum?
More Problems:
More Problems:


A 25,000kg asteroid is flying towards the Earth at a
speed of 1500.0m/s. How much Silly Putty would it
take to stop the asteroid if we launched it at a velocity
of -100.0m/s? Assume the Silly Putty sticks to the
asteroid.
A .005kg bullet hits a .500kg block of stationary wood.
If the bullet sticks to the block and they fly off together
at a velocity of 10.0m/s, then what was the original
velocity of the bullet before the collision?
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