Download Conservation of Momentum

Balancing the before and after
The law of conservation of momentum states
that, in the absence of an external force, the
momentum of a system remains unchanged.
The force or impulse that changes momentum
must be exerted on the object by something
outside the object.
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Molecular forces within a basketball have no effect
on the momentum of the basketball.
A push against the dashboard from inside does not
affect the momentum of a car.
These are internal forces. They come in balanced
pairs that cancel within the object.
The momentum before firing is zero. After firing,
the net momentum is still zero because the
momentum of the cannon is equal and opposite to
the momentum of the cannonball.
Momentum has both direction and magnitude. It is
a vector quantity.
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The cannonball gains momentum and the recoiling
cannon gains momentum in the opposite direction.
The cannon-cannonball system gains none.
The momenta of the cannonball and the cannon are
equal in magnitude and opposite in direction.
No net force acts on the system so there is no net
impulse on the system and there is no net change in
the momentum.
A 200-kg cannon fires a 10-kg cannonball with
a velocity of 150 m/s. Using the law of
conservation of momentum, calculate the
velocity of the cannon.
Momentum before firing equals the momentum
after firing.
net pbefore = net pafter
net (mv)before = net (mv)after
(200)(0) + (10)(0) = 200vcannon + 10(150)
0 = 200vcannon + 1500
-1500 = 200vcannon
-7.5 = vcannon
The cannon recoils at a rate of 7.5 m/s in
the opposite direction of the cannon ball.
The collision of objects clearly shows the
conservation of momentum.
http://www.youtube.com/watch?v=4IYDb6K5UF8
The net momentum before a collision is the sum of
all the momentum before the collision.
The net momentum after a collision is the sum of
all the momentum after the collision.
When objects collide without being permanently
deformed and without generating heat, the
collision is an elastic collision.
A collision in which the colliding objects become
distorted and generate heat during the collision is
an inelastic collision.
A cue ball with a mass of 170g collides with a
stationary 160g eight ball. If the velocity of the
cue ball was 4 m/s before the collision and 1.8
m/s after the collision, what is the velocity of the
eight ball after the collision?
before
after
m1v1 + m2v2 = m1v1 + m2v2
m1 = mass of cue ball (170g = 0.17 kg)
v1 = velocity of cue ball before collision (4.0 m/s)
m2 = mass of the eight ball (160g = 0.16kg)
v2 = velocity of the eight ball before collision (0 m/s)
v1 = velocity of cue ball after collision (1.8 m/s)
v2 = velocity of the eight ball after collision
(unkown)
m1v1 + m2v2 = m1v1 + m2v2
(0.17kg)(4.0m/s)+(0.16kg)(0m/s) = (0.17kg)(1.8m/s)+(0.16kg)(v2)
0.68 + 0 = 0.306 + .16v2
0.374 = .16v2
2.34 m/s = v2
The eight ball’s velocity after the collision is
2.34 m/s.
In an inelastic collision between two freight cars,
the momentum of the freight car on the left is
shared with the freight car on the right.
m1= 2000 kg
v
before
= 4 m/s
m1= 1000 kg
v
before
= 0 m/s
Vafter = ?
before
after
m1v1 + m2v2 = (m1 + m2)v
(2000kg)(4m/s) + (1000kg)(0m/s) = (2000kg+1000kg)v
8000 + 0 = 3000v
8/3 = v
The final velocity of the two freight cars is 2.67
m/s