1. Conservation of Momentum
Balancing the before and after

2. Conservation of momentum
The law of conservation of momentum states that, in the absence of an external force, the momentum of a system remains unchanged.

3. The force or impulse that changes momentum must be exerted on the object by something outside the object.
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.

4. The momentum before firing is zero
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.

5. Momentum has both direction and magnitude. It is a vector quantity.
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.

6. Example
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.

7. Momentum before firing equals the momentum after firing
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 = 200vcannon -7.5 = vcannon The cannon recoils at a rate of 7.5 m/s in the opposite direction of the cannon ball.

8. Elastic and inelastic collision
The collision of objects clearly shows the conservation of momentum.

9. 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.

10. Elastic collision
When objects collide without being permanently deformed and without generating heat, the collision is an elastic collision.

11. Inelastic collision
A collision in which the colliding objects become distorted and generate heat during the collision is an inelastic collision.

12. Example elastic 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?

13. before after m1v1 + m2v2 = m1v1 + m2v2 m1 = mass of cue ball (170g = 0
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 (unknown)

14. The eight ball’s velocity after the collision is 2.34 m/s.
m1v1 + m2v2 = m1v1 + m2v2
(0.17kg)(4.0m/s)+(0.16kg)(0m/s) = (0.17kg)(1.8m/s)+(0.16kg)(v2)
= v2
0.374 = .16v2
2.34 m/s = v2
The eight ball’s velocity after the collision is
2.34 m/s.

15. Example inelastic collision
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
m2= 1000 kg
v before = 4 m/s
v before = 0 m/s
Vafter = ?

16. The final velocity of the two freight cars is 2.67 m/s
before after
m1v1 + m2v2 = (m1 + m2)v
(2000kg)(4m/s) + (1000kg)(0m/s) = (2000kg+1000kg)v
= 3000v
8/3 = v
The final velocity of the two freight cars is m/s