Examples

1. Lets say we start with two balls with masses  and  having initial velocities  and  respectively. They collide and end up with velocities  and 

  

Before:

The total momentum  . The total energy is  .

After:

The total momentum  . The total energy is  .

Equating the total momentum and energy before and after the collision

  

Let's try to analyze the collision of two balls in one dimension.




Elastic Collision Sample Problem:

2. A 0.450kg ice puck moving east with a speed of 3.00m/s has a head on collision with a 0.900kg puck at initially at rest. Assume a perfect elastic collision, what will be the the speed and direction of each after the collision?

First step is to realize that in perfectly elastic collisions, kinetic energy and momentum are conserved. If we allow m1 and m2 represent the masses and u and v represent the velocities before and after respectively, we obtain the following two equations:

KE: 1/2m1u1^2 + 1/2m2u2^2 = 1/2m1v1^2 + 1/2m2v2^2

Momentum: m1u1 + m2u2 = m1v1 + m2v2

Since the second object is initially at rest we can simplify (u2 = 0)

KE: 1/2m1u1^2 = 1/2m1v1^2 + 1/2m2v2^2

Momentum: m1u1 = m1v1 + m2v2

Next lets work with the kinetic energy relation and isolate v1:

1/2m1v1^2 = 1/2m1u1^2 - 1/2m2v2^2

m1v1^2 = m1u1^2 - m2v2^2

v1^2 = m1u1^2/m1 - m2v2^2/m1

v1^2 = u1^2 - m2v2^2/m1

Lovely, now we need to get v2 in terms of v1 to solve. We do this using the momentum equation, and isolating v2 in terms of v1:

m1u1 = m1v1 +m2v2

m2v2 = m1u1 - m1v1

v2 = (m1u1 - m1v1)/m2

From this point on substitute in the following values that you're given:

u1 = 3m/s

m1 = 0.45 kg

m2 = 0.9 kg

You will have to isolate all the v1's on one side then finally solve. Algebraically its quite a pain...

Anyway the final solution you should obtain is:

v1 = -1 m/s (negative because its now travelling opposite direction, ie west)

v2 = 2 m/s (positive because it is now heading east)