A0.17kg ball rolls at 0.75m/s to the right on a frictionless surface and collides with a 0.17kg ball rolling to the left at 0.65m/s. (elastic collision) what is the final velocity of the second ball when the first is 0.5m/s to the left? what is the change in kinetic energy before and after?

Both momentum and kinetic energy are conserved in elastic collisions (assuming that this collision is perfectly elastic, meaning no net loss in kinetic energy)

To find the final velocity of the second ball you have to use the conversation of momentum:

*i is initial and f is final*

Δpi = Δpf

So the mass and velocity of each of the balls before and after the collision must be equal so

Let one ball be ball 1 and the other be ball 2

m₁ = 0.17kg

v₁i = 0.75 m/s

m₂ = 0.17kg

v₂i = 0.65 m/s

v₂f = 0.5

m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f

Since the mass of the balls are the same we can factor it out and get rid of the numbers below it so

m(v₁i + v₂i) = m(v₁f + v₂f)

The masses now cancel because we factored them out on both sides so if we divide mass over to another side the value will cancel out so

v₁i + v₂i = v₁f + v₂f

Now we want the final velocity of the second ball so we need v₂f

so...

(v₁i + v₂i) - v₁f = v₂f

Plug in the numbers now:

(0.75 + 0.65) - 0.5 = v₂f

v₂f = 0.9 m/s

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