In a billiards game, one player hits the cue ball towards another ball. The cue ball has a mass of 0.1kg and hits the other ball with a velocity of 2m/s. If the collision is completely elastic and the cue ball travels with a velocity of 0.8m/s after the collision, what is the mass and velocity of the other ball?

We assume the other ball is at rest initially, and that the collision is "head-on." That is, final velocities are in the same direction as the initial velocity of the cue ball. We further assume no energy loss, and that the balls do not rotate.

__

initial momentum

Assuming the ball being hit is initially at rest, it contributes no momentum to the system. The total momentum is ...

(0.1 kg)(2 m/s) = 0.2 kg·m/s

__

initial velocity difference

If the cue ball has velocity v1 and the hit ball has velocity v2, the initial difference of velocities is ...

v1 -v2 = 2 m/s -0 m/s = 2 m/s

__

final velocity difference

After the collision, we must have ...

v1 -v2 = -2 m/s . . . . the velocity difference changes sign

v2 = v1 +2 m/s = 0.8 m/s +2 m/s

v2 = 2.8 m/s

The velocity of the other ball is 2.8 m/s after the collision.

__

mass of the other ball

The total momentum after the collision is the same as before, so we must have ...

m1·v1 +m2·v2 = initial momentum

(0.1 kg)(0.8 m/s) +(m2)(2.8 m/s) = 0.2 kg·m/s

m2 = (0.2 kg·m/s -0.08 kg·m/s)/(2.8 m/s) = (0.12/2.8) kg = 3/70 kg

The mass of the other ball is 3/70 ≈ 0.04286 kg.