Question

A cue ball (mass = 0.170 kg) is at rest on a frictionless pool table. The...

A cue ball (mass = 0.170 kg) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick, which applies an impulse of +1.20 N · s to the ball. The ball then slides along the table and makes an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is struck.

? m/s

According to impulse momentum theorm,

I = mvf - mvi

For the cue ball, the initial velocity is 0.

When it got a +1.20 N s impulse, the velocity afterwards is

1.20 = 0.170 v - 0

v = 7.06 m/s

Therefore, the cue ball with 7.07 m/s make an elastic head-on collision with the second ball.

Since the two balls are equal masses, then after the collision, the second ball has the velocity of 7.06 m/s. The cueball would be at rest.

You can solve it by conservation of momentum and kineticenergy.

Denot v1 and v2 as the velocity of cueball and the second ball after the collision, respectively.

conservation of momentum:

mv = mv1 + mv2

conservation of energy

Solve both equations, we find v1 = 0, andv2 = v = 7.06 m/s

So the velocity is 7.06 m/s for the second ball right after the collision.