Question

A cue ball (mass = 0.160 kg) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick, which applies an impulse of +1.50 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.

Answer 1

We are given an impulse applied to the cue ball. We can interpret this impulse as the ball's change in momentum:

J=Δp

J=p_f − p_i

Initially, the cue ball is at rest, so p_i=0. Our equation becomes:

J = p_f − 0

1.50 Ns = p_f

So, the cue ball moves with a momentum of 1.50 kg.m/s. This is the total momentum of the system, and it will be conserved durin the collision with the second ball.

We can also use this momentum to find the velocity at which the cue ball moves:

And, next, we can use this velocity to find the ball's kinetic energy.

Now, let's consider the collision. This is an elastic collision, so momentum and kinetic energy are both conserved. We can express the momentum as:

The two balls have equal mass, so this equation simplifies to:

Now, let's consider energy conservation. The sum of each ball's kinetic energy after the collision must equal the total energy of the system:

We have an expression for the first ball's velocity, so let's substitute that into our expression.