physics elastic collision problem?

Question

A cue ball (mass = 0.165 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 intially at rest. find the velocity of the second ball just after it is struck.

Answer 1

More precisely, an elastic collision has four significant parts: initial strike, elastic deformation of both balls, elastic recovery of both balls, and separation. There are three points in this process where the velocities V1 and V2 are defined. At inital strike, V1 = 1.50 / 0.165, V2 = 0. At maximum deformation, V1 = V2 = 0.5 * 1.50 / 0.165. At separation V1 = 0, V2 = 1.50 / 0.165. These points are separated by nonzero times, so you need to decide which of them corresponds to "just after" the strike.

Answer 2

Velocity of cue ball after impulse is its momentum (1.50 N.s) gained from the impulse divided by the mass of cue ball (0.165 kg).

In an elastic head-on collision, all the momentum is transfered to the second ball. It has a velocity of 1.50/0.165 m/s = 9.09 m/s just after it is struck.