Question

A cue ball with a mass of 1.25 Kg moving at 3.0 m/s, east strikes the

stationary eight ball of equal mass. The cue ball moves back at 1.0 m/s
after the collision. What is the velocity of the eight ball after the collision?

Answer 1

4.0 m/s to the EAST

Step-by-step explanation:

This is a conservation of momentum problem. Total momentum before the collision must equal total momentum after the collision.

Momentum = P = mv

Assume positive direction is to the EAST, negative is to the WEST

v = velocity of 8-ball after the collision

(1.25 kg)(3.0 m/s) + (1.25 kg)(0 m/s) = (1.25 kg)(-1.0 m/s) + (1.25 kg)(v)

3.75 kg·m/s + 0 = -1.25 kg·m/s + (1.25 kg)(v)

(1.25 kg)(v) = 1.25 kg·m/s + 3.75 kg·m/s = 5.0 kg·m/s

v = (5.0 kg·m/s) / (1.25 kg) = 4.0 m/s

v = velocity of 8-ball after the collision = 4.0 m/s to the EAST