Question

In a pool game, the cue ball, which has an initial speed of 4.5 m/s, makes an elastic collision with the 8-ball, which is initially at rest. After the collision, the 8-ball moves at an angle θ = 39.7° with respect to the original direction of the cue ball, as shown in the figure. At what angle φ does the cue ball travel after the collision? Assume the pool table is frictionless and the masses of the cue ball and the 8-ball are equal. What is the speed of the cue ball after the collision? What is the speed of the 8-ball after the collision?

Answer 1

A)

Using law of conservation of momentum,

4.5 = Va cos(50.3) + Vb cos(39.7). ....... (1)

Va sin(50.3) = Vb sin(39.7) ....... (2)

From equation (1) and (2)

Va = 4.5/[cot(50.3) + sin(50.3)/tan(50.3)]

Va = 3.06 m/s

B)

V8 = 3.06 sin(50.3)/sin(39.7) = 3.69 m/s