Question

In a pool game, the cue ball, which has an initial speed of 2.5 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 theta = 14.3 degrees with respect to the original direction of the cue ball. Assume the pool table is frictionless and the masses of the cue ball and the 8-ball are equal. The cue ball travels at an angle of 90 degrees after the collision.

What is the speed of the cue ball and what is the speed of the 8 ball after collision?

Answer 1

here,

let the masses of ball be m

initial speed of cue ball , u1 = 2.5 m/s

theta1 = 14.3 degree

theta2 = 90 degree

let the final speed of cue ball and the 8-ball be v1 and v2

using conservation of momentum

m * u1 = m * v1 * ( cos(theta1) i + sin(theta2) j) + m * v2 * ( cos(theta2) i - sin(theta2) j)

2.5 i = v1 * ( cos(14.3) i + sin(14.3) j) + v2 * ( 0 - sin(90) j)

on compairing

2.5 = 0.969 * v1 => v1 = 2.58 m/s

and

v1 * sin(14.3) - v2 = 0

v2 = 2.58 * sin(14.3) = 0.64 m/s

the final speed of cue ball is 2.58 m/s

the final speed of 8-ball is 0.64 m/s