A cue ball initially moving at 2.5 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue ball's final speed is 1.2 m/s at an angle of? With respect to its original line of motion. Find the eight ball's speed after the collision. Assume an elastic collision (ignoring friction and rotational motion). Answer in units of m/s

1.3 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = total momentum after collision

m'u' + mu = m'v' + mv ... Equation 1

Where m' = mass of the cue ball, m = mass of the eight ball, u' = initial speed of the cue ball, u = initial speed of the eight ball, v' = final speed of the cue ball, v = final speed of the eight ball.

making v the subject of the equation,

v = (m'u'+mu-m'v') / m ... Equation 2

Given: u' = 2.5 m/s, u = 0 m/s (stationary), v' = 1.2 m/s

Let: m' = m = y kg

v = (2.5*y+0*y-1.2*y) / y

v = 2.5y-1.2y/y

v = 1.3y/y

v = 1.3 m/s

Thus the final speed of the eight ball = 1.3 m/s