Question

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

Answer 1

1.3 m/s

Step-by-step 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