Question

A bowling ball with a mass of 9kg is thrown down a lane with a constant speed of 3 m/s. The ball hits the 1.5kg pin, initially at rest, at the end of the lane. After the collision the pin moves with a speed of 5 m/s. How fast is the ball moving after the impact?​

Answer 1

Answer:

M1 V1 = M1 V2 + M2 V3    conservation of momentum

V2 = (M1 V1 - M2 V3) / M1      where V2 = speed of M1 after impact

V2 = (3 * 9 - 1.5 * 5) / 9  =   (27 - 7.5) / 9 = 2.17 m/s

Note: All speeds are in the same direction and have the same sign