Question

A ping pong ball (mass 0.0029 kg) strikes a bowling ball (mass 14 kg) head on with a speed of 89 m/s and bounces off elastically. What is the speed of the ping pong ball after impact? Ignore rotational motion.

Answer 1

let V1 and V2 be the velocities of the ping pong ball and bowling ball after collision and U1 = 89m/s and U2 =0 be the velocities of the ping pong ball and bowling ball before collision. and let M1 =0.0028kg and M2 = 14kg be their respective masses

applying conservation of the momentum :

totalr momentum before collision = total momentum after collision =

so ,

M1U1 + M2U2 = M1V1 +M2V2 ......(1)

applying conservation of the energy :

total energy before collision = total energy after collision

so, 1/2M1U12 +1/2 M2U22 =1/2 M1V12 +1/2M2V22 ......(2)

Now solving euation (1) and (2) we get

V1 = ((M1 - M2) / (M1 + M2))U1 = ((0.0029 - 14) / (0.0029 + 14))89 = 88.9631 m/s

so the speed of the ping pong ball after impact = 88.9631 m/s