A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm above a horizontal surface.

Question 40

A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm above a horizontal surface. It falls vertically and then bounces back up so that the maximum height reached by the top of the ball is 45 cm, as shown.

If the kinetic energy of the ball is 0.75 J just before it strikes the surface, what is its kinetic energy just after it leaves the surface?
A 0.36 J                       B 0.39 J                       C 0.40 J                       D 0.42 J

Reference: Past Exam Paper – June 2013 Paper 11 Q17

Solution:

Answer: B.


This is a tricky question. It is necessary to deal with the bottom of the ball at 72 cm at the start and 37 cm at the end.

From the conservation of energy, as the ball falls, its potential energy is being converted to kinetic energy.


Just before it strikes the surface, the kinetic energy is 0.75 J (all potential energy has been converted to KE). The bottom of the ball has zero PE and a KE of 0.75 J. This value of KE = 0.75 J is for the bottom of the ball. So, throughout our calculation, we need to consider the bottom of the ball instead of the top.

PE of bottom of the ball at the point of release = KE of bottom of ball just before striking the surface
mgh = KE
m × 9.81 × 0.72 = 0.75
Mass m of the ball = 0.11 kg


Now, for the rebound, all of the kinetic energy just after the ball leaves the surface is converted to potential energy at the maximum height reached. The maximum of height of the bottom of the ball after rebound is 45 – 8 = 37 cm.

KE just after leaving the surface = PE of bottom of ball at a height of 37 cm
KE = mgh
KE = 0.11 × 9.81 × 0.37 = 0.3854 J


Note that 0.75 J is the KE of the ball just before hitting the surface while 0.3854 J is the KE of the ball just after leaving the surface. This indicates that some energy (0.75 – 0.3854) J has been lost.