A rubber ball with mass 0.20 kg is dropped vertically from a height of 1.5 m above a floor. The ball bounces off of the floor, and during the bounce 0.60 J of energy is dissipated. What is the maximum height of the ball after the bounce

Given:

Mass = 0.2 kg

Height = 1.5 m

Potential energy of the drop, PE = m * g * h

= 0.2 x 9.81 x 1.5

= 2.94 J

After the drop, 0.6 J of energy is dissipated, so amount of energy left

= 2.94 - 0.6

= 2.34 J

This energy is eqal to thenew potential energy which is:

m * g * h = 2.34

0.2 * 9.81 * h = 2.34

= 2.34/1.962

= 1.19 m.

The maximum height of the ball after the bounce is 1.2 m

Explanation:

Potential Energy = mass * Height * acceleration of gravity

PE=mgh

= 0.2 x 9.8 x 1.5

P. E = 2.94 J

During bounce of ball 0.60 J of energy is lost. So

2.94 - 0.6 = 2.34 J

now new energy is 2.34

New P. E = mgh

2.34 = 0.2 x 9.8 x h

h = 2.34 / 0.2 x 9.8

h = 1.2 m

The maximum height of the ball after the bounce is 1.2 m