A small ball is dropped from a height of 20 m above a horizontal floor. After striking the floor, it rebounds to a maximum height of 5 m . The percentage of kinetic energy lost by the ball during the collision with the floor is:

Step-by-step solution

Let the mass of the ball be m . The kinetic energy of the ball just before striking the floor is equal to its initial potential energy: K_ i = mgh_ i = mg(20) The kinetic energy of the ball just after the rebound is equal to the potential energy at its maximum rebound height: K_ f = mgh_ f = mg(5) The loss in kinetic energy during the collision is: K = K_ i - K_ f = 20mg - 5mg = 15mg The percentage loss of kinetic energy is: Percentage loss = ( K K_ i ) 100 Percentage loss = ( 15mg 20mg ) 100 = 75 % Answer: 75 %