A ball of mass 20 g falls from a height of 45 m. It rebounds from the ground to a height of 40 m. Calculate :
the loss in kinetic energy on striking the ground. [g = 10 m s⁻²]
Topic: Loss of mechanical energy
Answer
Given: m = 0.02 kg, g = 10 m s⁻², rebound height = 40 m
Kinetic energy just BEFORE the impact = potential energy at the start = 9 J (from part (a)).
Kinetic energy just AFTER the impact = the potential energy it can reach on the rebound: = m g h′ = 0.02 × 10 × 40 = 8 J
Loss in kinetic energy on striking the ground: = 9 − 8 = 1 J
Loss in kinetic energy = 1 J
The rebound height tells you how much energy survived the bounce. The ball rises to 40 m instead of the original 45 m, so it left the ground with 8 J instead of the 9 J it arrived with — 1 J was lost in the impact, as heat and sound and in deforming the ball.
Work in ENERGIES rather than heights. Subtracting the heights (45 − 40 = 5) gives a distance, not an energy; you have to convert each height into its energy first.
This is why a real ball never bounces back to the height it was dropped from. A perfectly elastic collision would lose nothing and return to 45 m.