A body is dropped on ground from a height 'h₁' and after hitting the ground, it rebounds to a height 'h₂'. If the ratio of velocities of the body just before and after hitting ground is 4 , then percentage loss in kinetic energy of the body is x/4. The value of x is ____________.

Solution

Physics sir · 4 steps

Kinetic energy goes as v², so a speed ratio of 4 becomes an energy ratio of 16.

1. Speeds at the ground. Let v₁ be the speed just before impact and v₂ the speed just after. We are given v₁/v₂ = 4, so v₂ = v₁/4.
2. Compare kinetic energies. The mass does not change, so K₂/K₁ = (v₂/v₁)² = (1/4)² = 1/16. The body keeps only one-sixteenth of its kinetic energy.

   Percentage loss = (K₁ - K₂)/K₁ × 100 = (1 - 1/16) × 100 = 15/16 × 100 = 93.75%

3. Match the form. Set x/4 = 93.75, which gives x = 4 × 93.75 = 375.
4. So. The value of x is 375.

The common slip: 300. Treating kinetic energy as proportional to v instead of v² gives a 75% loss, and so x = 300, not 375.

The idea to keep. For any loss in kinetic energy at fixed mass, square the speed ratio first. Here e = v₂/v₁ and the fraction of KE lost is 1 - e².

Answer from the NTA final answer key. Solution by PrepWiser.