The momentum of a body is increased by 50 %. The percentage increase in the kinetic energy of the body is ___________ %.

Solution

Physics sir · 4 steps

Write kinetic energy in terms of momentum, K = p²/2m, and the 50% becomes a factor of 1.5 squared.

1. Link K and p. Since p = mv and K = 1/2mv², you can write K = p²/2m. The mass does not change, so K is proportional to p².
2. New momentum. A 50% increase makes the new momentum p' = 1.5 p.

   K'/K = (p'/p)² = (1.5)² = 2.25

3. Percentage change. The increase is (K' - K)/K × 100 = (2.25 - 1) × 100 = 125%.
4. So. The kinetic energy increases by 125%.

The common slip: 100. Doubling the percentage, as if a small-change approximation held, gives 100, not 125. That shortcut only works for changes of a few percent.

The idea to keep. For a fixed mass, K = p²/2m, so K goes as p². For a large change, square the actual ratio of momenta. Do not just double the percentage.

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