Question

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be

• A. \(500\%\)
• B. \(600\%\)
• C. \(6\%\)
• D. \(60\%\)

Answer 1

The Correct Option is A

Kinetic energy \( K \) is defined as \( K = \frac{P^2}{2m} \). This implies momentum \( P \) can be written as \( P = \sqrt{2mK} \). If the final kinetic energy \( K_f \) is 36 times the initial kinetic energy \( K_i \), such that \( K_f = 36 K_i \), then the final momentum \( P_f \) is \( P_f = \sqrt{2m \cdot 36K_i} = 6P_i \). The percentage increase in momentum is calculated as \( \text{Percentage increase} = \frac{P_f - P_i}{P_i} \times 100\% = \frac{6P_i - P_i}{P_i} \times 100\% = \frac{5P_i}{P_i} \times 100\% = 500\% \).