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

The kinetic energy \( K \) is given by:
\[K = \frac{P^2}{2m}\]
Therefore, momentum \( P \) can be expressed as:
\[P = \sqrt{2mK}\]
If the final kinetic energy \( K_f \) is 36 times the initial kinetic energy \( K_i \), we have:
\[K_f = 36 K_i\]
Thus, the final momentum \( P_f \) will be:
\[P_f = \sqrt{2m \cdot 36K_i} = 6P_i\]
The percentage increase in momentum is:
\[\text{Percentage increase} = \frac{P_f - P_i}{P_i} \times 100\%\]
\[= \frac{6P_i - P_i}{P_i} \times 100\%\]
\[= \frac{5P_i}{P_i} \times 100\% = 500\%\]