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

Answer: (A)

$500\%$

Answer 2

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:
$\text{Percentage increase} = \frac{P_f - P_i}{P_i} \times 100\%$
$= \frac{6P_i - P_i}{P_i} \times 100\%$
$= \frac{5P_i}{P_i} \times 100\% = 500\%$