Question 6 of 130
When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be:
Show answer and solution
Answer: (D)
Kinetic energy
If
So,
$\begin{aligned}
\% \text { increase in momentum } & =\frac{\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\
& =\frac{6 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\
& =500 \%
\end{aligned}$
If
So,
$\begin{aligned}
\% \text { increase in momentum } & =\frac{\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\
& =\frac{6 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\
& =500 \%
\end{aligned}$
