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

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Aicharya · Accepted Answer

Kinetic energy $(K)=\frac{P^2}{2 m}$ $\Rightarrow \mathrm{P}=\sqrt{2 \mathrm{mK}}$
If $K_f=36 K_i$ So, $\mathrm{P}_{\mathrm{f}}=6 \mathrm{P}_{\mathrm{i}}$ $\begin{aligned} \% \text { increase in momentum } & =\frac{\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\ & =\frac{6 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\ & =500 \% \end{aligned}$

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

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