If the kinetic energy of a moving body becomes four times of its initial kinetic energy, then the percentage change in its momentum will be

We know that,

Kinetic energy, $\mathrm{K}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}\Rightarrow \mathrm{p}\propto \sqrt{\mathrm{K}}$                        ...(i)

where, p = linear momentum of the body

and m = mass of the body.

Considering Eq. (i), we can write

$\frac{{\mathrm{p}}_2}{{\mathrm{p}}_1}=\frac{\sqrt{{\mathrm{K}}_2}}{\sqrt{{\mathrm{K}}_1}}$                      ...(ii)

According to question,

$\Rightarrow {\mathrm{K}}_2=4{\mathrm{K}}_1$                       …(iii)

Putting the value of K₂ in Eq. (ii), we get

$\begin{array}{l}\frac{p_2}{p_1}=\frac{\sqrt{4K_1}}{\sqrt{K_1}}\\ \Rightarrow \frac{p_2}{p_1}=2\\ \Rightarrow p_2=2p_1\end{array}$

$\begin{array}{c}\therefore \text{ Percentage change in momentum }=\frac{{\mathrm{p}}_2-{\mathrm{p}}_1}{{\mathrm{p}}_1}\times 100\\ =\frac{2{\mathrm{p}}_1-{\mathrm{p}}_1}{{\mathrm{p}}_1}\times 100=\frac{{\mathrm{p}}_1}{{\mathrm{p}}_1}\times 100=100\mathrm{\%}\end{array}$