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Q.
The momentum of a body is increased by $25\%$. The kinetic energy is increased by about
Work, Energy and Power
Solution:
Kinetic energy of the body is $K=\frac{p^{2}}{2m}$
where $p$ is the momentum and $m$ is the mass of a body respectively.
$\therefore K ∝ p^{2}$
When the momentum of a body is increased by $25\%$, its momentum will become
$p'=p+\frac{25}{100}p=\frac{125}{100}p=\frac{5}{4}p$
$\therefore \frac{K'}{K}=\frac{p'^{2}}{p^{2}}=\left(\frac{5}{4}\right)^{2}=\frac{25}{16} or K'=\frac{25}{16}K$
Percentage increase in the kinetic energy of the body
$=\frac{K'-K}{K}\times100=\frac{\left(25/16\right)K-K}{K}\times100$
$=\frac{9}{16}\times100=56\%$