Question

The momentum of a body is increased by $25\%$. The kinetic energy is increased by about

• A. $25\%$
• B. $5\%$
• C. $56\%$
• D. $38\%$

Answer 1

Answer: (C)

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.
$\therefore K\propto p^2$
When the momentum of a body is increased by $25\%$, its momentum will become
$p^{\prime }=p+\frac{25}{100}p=\frac{125}{100}p=\frac{5}{4}p$
$\therefore \frac{K^{\prime }}{K}=\frac{p^{\prime 2}}{p^2}={\left(\frac{5}{4}\right)}^2=\frac{25}{16}or{K}^{\prime }=\frac{25}{16}K$
Percentage increase in the kinetic energy of the body
$=\frac{K^{\prime }-K}{K}\times 100=\frac{\left(25/16\right)K-K}{K}\times 100$
$=\frac{9}{16}\times 100=56\%$