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)

$56\%$

Answer 2

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 ? 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%$