Question

If increase in linear momentum of a body is $50\%$, then change in its kinetic energy is

• A. 0.25
• B. 1.25
• C. 1.5
• D. 0.5

Answer 1

Answer: (B)

1.25

Answer 2

The relation between kinetic energy $K$, and momentum $p$, is $p=\sqrt{2 m K}$ where, $m$ is mass. Given, $p_{1} =p, p_{2}=p_{1}+50 \%$ of $p_{1}$ $p_{2}=p_{1}+\frac{p_{1}}{2}=\frac{3}{2} p_{1}=1.5 p_{1}$ $\therefore \frac{K_{1}}{K_{2}}=\frac{p_{1}^{2}}{p_{2}^{2}}$ $\Rightarrow K_{2} =\frac{p_{2}^{2}}{p_{1}^{2}} K_{1}$ $\Rightarrow K_{2} =\frac{(1.5)^{2}}{1} \times K=2.25 K$ $\therefore$ Change in K E$=2.25-1$ $=1.25=125 \%$