Question

If the momentum of a body increases by $50\%$, its kinetic energy will increase by

• A. 0.5
• B. 1
• C. 1.25
• D. 1.5

Answer 1

Answer: (C)

1.25

Answer 2

Relation between kinetic energy (KE) and momentum (p) is, $K E=\frac{p^{2}}{2 m}$ Initial kinetic energy $K E_{1}=\frac{\vec{p}_{2}^{2}}{2 m}$,Similarly, final kinetic energy $K E_{2}=\frac{\vec{p}_{1}^{2}}{2 m}$ $\therefore \frac{E_{2}-E_{1}}{E_{1}}=\frac{p_{2}^{2} / 2 m-p_{1}^{2} / 2 m}{p_{1}^{2} / 2 m}$ ie. Percentage increase in $KE =\frac{p_{2}^{2}-p_{1}^{2}}{p_{1}^{2}} \times 100$ Now, let, $p_{1}=100$ then $p_{2}=150$ So, $\%$ increase in KE $=\frac{(50)^{2}-(100)^{2}}{(100)^{2}} \times 100$ or $\%$ increase in $E=125$