Question

If the momentum of body is increased by $100\%$, then the percentage increase in the kinetic energy will be:

• A. 330%
• B. 225%
• C. 200%
• D. 300%

Answer 1

Answer: (D)

Kinetic energy is related to momentum through the relation
$K=\frac{p^2}{2m}$
or $\frac{K_2}{K_1}=\frac{p_2^2}{p_1^2}...(i)$
Given, $p_1=p,p_2=p+\frac{p\times 100}{100}=p+p=2p$,
$K_1=K$
$\therefore \frac{K_2}{K}=\frac{(2p{)}^2}{p^2}=\frac{4p^2}{p^2}$
Increase in kinetic energy
$\frac{K_2-K}{K}=\frac{4p^2-p^2}{p^2}=3$
Hence, percentage increase in kinetic energy
$\left(\frac{K_2-K}{K}\right)\times 100=(3\times 100)$
$=300\%$