Question

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

• A. 3.3
• B. 2.25
• C. 2
• D. 3

Answer 1

Answer: (D)

3

Answer 2

Kinetic energy is related to momentum through the relation
$K=\frac{p^{2}}{2 m}$
or $\frac{K_{2}}{K_{1}}=\frac{p_{2}^{2}}{p_{1}^{2}} \,\,\,...(i)$
Given, $p_{1}=p, p_{2}=p+\frac{p \times 100}{100}=p+p=2 p$,
$K_{1}=K$
$\therefore \frac{K_{2}}{K}=\frac{(2 p)^{2}}{p^{2}}=\frac{4 p^{2}}{p^{2}}$
Increase in kinetic energy
$\frac{K_{2}-K}{K}=\frac{4 p^{2}-p^{2}}{p^{2}}=3$
Hence, percentage increase in kinetic energy
$\left(\frac{K_{2}-K}{K}\right) \times 100=(3 \times 100)$
$=300\%$