Question

If the momentum of the body is increased by 100%, the percentage increase in the kinetic energy is

• A. 150 %
• B. 200 %
• C. 225 %
• D. 300 %

Answer 1

Answer: (D)

300 %

Here,

Let initially momentum $=p$

Final momentum$=p+p\ast \frac{100}{100}=2p$

$K{E}_i=\frac{p^2}{2m}$

$K{E}_f=\frac{4p^2}{2m}=4KE$

$\Delta KE=3KE$

$\%KE$ = $\frac{3KE}{KE}\times 100=300\%$

Option $\text{D}$ is the correct answer