Question

If momentum is increased by $20\%$, then kinetic energy increases by

• A. $48\%$
• B. $44\%$
• C. $40\%$
• D. $36\%$

Answer 1

Answer: (B)

The kinetic energy is given by $KE=\frac{p^2}{2m}$
So, $\Delta KE=\frac{2p\Delta p}{2m}=\frac{p\Delta p}{m}$
$\Rightarrow \frac{\Delta KE}{KE}=\frac{p\Delta p}{m}\times \frac{2m}{p^2}=\frac{2\Delta p}{p}$
Since momentum $p$ increases by $20\%$, so the final momentum becomes $1.2p$.
Hence, $K{E}_{\text{final }}=\frac{(1.2p{)}^2}{2m}=1.44\frac{p^2}{2m}=1.44KE$
So, $\%$ change in $KE=44\%$.