Ans : (a)
Momentum $p_2 = p_1+50 \% \: of\: p_1$
$ \Rightarrow p_2 = \large\frac{3}{2} \times p_1$
Since, $KE \sim v^2 \sim p^2$
Hence, $KE_2 = \large\frac{9}{4} KE_1$
So, $ \% $ increase in $KE$ is
$\large\frac{KE_2 – KE_1}{KE_1} \times100 =( { \large\frac{9p_1}{4} – p_1} ) \times100 = 125 \%$