Explanation

(c) The relation between linear momentum and kinetic energy is
${\ p}^{2} = 2mk$ ….(i)
But linear momentum is increased by 50%, then
$p^{'} = \frac{150}{100}p$
$p^{'} = \frac{3}{2}p$
Hence, $p^{'2} = 2mk'$
Or $\left( \frac{3}{2}p \right)^{2} = 2mk'$
Or $\frac{9}{4}p^{2} = 2mk'$ ….(ii)
On putting the value of $p^{2}$ from Eq. (i) in Eq. (ii)
$\frac{9}{4} \times 2mk = 2mk'$
Or $K^{'} = \frac{9}{4}k$
So, the increase in kinetic energy is
$\mathrm{\Delta}K = \frac{9}{4}k - k = \frac{5}{4}k$
Hence, percent increase in kinetic energy
$= \frac{(5/4)K}{K} \times 100\%$
$= \frac{5}{4} \times 100\% = 125\%$