Explanation

(b) Kinetic energy, $K = \frac{P^{2}}{2m}$
Where $P$ is the momentum and $m$ is the mass. When momentum is increased by 20%, then
$\rho^{'} = P + \frac{20}{100}P = 1.2P$
$\therefore K^{'} = \frac{(1.2P)^{2}}{2m} = \frac{1.44P^{2}}{2m} = 1.44K$
$K^{'} = K + 0.44K\ \Rightarrow \frac{K^{'} - K}{K} = 0.44$
Percentage increase in kinetic energy is
$\frac{K^{'} - K}{K} \times 100 = 0.44 \times 100 = 44\%$