If the linear momentum of a body is increased by 50%, the kinetic energy will be increased by ________.

We know that, Kinetic Energy of a body, K = $\frac{{\mathrm{p}}^2}{2\mathrm{m}}$
 

 Hence, 
$K_1$= $\frac{{p_1}^2}{2m}$ 

and 
$K_2$= $\frac{{p_2}^2}{2m}$

Given that,  $p_2$ = $\frac{150}{100}$ $p_1$   ---------- (1)

$$\begin{gathered} \frac{K_2}{K_1} =\frac{{p_2}^2}{{p_1}^2} \\ Or \\ \frac{K_2}{K_1} ={\left(\frac{p_2}{p_1}\right)}^2 \end{gathered}$$   

From Equation (1)

$\frac{K_2}{K_1} = {\left(\frac{150}{100}\right)}^{2 } = \frac{9}{4}$

$\% Change in Kinetic Energy = \frac{K_2 - K_1}{K_1} \times 100$
 

$\% Change in Kinetic Energy = \frac{(9/4) K_1 - K_1}{K_1} \times 100$ 

$\% Change in Kinetic Energy = \frac{9 - 4}{4} \times 100 = \frac{5}{4} \times 100$

$\% Change in Kinetic Energy = 125 \%$