If the linear momentum is increased by 50%, the kinetic energy will increase by (in %)

Let ${\mathrm{P}}_1=\mathrm{P}$ ,  ${\mathrm{P}}_2={\mathrm{P}}_1+50\mathrm{\%}$ of  ${\mathrm{P}}_1={\mathrm{P}}_1+\frac{{\mathrm{P}}_1}{2}=\frac{3{\mathrm{P}}_1}{2}$

$\mathrm{E}\propto {\mathrm{P}}^2 \Rightarrow \frac{{\mathrm{E}}_2}{{\mathrm{E}}_1}={\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)}^2={\left(\frac{3{\mathrm{P}}_1\text{}/\text{}2}{{\mathrm{P}}_1}\right)}^2=\frac{9}{4}$

$\Rightarrow {\mathrm{E}}_2=2.25\mathrm{E}={\mathrm{E}}_1+1.25{\mathrm{E}}_1$

$\therefore {\mathrm{E}}_2={\mathrm{E}}_1+125\mathrm{\%}\text{\hspace{0.33em}of }{\mathrm{E}}_1$ 

i.e. kinetic energy will increase by 125%.