In a collinear collision , a particle with an initial speed $v_o$ strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particle after collision, is

Momentum is conserved in all type of collisions. 

Final Kinetic energy is 50% more than initial kinetic energy

$\Rightarrow \frac{1}{2}m{v}_2^2+\frac{1}{2}m{v}_1^2=\frac{150}{100}\times \frac{1}{2}m{v}_0^2$

$\Rightarrow {\mathrm{v}}_2^2+{\mathrm{v}}_1^2=\frac{3}{2}{\mathrm{v}}_0^2\dots \dots .\left(\mathrm{i}\right)$

from momentum conservation gives,

$\begin{array}{l}m{v}_0=m{v}_1+m{v}_2\\ v_0=v_2+v_1\dots .\left(ii\right)\end{array}$

Squaring Eqs. (ii), we get

$\begin{array}{l}{\mathrm{v}}_1^2+{\mathrm{v}}_2^2+2{\mathrm{v}}_1{\mathrm{v}}_2={\mathrm{v}}_0^2\\ \Rightarrow 2{\mathrm{v}}_1{\mathrm{v}}_2=\frac{-v_0^2}{2} (\mathrm{from} \mathrm{eq} (\mathrm{i}\left)\right)\\ \therefore {\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)}^2={\left({\mathrm{v}}_1+{\mathrm{v}}_2\right)}^2-4{\mathrm{v}}_1{\mathrm{v}}_2\\ =2v_0^2\\ \text{ or }{\mathrm{v}}_{\mathrm{rel}}=\sqrt{2}{\mathrm{v}}_0\end{array}$