\(\begin{gathered}
  \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{3}{2}\left( {\frac{1}{2}mv_0^2} \right) \hfill \\
   \Rightarrow \,v_1^2 + v_2^2 = \frac{3}{2}v_0^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right) \hfill \\ 
\end{gathered} \)

From momentum conservation 

\(m{v_0} = m\left( {{v_1} + {v_2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {ii} \right)\)

Squarring both sides,

\(\begin{gathered}
  {\left( {{v_1} + {v_2}} \right)^2} = v_0^2 \hfill \\
   \Rightarrow \,v_1^2 + v_2^2 + 2{v_1}{v_2} = v_0^2 \hfill \\
  \,\,\,\,\,\,2{v_1}{v_2} = \frac{{v_0^2}}{2} \hfill \\
  {\left( {{v_1} - {v_2}} \right)^2} = v_1^2 + v_2^2 - 2{v_1}{v_2} = \frac{3}{2}v_0^2 + \frac{{v_0^2}}{2} \hfill \\ 
\end{gathered} \)

Solving we get relative velocity between the two particles

\({v_1} - {v_2} = \sqrt 2 {v_0}\)