Conservation of linear momentum gives
$m_{1} u_{1} +m_{2} u_{2} =m_{1} v_{1}+m_{2} v_{2}$
$\Rightarrow m_{2} v_{2} =\frac{3 u}{2}$
Conservation of kinetic energy gives
$\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}$
$=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}$
$\Rightarrow m_{2} v_{2}^{2}=\frac{15 u^{2}}{8}$
Hence, on solving Eqs. (i) and (ii), we get
$m_{2}=1.2\, kg$