Given, $m_{1}=1 \mathrm{~kg}$
$\begin{aligned}
&u_{1}=12 \mathrm{~ms}^{-1} \\
&m_{2}=2 \mathrm{~kg} \\
&u_{2}=-24 \mathrm{~ms}^{-1}
\end{aligned}$
(-ve sign implies that, it is travelling in the opposite direction)
$e=\frac{2}{3}$
As we know, coefficient of restitution, $e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}$.
$\begin{aligned}
&\frac{2}{3}=\frac{v_{2}-v_{1}}{12+24} \\
&\frac{2}{3}=\frac{v_{2}-v_{1}}{36}
\end{aligned}$
or $\quad v_{2}-v_{1}=24...(i)$
According to the law of conservation of momentum,
Initial momentum of the system = Final momentum of the system
$\Rightarrow \quad m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}$
Substituting the given values in the above relation, we get
$\begin{array}{ll}
& 1 \times 12+2 \times(-24)=1 \times v_{1}+2 v_{2} \\
\Rightarrow & 12-48=v_{1}+2 v_{2} \\
\Rightarrow & v_{1}+2 v_{2}=-36 ...(ii)
\end{array}$
Solving Eqs. (i) and (ii), we get
$\begin{aligned}
v_{2} &=-4 \mathrm{~ms}^{-1} \\
\Rightarrow \quad-4-v_{1} &=24 \\
\Rightarrow \quad \quad v_{1} &=-24-4 \quad[\because \text { from Eq. (i) }] \\
&=-28 \mathrm{~ms}^{-1}
\end{aligned}$