A \(1 \mathrm{~kg}\) ball moving at \(12 \mathrm{~ms}^{-1}\) collides with a \(2 \mathrm{~kg}\) ball moving in opposite direction at \(24 \mathrm{~ms}^{-1}\). If the coefficient of restitution is \(2 / 3\), then their velocities after the collision are

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
\(\Rightarrow m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}\)
Substituting the given values in the above relation, we get
\(1 \times 12+2 \times(-24)=1 \times v_{1}+2 v_{2} \)
\( \Rightarrow 12-48=v_{1}+2 v_{2} \)
\( \Rightarrow v_{1}+2 v_{2}=-36 ...(ii)\)
Solving Eqs. (i) and (ii), we get
\(v_{2} =-4 \mathrm{~ms}^{-1} \)
\( \Rightarrow -4-v_{1} =24 \)
\( \Rightarrow v_{1} =-24-4 [\because \text { from Eq. (i) }] \)
\( =-28 \mathrm{~ms}^{-1}\)