Ans.

On a horizontal plane, a mass $m_1$ hits another mass $m_2$. $u$ and zero are the velocities of the masses $m_1$ and $m_2$ respectively.

Let, $v_1$ and $v_2$ be their respective velocities after collision.

According to the conservation of linear momentum,

$m_1u+m_1v_1+m_2v_2$

If $e$ be the coefficient of restitution, then

$\displaystyle{e=\frac{v_2-v_1}{u}}$

After collision, the loss of kinetic energy is

Total kinetic energy before collision – Total kinetic energy after collision

= $\frac{1}{2}m_1u^2-\left(\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\right)$

= $\frac{1}{2(m_1+m_2)}\left[(m_1+m_2)m_1v_1^2+(m_1+m_2)m_2v_2^2\right]$

= $\frac{1}{2(m_1+m_2)}\left[(m_1^2u^2+m_1m_2u^2)-\left(m_1^2v_1^2+m_2^2v_2^2+m_1m_2(v_1^2+v_2^2)\right)\right]$

= $\frac{1}{2(m_1+m_2)}\left[(m_1^2u^2+m_1m_2u^2)-\left({(m_1v_1+m_2v_2)}^2-2m_1m_2v_1v_2+m_1m_2(v_1^2+v_2^2)\right)\right]$

= $\frac{1}{2(m_1+m_2)}\left[(m_1^2u^2+m_1m_2u^2)-\left({(m_1v_1+m_2v_2)}^2+m_1m_2{(v_1-v_2)}^2\right)\right]$

=$\frac{1}{2(m_1+m_2)}\left[(m_1^2u^2+m_1m_2u^2)-\left(m_1^2u^2+m_1m_2{(v_1-v_2)}^2\right)\right]$
[ using equation (1) ]

=$\frac{1}{2(m_1+m_2)}\left[m_1m_2u^2-m_1m_2{(v_1-v_2)}^2\right]$

=$\frac{1}{2(m_1+m_2)}\left[m_1m_2u^2-m_1m_2e^2u^2\right]$
[ using equation (2) ]

=$\displaystyle{\frac{1}{2}\frac{m_1m_2}{(m_1+m_2)}(1-e^2)u^2}$