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