Consider two point masses involved in a semi-elastic impact:
Suppose the two point masses are moving parallel to each other with the velocities $v_1\neq v_2$ (and no external forces/accelerations). Suppose that they collide at some point in time. I want to determine the energy "lost" (or dissipated to the environment) in this process. Let me denote the respective speeds after the collision with $v_1'$ and $v_2'$
Since the collision is semi-elastic, I will introduce a restitution coefficient $k\in[0,1]$ which defines $k=\dfrac{v_2'-v_1'}{v_1-v_2}$.
With this given, one can easily derive that \begin{align} v_1' &= \frac{m_1 v_1 + m_2 v_2 - m_2(v_1 - v_2)k}{m_1 + m_2},\\ v_2' &= \frac{m_1 v_1 + m_2 v_2 - m_1(v_2 - v_1)k}{m_1 + m_2}. \end{align} by using the conservation of momentum: $$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'.$$
The energy that is lost is then given by the difference of the kinetic energies before and after the collision: \begin{equation}\label * \tag{*} 2\cdot\Delta E=m_1v_1^2+m_2 v_2^2-m_1 (v_1')^2-m_2 (v_2')^2.\end{equation} I have seen in various sources that \begin{equation}\label{**} \tag{**}2\cdot\Delta E = \frac{m_1 \cdot m_2}{m_1 + m_2}(v_1 - v_2)^2 \cdot (1 - k^2).\end{equation}
How can I derive \eqref{**} from \eqref{*}? When I try to insert the expressions for $v_1'$ and $v_2'$ from above into $\eqref{*}$, then my calculations get discombobulated very quickly.
The easiest way is to change into the center-of-mass frame.
The kinetic energy contributions split into a center-of-mass component (all mass at CoM, moving at averaged velocity) plus the kinetic energy of relative motion: $$ KE_{total}=KE_{CoM}+KE_{rel} $$ The $KE_{CoM}$ is unchanged in the collision by conservation of momentum. The relative KE is equivalent to a reduced mass $\frac{m_1m_2}{m_1+m_2}$ moving at relative velocity $v_1-v_2$: $$ m_1\left(v_1-\frac{m_1v_1+m_2v_2}{m_1+m_2}\right)^2 +m_2\left(v_2-\frac{m_1v_1+m_2v_2}{m_1+m_2}\right)^2 =\left(\frac{m_1m_2}{m_1+m_2}\right)(v_1-v_2)^2 $$ So \begin{align*} 2\cdot KE_{rel,before}&=\frac{m_1m_2}{m_1+m_2}(v_1-v_2)^2\\ 2\cdot KE_{rel,after}&=\frac{m_1m_2}{m_1+m_2}(v_1'-v_2')^2. \end{align*} Subtracting and using $v_1'-v_2'=-k(v_1-v_2)$ then gives the result.