Loss in the Kinetic Energy during a Perfectly Inelastic and an Inelastic Head-on Collision

For a PERFECTLY INELASTIC head-on collision of masses $m_1$, $m_2$ with initial velocities $u_1$, $u_2$, the bodies move together afterwards with one common final velocity v; conservation of momentum, $m_1u_1+m_2u_2=(m_1+m_2)v$, immediately gives $v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$. Working out the loss in kinetic energy, $\Delta(KE)=(KE)_{\text{initial}}-(KE)_{\text{final}}$, and simplifying algebraically gives the compact result $\Delta(KE)=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(u_1-u_2)^2$. Since masses are always positive and $(u_1-u_2)^2$ is always non-negative, there is ALWAYS a loss of kinetic energy in a perfectly inelastic collision (zero only in the degenerate case of equal initial velocities, i.e. no real collision at all).

More generally, for an INELASTIC collision with an arbitrary coefficient of restitution e (using the momentum equation together with the definition of e), the final velocities work out to $v_1=\frac{m_1-em_2}{m_1+m_2}u_1+\frac{(1+e)m_2}{m_1+m_2}u_2$ and $v_2=\frac{m_2-em_1}{m_1+m_2}u_2+\frac{(1+e)m_1}{m_1+m_2}u_1$, and the kinetic energy loss generalises to $\Delta(KE)=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(u_1-u_2)^2(1-e^2)$. Since $e<1$ for any real collision, $(1-e^2)$ is always positive, so there is always SOME loss of KE in an inelastic collision; setting $e=0$ recovers the perfectly-inelastic maximum-loss result above, and setting $e=1$ correctly makes the loss vanish, recovering the elastic case. The quantity $\mu=\frac{m_1m_2}{m_1+m_2}$ appearing throughout these formulas is called the REDUCED MASS of the two-body system. …