Loss of kinetic energy in perfect inelastic collision

Even though momentum is exactly conserved in a perfectly inelastic collision, kinetic energy is not -- the sticking-together process necessarily dissipates some of it as heat, sound, and permanent deformation. This loss can be worked out exactly.

The total kinetic energy before collision is $KE_i = \tfrac12 m_1u_1^2 + \tfrac12 m_2u_2^2$, and after collision (both masses now moving together at the common velocity $v$) it is $KE_f = \tfrac12(m_1+m_2)v^2$. The loss is $\Delta Q = KE_i - KE_f$. Substituting the common-velocity formula $v=(m_1u_1+m_2u_2)/(m_1+m_2)$ from the previous section and simplifying (using $a^2-2ab+b^2=(a-b)^2$ on the resulting expression) collapses this difference into a single, elegant closed form:

$\boxed{\Delta Q = \frac{1}{2}\left(\frac{m_1m_2}{m_1+m_2}\right)(u_1-u_2)^2}$ …