Past exam of the physics 1 paper of the University of Oxford / 2019 / 2

Let $\mathbf{w}={\mathbf{v}}_1-{\mathbf{v}}_2$ be the relative velocity and $\mathbf{V}=(M_1{\mathbf{v}}_1+M_2{\mathbf{v}}_2)/(M_1+M_2)$ the centre-of-mass velocity. Inverting givesSubstitute into the sum of the two kinetic energies. The cross terms cancel, while the coefficient of $|\mathbf{w}{|}^2/2$ is the reduced mass $\mu =M_1{M}_2/(M_1+M_2)$. HenceThe first term is translation of the centre of mass; the second is the kinetic energy measured in the centre-of-momentum frame. For an isolated perfectly inelastic collision, the bodies finish with a common velocity, while their total momentum and thus $\mathbf{V}$ are unchanged. The kinetic energy converted to internal forms is thereforewhere $v$ is their initial relative speed. Total energy remains conserved; it is the translational-plus-relative kinetic energy that decreases.