Past exam of the physics 1 paper of the University of Oxford / 2014 / 1

Use nonrelativistic mechanics and let $\mathbf{V}=(m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2)/(m_1+m_2)$ be the constant centre-of-mass velocity. In the centre-of-mass frame, conservation of momentum gives $m_1{\mathbf{u}}_1+m_2{\mathbf{u}}_2=0$ before and after the elastic collision. Thus ${\mathbf{u}}_2=-(m_1/m_2){\mathbf{u}}_1$, and the kinetic energy isThe decomposition $T_{\mathrm{lab}}=T_{\ast }+\frac{1}{2}(m_1+m_2)V^2$ and constant centre-of-mass velocity show that the elastic collision preserves $T_{\ast }$, so the positive constant multiplying $u_1^2$ forces that speed to remain unchanged. The same momentum relation fixes the second speed:Only their directions can change; the two velocities remain antiparallel in this frame. This derivation does not require a head-on collision.

For equal masses, let the incoming laboratory frame velocity be $\mathbf{v}$ and the outgoing velocities be $\mathbf{a},\mathbf{b}$. Conservation of momentum and kinetic energy giveSquaring the first equation and subtracting the second yields $2\mathbf{a}\cdot \mathbf{b}=0$. HenceThe stated oblique collision ensures both outgoing speeds are nonzero. In the degenerate head-on case one particle can stop, and a direction cannot be assigned to its zero velocity.