2D Elastic Collision in the Centre-of-Mass Reference frame

Problem statement:

Two particles with different masses collide elastically. Show that in the centre-of-mass reference frame the final speeds of the two balls are the same as before the collision.

In the centre-of-mass reference frame the momenta of the two before have equal magnitudes: $$|\vec p_{1,i}| = |\vec p_{2,i}|$$ Layer 1 In the centre-of-mass reference frame.

Solution

Because this is an elastic collision we can write $$K_{\rm total,i} = K_{\rm total,f}$$ $$ \frac{p^2_{1,i}}{2m_1} + \frac{p^2_{2,i}}{2m_2} = \frac{p^2_{1,f}}{2m_1} + \frac{p^2_{2,f}}{2m_2}$$ Because we're in the c.m. reference frame, $|\vec p_{1,i}| = |\vec p_{2,i}|$ and $|\vec p_{1,f}| = |\vec p_{2,f}|$ $$ \frac{p^2_{1,i}}{2m_1} + \frac{p^2_{1,i}}{2m_2} = \frac{p^2_{1,f}}{2m_1} + \frac{p^2_{1,f}}{2m_2}$$ Let's get rid of the 2 in the denominator and factor out the $p$ terms. $$ \left( \frac{1}{m_1} + \frac{1}{m_2} \right) p^2_{1,i} = \left( \frac{1}{m_1} + \frac{1}{m_2} \right) p^2_{1,f}$$ $$ p^2_{1,i} =p^2_{1,f} $$ $$ (m_1v_{1,i})^2 = (m_1v_{1,f})^2 $$ $$ v^2_{1,i} =v^2_{1,f} $$ $$ |\vec v_{1,i}| =|\vec v_{1,f}| $$
N. Alberding, 2013