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:
$$|\vec p_{1,i}| = |\vec p_{2,i}|$$
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