Answer
The proof is below.
Work Step by Step
We know the following equation for elastic collisions:
$v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i}$
Thus, we find:
$v_{1f}=\frac{3m-m}{4m}v+\frac{2m}{4m}(-v)=0$
Momentum is conserved, so it follows:
$m(-v)+3mv=mv_f$
$v_f=2v$
Thus, the second mass rebounds at twice the speed.
