## Essential University Physics: Volume 1 (3rd Edition)

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.