Answer
$\frac{2}{5}v, \frac{7}{5}v$
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:
$v_{1f}=\frac{4m-m}{5m}v+\frac{2m}{5m}(2v)=\frac{7}{5}v$
We use conservation of momentum to obtain:
$2mv+4mv = \frac{7}{5}(4m)v+mv_f$
$v_f = \frac{2}{5}v$
