Answer
$2.31m/s$
$3.46m/s$
Work Step by Step
According to the law of conservation of energy
$\frac{Ks^2}{2}=\frac{m_av_{a,2}^2}{2}+\frac{m_bv_{b,2}^2}{2}$
This simplifies to:
$v_{b,2}^2=\frac{Ks^2}{m_b\cdot(\frac{m_b}{m_a}+1)}$
We plug in the known values to obtain:
$v_{b,2}^2=\frac{5000(0.04)}{15(\frac{15}{10}+1)}$
$\implies v_{b,2}=2.31m/s$
Now to determine the velocity of block A, we use the law of conservation of momentum
$m_av_{a,1}+m_bv_{b,1}=m_av_{a,2}+m_b v_{b,2}$
$\implies 0=m_av_{a,2}+m_bv_{b,2}$
$v_{a,2}=-\frac{m_b v_{b,2}}{m_a}$
We plug in the known values to obtain:
$v_{b,2}=-\frac{15(2.31)}{10}=3.46m/s$
