## Engineering Mechanics: Statics & Dynamics (14th Edition)

$2.31m/s$ $3.46m/s$
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: $v_{b,2}=-\frac{15(2.31)}{10}=3.46m/s$