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

$0.3m$
The maximum compression of the spring can be determined as: According to the law of conservation of momentum $m_av_{a,1}+m_bv_{b,1}=(m_a+m_b) v$ $\implies v=\frac{m_av_{a,1}+m_bv_{b,1}}{m_a+m_b}$ $\implies v=\frac{0+10(15)}{15+10}=6m/s$ Now according to the law of conservation of energy $\frac{Ks^2}{2}=\frac{(m_a+m_b) v^2}{2}$ $\implies s^2=\frac{(m_a+m_b) v^2}{K}$ We plug in the known values to obtain: $s^2=\frac{(15+10)36}{10000}$ $s^2=0.09m^2$ $\implies s=0.3m$