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

$481~m$
We can determine the required maximum compression of the spring as follows: According to the law of conservation of linear momentum $m_Av_{A_1}+m_Bv_{B_1}=(m_A+m_B)v$ We plug in the known values to obtain: $30\times 10^{3}(5.556)+15\times 10^3(-2.778)=(30\times 10^3+15\times 10^3)v$ This simplifies to: $v=2.778m/s$ Now, we use the law of conservation of energy $\frac{1}{2}m_Av_{A_1}^2+\frac{1}{2}m_Bv_{B_1}^2+\frac{1}{2}Ks_1^2=\frac{1}{2}m_Av^2+\frac{1}{2}m_Bv^2+\frac{1}{2}Ks_{max}^2$ We plug in the known values to obtain: $\frac{1}{2}(30\times 10^3)(5.556)^2+\frac{1}{2}(15\times 10^3)(2.778)^2+0=\frac{1}{2}(30\times 10^3)(2.778)^2+\frac{1}{2}(15\times 10^3)(2.778)^2+\frac{1}{2}(3\times 10^6)s_{max}^2$ This simplifies to: $s_{max}=0.4811mm=481.1m$