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

$9.52^{\circ}$
According to the law of conservation of momentum $\Sigma mv_1=\Sigma mv_2$ $\implies 0=m_A v_A+m_Bv_B$ $\implies v_A=v_B=v$ Now, according to the law of conservation of energy $\frac{1}{2}m_Av_{A_{\circ}}^2+\frac{1}{2}m_Bv_{B_{\circ}^2}+\frac{1}{2}KS^2=\frac{1}{2}mv_A^2+\frac{1}{2}m_Bv_B^2+0$ $\implies KS^2=m_Av^2+m_Bv^2$ $\implies (60)(0.3)^2=5v^2+5v^2$ This simplifies to: $v=0.735m/s$ Now, we can determine the required angle as $\frac{1}{2}mv^2+0=0+mgh(1-cos\theta)$ We plug in the known values to obtain: $\frac{1}{2}(5)(0.735)^2=5(9.81)(2)(1-cos\theta)$ $\implies \theta=\phi=9.52^{\circ}$