Answer
$9.52^{\circ}$
Work Step by Step
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:
$\frac{1}{2}(5)(0.735)^2=5(9.81)(2)(1-cos\theta)$
$\implies \theta=\phi=9.52^{\circ}$
