運動量保存の法則より \[ m_A v_A=m_A {v_A}' +m_B {v_B}' \] 水平方向 \[ m_A v_A=m_A {v_A}' \cos\theta_A +m_B {v_B}' \cos\theta_B \] 垂直方向 \[ 0=m_A {v_A}' \sin\theta_A - m_B {v_B}' \sin\theta_B\\ \] \[ m_A {v_A}' \sin\theta_A = m_B {v_B}' \sin\theta_B \] $m_A=m_B$より \[ {v_A}' \sin\theta_A={v_B}' \sin\theta_B \] \[ \therefore {v_B}'=\frac{\sin\theta_A}{\sin\theta_B}{v_A}' \] 水平方向でも$m_A=m_B$より \[\begin{align} v_A &={v_A}'\cos\theta_A+{v_B}'\cos\theta_B\\ &={v_A}'\cos\theta_A+\left(\frac{\sin\theta_A}{\sin\theta_B}\times {v_A}' \right)\cos\theta_B\\ &={v_A}' \left(\cos\theta_A +\frac{\sin\theta_A}{\sin\theta_B}\times \cos\theta_B \right)\\ &={v_A}'\left(\cos\theta_A+ \frac{\sin\theta_A}{\tan\theta_B} \right) \end{align}\] \[\begin{align} {v_A}'=\frac{v_A}{\cos\theta_A+ \frac{\sin\theta_A}{\tan\theta_B}} &=\frac{6.0\,\rm{m/s}}{\cos30^{\circ} + \frac{\sin30^{\circ}}{\tan60^{\circ}}} \\ &=\frac{6.0\,\rm{m/s}}{\frac{\sqrt{3}}{2}+ \frac{\frac{1}{2}}{\sqrt{3}}}\\ &=\frac{6}{\frac{\sqrt{3}}{2}+\frac{1}{2\sqrt{3}}}\\ &=\frac{6\times 2\sqrt3}{\sqrt3\times \sqrt3+1}\\ &=\frac{12\sqrt{3}}{4}\\ &=3\sqrt3\ \rm{m/s}\\ &=5.196\ \rm{m/s} \end{align}\] \[ \therefore {v_A}'=5.20\ \rm{m/s} \] \[\begin{align} {v_B}'=\frac{\sin\theta_A}{\sin\theta_B}{v_A}' &=\frac{\sin30^\circ}{\sin60^\circ}\times3\sqrt3\,\rm{m/s}\\ &=\frac{\frac{1}{2}}{\frac{\sqrt3}{2}}\times3\sqrt3\,\rm{m/s}\\ &=\frac{1}{\sqrt{3}}\times3\sqrt{3}\\ &=3 \end{align}\] \[ \therefore {v_B}'=3.00\ \rm{m/s} \]