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
\[\begin{align}
v_A
&={v_A}'\cos\theta_A+{v_B}'\cos\theta_B\\
&={v_A}'\cos\theta_A+\left(\frac{\sin\theta_A}{\sin\theta_B}\times {v_A}' \right)\cos\theta_B\\
&={v_A}' \left(\cos\theta_A +\frac{\sin\theta_A}{\sin\theta_B}\times \cos\theta_B \right)\\
&={v_A}'\left(\cos\theta_A+ \frac{\sin\theta_A}{\tan\theta_B} \right)
\end{align}\]
\[\begin{align}
{v_A}'=\frac{v_A}{\cos\theta_A+ \frac{\sin\theta_A}{\tan\theta_B}}
&=\frac{6.0\,\rm{m/s}}{\cos30^{\circ} + \frac{\sin30^{\circ}}{\tan60^{\circ}}} \\
&=\frac{6.0\,\rm{m/s}}{\frac{\sqrt{3}}{2}+ \frac{\frac{1}{2}}{\sqrt{3}}}\\
&=\frac{6}{\frac{\sqrt{3}}{2}+\frac{1}{2\sqrt{3}}}\\
&=\frac{6\times 2\sqrt3}{\sqrt3\times \sqrt3+1}\\
&=\frac{12\sqrt{3}}{4}\\
&=3\sqrt3\ \rm{m/s}\\
&=5.196\ \rm{m/s}
\end{align}\]
\[
\therefore
{v_A}'=5.20\ \rm{m/s}
\]
\[\begin{align}
{v_B}'=\frac{\sin\theta_A}{\sin\theta_B}{v_A}'
&=\frac{\sin30^\circ}{\sin60^\circ}\times3\sqrt3\,\rm{m/s}\\
&=\frac{\frac{1}{2}}{\frac{\sqrt3}{2}}\times3\sqrt3\,\rm{m/s}\\
&=\frac{1}{\sqrt{3}}\times3\sqrt{3}\\
&=3
\end{align}\]
\[
\therefore
{v_B}'=3.00\ \rm{m/s}
\]