5. In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=30^{\circ}, A C=1, M$ is the midpoint of $A B$, fold $\triangle A C M$ along $C M$, so that the distance between points $A$ and $B$ is $\sqrt{2}$, then the distance from point $A$ to the plane $B C M$ equals $\qquad$ .
Draw a perpendicular line $A^{\prime} D$ from $A^{\prime}$ to $C M$, intersecting $A B$ at point $E$, and connect $A D, A E$. Then $\triangle A^{\prime} C M$ is an equilateral triangle $\Rightarrow A^{\prime} D=A D=\frac{\sqrt{3}}{2}, A^{\prime} C=1$
$\Rightarrow C E=\frac{\sqrt{3}}{3}, A^{\prime} E=\frac{2 \sqrt{3}}{3} \Rightarrow D E=\frac{\sqrt{3}}{6}$.
Given $A C=1, B C=\sqrt{3}, A B=\sqrt{2} \Rightarrow A C \perp A B, A E \perp B C$, and $A E=\frac{\sqrt{6}}{3}$.
Thus $A E \perp D E \Rightarrow A E \perp$ plane $B C M$. Therefore, the distance from point $A$ to plane $B C M$ is $\frac{\sqrt{6}}{3}$.