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 $B C M$ is $\qquad$
5. $\frac{\sqrt{6}}{3}$.

As shown in Figure 1, take the midpoint $D$ of $CM$, and connect $AD$.
It is easy to see that
$AD \perp CM$.
In $\triangle BCM$,

draw $DE \perp CM$ intersecting $BC$ at point $E$, and connect $AE$, knowing that $AE \perp CM$.
Then $AD=\frac{\sqrt{3}}{2}, DE=CD \tan 30^{\circ}=\frac{\sqrt{3}}{6}$,
$$
CE=2DE=\frac{\sqrt{3}}{3} \text{. }
$$

In $\triangle ABC$, we have
$$
\begin{array}{l}
AC=1, AB=\sqrt{2}, BC=\sqrt{3} \\
\Rightarrow AC^{2}+AB^{2}=BC^{2} \\
\Rightarrow \angle BAC=90^{\circ} .
\end{array}
$$

Also, $AC^{2}=1=CE \cdot BC$, by the projection theorem, we know $AE \perp BC, AE=\frac{AC \cdot AB}{BC}=\frac{\sqrt{6}}{3}$.
Since $AE \perp BC, AE \perp CM$, thus, $AE \perp$ plane $BCM$.
Therefore, the distance from point $A$ to plane $BCM$ is $AE=\frac{\sqrt{6}}{3}$.