Example 2 In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of $A B$. Fold $\triangle A C M$ along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. Find the volume of the tetrahedron $A-B C M$.
Solution: The figures before and after folding are shown in Figure 3 and Figure 4, respectively. By comparing the two figures, we can see that in Figure 4, $A C=2, A M=2, B C=2 \sqrt{3}, \angle A C M=60^{\circ}, \angle B C M=30^{\circ}$.
Take the midpoint $D$ of $C M$, and connect $A D$.
In $\triangle B C M$, draw $D E \perp C M$ intersecting $B C$ at $E$, and connect $A E$. Then
$$
\begin{array}{l}
A D=A C \sin 60^{\circ}=\sqrt{3}, \\
D E=C D \tan 30^{\circ}=\frac{\sqrt{3}}{3}, \\
C E=2 D E=\frac{2 \sqrt{3}}{3} .
\end{array}
$$

In $\triangle A B C$, since $A C^{2}+A B^{2}=B C^{2}$, then $\angle B A C=90^{\circ}$. Therefore, $\cos \angle A C E=\frac{\sqrt{3}}{3}$.
In $\triangle A C E$, by the cosine rule, we get $A E^{2}=\frac{8}{3}$.
Calculations show that
$$
A E^{2}+C E^{2}=A C^{2}, A E^{2}+D E^{2}=A D^{2} \text {. }
$$

Thus, $A E \perp B C, A E \perp D E$.
Therefore, $A E \perp$ plane $B C M$. Hence,
$$
\begin{array}{l}
V_{A-B C M}=\frac{1}{3} A E \cdot S_{\triangle B C M} \\
=\frac{1}{3} \times \frac{2 \sqrt{6}}{3} \times \frac{1}{2} \times 2 \times 2 \sin 60^{\circ}=\frac{2 \sqrt{2}}{3} .
\end{array}
$$