Example 13 (1998 National High School Mathematics League Question) In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=30^{\circ}, A C=2, M$ is the midpoint of $A B$. $\triangle A C M$ is folded along $C M$, making the distance between $A$ and $B$ equal to $2 \sqrt{2}$. At this point, the volume of the tetrahedron $A-B C M$ is $\qquad$ .
Solve: Fill in $\frac{2}{3} \sqrt{2}$. Reason: To find the volume of the tetrahedron after folding, the key is to determine the height of the tetrahedron. It can be observed that after folding, $\triangle A B C$ is a right triangle $(A C=2, B C=2 \sqrt{3}, A B=2 \sqrt{2})$, which helps solve the problem.

The folded tetrahedron is shown in Figure 5-12. 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,
$$
A D=2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}, D E=C D \cdot \tan 30^{\circ}=1 \cdot \frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{3}, C E=
$$
$2 D E=\frac{2 \sqrt{3}}{3}$. In $\triangle A B C$, $A C=2, A B=2 \sqrt{2}, B C=2 \sqrt{3}$, so
$A C^{2}+A B^{2}=B C^{2}$, thus $\angle B A C=90^{\circ}$. In $\triangle A C E$, $A E^{2}=A C^{2}+$
$C E^{2}-2 A C \cdot C E \cos \angle A C E=4+\frac{4}{3}-2 \cdot 2 \cdot \frac{2}{3} \cdot \sqrt{3} \cdot \frac{2}{2 \sqrt{3}}=$
$\frac{8}{3}$, so $A E^{2}+C E^{2}=\frac{8}{3}+\frac{4}{3}=4=A C^{2}$, and $A E^{2}+D E^{2}=\frac{8}{3}+\frac{1}{3}=3=A D^{2}$, therefore $A E \perp B C, A E \perp$ $D E$, hence $A E \perp$ plane $B C M$. Therefore, $V_{A-B C M}=\frac{1}{3} \cdot A E \cdot S_{\triangle H C M}=\frac{1}{3} \cdot \frac{2 \sqrt{6}}{3} \cdot \frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{\sqrt{3}}{2}=\frac{2 \sqrt{2}}{3}$.