Example 4 In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of $A B$, and $\triangle A C M$ is folded along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. At this moment, the volume of the tetrahedron $A-B C M$ is $\qquad$ .
Analysis: The first difficulty is how to make a more intuitive three-dimensional figure? Before folding, $M A=M B$ $=M C=A C=2$, these lengths will not change after folding into a pyramid. From the basic conclusion 5, we know that the projection of point $M$ on plane $A B C$ after folding is the circumcenter of $\triangle A B C$; also, $B C=2 \sqrt{3}$, and after folding, $A B=2 \sqrt{2}$, so $\angle B A C=90^{\circ}$ after folding. Therefore, the circumcenter of $\triangle A B C$ is the midpoint of its hypotenuse $B C$, i.e., the projection of $M$ on plane $A B C$ is the midpoint $D$ of $B C$, which inspires us to make Figure 5. Moreover, $\angle C B M=30^{\circ}$ will not change after folding, thus,
$$
M D=\frac{1}{2} M B=1, S_{\triangle A B C}=\frac{1}{2} A C \cdot A B=2 \sqrt{2} \text {, }
$$

Therefore, $V_{M-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot M D=\frac{2 \sqrt{2}}{3}$.