15. (1998 National High School Competition Question) In $\triangle ABC$, $\angle C=90^{\circ}, \angle B=30^{\circ}, AC=2, M$ is the midpoint of $AB$. $\triangle ACM$ is folded along $CM$, making the distance between $A$ and $B$ equal to $2\sqrt{2}$. At this point, the volume of the tetrahedron $A-BCM$ is $\qquad$.
15. Fill in $\frac{2 \sqrt{2}}{3}$. Reason: The tetrahedron $A-B C M$ after folding is shown in the figure on the right. 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 $A D=2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}, D E=C D \cdot \tan 30^{\circ}=1 \cdot \frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{3}, C E=$ $2 D E=\frac{2}{3} \sqrt{3}$. In $\triangle A B C$, $A C=2, A B=2 \sqrt{2}, B C=2 \sqrt{3}$, so $A C^{2}+A B^{2}=B C^{2}$, therefore $\angle B A C=90^{\circ}$. In $\triangle A C E$, $A E^{2}=A C^{2}+C E^{2}-2 A C \cdot C E \cdot \cos \angle A C E=4+$ $\frac{4}{3}-2 \cdot 2 \cdot \frac{2}{3} \sqrt{3} \cdot \frac{2}{2 \sqrt{3}}=\frac{8}{3}$, so $A E^{2}+C E^{2}=\frac{8}{3}+\frac{4}{3}=4=A C^{2}$. Also, $A E^{2}+D E^{2}=\frac{8}{3}+\frac{1}{3}=3=$ $A D^{2}$, thus $A E \perp B C, A E \perp D E$, hence $A E \perp$ plane $B C M$, so $V_{A-B C M}=\frac{1}{3} A E \cdot S_{\triangle B C M}=\frac{2}{3} \sqrt{2}$.