Example 2 In the tetrahedron $ABCD$, let $AB=1, CD=\sqrt{3}$, the distance between the lines $AB$ and $CD$ is 2, and the angle between them is $\frac{\pi}{3}$. Find the volume of the tetrahedron $ABCD$.
Solution 1 As shown in Figure $16-2$, draw $B E \Perp C D$, and connect $A E, B E, D E$. Then quadrilateral $B C D E$ is a parallelogram.
Since the angle formed by the skew lines $A B$ and $C D$ is $60^{\circ}$, we have $\angle A B E=60^{\circ}$
Thus: $S_{\triangle A B E}=\frac{1}{2} A B \cdot B E \sin 60^{\circ}=\frac{3}{4}$
Since $C D / / B E$, it follows that $C D / /$ plane $A B E$.
Therefore, the distance $h$ from point $D$ to plane $A B E$ is equal to the distance between lines $A B$ and $C D$, i.e., $h=2$. Hence $V_{A-A C D}=V_{A-B l D:}=V_{D-A B E}=\frac{1}{3} \times \frac{3}{4} \times 2=\frac{1}{2}$
Solution 2 As shown in Figure $16-3$, take the midpoints $P, Q, R, S$ of $B C, C A, A D, D B$ respectively, then quadrilateral $P Q R S$ is a parallelogram with $\angle Q P S=60^{\circ}$,
and $P Q=\frac{1}{2} A B=\frac{1}{2}, P S=\frac{1}{2} C D=\frac{\sqrt{3}}{2}$
$A B / /$ plane $P Q R S$, and $C D / /$ plane $P Q R S$
Let the distances from $A B, C D$ to plane $P Q R S$ be $h_{1}, h_{2}$ respectively, then $h_{1}+h_{2}=2$. By the formula for the volume of a frustum, we get
$$
\begin{aligned}
V_{\text {ASK U }} & =V_{\text {AS-PQK: }}+V_{(T) \text {-YRSS }}=\frac{1}{3}\left(h_{1}+h_{2}\right) \cdot S_{\text {IQRS }} \\
& =\frac{1}{3} \times 2 \times \frac{3}{4}=\frac{1}{2}
\end{aligned}
$$