5. Given a spatial quadrilateral $ABCD$ with diagonals $AC$ $=10 \, \text{cm}, BD=6 \, \text{cm}, M, N$ are the midpoints of $AB, CD$ respectively. If the angle formed by the skew lines $AC, BD$ is $60^{\circ}$, then the length of $MN$ is $\qquad$
5. $\sqrt{19} \mathrm{~cm}$ or $7 \mathrm{~cm}$.

As shown in Figure 6, take the midpoint $E$ of $B C$, and connect $M E$ and $N E$. By the Midline Theorem of a triangle, we have
$$
\begin{array}{l}
M E=\frac{1}{2} A C=5, \\
N E \perp \frac{1}{2} B D=3 .
\end{array}
$$

By the definition of the angle formed by skew lines (not greater than a right angle), the angle between $M E$ and $E N$ is the angle formed by the skew lines $A C$ and $B D$, thus, $\angle M E N=120^{\circ}$ or $60^{\circ}$.
Applying the Law of Cosines in $\triangle M E N$, we get
$$
\cos \angle M E N=\frac{M E^{2}+E N^{2}-M N^{2}}{2 M E \cdot E N},
$$

which means $\pm \frac{1}{2}=\frac{5^{2}+3^{2}-M N^{2}}{2 \times 5 \times 3}$.
Solving for $M N^{2}$, we get $M N^{2}=34 \pm 15$.
Therefore, $M N=7 \mathrm{~cm}$ or $\sqrt{19} \mathrm{~cm}$.