5. As shown in Figure 3, the radius of the base of the cone $R=3 \mathrm{dm}$, the slant height $l=5 \mathrm{dm}$, $AB$ is the diameter of the base, $C$ is a point on the circumference of the base, $\angle COB=150^{\circ}$, $D$ is a point on $VB$, $VD$ $=\sqrt{7} \mathrm{dm}$. There is an ant that crawls from point $C$ to $D$ along the surface of the cone. The shortest distance the ant can crawl is ( ) dm.
(A) $3 \sqrt{2}$
(B) $4 \sqrt{2}$
(C) $\frac{15}{2}$
(D) $2 \sqrt{7}$
5.B.

As shown in Figure 7, if the cone is cut along the generatrix $V C$ and then flattened onto the plane $V A B$, point $C$ will land on $C^{\prime}$. Therefore, the shortest path that the ant should take is the straight line $C^{\prime} D$.

According to the arc length formula of a sector, in the base sector $O B C$, $\overparen{B C}=\frac{\alpha_{1} \pi R}{180}$. After the lateral surface is unfolded, in the sector $V B C^{\prime}$, $\overparen{B C^{\prime}}=\frac{\alpha_{2} \pi l}{180}$.
Since $\overparen{B C}=\overparen{B C^{\prime}}$, we have $\alpha_{2}=\frac{R}{l} \alpha_{1}$.
Given $\alpha_{1}=\angle C O B=150^{\circ}, R=3, l=5$, then $\alpha_{2}=\frac{3}{5} \times 150^{\circ}=90^{\circ}$.
Thus, by the Pythagorean theorem,
$$
C^{\prime} D=\sqrt{25+7}=4 \sqrt{2} \text {. }
$$