5. In a regular tetrahedron \( P-ABC \), \( AB=1, AP=2 \). A plane \(\alpha\) passing through \( AB \) bisects its volume. Then the cosine of the angle formed by the edge \( PC \) and the plane \(\alpha\) is \(\qquad\)
Let the midpoints of $AB$ and $AC$ be $K$ and $M$, respectively. It is easy to prove that the plane $ABM$ is the plane $\alpha$.
By the median length formula, we know $AM^2 = \frac{3}{2}$.
Thus, $KM = \sqrt{AM^2 + AK^2} = \sqrt{\frac{3}{2} - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{5}}{2}$.
It is also easy to see that the projection of line $PC$ onto plane $\alpha$ is line $MK$, and $CM = 1$, $KC = \frac{\sqrt{3}}{2}$. Therefore,
$$
\cos \angle KMC = \frac{KM^2 + MC^2 - KC^2}{2 \cdot KM \cdot MC} = \frac{3 \sqrt{5}}{10}.
$$

Thus, the cosine of the angle between the edge $PC$ and the plane $\alpha$ is
$$
\text{ } \frac{3 \sqrt{5}}{10} \text{. }
$$