C

Step 1: In an elastic collision, both momentum and kinetic energy are conserved.

Step 2: Let the mass of each particle be m. Before the collision, particle P has a momentum of mv (moving with speed v), and particle Q is at rest (momentum = 0). Hence, total initial momentum = mv.

Step 3: After the collision, let particle P have a velocity of u (unknown) and particle Q have a velocity of v (which we will prove). The total final momentum will then be mu + mv. According to conservation of momentum:
\( mv = mu + mv \)
Thus we find that \( u + v = v \) implies \( u = 0 \). This means particle P comes to rest after the collision.

Step 4: By conservation of kinetic energy, since kinetic energy before collision = kinetic energy after collision, we also have:
\( \frac{1}{2}mv^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 \)
which simplifies to determine that Q must move with speed v.

Step 5: Therefore, in the end, particle P comes to rest, and particle Q moves forward with speed v.
Thus, the correct answer is C.