Solution

First, let's consider the action of the force on the 1st particle. By condition, it is known that it began to move in the opposite direction to the original one. Consequently, the force acted against the particle's motion. Then the impulse of the force in the direction of the force's action (x-axis) is:
$$F \cdot \Delta t = m \cdot 2v - (-mv) = 3mv$$

Now let's consider the motion of the second particle. The force acts perpendicularly to its motion; therefore, the velocity will remain $2v$ in the direction of its original motion. On the other hand, the acting force will impart a velocity to the particle of mass $2m$ in the direction of its action. The impulse of the force:
$$F \cdot \Delta t = 2mv_x - 0 = 2mv_x$$

After transformations we obtain:
$$2mv_x = 3mv$$
From which it follows that:
$$v_x = \frac{3}{2}v$$

We conclude: the second particle has its original velocity $2v$ in the direction of its original motion and $v_x = \frac{3}{2}v$ acquired under the action of the force — directed perpendicular to the velocity $2v$.

We use the Pythagorean theorem to find the required velocity:
$$V = \sqrt{\frac{9}{4}v^2 + 4v^2} = \frac{5}{2}v$$

Note: the velocity vector of the second particle has a direction that can be determined from the relation $\tan\alpha = \frac{v_x}{2v} = \frac{3}{4}$, which gives an angle of $36.87^\circ$ to its original direction.

Alternative solution:

Change in momentum of the first particle: $m(2v - (-v)) = 3mv$
The same change in momentum will occur for the second particle since $\Delta p = F\Delta t$.
Since the initial momentum of the second particle is perpendicular to the first, we use the Pythagorean theorem for the addition of momenta:
$$p_2 = \sqrt{(4mv)^2 + (3mv)^2} = 5mv$$
Then its velocity:
$$v_2 = p_2 / 2m = 2.5v$$