A

Step 1: Understand the parameters of the problem.
We have two balls: one with mass 3m (which we will call ball A) moving with a velocity v, and another with mass m (ball B) that is stationary.

Step 2: Apply the principles of elastic collision.
In an elastic collision, both momentum and kinetic energy are conserved.

Step 3: Write down the conservation of momentum:
The total momentum before the collision = The total momentum after the collision.
Therefore, we have:
$$3m imes v + m imes 0 = 3m imes v_A + m imes v_B$$
$$3mv = 3mv_A + mv_B$$
Let's simplify this:
$$3v = 3v_A + v_B ag{1}$$

Step 4: Write down the conservation of kinetic energy:
The total kinetic energy before the collision = The total kinetic energy after the collision.
Thus, we have:
$$\frac{1}{2} (3m)v^2 + 0 = \frac{1}{2}(3m)v_A^2 + \frac{1}{2}m v_B^2$$
Simplifying this we get:
$$3v^2 = 3v_A^2 + v_B^2 ag{2}$$

Step 5: Solve the equations (1) and (2).
From equation (1), we can express v_B in terms of v_A:
$$v_B = 3v - 3v_A ag{3}$$

Substitute (3) into (2):
$$3v^2 = 3v_A^2 + (3v - 3v_A)^2$$
Expand the right side:
$$3v^2 = 3v_A^2 + (9v^2 - 18vv_A + 9v_A^2)$$
Combine like terms:
$$0 = 6v_A^2 - 18vv_A + 6v^2$$
Divide the whole equation by 6:
$$0 = v_A^2 - 3vv_A + v^2$$
This is a quadratic equation in v_A. We can use the quadratic formula:
$$v_A = \frac{3v \pm \sqrt{(3v)^2 - 4 imes 1 imes v^2}}{2 imes 1}$$
$$= \frac{3v \pm \sqrt{9v^2 - 4v^2}}{2} = \frac{3v \pm \sqrt{5v^2}}{2}$$
$$= \frac{3v \pm \sqrt{5}v}{2}$$
Thus, there are two potential solutions:
$$v_A = \frac{(3 + \sqrt{5})v}{2} ext{ or } v_A = \frac{(3 - \sqrt{5})v}{2}$$
Since we want the velocity after collision for ball A and knowing that it cannot exceed ball A's initial velocity, we select:
$$v_A = \frac{(3 - \sqrt{5})v}{2}$$

Step 6: Substitute back to find v_B.
Using equation (3):
$$v_B = 3v - 3 \left(\frac{(3 - \sqrt{5})v}{2}\right) = 3v - \frac{(9 - 3\sqrt{5})v}{2}$$
$$= \frac{6v}{2} - \frac{(9 - 3\sqrt{5})v}{2}$$
$$= \frac{(6 - 9 + 3\sqrt{5})v}{2} = \frac{(-3 + 3\sqrt{5})v}{2}$$

Thus, the final velocities are:
(i) The velocity of ball of mass m (ball B) is: $$v_B = \frac{(3\sqrt{5} - 3)v}{2}$$
(ii) The velocity of ball of mass 3m (ball A) is: $$v_A = \frac{(3 - \sqrt{5})v}{2}$$.
Therefore, the correct answers are:
V_A = (3 - √5)v/2
V_B = (3√5 - 3)v/2.