The correct answer is: C. Bears a constant ratio to their velocity of approach.
Newton’s law of collision of elastic bodies states that when two moving bodies collide each other, the velocities of separation are in the same ratio as the velocities of approach, and the relative speed of separation is equal to the relative speed of approach before the collision.
In other words, if two objects are moving towards each other with velocities $v_1$ and $v_2$, and they collide elastically, then after the collision they will be moving away from each other with velocities $v_1’$ and $v_2’$, such that:
$$\frac{v_1′}{v_2′} = \frac{v_1}{v_2}$$
and
$$v_1′ + v_2′ = v_1 + v_2$$
This law can be derived from the conservation of momentum and the conservation of energy.
Option A is incorrect because the velocities of separation are not directly proportional to the velocities of approach.
Option B is incorrect because the velocities of separation are not inversely proportional to the velocities of approach.
Option D is incorrect because the velocities of separation are not equal to the sum of the velocities of approach.