A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 m s$$^{-1}$$ with respect to the man. The speed of the man with respect to the surface is:

We begin by noting that the surface is perfectly smooth, so there is no external horizontal force acting on the combined system of the man and his son. Because of this, the total linear momentum of the system in the horizontal direction must remain constant. This is simply an application of the Law of Conservation of Linear Momentum, which states:

$$\text{Total initial momentum} \;=\;\text{Total final momentum}.$$

Initially, both the man and the son are at rest, so their initial momenta are zero. Hence, the total initial momentum of the system is

$$0 + 0 = 0.$$

Now the man pushes the son. Let us choose the direction in which the son eventually moves as the positive direction. We define:

$$v_s = \text{speed of the son with respect to the ground}$$ $$v_m = \text{speed of the man with respect to the ground}.$$

Because they move apart while facing each other, the man travels in the opposite (negative) direction to the son. Therefore, in our chosen sign convention, the man's final momentum is negative. Applying the conservation of momentum, we write

$$\bigl(\text{mass of son}\bigr)\,(+v_s) \;+\;\bigl(\text{mass of man}\bigr)\,(-v_m) \;=\;0.$$

Substituting their masses, we have

$$20\,v_s \;-\;50\,v_m \;=\;0.$$

Solving this linear equation for $$v_m$$ gives

$$50\,v_m = 20\,v_s,$$ $$v_m = \frac{20}{50}\,v_s,$$ $$v_m = 0.4\,v_s.$$

Next, we use the information about their relative speed. The son moves away from the man at a speed of $$0.70\;\text{m s}^{-1}$$ with respect to the man. The relative speed of two bodies moving in opposite directions equals the sum of their individual speeds with respect to the ground. Therefore, we write

$$v_{\text{relative}} = v_s + v_m.$$

Substituting the given relative speed and the relationship $$v_m = 0.4\,v_s$$ obtained above, we get

$$0.70 = v_s + v_m = v_s + 0.4\,v_s = 1.4\,v_s.$$

Now we solve for $$v_s$$:

$$v_s = \frac{0.70}{1.4} = 0.50\;\text{m s}^{-1}.$$

Substituting this back into $$v_m = 0.4\,v_s$$ gives

$$v_m = 0.4 \times 0.50 = 0.20\;\text{m s}^{-1}.$$

Because we assigned the man's motion opposite to the son's as the negative direction, the magnitude of his speed with respect to the surface is simply $$0.20\;\text{m s}^{-1}.$$

Hence, the correct answer is Option A.