Q 11-04-193JEE MainJEE Main 2019 (12 Apr, Shift 1)Easy
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 $0.70\ \text{m s}^{-1}$ with respect to the man. The speed of the man with respect to the surface is:
Answer: (A) $0.20\ \text{m s}^{-1}$
Momentum is conserved: $50v_m = 20v_s$, with the two moving in opposite directions and $v_s + v_m = 0.70$.
$$v_m = \frac{20}{70}\times0.70 = 0.20\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics