A man (mass = 50 k g ) and his son (mass = 20 k g ) 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:

Step-by-step solution

By conservation of linear momentum, P → i = P → f Where P → i and P → f are initial and final momentum of system (boy + man) Now, 0 = m b v → b + m m v → m ......(i) v b = velocity of boy w.r.t. ground v m = velocity of man w.r.t. ground v → bm = velocity of boy w.r.t. man So, v → bm = v → b - v → m or v → b = v → bm + v → m or v → b = 0 . 7 + v → m ........(ii) From equation (i) and (ii), 0 = 20 0 . 7 + v → m + 50 &#160