A man of mass M stands at one end of plank of length L which lies at rest on a frictionless surface. The man walks to the other end of the plank. If the mass of the plank is `(M/3)`, the distance that the man move relative to the ground is

Explanation

The net external force on the system (man + plank) is zero (their weights are balanced by the normal reaction on the sled and the ice is frictionless). The centre of mass of the system is to be at same position during their movements. Obviously, the plank will shift towards left.
Let the displacement of plank relative to ground = x
The displacement of man relative to ground `= - (L-x)`
Moments of masses about C.M. = 0
`Mx-m(L-x)=0`
`rArr Mx = m(L-x)`
`rArr x = (mL)/(m+M)`
As a shortcut remember the distance moved by bodies w.r.t. ground is in the inverse ratio of their masses.