A block of mass `M` is tied to one end of a massless rope. The other end of the rope is in the hands of a man of mass `2M` as shown in Fig. The block and the man art resting on a rough wedge of mass `M`. The whole system is resting on a smooth horizontal surface. The man starts walking towards right while holding the rope in his hands. Pulley is massless and frictionless. Find the displacement of the wedge when the block meets the pulley. Assume wedge is sufficiently long so that man does not fall down.

Options

A. `1/2 m` towards right
B. `1/2 m` towards left
C. The wedge does not move at all
D. `1 m` towards left

Answer

B

Explanation

Let `x` be displacement of wedge w,.er.t ground `y` be displacement of man w.r.t ground and `z` be diaplacement of block w.r.t ground. As the length of string will remain same, so
`(l_(1)+z-x)+(l_(2)+y-x)=l_(1)+l_(2)`
`implies z+y=2x`…………i

Also if the pulley and the blcok met, then we can write
`x=z+2` ..............ii
`/_\x_(cm)=0implies(Mx+2My+Mz)/(4M)=0`
`implies x+2y+z=0`.............iii
From eqn i, ii and iii
`x=(-1)/2m,y=3/2m,z=(-5)/2m`