In the arrangement shown in fig. the block of mass `m=2kg` lies on the wedge of mass `M=8kg`. The initial acceleration of the wedge, if the surfaces are smooth, is

Options

A. `(sqrt(3)g)/(23) ms^(-2)`
B. `(3sqrt(3)g)/(23) ms^(-2)`
C. `(3g)/(23) ms^(-2)`
D. `(g)/(23) ms^(-2)`

Answer

B

Explanation

If initial acceleration of M towards right is A, then we can shown that acceleration of m w.r.t. M down the incline is

`a=A(1+cos theta)(3A)/(2) ( because theta = 60^(@))`
FBD of block m (w.r.t. M) is shown below:
FBD of M
Equartion of motion:
for m: `mg sqrt(3)/(2) + mA xx 1/2 -T =m 3/2 A` ..(i)
`N+mA (sqrt(3))/(2)=mg 1/2` ...(ii)
For M: `T+N(sqrt(3))/(2)=MA` ..(iii)
From eqs. (i), (ii) and (iii) `A=(3sqrt(3)g)/(23) ms^(-1)`.