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

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

$\mathrm{a}=\mathrm{A}(1+\cos \mathrm{\theta })=\frac{3\mathrm{A}}{2}\left(\because \mathrm{\theta }={60}^{\circ }\right)$

FBD of block m (w.r.t. M) is shown below:

FBD of M (figure) Equation of motion: 

$\mathrm{For} \mathrm{m} : \mathrm{mg}\frac{\sqrt{3}}{2}+\mathrm{mA}\times \frac{1}{2}-\mathrm{T}=\mathrm{m}\frac{3}{2}\mathrm{A} ...\left(\mathrm{i}\right)$

$\mathrm{N}+\mathrm{mA}\frac{\sqrt{3}}{2}=\mathrm{mg}\frac{1}{2} ...\left(\mathrm{ii}\right)$

$\text{ For }\mathrm{M} : \mathrm{T}+\mathrm{N}\frac{\sqrt{3}}{2}=\mathrm{MA} ...\left(\mathrm{iii}\right)$

From Eqs. (i), (ii), and (iii)  $\mathrm{A}=\frac{3\sqrt{3}\mathrm{g}}{23}{\mathrm{ms}}^{-2}$