Centre of mass

Assume that the block of mass $M$ moves with an acceleration $A$ in negative $x$ direction , and the block of mass $m$ has a relative acceleration $a_r$ with respect to the block along the incline.

Now, since there is no external force, total force along horizontal direction can be equated to $0$ : $\implies MA+m(A-a_r \cos\theta) =0$

and writing the net force equation on the block of mass $m$ along the incline yields: $mg \sin \theta=m(a_r-A\cos\theta)$

Which on solving gives $A=\dfrac{mg\sin\theta\cos\theta}{M+m\sin^2 \theta} \\ a_r=\dfrac{(M+m)g\sin^2 \theta}{M+m\sin^2 \theta}$

Now , acceleration vectors of both the masses can be given by $\overrightarrow{A}=-\dfrac{mg\sin \theta \cos \theta }{M+m\sin^2 \theta} \hat{i} \\ \overrightarrow{a}=(a_r \cos\theta -A ) \hat{i} - a_r \sin\theta \hat{j}$ .

Note that acceleration of centre of mass will be given by $\overrightarrow{A}_{COM}=\dfrac{m\overrightarrow{a}+M\overrightarrow{A}}{M+m}$ , and it can be noted upon by putting the values that acceleration of centre of mass remains in the vertical direction which was indeed true as there are no forces acting in the horizontal direction and thus centre of mass follows a straight line path downwards.

By substitutions, we get $\overrightarrow{A}_{COM}=-\dfrac{mg\sin^2 \theta}{M+m\sin^2\theta} \hat{j}$ .

Now, we can calculate the value of $g$ by using the value of acceleration of $M$ , which comes out to be $5 ms^{-2}$ downwards.

Now, plugging in all the values, we obtain $a_{com}=\boxed{1 ms^{-2}}$ .

a(com)along x dirn will be zero . let acch of block be A. resolve A along x and y axis. A(com) in x =4(A/2-1)+8(-1)=0 A=6. a(com) in y = 4*3/12=1