Let us draw free body diagram of each body and fix the coordinate system, as shown in the figure. After analysing the motion of $M$ and $m$ on the basis of force diagrams, let us draw the kinematical diagram for accelerations (Fig.). As the length of threads are constant so, $d s_{m M}=d s_{M}$ and as $\vec{v}_{m M}$ and $\vec{v}_{M}$ do not change their directions that why
$\mathbf{4 5}$
As $\vec{w}_{m}=\vec{w}_{m M}+\vec{w}_{M}$
so, from the triangle law of vector addition $w_{m}=\sqrt{2} w$
From the Eq. $F_{x}=m w_{x}$, for the wedge and block :
$T-N=M w$
and $\quad N=m w$
Now, from the Eq. $F_{y}=m w_{y}$, for the block $m g-T-k N=m w$
Simultaneous solution of Eqs. (2), (3) and (4) yields :
$m$$$
w=\frac{m g}{(k m+2 m+M)}=\frac{g}{(k+2+M / m)}
$$
Hence using Eq. (1)
$$
w_{m}=\frac{g \sqrt{2}}{(2+k+M / m)}
$$