A block of mass m slides down on an inclined plane of a wedge of same mass m (see fig.). Friction is absent everywhere. Which of the following statement(s) is/are correct?

Let a be the acceleration of wedge leftwards and $a_r$ be the relative acceleration of block down the plane. Then, absolute acceleration of block in horizontal direction will be $\left({\text{a}}_r\text{\hspace{0.17em}cosθ\hspace{0.17em}-\hspace{0.17em}a}\right)$rightwards.${\overline{\text{F}}}_{\text{horizontal\hspace{0.17em}\hspace{0.17em}}}\text{=\hspace{0.17em}0\hspace{0.17em}\hspace{0.17em}}\Rightarrow {\overline{\text{a}}}_{\text{horizontal\hspace{0.17em}}}\text{\hspace{0.17em}=\hspace{0.17em}0}$

$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}a}}_r\text{\hspace{0.17em}cosθ\hspace{0.17em}-\hspace{0.17em}a\hspace{0.17em}=\hspace{0.17em}a}$

or    ${\text{2a\hspace{0.17em}=\hspace{0.17em}a}}_r\text{\hspace{0.17em}cosθ\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{...}\left(\text{1}\right)$

FBD of wedge             FBD of block w.r.t. wedge

$\text{N\hspace{0.17em}sin\hspace{0.17em}θ\hspace{0.17em}=\hspace{0.17em}ma\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{...}\left(\text{ii}\right)$

Net force on block perpendicular to plane is zero.

So, $\text{N\hspace{0.17em}+\hspace{0.17em}masin\hspace{0.17em}θ\hspace{0.17em}=\hspace{0.17em}mgcos\hspace{0.17em}θ\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{...}\left(\text{iii}\right)$

On solving eqs. (i), (ii) and (iii), we get

${\text{a}}_{\text{r}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{2g\hspace{0.17em}sinθ}}{{\text{1+sin}}^{\text{2}}\text{θ}}$

For block, ${\text{a}}_{\text{y}}{\text{\hspace{0.17em}=\hspace{0.17em}a}}_{\text{r}}\text{\hspace{0.17em}sinθ\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{2g\hspace{0.17em}sin}}^{\text{2}}\text{θ}}{{\text{1+sin}}^{\text{2}}\text{θ}}$

${\text{a}}_{\text{COM}}\text{=}\frac{\text{m}\left(\text{0}\right)\text{+m}\left[\frac{{\text{2g\hspace{0.17em}sin}}^{\text{2}}\text{θ}}{{\text{1+sin}}^{\text{2}}\text{θ}}\right]}{\text{m\hspace{0.17em}+m}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{g\hspace{0.17em}sin}}^{\text{2}}\text{θ}}{{\text{1+sin}}^{\text{2}}\text{θ}}$