Block A slides down the wedge of same mass. All surfaces are smooth. The angle of inclination of wedge is θ with horizontal. Match the following columns:

Let a be the acceleration of wedge towards left and ar be the relative acceleration of the block down the planeSo, arcos⁡θ−a=a2a=arcos⁡θFor block A,N+masin⁡θ=mgcos⁡θFor wedge,Nsin⁡θ=maFrom equations (ii), (iii) and (i)ar=2gsin⁡θ1+sin2⁡θand a=gsinθcosθ1+sin2θAbsolute acceleration of block A wrt ground is given bya→A=(arcosθ−a)i^−(arsinθ)j^ ⇒a→A=(gsinθcosθ1+sin2θ)i^−(2gsin2θ1+sin2θ)j^ ⇒|a→A|=gsinθ1+sin2θ(1+3sin2θ)ay = acceleration of block in vertical downward direction=arsin⁡θ=2gsin2⁡θ1+sin2⁡θacom=may+02m=ay2=gsin2⁡θ1+sin2⁡θNet force on system in horizontal direction is zero. So, acceleration of COM in horizontal direction is zero.