JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Laws of Motion

All contacts are frictionless. Let the wedge recoil with acceleration $A$ and the block accelerate at $a_r$ relative to the wedge down the $30^{\circ }$ face. The horizontal push on the wedge gives $N\sin \theta =MA$, and for the block, resolving along and perpendicular to the incline in the wedge frame gives $mg\sin \theta +mA\cos \theta =m{a}_r$ and $N=mg\cos \theta -mA\sin \theta$. Eliminating $N$ and $A$ yields $a_r=\frac{(M+m)g\sin \theta }{M+m{\sin }^2\theta }$. With $\theta =30^{\circ }$, $M=16$, $m=8$: $a_r=\frac{24g(1/2)}{16+8(1/4)}=\frac{12g}{18}=\frac{2g}{3}$.