A triangular wedge of mass M lies on a smooth horizontal table with half of its base projecting out of the edge of the table. A block of mass m is kept at the top of the smooth incline surface of the wedge and the system is let go, Find the maximum value of $\frac{M}{m}$ for which the block will land on the table. Take $\theta ={60}^0.$

 

Let $a_0=$ acceleration of the wedge
$a=$ acceleration of the block relative to the wedge. We consider motion of the block in reference attached to the wedge
 
$N+m{a}_0\sin \theta =mg\cos \theta \mathrm{.........}\left(1\right)$ 
$m{a}_0\cos \theta +mg\sin \theta =ma\Rightarrow a_0\cos \theta +g\sin \theta =a\mathrm{.........}\left(2\right)$ 
For motion of the wedge
 
 $N=\sin \theta =M{a}_0\mathrm{...........}\left(3\right)$
Eliminating N between (1) and (3)
$a_0=\frac{mg\sin \theta \cos \theta }{(M+m{\sin }^2\theta )}$ 
Putting this in (2) gives
$a=\frac{(M+m)g\sin \theta }{(M+m{\sin }^2\theta )}$ 
$\therefore \frac{a}{a_0}=\frac{M+m}{m}$ 
For the block to remain on the table, it is required that by the time displacement of the block relative to the wedge becomes equal to “S”, the horizontal displacement of the wedge must have become larger than
 
$\Rightarrow \frac{1}{2}a{t}^2=S$ ……..(1) and $\Rightarrow \frac{1}{2}{a}_0{t}^2\ge \frac{L}{2}\mathrm{.........}\left(2\right)$  
For the limiting case let’s take the ratio $[\left(1\right)÷\left(2\right)]$
$\frac{a}{a_0}=\frac{2S}{L}\Rightarrow \frac{M+m}{m}=\frac{2S}{L}$ 
$\frac{M}{m}+1=\frac{2}{\cos \theta }=4[\cos {60}^0=\frac{1}{2}]$ 
$\therefore \frac{M}{m}=3$