Figure shows a wedge A of mass 6m smooth semicircular groove radius a = 8.4m placed on a smooth horizontal surface. A small block B of mass m is released from a position in groove where its radius is horizontal. Find the speed (in ms^-1 ) of bigger block when smaller block reaches its bottommost position

Let us assume smaller block is moving with speed v₁ relative to bigger block when it reaches the bottom most position

And at this instant let us assume bigger block is moving at v₂  

Then using conservation of momentum in horizontal direction we have 

$6{\mathrm{mv}}_2=\mathrm{m}\left({\mathrm{v}}_1-{\mathrm{v}}_2\right).$.....(i)

Now using energy conservation
$\mathrm{mga}=\frac{1}{2}\mathrm{m}{\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)}^2+\frac{1}{2}\left(6\mathrm{m}\right){\mathrm{v}}_2^2$......(ii)

Solving (i) and (ii), we get $2\mathrm{ga}=36{\mathrm{v}}_2^2+6{\mathrm{v}}_2^2$

${\mathrm{v}}_2=\sqrt{\frac{\mathrm{ga}}{21}}=\sqrt{\frac{10\times 8.4}{21}}=2{\mathrm{ms}}^{-1}$