The figure from below shows a block which is shaded with blue color and whose mass is $9\,kg$ and of radius equal to $2\,m$ is put over a frictionless table. Then a small block labeled $A$ is put on $A$ and released so it slides in the circular cavity of the bigger block, there is no friction between them. Find the speed of the block in $\frac{m}{s}$ when the block reaches the lowest point of the half circle cavity.

The alternatives given are as follows:

$\begin{array}{ll}

1.&2\,\frac{m}{s}\\

2.&3\,\frac{m}{s}\\

3.&4\,\frac{m}{s}\\

4.&6\,\frac{m}{s}\\

5.&9\,\frac{m}{s}\\

\end{array}$

I'm lost at this problem. How should I exactly relate the conservation of momentum?.

I recall that the energy will be preserved as follows:

$E_u=E_k$

$mgR=\frac{1}{2}mv^2$

$\frac{mv^2}{R}=mg$

$v^2=Rg$

But if I equate these two expressions I will get to an inconsistency. What part did I got wrong?. Can someone help me here please?.