Question

A small cube of mass 'm' slides down a circular path of radius 'R' formed from a large block of mass 'M' as shown in figure 'M' rests on a table and both blocks move without friction. The blocks are initially at rest and 'm' starts from the top of the path. Find the velocity 'v' of the cube as it leaves the block. Initially the line joining m and the centre is horizontal.

Answer 1

$\begin{array}{c}wehave\\ mu=mv--\left(1\right)\\ wall=\Delta k\\ \Rightarrow mgR=\frac{1}{2}m{v}^2+\frac{1}{2}m{u}^2\\ \Rightarrow mgR=\frac{1}{2}mvH2+\frac{1}{2}\times {\left(\frac{mv}{M}\right)}^2\times M\\ \Rightarrow mgR=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{m^2{v}^2}{M}\\ \Rightarrow gR=\frac{v^2}{2}+\frac{m}{M}\frac{v^2}{2}\\ \Rightarrow \frac{2gR}{1+\frac{m}{M}}=v^2\\ \Rightarrow v=\sqrt{\frac{2gR}{1+\frac{m}{M}}}\\ whichistherequiredanswer.\end{array}$