Question

Figure shows a wedge A of mass 6 m smooth semicircular groove of radius a $=$ 8.4 m 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

• A. $3m/s$
• B. $2m/s$
• C. $7m/s$
• D. $4m/s$

Answer 1

The correct option is B $2m/s$

Let $v_1$ be the velocity of small block B relative to wedge A when block B is at the bottom-most position and at the same instant let $v_2$ be the velocity of wedge A

To find the speed of wedge A i.e $v_2$ we apply conservation laws as follows :

Applying the law of conservation of momentum along the horizontal direction

Next to apply the law of conservation of energy

solving equation (i) and (ii)

Hence the speed of the bigger block when the smaller block reaches its bottom position is 2m/s