A small ball `(` uniform solid sphere `)` of mass `m` is released from the top of a wedge of the same mass `m`. The wedge is free to move on a smooth horizontal surface. The ball rolld without sliding on the wedge. The required height of the wedge are mentioned in the figure.

The speed of the wedge when the ball is just going to leave the wedge at point `'P'` of the wedge is

Options

A. `sqrt((5gh)/(9))`
B. `sqrt(gh)`
C. `sqrt((5gh)/(6))`
D. None of these

Answer

A

Explanation

`mgh=(1)/(2)mv^(2)+(1)/(2)mv^(2)+(1)/(2).(2)/(5)mr^(2)((2v)/(r))^(2)`
`=(1)/(2)mv^(2)[1+1+(8)/(5)]=(1)/(2)mv^(2)(18)/(5)=(9mv^(2))/(5)`
`rArr v=sqrt((5)/(9)gh)`