In the figure shown, a ring `A` is initially rolling without sliding with a velocity `v`, on the horizontal surface of the body `B` (of same mass as `A`). All surfaces arc smooth. `B` has no initial velocity. What will be the maximum height reached by `A` on `B`?

Options

A. `(3v^(2))/(4g)`
B. `(v^(2))/(4g)`
C. `(v^(2))/(2g)`
D. `(v^(2))/(3g)`

Answer

B

Explanation

When the ring is at the maximum height, the wedge and the ring have the same horizontal component of velocity. As all the surfaces are as smooth, in the sabsence of friction between the ring and the wedge surface, angular velocity of the ring remains constant.
For conservatioin of mehanical energy we get
`1/2mv^(2)+1/2Iomega^(2)=1/2mv^'(2)+1/2Iomega^(2)+1/2mv'^(2)+mgh`
Where `v'` is final common velocity
`v'=v/2` (from, conservation of momentum) and `h=(v^(2))/(4g)`