In the figure shown, a ring `A` is initially rolling without sliding with a velocity `1`, 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 maximum height, the wedge and ring have same x component of velocity. In the absence of friction between the ring and the wedge surface, angular velocity of the ring remains constant.
For conservation of energy, we get: `1/2 mv^(2) + 1/2I omega^(2) =1/2mv^(2) + 1/2Iomega^(2) +1/2mv^(2) = mgh`
When, `v'` is final common velocity, `v'= v/2` (from conservation of momentum)
Solve to get `h=v^(2)/(4g)`