A block of mas `m_(1)` is projected with a velocity `v_(0)` so that it climbs onto the smooth wedge of mass `m_(2)`. If the block does not level the wedge, find the maximum height attained by the block.

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)`