A block of mass `m` is pushed towards a movable wedge of mass `nm` and height `h`, with a velocity `u`. All surfaces are smooth. The minimum value of `u` for which the block will reach the top of the wedge is

Options

A. Block will reach top of the wedge if `u=sqrt(2gh(1-(1)/(n)))`
B. Block will reach top of the wedge if `u=sqrt(2gh(1+(1)/(n)))`
C. If the block overshoots `P`, the angle of projectile is `alpha`.
D. If the block overshoots `P`, the angle of projectile is less than `alpha`.

Explanation

When the particle just reaches the top of the wedge
`mu=(m+n)v`…..(`1`)
`u=sqrt(2gh(1+(1)/(n)))`
Angle of projection as observed by the ground will be loss than `alpha`
`vecv_(block//ground)=vecv_(block//"wedge")+vecv_(block//ground)`