A small ball of mass `m` is projected with a minimum horizontal velocity `v_(0)` on a smooth wedge of mass `M` so that it will reach the highest point of the wedge. Find the value of `v_(0)`

Explanation

if the ball reaches at point `P`, the velocity of the ball with respect to wedge should be `sqrt(gR)`
using work energy theorem from centre of mass frame at `A` and the highest point `P` ltbr `W_(ext)+W_(int)+0=(/_\K)-(cm)`
`W_("gravity")=(/_\K)_(CM)`
`-mg(2R)=[1/2muv_(rel)^(2)]_("final")-[1/2muv_(rel)^(2)]_("initial")`
`-mg2R=1/2mu(V+v)^(2)-1/2muv_(0)^(2)`..............i
`mu=((mM)/(m+M))` and `v=sqrt(gR)`
As there is no external forces acting on the system in horizontal direction. the linear momentum of the system should be conserved.
i.e. `mv_(0)=m(V-v)+MV`
`V=(m(v_(0)+v))/((m+M))` ............ii
from eqn i and ii we get
`implies v_(0)=sqrt((5+(4m)/M))gR`