Potential energy of a particle of mass `m`, depends on distance `y` from line `AB` according to given relation `U = (K)/(sqrt(y^(2) + a^(2))`, where `K` is a positive constant. A particle of mass `m` is projected from `y = sqrt(3)` towards line `AB`. (perpendicular to it) then minimum velocity so that it connot return to its initial point is `sqrt((K)/(aNm))`, calculate `N`.

Answer

1

Explanation

`U_(i) + k_(i) = U_(f) + k_(f)`
`(K)/(2a) + (1)/(2)mV_(0)^(2) = (K)/(a)`
`(mV_(0)^(2))/(2) = (K)/(2a) rArr v_(0) = sqrt((K)/(am))`