The potential energy `U`(in `J`) of a particle is given by `(ax + by)`, where `a` and `b` are constants. The mass of the particle is `1 kg` and `x` and `y` are the coordinates of the particle in metre. The particle is at rest at `(4a, 2b)` at time `t = 0`.
Find the speed of the particle when it crosses x-axis

Options

A. `2 sqrt(a^(2) + b^(2))`
B. `sqrt(a^(2) + b^(2))`
C. `(1)/(2) sqrt(a^(2) + b^(2))`
D. `sqrt(((a^(2)+b^(2)))/(2))`

Answer

A

Explanation

`vec (a) = vec(F)//m =(1)/(m) ((del U)/(del X) hat i + (del U)/(del Y) hat j)`
`= -(a hat i + b j)` since, `m =1 kg`.