A particle is moving with momentum P and kinetic K in a gravity free space. A constant force F (directed perpendicular to the initial momentum) now acts on the particle for time t. Final kinetic energy of the particle will be

Options

A. `K [1 + ((Ft)/(P))^(2)]`
B. `K [1 + 2 ((Ft)/(2))^(2)]`
C. `K sqrt(1 + ((Ft)/(P))^(2))`
D. `K sqrt(1 + ((Ft)/(P))^(4))`

Answer

A

Explanation

`a_(t) = (F)/(m)`
Integrating, we get `v_(t) = (F)/(m) t`
As `(p^(2))/(2m) = K :. V_(t) = (F)/(p^(2)) xx 2 Kt`
Now, `K_(f) = (1)/(2) mv^(2) = (1)/(2) m ((f^(2)4K^(2)t^(2))/(p^(4)) + v^(2))`
`= (2F^(2)K^(2)t^(2)m)/(p^(4)) + K = K [(2KF^(2)t^(2)m)/(p^(2)) + 1]`
`= K [1 + (2m)/(p^(2) xx p^(2)) xx (F^(2)t^(2))/(p^(2))]`