A particle moves in the `xy` plane under the influence of a force such that its linear momentum is `vecP(t) = A [haticos(kt)-hatjsin(kt)]`, where `A` and `k` are constants. The angle between the force and momentum is

Options

A. `0^@`
B. `30^@`
C. `40^@`
D. `90^@`

Answer

D

Explanation

`vecF = (dvecP)/(dt) = -kA sin (kt) hati - kA cos (kt)hatj`
`vecP = A cos (kt) hati - A sin (kt) hatj`
Now `vecF cdot vecP = 0`
Therefore, angle between `vecF and vecP` should be `90^@`.