F particle moves in x-y plane under the action of force `vecF` such that the value of its linear momentum `(vecp)` at any instant t is `p_x=2` cos t and py=2 sin t. The angle `theta` between `vecF vecP` at a given time t will be-

Options

A. `90^(@)`
B. `0^(@)`
C. `180^(@)`
D. `30^@`

Answer

A

Explanation

`p=sqrt(px^(2)+py^(2))=sqrt(4 cos^(2)t+4 sin^(2)t)=2` (constant), As p remains constant, `vecF` acts at right angle to `vecp`.