A particle moves in X-Y plane under the action of forces F such that the values of linear momentum 'p' at any times is `p_(x) = 2 cos t` and `p_(y) = 2 sin t` . The angle between `vec(F)` and at the time t will be:

Options

A. 0°
B. 30°
C. 90°
D. 180°

Answer

C

Explanation

Here `vec(p)=vec(p_(x))hati + vec(p_(y))hatj=2costhati+2sinthatj`
Now `vec(F)=(dvecp)/(t)=-2sinthati+2costhatj`. The angle will be
`costheta=(vec(F).vec(p))/(Fp)`
`costheta=(-4sintcost + 4sintcost)/(Fxxp)`
`theta=90^@`