Question

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

Answer 1

Momentum of the body `vecP=2cos thati+2sinthatj`
So force on the body is `vecF=(dvecp)/(dt)=-2sin t hati_2cos thatj`
If angle between `vecF` and `becp` is `theta` then
`costheta=(vecF.vecp)/(Fp)=(-4cos t sin t+4cos t sin t)/(Fp)=0` so `theta=90^(@)`