A particle moves in the xy-plane under the action of a force F such that the components of its linear momentum p at any time t are `p_(x)=2cost`, `p_(y)=2sint`. The angle between F and p at time t is

Options

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

Answer

A

Explanation

Given that `vec(P)=P_(x)hatj+P_(y)hatj=2costhati+2sinthatj`
:. `vec(F)=(dvec(p))/(dt)=-2sinthati+2costhatj`
Now, `vec(F).vec(p)=0` i.e., angle between `vec(F)` and `vec(p)` is `90^(@)` .