Question

A particle moves in the XY plane under the actionof force F vector such that the value of its linear moment P vector at any time t is P vector = 2costi^+2sintj^. Find the angle between F vector And P vector at = 4 sec.

Answer 1

Position vector r =  x i + y j  in the x y plane.
    force vector  F = m a = m dv/dt = d P / d t
    momentum vector  P = m v
           P = 2 cos t  i + 2 Sin t  j
           F = - 2 sin t  i + 2 Cos t  j

Angle of vector P with x axis:  Tan Ф1 = 2 Sin t / (2 Cos t) = Tan t
Angle of vector F with x axis :  Tan Ф2 = 2 Cos t / (- 2 sin t) = - Cot t
         = - tan (π/2 - t) = tan [π - (π/2 - t) ] = tan (π/2 + t)   or    tan (3π/2 - t)

Angle between P and F = Ф2 - Ф1
Tan (Ф₂ - Ф₁) = - [tan t + Cot t] / [ 1 - tan t  * Cot t ]
                     = infinity
So angle  Ф₂ - Ф₁ = π/2

The momentum and force vectors are perpendicular to each other.

This is the case of uniform circular motion. F is the centripetal force.