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 px=2cost and py=2sint. The angle between F and p at the time t will be: (A) θ = 0°(B) θ = 30° (C) θ = 90°(D) θ = 180°

A particle moves in X-Y plane under the action of forces F such that

1.	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 px=2cost and py=2sint. The angle between F and p at the time t will be: (A) θ = 0°(B) θ = 30° (C) θ = 90°(D) θ = 180°
Answer» From Newtons second LAW, Force F is nothing but rate of change of momentum. Hence, F = dP/dt Given, Px = 2cost & PY = 2sint => Fx = -2sint & Fy = 2cost Now to find the angle between the force vector and the momentum vector, let us first find the dot product of these two VECTORS. F.P = FxPx + FyPy = 2cost-2sint + 2sint2cost = -4costsint + 4costsint = 0 ALSO, F.P = \|F\|\|P\|cosθ => cosθ = 0 => θ = 90 degrees. Hence, the angle between force and momentum vectors is 90 degrees

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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 px​=2cost and py​=2sint. The angle between F and p​ at the time t will be: (A) θ = 0°(B) θ = 30° (C) θ = 90°(D) θ = 180°​

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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 px=2cost and py=2sint. The angle between F and p at the time t will be: (A) θ = 0°(B) θ = 30° (C) θ = 90°(D) θ = 180°