A particle moves in the x-y plane under the influence of a force such that its linear momentum is p→(t)=A[i^cos(kt)−j^sin(kt)] , where A and k are constants. Angle between the force and the momentum is:
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a
0°
b
30°
c
45°
d
90°
answer is D.
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Detailed Solution
F→=dp→dt=Ak[−i^sinkt−j^coskt]Let angle between F→ and p→ be θ , thencosθ=F→⋅p→FpBut F→⋅p→=0 so cosθ=0⇒θ=90∘