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A particle moves in the x-y plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (kt) - \hat{j} \sin (kt)]$ , where A and k are constants. Angle between the force and the momentum is:

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Correct option is D

F→=dp→dt=Ak[−i^sin⁡kt−j^cos⁡kt]Let angle between F→ and p→ be θ , thencos⁡θ=F→⋅p→FpBut F→⋅p→=0 so cos⁡θ=0⇒θ=90∘

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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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Ready to Test Your Skills?

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A particle moves in the x-y plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (kt) - \hat{j} \sin (kt)]$ , where A and k are constants. Angle between the force and the momentum is: