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IIT-JEE 2007 · Paper 2, Physics question 3

A particle moves in the X-Y plane under the influence of a force such that its linear momentum is p⃗(t)=A[i^cos⁡(kt)−j^sin⁡(kt)]\vec p(t)=A\left[\hat i\cos(kt)-\hat j\sin(kt)\right], where AA and kk are constants. The angle between the force and the momentum is

Choose one option
Answer and worked solution

Official answer: D

Constant magnitude. ∣p⃗∣=A|\vec p|=A for all tt, so p⃗⋅p⃗\vec p\cdot\vec p is constant and p⃗⋅dp⃗dt=0\vec p\cdot\dfrac{d\vec p}{dt}=0.

Check. F⃗=dp⃗dt=−Ak[i^sin⁡kt+j^cos⁡kt]\vec F=\dfrac{d\vec p}{dt}=-Ak\left[\hat i\sin kt+\hat j\cos kt\right], and F⃗⋅p⃗=−A2k(sin⁡ktcos⁡kt−cos⁡ktsin⁡kt)=0\vec F\cdot\vec p=-A^2k(\sin kt\cos kt-\cos kt\sin kt)=0.

(D) 90°

Topics:Newton's lawsUniform circular motion
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