JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Self Evaluation Test - Laws of Motion

  • question_answer
    A particle moves in the X-Y plane under the influence of 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. The angle between the force and the momentum is

    A) \[0{}^\circ \]

    B) \[30{}^\circ \]  

    C) \[45{}^\circ \]

    D) \[90{}^\circ \]

    Correct Answer: D

    Solution :

    [d] \[\vec{p}\left( t \right)=A\left[ \hat{i}cos\left( kt \right)-\hat{j}\sin \left( kt \right) \right]\] \[F=\frac{d\vec{p}}{dt}=Ak[-\hat{i}cos(kt)-\hat{j}\sin (kt)]\] Here, \[\vec{F}.\vec{P}=0\]    but \[\vec{F}.\vec{P}=Fp\cos \theta \] \[\therefore \cos \theta =0\Rightarrow \theta =90{}^\circ \]


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