JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    A particle moves in the xy-plane under the action of a force F such that the components of its linear momentum p at any time t are \[{{p}_{x}}=2\cos t\], \[{{p}_{y}}=2\sin t\]. The angle between F and p  at time t is [MP PET 1996; UPSEAT 2000]

    A)             \[90{}^\circ \]

    B)             \[0{}^\circ \]

    C)             \[180{}^\circ \]         

    D)             \[30{}^\circ \]

    Correct Answer: A

    Solution :

                    Given that \[\vec{p}={{p}_{x}}\hat{i}+{{p}_{y}}\hat{j}=2\cos t\ \hat{i}+2\sin t\ \hat{j}\]             \ \[\vec{F}=\frac{d\vec{p}}{dt}=-2\sin t\ \hat{i}+2\cos t\ \hat{j}\]             Now, \[\vec{F}.\vec{p}=0\] i.e. angle between \[\vec{F}\ \text{and }\vec{p}\ \]is 90°.        


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