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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Topic Test - Newton's Laws Of Motion

  • question_answer
    A particle moves in the \[x-y\] plane under the action of a force \[\overrightarrow{F}\] such that the value of its linear momentum \[\overrightarrow{P}\] at any time \[t\] is \[{{P}_{x}}=2\,\cos t,\] \[{{P}_{y}}=2\sin t\]. The angle \[\theta \] between \[\overrightarrow{F}\] and \[\overrightarrow{P}\] at a given time t will be

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

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

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

    D) \[30{}^\circ \]

    Correct Answer: A

    Solution :

    [a] \[P=\sqrt{P_{x}^{2}+P_{y}^{2}}=\sqrt{{{(2\,\cos t)}^{2}}+{{(2\,\sin t)}^{2}}}=2\] If m is the mass of the body, then kinetic energy \[=\frac{{{P}^{2}}}{2m}=\frac{{{(2)}^{2}}}{2m}=\frac{2}{m}\] Since kinetic energy does not change with time, both work done and power are zero. Now, power \[=Fv\,\cos \theta =0\] As         \[F\ne 0,v\ne 0\] \[\therefore \]      \[\cos \theta =0\] or  \[\theta =90{}^\circ \] As direction of \[\vec{p}\] is same that of \[\vec{v},\] \[(\because \overrightarrow{P}=m\overrightarrow{v}),\] hence angle between \[\overrightarrow{F}\] and \[\overrightarrow{P}\] is equal to \[90{}^\circ \].


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