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JEE Main & Advanced Physics Vectors Question Bank Multiplication of Vectors

  • 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 anytime 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   [MNR 1991; UPSEAT 2000]

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

    B)                             \[\theta =30{}^\circ \]

    C)                 \[\theta =90{}^\circ \]

    D)                             \[\theta =180{}^\circ \]

    Correct Answer: C

    Solution :

                        \[{{P}_{x}}=2\cos t\], \[{{P}_{y}}=2\sin t\] \\[\vec{P}=2\cos t\ \hat{i}+2\sin t\ \hat{j}\]                 \[\vec{F}=\frac{d\vec{P}}{dt}=-2\sin t\ \hat{i}+2\cos t\ \hat{j}\]                 \[\vec{F}.\vec{P}=0\]\\[\theta =90{}^\circ \]


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