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JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Graphical Questions

  • question_answer
    A particle of mass m moving with a velocity u makes an elastic one dimensional collision with a stationary particle of mass m establishing a contact with it for extremely small time T. Their force of contact increases from zero to F0 linearly in time \[\frac{T}{4}\], remains constant for a further time \[\frac{T}{2}\] and decreases linearly from F0 to zero in further time \[\frac{T}{4}\] as shown. The magnitude possessed by F0 is

    A)                         \[\frac{mu}{T}\]          

    B)                           \[\frac{2mu}{T}\]            

    C)                         \[\frac{4mu}{3T}\]              

    D)                         \[\frac{3mu}{4T}\]

    Correct Answer: C

    Solution :

                    Change in momentum = Impulse             = Area under force-time graph             \ \[mv=\]Area of trapezium               Þ \[mv=\frac{1}{2}\left( T+\frac{T}{2} \right)\ {{F}_{0}}\]             Þ \[mv=\frac{3T}{4}{{F}_{0}}\]Þ\[{{F}_{0}}=\frac{4mu}{3T}\]            


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