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BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    A particle of mass m moving with velocity u makes an elastic one dimensional collision with a stationary particle of mass m. They are in contact for a short time T. Their force of interaction increases from zero to \[{{F}_{0}}\]linearly in time \[\frac{T}{2}\] and decreases linearly to zero in further time \[\frac{T}{2}\] (shown in figure). The magnitude of \[{{F}_{0}}\]is

    A)  mu/2 T                                

    B)  mu/T

    C)  2mu/T                 

    D)         None of these

    Correct Answer: C

    Solution :

    In elastic one dimensional collision particle rebounds with same speed in opposite direction. So, change in momentum \[=2\text{ }mu\] But, impulse \[=F\times T=\] Change in momentum \[\Rightarrow \]               \[{{F}_{0}}\times T=2mu\] or            \[{{F}_{0}}=\frac{2mu}{T}\]


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