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BHU PMT BHU PMT Solved Paper-2004

  • question_answer
    A ball of mass 150g moving with an acceleration 20 m/s2 is hit by a force, which acts on it for 0.1 s. The impulsive force is :                                                                                                                                           [BHU PMT-2004]

    A)  0.5 N-s                

    B)  0.1 N-s

    C)  0.3 N-s                                

    D)  1.2 N-s

    Correct Answer: C

    Solution :

                     If a constant force \[\overset{\to }{\mathop{F}}\,\] is applied a body for a short interval of time \[\Delta t\], then the impulse of this force will be \[\overset{\to }{\mathop{F}}\,\times \Delta \,t\]. Also from Newton?s law \[\overset{\to }{\mathop{F}}\,=m\overset{\to }{\mathop{a}}\,\] Where \[m\]is mass and a is acceleration. Given, \[m=150\,\,g=150\times {{10}^{-3}}\,kg,\]\[a=20\,\,m/{{s}^{2}}\]                 \[\therefore \]                  \[F=0.15\times 20=3\,N\] Impulse\[=F\Delta \,t=3\times 0.1=0.3\,N-s\] Note: Impulse is a vector quantity, its direction is the same as that of the force.


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