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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    n small balls each of mass m impinge elastically each second on a surface with velocity u. The force experienced by the surface will be               [MP PMT/PET 1998; RPET 2001; BHU 2001; MP PMT 2003]

    A)             mnu    

    B)             2 mnu

    C)             4 mnu

    D)             \[\frac{1}{2}\,mnu\]

    Correct Answer: B

    Solution :

                    \[\vec{F}=\frac{d\vec{p}}{dt}=\] Rate of change of momentum As balls collide elastically hence, rate of change of momentum of ball = \[n[mu-(mu)]\]= \[2mnu\] i.e. \[F=2mnu\]


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