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JEE Main & Advanced Physics Rotational Motion Question Bank Mock Test - Centre of Mass, Conservation of Linear Momentum and Collisions

  • question_answer
    In a gravity free space, a man of mass \[M\] standing at a height \[h\] above the floor, throws a ball of mass m straight down with a speed \[u\]. When the ball reaches the floor, the distance of the man above the floor will be

    A)  \[h(1+m/M)\]

    B)  \[h(2-m/M)\]

    C)  \[2h\]

    D)  a function of \[m\], \[M\], \[h\] and \[u\]

    Correct Answer: A

    Solution :

    [a]  As in gravity free space displacement of centre of mass of man and ball system should not move. If displacement of the ball be h then the displacement of man in upward direction. \[mh=Mh'\Rightarrow h'=\frac{mh}{M}\]              (i) Hence the position of man from ground \[H=h+h'=h=h\left[ 1+\frac{m}{M} \right]\]


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