Punjab Medical Punjab - MET Solved Paper-2001

  • question_answer
    A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by \[1\,\,m/s\] so as to have same K.E. as that of the boy. The original speed of the man will be

    A) \[\sqrt{2}\,\,m/s\]                         

    B) \[(\sqrt{2}-1)\,\,m/s\]

    C) \[\frac{1}{(\sqrt{2}-1)}\,\,m/s\]

    D) \[\frac{1}{\sqrt{2}}\,\,m/s\]

    Correct Answer: C

    Solution :

    Suppose mass and speed of man is \[M\] and \[V\] respectively. Suppose the speed of the boy is \[v\] Then.                 \[\frac{1}{2}M{{V}^{2}}=\frac{1}{2}\left[ \frac{1}{2}\cdot \left( \frac{M}{2}\cdot  \right){{v}^{2}} \right]\]                ? (1)                 \[\frac{1}{2}M{{(V+1)}^{2}}=\frac{1}{2}\left( \frac{M}{V} \right){{v}^{2}}\]           ... (2) Dividing equation (1) by equation (2) we obtain                 \[\frac{{{V}^{2}}}{{{(V+1)}^{2}}}=\frac{1}{2}\]    or            \[\frac{V}{(V+1)}=\frac{1}{\sqrt{2}}\]


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