JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Conservation of Energy and Momentum

  • 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 1m/s so as to have same K.E. as that of the boy. The original speed of the man will be                                                [Pb. PMT 2001]

    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 :

                        Let m = mass of boy, M = mass of man v = velocity of boy, V = velocity of man \[\frac{1}{2}M{{V}^{2}}=\frac{1}{2}\left[ \frac{1}{2}m{{v}^{2}} \right]\]                  ?..(i) \[\frac{1}{2}M{{(V+1)}^{2}}=1\left[ \frac{1}{2}m{{v}^{2}} \right]\]            ?..(ii) Putting \[m=\frac{M}{2}\] and solving \[V=\frac{1}{\sqrt{2}-1}\]


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