AMU Medical AMU Solved Paper-2015

  • question_answer
    A man racing with his son has half the kinetic energy of the son, who has half the mass of the father. The man speeds up by 1 m/s and has the same kinetic energy as the son. What was the original speed of the man?

    A)  4.8 m/s

    B)  3.6 m/s

    C)  2.4 m/s

    D)  1.2 m/s

    Correct Answer: C

    Solution :

                    Let the velocity of father \[={{v}_{1}}\] According to the question,                 \[{{v}_{i}}=\frac{1}{2}{{K}_{son}}\] Final velocity of the father                 \[{{v}_{t}}=({{v}_{1}}+1.0)\,\,m/s\] Then,    \[{{K}_{t}}={{K}_{son}}\]                 \[{{K}_{i}}=\frac{1}{2}{{K}_{t}}\]                 \[{{K}_{i}}=\frac{1}{2}\left[ \frac{1}{2}m\,{{({{v}_{i}}+1.0)}^{2}} \right]\]                 \[{{K}_{i}}=\frac{1}{4}m\,(v_{i}^{2}+1+2{{v}_{i}})\]                 \[\frac{1}{2}mv_{i}^{2}=\frac{1}{4}m\,(v_{i}^{2}+1+2{{v}_{i}})\]                 \[v_{i}^{2}=\frac{v_{i}^{2}}{2}+\frac{1}{2}+{{v}_{i}}\]                 \[v_{i}^{2}-{{v}_{i}}-\frac{1}{2}=0\]                 \[v_{i}^{2}-2{{v}_{i}}-1=0\]                                 \[{{v}_{i}}=2.4\,m/s\]


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