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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    Two bodies of masses 1 kg and 2 kg have equal momentum. Then, the ratio of their kinetic energies is:

    A)  \[2:1\]                                 

    B)  \[3:1\]

    C)  \[1:3\]                                 

    D)  \[1:1\]

    Correct Answer: A

    Solution :

    Kinetic energy of a body                 \[E=\frac{1}{2}m{{\upsilon }^{2}}=\frac{1}{2m}\,{{(m\upsilon )}^{2}}\]                 \[=\frac{{{p}^{2}}}{2m}\]                              \[(\because \,\,p=m\upsilon )\] or            \[E\propto \frac{1}{m}\]                               [\[p=\]constant (given)] \[\therefore \]  \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{m}_{2}}}{{{m}_{1}}}\]                 \[=\frac{2}{1}=2:1\]


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