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VMMC VMMC Medical Solved Paper-2005

  • question_answer
    A 30 g bullet travelling initially at500 m/s penetrates 12 cm in to wooden block. The average force exerted will be:

    A) 31250 N

    B)                        41250 N               

    C)        31750 N               

    D)        30450 N

    Correct Answer: A

    Solution :

    From kinetic energy relation \[\frac{1}{2}m{{\upsilon }^{2}}=F.s\] or            \[F=\frac{m{{\upsilon }^{2}}}{2s}\]                 \[=\frac{30\times {{10}^{-3}}\times {{(500)}^{2}}}{2\times 12\times {{10}^{-2}}}\] \[=31250\,N\]


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