JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    A machine gun fires a bullet of mass 40 g with a velocity\[1200\text{ }m{{s}^{-1}}\]. The man holding it, can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most? [AIEEE 2004]

    A) One

    B) Four

    C) Two

    D) Three

    Correct Answer: D

    Solution :

    [d] The force is measured as the time rate of change of momentum we have to remembered that momentum and force both are vector quantity.
    The force exerted by machine gun on man's hand in firing a bullet = Change in momentum per second on a bullet or rate of change of momentum
    \[=\left( \frac{40}{1000} \right)\times 1200\]
    The force exerted by man on machine gun= Force exerted on man by machine gun = 48 N
    Hence, number of bullets fired\[=\frac{144}{48}=3.\]

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