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Center of Mass question

2004 · Shift 0 · Q162
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Center of Mass question

2004 · Shift 0 · Q162

JEE MainPhysicsCenter of MassMCQ+4 / −1
A machine gun fires a bullet of mass 40g40g40g with a velocity 1200ms−1.1200m{s^{ - 1}}.1200ms−1. The man holding it can exert a maximum force of 144N144N144N on the gun. How many bullets can he fire per second at the most?
  1. A
    Two
  2. B
    Four
  3. C
    One
  4. D
    Three
View written solutionFree

Correct answer: D

  1. Given data

    • Mass of one bullet: m=40 g=0.04 kgm = 40\text{ g} = 0.04\text{ kg}m=40 g=0.04 kg
    • Speed of bullet: v=1200 m s−1v = 1200\text{ m s}^{-1}v=1200 m s−1
    • Maximum force man can exert: F=144 NF = 144\text{ N}F=144 N

  2. Momentum given to one bullet

    The momentum of one bullet is p=mv=0.04×1200=48 kg m s−1.p = mv = 0.04 \times 1200 = 48\text{ kg m s}^{-1}.p=mv=0.04×1200=48 kg m s−1.

  3. Relation between force and rate of change of momentum

    If nnn bullets are fired per second, then total momentum carried away per second is n×48.n \times 48.n×48.

    This equals the recoil force on the gun: F=48n.F = 48n.F=48n.

  4. Use the maximum force condition

    48n=14448n = 14448n=144 n=14448=3.n = \frac{144}{48} = 3.n=48144​=3.

  5. Conclusion

    The maximum number of bullets that can be fired per second is 3.\boxed{3}.3​.

  6. Option check

    • A: Two   →\;\to→ incorrect
    • B: Four   →\;\to→ incorrect
    • C: One   →\;\to→ incorrect
    • D: Three   →\;\to→ correct

Therefore, the correct option is D.

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