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Center of Mass question

2023 · 30 Jan · Shift 2 · Q50
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Center of Mass question

2023 · 30 Jan · Shift 2 · Q50

JEE MainPhysicsCenter of MassMCQ+4 / −1
A machine gun of mass 10 kg10 \mathrm{~kg}10 kg fires 20 g20 \mathrm{~g}20 g bullets at the rate of 180 bullets per minute with a speed of 100 m s−1100 \mathrm{~m} \mathrm{~s}^{-1}100 m s−1 each. The recoil velocity of the gun is
  1. A
    0.02 m/s0.02 \mathrm{~m} / \mathrm{s}0.02 m/s
  2. B
    1.5 m/s1.5 \mathrm{~m} / \mathrm{s}1.5 m/s
  3. C
    2.5 m/s2.5 \mathrm{~m} / \mathrm{s}2.5 m/s
  4. D
    0.6 m/s0.6 \mathrm{~m} / \mathrm{s}0.6 m/s
View written solutionFree

Correct answer: D

  1. Given data
  • Mass of gun: M=10 kgM = 10\,\text{kg}M=10kg
  • Mass of each bullet: m=20 g=0.02 kgm = 20\,\text{g} = 0.02\,\text{kg}m=20g=0.02kg
  • Rate of firing: 180180180 bullets/minute
  • Speed of each bullet: v=100 m s−1v = 100\,\text{m s}^{-1}v=100m s−1
  1. Convert firing rate into bullets per second
180 bullets/min=18060=3 bullets/s180\ \text{bullets/min} = \frac{180}{60} = 3\ \text{bullets/s}180 bullets/min=60180​=3 bullets/s
  1. Mass of bullets fired per second

Each bullet has mass 0.02 kg0.02\,\text{kg}0.02kg, so in 1 second:

m˙=3×0.02=0.06 kg/s\dot m = 3 \times 0.02 = 0.06\,\text{kg/s}m˙=3×0.02=0.06kg/s
  1. Rate of momentum carried away by bullets

Momentum of bullets fired per second:

F=m˙ v=0.06×100=6 NF = \dot m\, v = 0.06 \times 100 = 6\,\text{N}F=m˙v=0.06×100=6N

This is the backward recoil force on the gun.

  1. Find recoil velocity

If the gun recoils with speed VVV, then using momentum balance per second:

MV=(mass fired per second)×vMV = (\text{mass fired per second}) \times vMV=(mass fired per second)×v

So,

10V=0.06×100=610V = 0.06 \times 100 = 610V=0.06×100=6 V=610=0.6 m/sV = \frac{6}{10} = 0.6\,\text{m/s}V=106​=0.6m/s
  1. Match with options
V=0.6 m/s\boxed{V = 0.6\,\text{m/s}}V=0.6m/s​

So the correct option is D.

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