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Center of Mass question

2023 · 11 Apr · Shift 1 · Q60
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Center of Mass question

2023 · 11 Apr · Shift 1 · Q60

JEE MainPhysicsCenter of MassMCQ+4 / −1
An average force of 125 N125 \mathrm{~N}125 N is applied on a machine gun firing bullets each of mass 10 g10 \mathrm{~g}10 g at the speed of 250 m/s250 \mathrm{~m} / \mathrm{s}250 m/s to keep it in position. The number of bullets fired per second by the machine gun is :
  1. A
    25
  2. B
    50
  3. C
    5
  4. D
    100
View written solutionFree

Correct answer: B

  1. Given data

    • Average force on the gun: F=125 NF = 125\,\text{N}F=125N
    • Mass of each bullet: m=10 g=0.01 kgm = 10\,\text{g} = 0.01\,\text{kg}m=10g=0.01kg
    • Speed of each bullet: v=250 m/sv = 250\,\text{m/s}v=250m/s
    • Let the number of bullets fired per second be nnn.

  2. Concept used The average recoil force equals the rate of change of momentum of the bullets: F=n⋅mvF = n \cdot m vF=n⋅mv because each bullet carries momentum mvmvmv, and if nnn bullets are fired every second, momentum carried per second is nmvn mvnmv.

  3. Substitute the values 125=n⋅0.01⋅250125 = n \cdot 0.01 \cdot 250125=n⋅0.01⋅250

  4. Simplify 0.01⋅250=2.50.01 \cdot 250 = 2.50.01⋅250=2.5 So, 125=2.5n125 = 2.5n125=2.5n

  5. Solve for nnn n=1252.5=50n = \frac{125}{2.5} = 50n=2.5125​=50

  6. Check options

    • A: 25
    • B: 50
    • C: 5
    • D: 100

Hence, the correct answer is Option B.

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