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Center of Mass question

2021 · 22 Jul · Shift 2 · Q56
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  5. /2021 · 22 Jul · Shift 2 · Q56

Center of Mass question

2021 · 22 Jul · Shift 2 · Q56

JEE MainPhysicsCenter of MassMCQ+4 / −1
A bullet of '4 g' mass is fired from a gun of mass 4 kg. If the bullet moves with the muzzle speed of 50 ms −-− 1, the impulse imparted to the gun and velocity of recoil of gun are :
  1. A
    0.2 kg ms −-− 1, 0.1 ms −-− 1
  2. B
    0.4 kg ms −-− 1, 0.05 ms −-− 1
  3. C
    0.2 kg ms −-− 1, 0.05 ms −-− 1
  4. D
    0.4 kg ms −-− 1, 0.1 ms −-− 1
View written solutionFree

Correct answer: C

  1. Given data
  • Mass of bullet: m=4 g=0.004 kgm = 4\text{ g} = 0.004\text{ kg}m=4 g=0.004 kg
  • Mass of gun: M=4 kgM = 4\text{ kg}M=4 kg
  • Muzzle speed of bullet: v=50 m s−1v = 50\text{ m s}^{-1}v=50 m s−1
  1. Impulse imparted to the gun

The bullet acquires momentum

p=mv=0.004×50=0.2 kg m s−1p = mv = 0.004 \times 50 = 0.2\text{ kg m s}^{-1}p=mv=0.004×50=0.2 kg m s−1

By Newton's third law and conservation of momentum, the gun receives an equal and opposite momentum.

So, the impulse imparted to the gun is

J=0.2 kg m s−1J = 0.2\text{ kg m s}^{-1}J=0.2 kg m s−1

  1. Velocity of recoil of the gun

If recoil velocity is VVV, then momentum of gun is

MV=0.2MV = 0.2MV=0.2

Thus,

V=0.24=0.05 m s−1V = \frac{0.2}{4} = 0.05\text{ m s}^{-1}V=40.2​=0.05 m s−1

  1. Match with options
  • Impulse imparted to gun = 0.2 kg m s−10.2\text{ kg m s}^{-1}0.2 kg m s−1
  • Recoil velocity of gun = 0.05 m s−10.05\text{ m s}^{-1}0.05 m s−1

This matches Option C.

  1. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

So, the derived answer agrees with the stored correct answer.

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