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MCQEasyJEE 2023Impulse & Momentum

JEE Physics 2023 Question with Solution

A machine gun of mass 10 kg10 \, \text{kg} fires 20 g20 \, \text{g} bullets at the rate of 180180 bullets per minute with a speed of 100 m/s100 \, \text{m/s} each. The recoil velocity of the gun is:

  • A

    0.02 m/s0.02 \, \text{m/s}

  • B

    2.5 m/s2.5 \, \text{m/s}

  • C

    1.5 m/s1.5 \, \text{m/s}

  • D

    0.6 m/s0.6 \, \text{m/s}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

  • Mass of the gun = 10 kg10 \, \text{kg}
  • Mass of each bullet = 20 g=0.02 kg20 \, \text{g} = 0.02 \, \text{kg}
  • Speed of each bullet = 100 m/s100 \, \text{m/s}
  • Rate of fire = 180180 bullets per minute = 33 bullets per second

Find: The recoil velocity of the gun.

Use the principle of conservation of momentum. Initially, the system is at rest, so total initial momentum is zero.

Momentum balance:

0=Mvgun+(mv)R0 = M v_{\text{gun}} + (m v)R

Substituting the given values:

0=10 vgun+(0.02×100)×30 = 10 \, v_{\text{gun}} + (0.02 \times 100) \times 3

So,

10 vgun=−610 \, v_{\text{gun}} = -6 vgun=−610=−0.6 m/sv_{\text{gun}} = -\frac{6}{10} = -0.6 \, \text{m/s}

Therefore, the recoil velocity of the gun is 0.6 m/s0.6 \, \text{m/s} opposite to the direction of the bullets. The solution concludes this value but labels the correct option as A, whereas the listed option containing 0.6 m/s0.6 \, \text{m/s} is D.

Common mistakes

  • Using 180180 directly as bullets per second is incorrect because the rate is given per minute. Convert it to 33 bullets per second before applying momentum balance.

  • Not converting bullet mass from 20 g20 \, \text{g} to 0.02 kg0.02 \, \text{kg} leads to a wrong momentum value. Always use SI units consistently.

  • Ignoring the negative sign in recoil velocity is a conceptual error. The negative sign indicates that the gun moves opposite to the bullets; the asked recoil speed is the magnitude 0.6 m/s0.6 \, \text{m/s}.

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