A bullet of mass 10 g is horizontally fired with velocity 300 \text{ m s}^{-1} from a pistol of mass 1 kg. What is the recoil velocity of the pistol?

  1. A. 0.3 \text{ m s}^{-1}
  2. B. 3 \text{ m s}^{-1}
  3. C. -3 \text{ m s}^{-1} ✓
  4. D. -0.3 \text{ m s}^{-1}

Correct Answer: C. -3 \text{ m s}^{-1}

Explanation

By the principle of conservation of momentum, m_{1}v_{1} + m_{2}v_{2} = 0. Substituting the values: (0.01 \text{ kg})(300 \text{ m s}^{-1}) + (1 \text{ kg})(v_{2}) = 0, giving v_{2} = -3 \text{ m s}^{-1}.

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