A bullet of mass 20 g is fired from a gun of mass 2 kg with velocity 300 m/s. The recoil velocity of the gun is: MCQ with Answer and Explanation

By conservation of momentum: initial momentum = 0 (system at rest). Final momentum: m_bullet×v_bullet + m_gun×v_gun = 0. Thus (0.02 kg)(300 m/s) + (2 kg)(v_gun) = 0 ⇒ 6 + 2v_gun = 0 ⇒ v_gun = -3 m/s. Magnitude is 3 m/s; negative sign indicates opposite direction to bullet. This demonstrates momentum conservation in isolated systems. Memory tip: Recoil velocity = -(m_bullet/m_gun)×v_bullet. Such numerical problems test application of conservation laws, frequently appearing in competitive exams with varying mass ratios and velocities.