A 0.02 kg bullet is fired with a speed of 300 m s⁻¹ from a 4 kg gun. The recoil speed of the gun is:
A3.0 m s⁻¹
B0.67 m s⁻¹
C15 m s⁻¹
D1.5 m s⁻¹
Answer & Solution
Correct answer: D. 1.5 m s⁻¹
Momentum is conserved for the gun+bullet system: m_b v_b = m_g v_g ⟹ v_g = (0.02×300)/4 = 1.5 m s⁻¹.
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