State the law of conservation of linear momentum. Derive it from Newton's third law and explain the recoil of a gun. A 3 kg gun fires a 15 g bullet at 400 m s⁻¹; find its recoil velocity.

1. 1 · State the law. If no net external force acts on a system, its total linear momentum stays constant.
2. 2 · Derive it.
   • Bodies 1 and 2 interact for a time Δt; by the third law F₁₂ = −F₂₁.
   • By the second law, F₁₂ = Δp₁/Δt and F₂₁ = Δp₂/Δt.
   • So Δp₁ + Δp₂ = 0: the total momentum does not change.
3. 3 · Draw the gun and bullet. [Gun of mass M with an arrow V pointing left; bullet of mass m with an arrow v pointing right; write 'before firing: total momentum = 0']
4. 4 · Explain recoil. Before firing, total momentum = 0. After firing, mv + MV = 0, so V = −(m/M)v. The minus sign means the gun moves backward.
5. 5 · Solve the numerical. V = −(0.015 kg × 400 m s⁻¹) ÷ 3 kg = −6 ÷ 3 = −2 m s⁻¹. The gun recoils at 2 m s⁻¹.
6. 6 · Link to real life. A soldier presses the rifle butt firmly to the shoulder, so the gun and body recoil together, more slowly. Rockets fly by the same law.
7. 7 · Close in one line. Momentum lost by one part of an isolated system is gained by another.

Scoring tip: Write the 'before' and 'after' momentum on separate lines. Keep the sign of V, then say what the minus sign means.