Conservation of linear momentum: recoil, collisions, explosions and rockets
In an isolated system (no net external force) the total momentum stays constant. This explains gun recoil, explosions and how rockets fly.
The law
A system is a group of bodies that act on each other. If no outside body exerts a net force on it, it is an isolated or closed system. In such a system the vector sum of the momenta stays the same. The momentum of each body may change, in size, direction or both. But what one body gains, another loses. Internal forces change individual momenta, never the total. The law holds for collisions, explosions, nuclear reactions and radioactive decay.
Derivation: conservation of momentum from Newton's laws
- 1Two bodies 1 and 2 interact for a time Δt. No external force acts on them.
- 2Third law: the force on body 1 by body 2 is F₁₂, and the force on body 2 by body 1 is F₂₁ = −F₁₂.
- 3Second law: F₁₂ = Δp₁/Δt and F₂₁ = Δp₂/Δt.
- 4So Δp₁/Δt = −Δp₂/Δt, which gives Δp₁ + Δp₂ = 0.
- 5The total momentum p₁ + p₂ therefore does not change.
- 6★So, the total momentum of an isolated system is conserved.
★ key step
Worked example: recoil of a rifle
A 4 kg rifle fires a 20 g bullet at 400 m s⁻¹. Find the recoil velocity of the rifle.
- Before firing, the rifle and bullet are at rest. Total momentum = 0.
- After firing: mv + MV = 0, with m = 0.020 kg, v = 400 m s⁻¹, M = 4 kg.
- Bullet's momentum = 0.020 × 400 = 8 kg m s⁻¹.
- So 8 + 4V = 0, giving V = −8 ÷ 4 = −2 m s⁻¹.
- The minus sign means the rifle moves opposite to the bullet.
The rifle recoils at 2 m s⁻¹, backward. It is slow because its mass is 200 times the bullet's
Worked example: two trolleys that stick together
A 3 kg trolley moving at 4 m s⁻¹ runs into a 1 kg trolley at rest. They couple together. Find their common velocity.
- Momentum before = (3 kg × 4 m s⁻¹) + (1 kg × 0) = 12 kg m s⁻¹.
- Momentum after = (3 + 1) kg × v = 4v.
- Conservation of momentum: 4v = 12, so v = 3 m s⁻¹.
The coupled trolleys move on at 3 m s⁻¹ in the original direction
Explosions and rockets
- Explosion: a cracker or a shell at rest has zero momentum. After it bursts, the momenta of all the pieces still add up to zero. Two equal pieces fly apart with equal speeds in opposite directions.
- Rocket: burning fuel makes hot gas at very high pressure. The gas rushes out of the nozzle backward with a large momentum. The rocket gains an equal momentum forward.
- If m kg of gas leaves each second at speed v, the gas carries away momentum mv each second. So the rocket gains momentum mv each second. For a rocket of mass M this gain is Ma, so its acceleration is a = mv/M (ignoring gravity, as your NIOS lesson does). The lesson writes a = −mv/M; the minus sign only shows that the rocket moves opposite to the gas.
- A rocket does not need air to push against. It works in outer space too.
- A balloon released with its neck open shoots forward as the air rushes out backward.
Worked example: acceleration of a rocket
A rocket of mass 5000 kg throws out 50 kg of gas every second at a speed of 2000 m s⁻¹. Find its acceleration. Ignore gravity.
- Momentum given to the gas each second = mv = 50 kg s⁻¹ × 2000 m s⁻¹ = 1.0 × 10⁵ kg m s⁻².
- This is the forward force (thrust) on the rocket: 1.0 × 10⁵ N.
- a = mv/M = 1.0 × 10⁵ N ÷ 5000 kg = 20 m s⁻².
The rocket accelerates at 20 m s⁻², forward (opposite to the escaping gas)
ISRO's rockets rise from Sriharikota because hot gases rush out backward at great speed. The kick of a rifle on the shoulder and a Diwali rocket's flight follow the same law.
3 marks: “Two bodies interact only with each other. Using Newton's second and third laws, show that their total momentum does not change. Name the law this proves.” Third law: F₁₂ = −F₂₁. Second law: F₁₂ = Δp₁/Δt and F₂₁ = Δp₂/Δt. So Δp₁/Δt = −Δp₂/Δt, giving Δp₁ + Δp₂ = 0: the total momentum is constant. This is the law of conservation of linear momentum.
2 marks: “A 5 kg gun fires a 25 g bullet at 300 m s⁻¹. Find the recoil velocity of the gun.” 0 = mv + MV. Bullet's momentum = 0.025 × 300 = 7.5 kg m s⁻¹, so V = −7.5 ÷ 5 = −1.5 m s⁻¹. The gun recoils at 1.5 m s⁻¹, backward.
2 marks: “How does a rocket move forward in space, where there is no air?” Gases from the burning fuel are thrown backward at high speed and carry momentum backward. By conservation of momentum, the rocket gains an equal momentum forward. No air is needed.
Before = after: total momentum of an isolated system never changes (gun back, bullet forward; gas back, rocket forward).