Explosions and Recoil

An explosion is the opposite of a perfectly inelastic collision: a single body (or a pair of bodies at rest with respect to each other) suddenly splits into two or more fragments that fly apart. Energy is released — usually from chemical, nuclear, or stored elastic potential energy — while momentum remains conserved.

This final lesson of OCR Module 3.5 applies everything from Lessons 4–9 to the archetypal explosion and recoil situations: a gun firing a bullet, a rocket propelling itself by ejecting exhaust, a radioactive nucleus undergoing alpha decay, and two spring-loaded trolleys pushing each other apart. It also brings together the conceptual ideas about force, momentum, and energy that this module has developed.

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The Principle

For a closed system with no external resultant force, the total momentum is constant (Lesson 6). Therefore, if a body starts at rest and explodes:

Σp(before) = Σp(after) 0 = Σm_i v_i

The sum of final momenta is zero. This immediately gives you the direction and speed of one fragment if you know the others.

Key point: kinetic energy is not conserved in an explosion — it increases, because potential energy (chemical, elastic, nuclear) is converted into kinetic energy. This is the opposite of an inelastic collision, where KE decreases.

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Diagram — Collision vs Explosion

flowchart LR
  subgraph Collision
    A1[Two bodies moving] --> A2[Interaction]
    A2 --> A3[One or two bodies moving]
  end
  subgraph Explosion
    B1[Single body at rest] --> B2[Release of stored energy]
    B2 --> B3[Two or more bodies moving apart]
  end

In both processes, momentum is conserved. In a collision, KE generally decreases (inelastic) or stays the same (elastic). In an explosion, KE increases, because stored PE has been converted into KE.

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Worked Example 1 — Classic Recoil From Rest

A 5.0 kg object at rest explodes into two fragments: a 1.0 kg fragment moving east at 30 m s⁻¹, and a 4.0 kg fragment moving in some unknown direction. Find the velocity of the 4.0 kg fragment.

Before: p_total = 0. After: 0 = (1.0)(+30) + (4.0)(v) ⇒ v = −30 / 4.0 = −7.5 m s⁻¹

So the 4.0 kg fragment moves at 7.5 m s⁻¹ west. Notice that the speeds are inversely proportional to the masses: the 1/4 mass ratio gives a 4/1 speed ratio. This is the signature of all "explosion from rest" calculations.

Kinetic energy released:

• Before: 0 J
• After: ½ × 1.0 × 900 + ½ × 4.0 × 56.25 = 450 + 112.5 = 562.5 J

All 562.5 J came from the potential energy stored in whatever made the object explode (chemical propellant, compressed spring, etc.).

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Worked Example 2 — Spring-Loaded Trolleys

Two trolleys, of masses 2.0 kg and 3.0 kg, are at rest on a smooth horizontal track. A compressed spring between them is released and they fly apart. The 2.0 kg trolley is observed to move at 1.5 m s⁻¹. Find (a) the velocity of the 3.0 kg trolley and (b) the energy that was stored in the spring.

Momentum conservation:

• Before: 0
• After: (2.0)(+1.5) + (3.0)(v) = 0
• v = −3.0 / 3.0 = −1.0 m s⁻¹ (opposite direction to the 2.0 kg trolley)

Energy check:

• KE after: ½ × 2.0 × 2.25 + ½ × 3.0 × 1.0 = 2.25 + 1.50 = 3.75 J

The spring originally stored 3.75 J of elastic potential energy, which has all been converted into the kinetic energy of the two trolleys (assuming the spring is massless and no sound/heat is produced). The heavier trolley moves more slowly, consistent with momentum conservation.

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Worked Example 3 — Gun and Bullet

A gun of mass 4.0 kg fires a 0.020 kg bullet horizontally at a muzzle velocity of 400 m s⁻¹. Find the recoil velocity of the gun and the total kinetic energy released by the cartridge (assume the gun is free to recoil with no shoulder or mount).

Momentum conservation:

• 0 = 0.020 × 400 + 4.0 × v_gun
• v_gun = −8.0 / 4.0 = −2.0 m s⁻¹

The gun recoils at 2.0 m s⁻¹.

Kinetic energy released:

• KE_bullet = ½ × 0.020 × 400² = 1600 J
• KE_gun = ½ × 4.0 × 4.0 = 8.0 J
• Total = 1608 J

Energy distribution: 1600/1608 = 99.5% of the energy goes to the bullet, only 0.5% to the gun. This is a general result for explosions: the lighter fragment gets almost all the kinetic energy, because KE = p²/(2m) and the two fragments have the same magnitude of momentum. This is why bullets are dangerous and gun recoil is manageable.

Derivation: For an explosion from rest, |p₁| = |p₂| = p. KE_i = p² / (2 m_i). So KE_1 / KE_2 = m_2 / m_1 — the KE ratio is the reciprocal of the mass ratio. Lighter fragment, more kinetic energy.

In a real gun, the shooter's shoulder absorbs some of the recoil, making the effective "mass" of the gun + body much larger and further reducing the recoil velocity. But the bullet's muzzle velocity is essentially unchanged.

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Worked Example 4 — Alpha Decay of Radium

A stationary ²²⁶Ra nucleus (mass 226 u) decays into a ²²²Rn nucleus (mass 222 u) and an alpha particle (mass 4 u). The alpha particle flies off at 1.5 × 10⁷ m s⁻¹. Find the recoil velocity of the radon nucleus and the kinetic energy of each product.

(1 u = 1.66 × 10⁻²⁷ kg)

Momentum conservation:

• 0 = 4 × 1.5 × 10⁷ + 222 × v_Rn
• v_Rn = −6.0 × 10⁷ / 222 = −2.70 × 10⁵ m s⁻¹

The radon nucleus recoils at 2.70 × 10⁵ m s⁻¹ in the opposite direction — about 56 times slower than the alpha, matching the mass ratio 222/4.

Kinetic energies:

• Alpha: ½ × 4 × 1.66 × 10⁻²⁷ × (1.5 × 10⁷)² = ½ × 6.64 × 10⁻²⁷ × 2.25 × 10¹⁴ = 7.47 × 10⁻¹³ J
• Radon: ½ × 222 × 1.66 × 10⁻²⁷ × (2.70 × 10⁵)² = ½ × 3.685 × 10⁻²⁵ × 7.29 × 10¹⁰ = 1.34 × 10⁻¹⁴ J
• Total: 7.60 × 10⁻¹³ J ≈ 4.75 MeV

About 98.2% of the kinetic energy goes to the alpha particle and 1.8% to the radon nucleus, consistent with the inverse-mass result above. The total 4.75 MeV is the released nuclear binding energy for this decay — a figure that can be looked up in any nuclear physics textbook.

This is why alpha particles are so energetic (a few MeV) and why the "daughter" nucleus barely moves: momentum conservation forces the light fragment to take nearly all the KE.

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Rocket Propulsion — The Continuous Explosion

A rocket is really a continuous explosion: it ejects exhaust gas backwards, and by momentum conservation, the rocket accelerates forwards. For A-Level, the key qualitative idea is:

Thrust = v_exhaust × (dm/dt)

where:

• v_exhaust is the speed of the exhaust relative to the rocket (typically 2–4 km s⁻¹ for chemical rockets),
• dm/dt is the mass flow rate of the exhaust (in kg s⁻¹).

The full rocket equation (Tsiolkovsky's equation) is beyond A-Level, but OCR does expect candidates to understand qualitatively that:

1. The rocket works in a vacuum (unlike a jet engine, which needs atmospheric oxygen).
2. Increasing the exhaust velocity or the mass flow rate increases the thrust.
3. As fuel burns, the remaining mass of the rocket decreases, so its acceleration (for a given thrust) increases.

Simple Quantitative Example

A rocket burns fuel at 50 kg s⁻¹ with exhaust velocity 2500 m s⁻¹. Find the thrust.

Thrust = 50 × 2500 = 125 000 N = 125 kN