Explosions and Recoil

Spec mapping: OCR H556 Module 3.3 — Explosions and recoil as applications of momentum conservation; distribution of kinetic energy between fragments; qualitative application to rocket propulsion and radioactive decay. (Refer to the official OCR H556 specification document for exact wording.)

An explosion is, in the language of momentum, 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 at some speed. Energy is released from a stored internal source — chemical, nuclear, or elastic potential — and converted into the kinetic energy of the fragments. Crucially, momentum is still conserved throughout, just as it is in any closed-system interaction, because the internal forces between fragments are still Newton-III action-reaction pairs.

This final lesson of OCR Module 3.3 applies everything built up in Lessons 1–9 to four archetypal scenarios: a gun firing a bullet (mechanical explosion), a rocket propelling itself by ejecting exhaust (continuous explosion), a radioactive nucleus undergoing alpha decay (nuclear explosion), and two spring-loaded trolleys pushing each other apart (controlled mechanical explosion). It also brings together the conceptual ideas about force, momentum, and energy that this module has developed, and points forward to applications throughout Year 13 (nuclear physics, special relativity, kinetic theory).

Key Definition: An explosion is an interaction in which a closed system at rest (or moving as a single rigid body) suddenly splits into two or more parts that fly apart. The internal stored energy (chemical, nuclear, elastic) is released as kinetic energy. Total linear momentum is conserved (vector sum unchanged); total kinetic energy increases from zero (or near zero) by the amount of internal PE released.

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The Principle

For a closed system with no external resultant force, the total linear momentum is constant (Lesson 6). For an explosion of an initially stationary body:

$\sum \vec{p}_{\text{before}} = \vec{0} \quad \Longrightarrow \quad \sum_i m_i \vec{v}_i = \vec{0}$∑p​before​=0⟹∑i​mi​vi​=0

The sum of the final momentum vectors is zero. This single constraint is the most powerful starting point of every explosion problem. In 1-D it gives the speed and direction of one fragment if you know the others; in 2-D it gives two simultaneous equations (one per component).

Key contrast with collisions: in a collision, KE either stays the same (elastic) or decreases (inelastic). In an explosion, KE increases — it starts at zero (or near-zero) and ends at the total final-fragment KE. This extra KE comes from the release of internal potential energy (chemical, nuclear, or elastic), which is converted into the macroscopic kinetic energy of the fragments. Total energy is, of course, still conserved (First Law of Thermodynamics).

Common Exam Mistake: Trying to apply kinetic-energy conservation to find the velocities of explosion fragments. KE is not conserved in an explosion — it is the output of the problem, not a constraint. Always use momentum first, then compute KE as a follow-on.

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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 PE]
    B2 --> B3[Two or more bodies moving apart]
  end

In both processes momentum is conserved. The difference is in the energy flow: a collision sends macroscopic KE → heat/sound/deformation; an explosion sends internal PE → macroscopic KE.

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Worked Example 1 — Classic Recoil From Rest

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

Momentum (1-D, east positive): $0 = (1.0)(+30) + (4.0)(v) \Rightarrow v = -30/4.0 = -7.5$0=(1.0)(+30)+(4.0)(v)⇒v=−30/4.0=−7.5 m s$^{-1}$−1.

So the $4.0$4.0 kg fragment moves at $7.5$7.5 m s$^{-1}$−1 west. Notice the speeds are inversely proportional to the masses ($1:4$1:4 mass ratio gives $4:1$4:1 speed ratio in opposite directions). This is the signature of every "explosion from rest" calculation.

Kinetic energy released:

• Before: $0$0 J (system at rest).
• After: $\frac{1}{2}(1.0)(900) + \frac{1}{2}(4.0)(56.25) = 450 + 112.5 = 562.5$21​(1.0)(900)+21​(4.0)(56.25)=450+112.5=562.5 J.

All $562.5$562.5 J have come from the potential energy stored in whatever caused the explosion (chemical propellant, compressed spring, internal fuel). Energy distribution: $450/562.5 = 80\%$450/562.5=80% goes to the lighter (faster) fragment, $20\%$20% to the heavier (slower) one. This is the fundamental energy-asymmetry of explosions, derived below.

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The "Lighter Fragment Gets Most KE" Result

For an explosion from rest into two fragments, the magnitudes of momentum are equal and opposite: $|p_1| = |p_2| = p$∣p1​∣=∣p2​∣=p. The kinetic energy of each is then:

$\text{KE}_i = \frac{p_i^2}{2 m_i} \quad \Rightarrow \quad \frac{\text{KE}_1}{\text{KE}_2} = \frac{m_2}{m_1}$KEi​=2mi​pi2​​⇒KE2​KE1​​=m1​m2​​

The KE ratio is the reciprocal of the mass ratio — the lighter fragment gets more KE. In the limiting case $m_1 \ll m_2$m1​≪m2​, the heavy fragment gets nearly zero KE (because $p^2/(2m_2) \to 0$p2/(2m2​)→0 as $m_2$m2​ grows), and essentially all the released energy goes to the light fragment.

This is why bullets are dangerous and gun recoil is manageable; why alpha particles are MeV-energetic but the daughter nucleus barely moves; why a firework shell's small fragments fly at hundreds of metres per second; and why the ejected rocket exhaust gas carries kilojoules of KE for every joule given to the rocket itself.

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Worked Example 2 — Spring-Loaded Trolleys

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

(a) Momentum: Before $= 0$=0. After $= (2.0)(+1.5) + (3.0)(v) = 0 \Rightarrow v = -1.0$=(2.0)(+1.5)+(3.0)(v)=0⇒v=−1.0 m s$^{-1}$−1 (opposite direction to the $2.0$2.0 kg trolley).

(b) Spring PE → KE:

$E_{\text{spring}} = \text{KE}_{\text{after}} = \tfrac{1}{2}(2.0)(1.5)^2 + \tfrac{1}{2}(3.0)(1.0)^2 = 2.25 + 1.50 = 3.75 \text{ J}$Espring​=KEafter​=21​(2.0)(1.5)2+21​(3.0)(1.0)2=2.25+1.50=3.75 J

The spring originally stored $3.75$3.75 J of elastic PE, all converted to KE of the two trolleys (assuming a massless spring and no energy lost to sound or to compression/decompression of the spring material). The heavier trolley moves more slowly, exactly as momentum conservation requires.

This setup is a standard PAG 6 air-track experiment: two trolleys with a spring-loaded release between them. The energy conservation can be tested by computing $E_{\text{spring}}$Espring​ from the trolley speeds and comparing with the spring constant $\times$× compression.

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Worked Example 3 — Gun and Bullet

A gun of mass $4.0$4.0 kg fires a $0.020$0.020 kg bullet horizontally at a muzzle velocity of $400$400 m s$^{-1}$−1. Find the recoil velocity of the gun, the total kinetic energy released by the cartridge (assume the gun is free to recoil), and the energy fraction going to the bullet vs the gun.

Momentum: $0 = (0.020)(400) + (4.0)\,v_{\text{gun}} \Rightarrow v_{\text{gun}} = -8.0/4.0 = -2.0$0=(0.020)(400)+(4.0)vgun​⇒vgun​=−8.0/4.0=−2.0 m s$^{-1}$−1.

The gun recoils at $2.0$2.0 m s$^{-1}$−1 (opposite to the bullet).

Kinetic energies:

• $\text{KE}_{\text{bullet}} = \frac{1}{2}(0.020)(400)^2 = 1600$KEbullet​=21​(0.020)(400)2=1600 J.
• $\text{KE}_{\text{gun}} = \frac{1}{2}(4.0)(2.0)^2 = 8.0$KEgun​=21​(4.0)(2.0)2=8.0 J.
• Total chemical energy released $= 1608$=1608 J.

Energy distribution: $1600/1608 = 99.5\%$1600/1608=99.5% goes to the bullet, only $0.5\%$0.5% to the gun. This is exactly what the inverse-mass rule predicts: ratio of bullet mass to gun mass is $0.020/4.0 = 0.005$0.020/4.0=0.005, so the gun gets $0.5\%$0.5% of the energy and the bullet gets $99.5\%$99.5%.

In a real-world rifle, the shooter's shoulder transfers the gun's recoil impulse into the shooter's whole body. The effective "recoiling mass" becomes the shooter + gun (∼70 kg), so the recoil velocity is reduced to $v_{\text{recoil}} \approx 2.0 \times 4.0/70 = 0.11$vrecoil​≈2.0×4.0/70=0.11 m s$^{-1}$−1 — barely perceptible. The bullet's muzzle velocity is essentially unchanged, of course; the bullet doesn't "know" about the shooter's mass during the brief firing time.

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Worked Example 4 — Alpha Decay of Radium-226

A stationary $^{226}$226Ra nucleus (mass $226$226 u) decays into a $^{222}$222Rn nucleus (mass $222$222 u) and an alpha particle (mass $4$4 u). The alpha particle flies off at $1.5 \times 10^7$1.5×107 m s$^{-1}$−1. Find the recoil velocity of the radon nucleus and the kinetic energy of each product. ($1$1 u $= 1.66 \times 10^{-27}$=1.66×10−27 kg.)

Momentum: $0 = 4 \times 1.5 \times 10^7 + 222 \times v_{\text{Rn}}$0=4×1.5×107+222×vRn​, so $v_{\text{Rn}} = -6.0 \times 10^7 / 222 = -2.70 \times 10^5$vRn​=−6.0×107/222=−2.70×105 m s$^{-1}$−1.

The radon nucleus recoils at $2.70 \times 10^5$2.70×105 m s$^{-1}$−1, about $56\times$56× slower than the alpha — matching the mass ratio $222/4 = 55.5$222/4=55.5.

Kinetic energies:

• Alpha: $\frac{1}{2}(4 \times 1.66 \times 10^{-27})(1.5 \times 10^7)^2 = \frac{1}{2}(6.64 \times 10^{-27})(2.25 \times 10^{14}) = 7.47 \times 10^{-13}$21​(4×1.66×10−27)(1.5×107)2=21​(6.64×10−27)(2.25×1014)=7.47×10−13 J.
• Radon: $\frac{1}{2}(222 \times 1.66 \times 10^{-27})(2.70 \times 10^5)^2 = \frac{1}{2}(3.685 \times 10^{-25})(7.29 \times 10^{10}) = 1.34 \times 10^{-14}$21​(222×1.66×10−27)(2.70×105)2=21​(3.685×10−25)(7.29×1010)=1.34×10−14 J.
• Total: $\approx 7.60 \times 10^{-13}$≈7.60×10−13 J $\approx 4.75$≈4.75 MeV (using $1$1 eV $= 1.6 \times 10^{-19}$=1.6×10−19 J).

About $98.2\%$98.2% of the kinetic energy goes to the alpha particle and only $1.8\%$1.8% to the radon — consistent with the inverse-mass rule ($4/226 = 1.77\%$4/226=1.77%). The total $4.75$4.75 MeV is the released nuclear binding energy for this decay (specifically, $Q$Q-value).

This is why alpha particles emerge with sharply-defined, characteristic MeV-scale energies (you can spectroscopically identify the parent isotope from the alpha energy alone), and why the daughter nucleus barely moves — momentum conservation forces the light fragment to take nearly all the KE. The radon recoil energy is only $0.014$0.014 MeV — easily lost as thermal motion inside the source.

This same physics underpins ionising-radiation dosimetry, mass spectrometry of nuclear products, and the workings of an alpha-particle smoke detector.

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Rocket Propulsion — The Continuous Explosion

A rocket is best understood as a continuous explosion: it ejects exhaust gas backwards, and by momentum conservation the rocket accelerates forwards. The qualitative A-Level idea is simple:

$\text{Thrust} = v_{\text{exhaust}} \times \frac{dm}{dt}$Thrust=vexhaust​×dtdm​

where $v_{\text{exhaust}}$vexhaust​ is the exhaust speed relative to the rocket (typically $2$2–$4$4 km s$^{-1}$−1 for chemical rockets), and $dm/dt$dm/dt is the mass flow rate of exhaust (kg s$^{-1}$−1). This is a direct consequence of Newton II in the form $F = dp/dt$F=dp/dt applied to the rocket-plus-exhaust system.

The full Tsiolkovsky rocket equation ($\Delta v = v_{\text{exhaust}} \ln(m_0/m_f)$Δv=vexhaust​ln(m0​/mf​)) is beyond A-Level, but OCR does expect candidates to understand qualitatively that:

1. The rocket works in a vacuum — unlike a jet engine, it doesn't need atmospheric oxygen because it carries its own oxidiser. Saturn V and SpaceX Falcon both work in space.
2. Increasing the exhaust velocity or the mass flow rate increases the thrust.
3. As fuel burns, the rocket's remaining mass decreases, so its acceleration (for fixed thrust) increases.

Worked Example — Rocket Thrust

A rocket burns fuel at $50$50 kg s$^{-1}$−1 with exhaust velocity $2500$2500 m s$^{-1}$−1. Find the thrust.

$\text{Thrust} = (50)(2500) = 125\,000 \text{ N} = 125 \text{ kN}$Thrust=(50)(2500)=125000 N=125 kN

If the rocket has an instantaneous mass of $20\,000$20000 kg, the upward acceleration from thrust alone is $125\,000 / 20\,000 = 6.25$125000/20000=6.25 m s$^{-2}$−2. Against gravity ($9.81$9.81 m s$^{-2}$−2 downward), the net upward acceleration is $6.25 - 9.81 = -3.56$6.25−9.81=−3.56 m s$^{-2}$−2 — so this rocket cannot yet lift off. It needs more thrust or less mass. As fuel burns, mass decreases and the net acceleration becomes positive — typically around the $0.6$0.6-mass-fraction mark, the rocket "breaks" free of the ground in the SpaceX/Saturn V launch profile.

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Worked Example 5 — Three-Body Explosion in 2-D

A $6.0$6.0 kg stationary bomb explodes into three fragments of equal mass ($2.0$2.0 kg each). Fragment A flies east at $10$10 m s$^{-1}$−1. Fragment B flies north at $10$10 m s$^{-1}$−1. Find the velocity of fragment C and the total energy released.

Momentum conservation, components:

• $p_x$px​: $0 = (2.0)(10) + (2.0)(0) + (2.0)\,v_{Cx} \Rightarrow v_{Cx} = -10$0=(2.0)(10)+(2.0)(0)+(2.0)vCx​⇒vCx​=−10 m s$^{-1}$−1 (west).
• $p_y$py​: $0 = (2.0)(0) + (2.0)(10) + (2.0)\,v_{Cy} \Rightarrow v_{Cy} = -10$0=(2.0)(0)+(2.0)(10)+(2.0)vCy​⇒vCy​=−10 m s$^{-1}$−1 (south).

Magnitude: $|\vec{v}_C| = \sqrt{100 + 100} = \sqrt{200} = 14.14$∣vC​∣=100+100​=200​=14.14 m s$^{-1}$−1.

Direction: $45°$45° south of west (because $v_{Cx}$vCx​ and $v_{Cy}$vCy​ are both negative and equal magnitude).

The three velocity vectors close into a triangle (drawn tip-to-tail). Three momentum vectors of magnitudes $20$20, $20$20, $28.28$28.28 kg m s$^{-1}$−1 with angles $0°, 90°, 225°$0°,90°,225° from east respectively — sum to zero, as required.

Energy released:

• $\text{KE}_A = \frac{1}{2}(2.0)(100) = 100$KEA​=21​(2.0)(100)=100 J.
• $\text{KE}_B = \frac{1}{2}(2.0)(100) = 100$KEB​=21​(2.0)(100)=100 J.
• $\text{KE}_C = \frac{1}{2}(2.0)(200) = 200$KEC​=21​(2.0)(200)=200 J.
• Total $= 400$=400 J of chemical energy released.

Fragment C has double the KE of the others because it has $\sqrt{2}$2​ times the speed (KE $\propto v^2$∝v2). The momenta sum to zero as vectors, but the KE values do not sum to anything special — they reflect the asymmetric mass-and-direction distribution.

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