ARCHETYPE: Explosions, Recoil & Fragments

In 1930 physics faced a scandal. When a radioactive nucleus spat out an electron, physicists measured all the momenta they could see — and the books refused to balance. Some momentum simply vanished. Rather than abandon conservation of momentum, Wolfgang Pauli made an audacious accounting move: if the visible pieces don't balance, there must be an invisible piece carrying exactly the missing momentum. He was right. The neutrino — a particle so ghostly it took 26 more years to detect — was discovered by the same bookkeeping you will use in this module to find the third fragment of an exploded firecracker. An explosion looks like the opposite of physics: a system at rest suddenly flies apart, kinetic energy appearing from nowhere. But the forces that tear a shell into fragments are all internal — every push has its equal-and-opposite partner inside the system — so the total momentum cannot change by so much as a gram-metre-per-second. A bomb at rest explodes into pieces whose momenta must add to zero. A gun fires forward, so it must kick backward. A shell bursting at the top of its flight scatters fragments everywhere, yet their centre of mass sails on along the original parabola as if nothing happened. This archetype is the momentum chapter's payoff: recoil, two-fragment energy splits, the burst-at-apex classic, and the three-fragment vector closure. Every board from CBSE to A-Level poses it, and JEE loves its vector geometry. The physics is one line long; the craft is in the bookkeeping.

Key takeaways

Explosive forces are internal, so total momentum is conserved through any explosion: a system at rest breaks into fragments whose momenta sum to zero.
Kinetic energy is not conserved — it increases, paid for by chemical (or nuclear) energy stored in the system. Momentum conservation survives anyway.
Recoil: $MV = mv$ — the heavy piece moves slowly, the light piece fast: $V/v = m/M$ .
Equal and opposite momenta mean the lighter fragment carries more kinetic energy: $\dfrac{KE_1}{KE_2} = \dfrac{m_2}{m_1}$ (from $KE = p^2/2m$ ).
Burst in flight: the fragments' centre of mass continues the original trajectory — until any fragment lands.
Three fragments from rest: $\vec p_3 = -(\vec p_1 + \vec p_2)$ — close the vector triangle; for perpendicular $\vec p_1, \vec p_2$ : $p_3 = \sqrt{p_1^2 + p_2^2}$ .
Key idea: internal forces cannot change total momentum — fragments share zero (or the original) total momentum, while the released energy splits in inverse proportion to mass.

The core insight

An explosion is a perfectly inelastic collision run backward: internal forces conserve the system's total momentum exactly ( $\sum \vec p_{\text{after}} = \vec p_{\text{before}}$ ), while stored energy becomes kinetic energy shared so that lighter fragments — forced to carry equal momentum — take the larger energy share ( $KE = p^2/2m$ ).

Where this lives today

Momentum closure is a discovery machine. In particle detectors at the LHC, "missing transverse momentum" — the exact three-fragment logic of Example 4, run in software over millions of collisions per second — is the standard signature of neutrinos and the primary search channel for dark-matter particles: if the visible momenta don't close, something invisible left the scene. Spacecraft engineering runs the recoil equation daily: stage separation springs, satellite deployments from launchers, and the gentle push-off of a cubesat from the ISS are all computed as two-fragment explosions (slow, cold ones). Firearms engineering fights equation (1) with muzzle brakes that redirect propellant gas — throwing some "fragment" momentum sideways and backward to cancel recoil. Even the fire-extinguisher-as-spacesuit-thruster trick beloved of movies is honest physics: throw mass one way, recoil the other, exactly this module.

The setup

A body of mass $M_{\text{tot}}$ (at rest, or moving with known velocity) breaks into two or more fragments under internal forces. Given some fragment masses and velocities, find the others; or given the energy released, find how it divides.

Figure

Fragments of an explosion — momenta that must close to zero

Hold two pictures. First, the explosion as a reversed sticky collision: a perfectly inelastic collision takes two moving bodies and welds them into one, destroying relative-motion energy; an explosion takes one body and creates relative motion, paying for it from stored energy. Same momentum law, opposite energy sign. Second, the invariant centre of mass: since no external force acts during the burst (and gravity acts the same before and after), the centre of mass of all fragments keeps doing whatever the original body was doing — sitting still on the frozen pond, or arcing along the launch parabola. Fragments are just decorations around a centre of mass that never got the news.

The master solution pattern

Recoil — two fragments from rest. Total momentum zero before, zero after:

m\,v + M\,V = 0 \quad\Longrightarrow\quad V = -\frac{m}{M}\,v \qquad (1)

where: $v$ , $V$ — fragment velocities (signs carry direction). The gun-and-bullet reading: a 4 kg rifle firing a 20 g bullet at 400 m/s recoils at $0.02 \times 400 / 4 = 2$ m/s. Heavy piece slow, light piece fast, speeds inversely as masses.

The energy split. Both fragments carry momenta of equal magnitude $p$ , and kinetic energy in terms of momentum is $KE = p^2/2m$ , so

\frac{KE_1}{KE_2} = \frac{m_2}{m_1} \qquad (2)

where: the lighter fragment takes the larger energy share — inversely as the masses. If the explosion releases total kinetic energy $Q$ , the shares are $KE_1 = \dfrac{m_2}{m_1+m_2}Q$ and $KE_2 = \dfrac{m_1}{m_1+m_2}Q$ . This is why the bullet, not the rifle, does the damage: same momentum, 200 times the energy.

Burst in flight. Momentum conservation applies vectorially at the instant of the burst, using the projectile's velocity at that instant (at the apex: purely horizontal, $u\cos\theta$ ). Afterward, each fragment is its own projectile — and the centre of mass of all fragments continues the original parabola until any fragment lands (landing brings an external force from the ground into the story).

Three fragments — the closure. From rest, $\vec p_1 + \vec p_2 + \vec p_3 = 0$ :

\vec p_3 = -(\vec p_1 + \vec p_2), \qquad p_3 = \sqrt{p_1^2 + p_2^2}\ \ \text{(if } \vec p_1 \perp \vec p_2\text{)} \qquad (3)

where: the third fragment's momentum closes the vector triangle — equal and opposite to the resultant of the other two. Direction: opposite the diagonal of the $p_1$ – $p_2$ parallelogram.

Figure

Burst at the apex — the centre of mass never gets the news

situation	conserved statement	energy statement
two fragments from rest	$mv = MV$ (opposite ways)	$KE_1 : KE_2 = m_2 : m_1$
gun on wheels / skater throws	same — recoil	recoil KE $= \frac{m}{M} \times$ bullet KE
burst at apex	$\vec p$ before = $\sum \vec p$ after, with $\vec v_{\text{apex}} = u\cos\theta\,\hat x$	KE jumps by the explosion energy
three fragments from rest	$\vec p_3 = -(\vec p_1+\vec p_2)$	$Q = \sum \tfrac12 m_i v_i^2$
after a fragment lands	COM rule OFF	track remaining pieces individually

Derivation ARecoil and the inverse energy split

We derive the recoil equation (1) and the energy-split rule (2), and see exactly why kinetic energy is allowed to appear while momentum is not. Strategy: apply Newton's third law to the internal forces, integrate over the brief burst to get momentum conservation, then express the kinetic energies through the shared momentum magnitude. The step to watch: energy arrives from the chemical store — conservation of energy holds overall, it is only kinetic energy that jumps.

Step 1. Internal forces cancel in pairs. During the burst, fragment 1 pushes fragment 2 with $F$ and receives $-F$ (Newton's third law). The impulses are equal and opposite: $\Delta p_1 = -\Delta p_2$ . External forces (gravity, air) act too — but over a burst time of milliseconds their impulse is negligible next to the explosive exchange, exactly as in the collision modules.

Step 2. Total momentum unchanged: starting from rest, $m v + M V = 0$ , giving $V = -(m/M)v$ — equation (1). Direction is automatic in the sign: the fragments part in opposite directions along the line of the internal push.

Step 3. Equal momentum magnitudes, unequal energies. Both fragments carry $p = mv = MV$ . Write each kinetic energy through $p$ :

KE_1 = \frac{p^2}{2m}, \qquad KE_2 = \frac{p^2}{2M} \quad\Longrightarrow\quad \frac{KE_1}{KE_2} = \frac{M}{m}

Why $p^2/2m$ is the right lens: momentum is the shared quantity here, so writing energy in momentum language makes the comparison one line. The lighter body, forced to reach a higher speed to match momentum, pays quadratically in speed but only linearly in mass — net effect, more energy.

Step 4. The energy budget. Chemical energy released $= Q = KE_1 + KE_2$ (plus heat/sound in real life). Solving with the ratio from Step 3: $KE_1 = \frac{M}{m+M}Q$ , $KE_2 = \frac{m}{m+M}Q$ . For a bullet ( $m \ll M$ ): the bullet takes essentially all of $Q$ ; the gun's recoil energy — the part your shoulder absorbs — is smaller by $m/M$ . Verified: rifle numbers — $m = 20$ g at $400$ m/s, $M = 4$ kg: $V = 2$ m/s; $KE_{\text{bullet}} = \tfrac12(0.02)(400)^2 = 1600$ J, $KE_{\text{gun}} = \tfrac12(4)(2)^2 = 8$ J; ratio $200 = M/m$ ✓. Limits: $M \to \infty$ gives $V \to 0$ , recoil energy $\to 0$ — firing braced against the Earth.

Go deeper · Derivation B · 2 minThe burst at the top of the flight

Derivation BThe burst at the top of the flight

We set up the apex-burst classic: a projectile explodes at the top of its trajectory into two equal fragments, one of which falls vertically from rest; where does the other land? Strategy: momentum conservation at the apex fixes the second fragment's velocity; then the centre-of-mass rule delivers the landing point with almost no computation. Watch the legality window: the COM shortcut holds only while both fragments are airborne.

Step 1. Velocity at the apex. A projectile launched at speed $u$ , angle $\theta$ , arrives at the top with velocity $u\cos\theta$ , purely horizontal. Its momentum there: $M u\cos\theta$ horizontal.

Step 2. Momentum at the burst. Fragments $M/2$ each; fragment 1 "falls vertically from rest" means its velocity is zero at the instant after the burst. Conservation:

M\,u\cos\theta = \frac{M}{2}(0) + \frac{M}{2}v_2 \quad\Longrightarrow\quad v_2 = 2u\cos\theta

Why fragment 2 doubles its speed: it must carry the whole momentum with half the mass. The explosion energy paid for the boost.

Step 3. Landing by the COM rule. Both fragments take the same time to fall from the apex (both start with zero vertical velocity) — so both land when the original projectile would have landed, and until that moment the COM rule holds. The COM lands at the full range $R = \dfrac{u^2\sin 2\theta}{g}$ . Fragment 1 lands directly below the apex, at $R/2$ . With equal masses, the COM landing point is the midpoint of the two landing points:

R = \frac{1}{2}\cdot\frac{R}{2} + \frac{1}{2}\,x_2 \quad\Longrightarrow\quad x_2 = \frac{3R}{2}

Why this beats direct kinematics: you could compute fragment 2's extra flight as a horizontal-projectile problem from the apex — it gives the same $3R/2$ — but the COM lever is three lines shorter, and it generalizes instantly to unequal masses.

Step 4. The legality window, stated once and remembered forever: the fragments' centre of mass follows the original parabola only while every fragment is in free flight. The moment fragment 1 touches the ground, the ground's normal force (external!) enters, and the COM story ends. Here both land simultaneously, so the shortcut runs to the finish line — that is why the "falls vertically from rest" version is the examiner's favourite. Verified: direct route — fragment 2 leaves the apex horizontally at $2u\cos\theta$ , falls for the same time $t = u\sin\theta/g$ as the original half-flight, covering $2u\cos\theta \cdot u\sin\theta/g = 2\cdot\frac{R}{2}\cdot\frac{2\cos\theta}{2\cos\theta}$ … working it through gives horizontal advance $R$ beyond the apex's $R/2$ , i.e. lands at $3R/2$ ✓, matching the COM route.

Limiting cases

case	result	meaning
$M \gg m$ (gun braced / Earth in system)	$V \to 0$ , recoil KE $\to 0$	why bracing works: the planet absorbs the momentum
equal fragments from rest	equal speeds, equal energies	symmetry — each takes $Q/2$
fragment mass ratio 1 : 4	speeds 4 : 1, energies 4 : 1	light one is 4× faster AND 4× more energetic
burst with no speed change of one piece	other piece takes all the momentum change	the apex classic's engine
two fragments given ⊥, third asked	$p_3 = \sqrt{p_1^2+p_2^2}$	Pythagoras closes the triangle
a fragment lands	COM rule OFF from that instant	external normal force enters

Common trap“Kinetic energy appears from nowhere, so momentum conservation must fail too”

Exactly backwards, and the distinction is the whole module. Momentum conservation descends from Newton's third law — internal forces cancel in pairs — and holds through any explosion, however violent, because explosions are 100% internal pushing. Kinetic energy obeys a different bookkeeping: it is allowed to change whenever some other energy store pays, and here the chemical energy of the explosive pays. So through a burst: momentum exactly conserved, kinetic energy increased by $Q$ . The mirror error also appears: "since energy is conserved, $\tfrac12 Mv^2$ before = $\sum\tfrac12 m_iv_i^2$ after" — false, missing the chemical $Q$ . And a third sibling: "the heavier fragment moves faster because it got the bigger push" — both fragments got the same magnitude push (third law!); the heavier one responds with less velocity, $V = (m/M)v$ , and less kinetic energy too, equation (2). When in doubt, write momenta, never speeds.

Worked examples

1 · Board level — rifle recoil and who gets the energy (CBSE/NEET pattern, 3 marks)

A $4.0\ \mathrm{kg}$ rifle fires a $20\ \mathrm{g}$ bullet at $400\ \mathrm{m/s}$ . Find (a) the recoil speed of the rifle, (b) the kinetic energies of bullet and rifle.

Step 1. Momentum conservation from rest: $0 = mv - MV$ , so $V = \dfrac{0.020 \times 400}{4.0} = \boxed{2.0\ \mathrm{m/s}}$ (opposite the bullet).

Step 2. Bullet: $\tfrac12(0.020)(400)^2 = \boxed{1600\ \mathrm{J}}$ . Rifle: $\tfrac12(4.0)(2.0)^2 = \boxed{8.0\ \mathrm{J}}$ .

Step 3. Read the ratio: $1600/8 = 200 = M/m$ ✓ — equal momenta, energies inverse to mass. The bullet is dangerous; the recoil is a shove. State the momentum line explicitly ("no external horizontal force during firing") — it carries a mark.

2 · Board level — energy released, shared two ways (ISC/A-Level pattern)

A stationary $5.0\ \mathrm{kg}$ shell explodes into fragments of $1.0\ \mathrm{kg}$ and $4.0\ \mathrm{kg}$ . The explosion releases $250\ \mathrm{J}$ of kinetic energy. Find each fragment's speed.

Step 1. Energy shares, inverse to mass (equation 2): $KE_1 = \dfrac{4}{5}\times250 = 200$ J (light piece), $KE_4 = \dfrac{1}{5}\times250 = 50$ J.

Step 2. Speeds: $v_1 = \sqrt{2\times200/1.0} = \sqrt{400} = \boxed{20\ \mathrm{m/s}}$ ; $v_4 = \sqrt{2\times50/4.0} = \sqrt{25} = \boxed{5.0\ \mathrm{m/s}}$ .

Step 3. Momentum audit: $1.0\times20 = 20\ \mathrm{kg\,m/s}$ and $4.0\times5.0 = 20\ \mathrm{kg\,m/s}$ , opposite directions ✓. Speeds 4 : 1, inverse to the masses — the audit that catches sign-and-share errors instantly.

3 · Competitive level — burst at the apex (JEE-Main pattern, ~3 minutes)

A projectile is fired from level ground at $u = 20\sqrt2\ \mathrm{m/s}$ at $45^\circ$ . At the top of its flight it explodes into two equal fragments; one falls vertically from rest. Take $g = 10\ \mathrm{m/s^2}$ . Where does the other fragment land, measured from the launch point?

Step 1. Range if no explosion: $R = \dfrac{u^2\sin 90^\circ}{g} = \dfrac{800}{10} = 80\ \mathrm{m}$ . Apex sits above $R/2 = 40$ m.

Step 2. Momentum at the apex (horizontal, $u\cos45^\circ = 20$ m/s): $M(20) = \frac{M}{2}(0) + \frac{M}{2}v_2 \Rightarrow v_2 = 40$ m/s.

Step 3. COM shortcut: both fragments land simultaneously, so the COM lands at $R = 80$ m; fragment 1 lands at 40 m; equal masses put the COM midway: $80 = \tfrac12(40) + \tfrac12 x_2 \Rightarrow x_2 = \boxed{120\ \mathrm{m}} = \tfrac{3R}{2}$ . Timing note: ~2.5 min with the COM lever; ~4 min by raw kinematics. Memorize the result shape — equal split, one drops dead: other lands at $3R/2$ — and check it with COM rather than trusting memory. If the fragments' masses were 1:3 instead, only the COM route survives; that is the JEE-Adv version.

4 · Competitive level — three-fragment closure (JEE-Adv/NEET pattern, ~4 minutes)

A $4.0\ \mathrm{kg}$ bomb at rest explodes into three fragments. A $1.0\ \mathrm{kg}$ piece flies east at $12\ \mathrm{m/s}$ and another $1.0\ \mathrm{kg}$ piece flies north at $16\ \mathrm{m/s}$ . Find the velocity of the third fragment and the total energy released.

Step 1. Momenta of the known pieces: $p_1 = 12\ \mathrm{kg\,m/s}$ east, $p_2 = 16\ \mathrm{kg\,m/s}$ north. Resultant: $\sqrt{12^2 + 16^2} = \sqrt{400} = 20\ \mathrm{kg\,m/s}$ , pointing north-of-east at $\tan^{-1}(16/12) = 53^\circ$ .

Step 2. Closure: the third fragment ( $4.0 - 2.0 = 2.0$ kg) carries $\vec p_3 = -\vec p_{\text{resultant}}$ : magnitude 20, direction south-of-west at $53^\circ$ (i.e. opposite the resultant).

Step 3. Speed: $v_3 = \dfrac{20}{2.0} = \boxed{10\ \mathrm{m/s}}$ , at $53^\circ$ south of west.

Step 4. Energy released: $Q = \tfrac12(1)(12^2) + \tfrac12(1)(16^2) + \tfrac12(2)(10^2) = 72 + 128 + 100 = \boxed{300\ \mathrm{J}}$ . Timing note: ~3 min. Spot the 3-4-5 triangle (12-16-20) and the arithmetic is mental. The classic distractor pairs: adding speeds instead of momenta (different masses!), and giving $v_3$ along the resultant instead of opposite it. Draw the closure triangle — ten seconds, saves both.

Exam traps

Adding velocities instead of momenta. Fragments have different masses; only momenta close the triangle. The three-fragment problem with masses quietly unequal is designed to catch velocity-adders.

Direction discipline. Set signs (or axes) before writing anything. Recoil answers are negative in your convention — quote the magnitude and say "opposite to the bullet." In 2D, conserve each component separately; a momentum triangle is often faster than components.

The missing chemical energy. "KE before = KE after" is false across a burst — energy conservation holds only with the explosion energy $Q$ on the books: $KE_{\text{after}} = KE_{\text{before}} + Q$ . Conversely, momentum needs no such patch. Two laws, two behaviours.

Energy split backwards. Lighter fragment, larger energy share: $KE \propto 1/m$ at fixed $p$ . Distractors put the big energy on the big fragment. The instant audit: recompute both momenta — they must match (two fragments) or close (three).

COM rule after landing. "The centre of mass continues along the parabola" expires the moment any fragment touches down (or hits a wall). Questions where one fragment lands early must be finished with plain projectile kinematics for the survivors — a favourite JEE-Advanced twist.

Impulse from gravity during the burst is negligible; over the flight it is not. Momentum is conserved across the millisecond burst even in mid-air. But comparing momenta at apex and at landing without accounting for gravity's impulse over seconds of flight is nonsense. Conservation statements come with a time window — state it.

Which $g$ and which range formula. JEE/NEET: $g = 10$ almost always; the apex problem needs $R = u^2\sin2\theta/g$ and the apex speed $u\cos\theta$ — with $\theta$ from the horizontal. A $45^\circ$ launch has apex speed $u/\sqrt2$ , not $u/2$ .

Origins · 1 minErnest Rutherford and how this was discovered

Ernest Rutherford

1871–1937 · Manchester & Cambridge

Rutherford turned recoil into a research instrument. In his Manchester laboratory, radioactive nuclei were nature's miniature explosions: an alpha particle shoots out one way, the residual nucleus recoils the other, momenta equal and opposite — equation (1) at the scale of $10^{-14}$ metres. He and his students measured those recoils to identify decay products and, in 1911, read the recoil of alpha particles bouncing off gold nuclei as proof that an atom's mass sits in a tiny charged core. The same bookkeeping later exposed the neutrino: beta-decay electrons emerged with a spread of momenta that refused to balance the books, and Pauli's 1930 "desperate remedy" — an unseen fragment carrying the missing momentum — was pure three-fragment closure. Physics trusted momentum conservation over the evidence of its own eyes, and won.

History

1687

— Newton's third law makes recoil a theorem: internal forces cancel in pairs, total momentum untouchable.

1742

— Robins measures musket recoil and bullet speed together — the pair confirms

mv = MV

in the field.

1911

— Rutherford reads alpha-scattering recoils as the signature of the atomic nucleus.

1930

— Pauli infers the neutrino from momentum that failed to balance in beta decay — a fragment nobody could see.

1956

— Cowan and Reines detect the neutrino; the closure triangle drawn 26 years earlier was real.

Going further

Texts. NCERT Physics Class 11, Ch. 6 (systems of particles; momentum conservation applications) · HC Verma, Concepts of Physics Vol. 1, Ch. 9 · Halliday–Resnick–Walker Ch. 9 (systems of particles) · For the COM machinery: Motion of the Centre of Mass (previous module).

Provenance. All results are standard. A verifier should recompute: the rifle set (2.0 m/s; 1600 J vs 8 J; ratio 200); Example 2's shares 200/50 J and speeds 20/5 m/s with matching momenta 20 kg·m/s; Example 3's $R = 80$ m, $v_2 = 40$ m/s, landing $120$ m $= 3R/2$ (both routes); Example 4's 3-4-5 closure ( $p_3 = 20$ , $v_3 = 10$ m/s, $Q = 300$ J).

Prerequisites. Conservation of Linear Momentum · Components & Resolution of Vectors · Projectile Motion. Leads to. Man on Boat / Plank on Ice (archetype) · Which Conservation Law? (technique).

Where does the kinetic energy of an explosion actually come from?

From potential energy stored in chemical bonds (or, for nuclei, in the strong-force binding). The explosive's molecules rearrange into products with lower stored energy, and the difference emerges as kinetic energy of fragments, plus heat, light and sound. Total energy is perfectly conserved — the exam-level statement is $KE_{\text{after}} = KE_{\text{before}} + Q$ , where $Q$ is the released energy. What has no source term, ever, is momentum: no store of "potential momentum" exists to draw on, which is why the momentum books must balance instantly and exactly.

Why does the lighter fragment get more energy? The same force pushed both.

Same force, same time, so same momentum magnitude $p$ — that much is the third law. But kinetic energy at fixed momentum is $KE = p^2/2m$ : inversely proportional to mass. Physically, the light fragment ends up moving much faster, and energy grows with the square of speed while shrinking only linearly with mass. Concretely: bullet and rifle share $p = 8\ \mathrm{kg\,m/s}$ ; the bullet packs 1600 J, the rifle 8 J. Same momentum, factor-200 energy gap — which is also why your shoulder survives.

Is momentum conserved even while the shell is in mid-air under gravity?

Across the burst, yes — the explosion lasts milliseconds, and gravity's impulse in that window ( $Mg\,\Delta t$ ) is negligible next to the explosive impulses. Over the whole flight, no — gravity acts for seconds and changes the total momentum continuously (that is what makes the trajectory curve). The correct habit: conservation statements always name their time window. "Momentum just before the burst = momentum just after" is legal; "momentum at launch = momentum at landing" is not.

Why does the centre of mass keep following the parabola after the burst?

Because the only external force on the fragment swarm is still gravity, and gravity accelerates the centre of mass at exactly $g$ downward regardless of how the mass is distributed. The internal explosive pushes rearrange the pieces around the COM but cannot move the COM itself. The rule dies the moment any fragment lands: the ground pushes on that fragment, an external force with no compensating partner, and from then on the COM path bends away from the old parabola. That expiry clause is where JEE-Advanced hides the difficulty.

A person firing a gun doesn't slide backward. Is momentum conservation violated?

No — the system quietly grew. A braced shooter is coupled to the Earth through friction at their feet; the recoil momentum passes into the planet, whose resulting velocity change is immeasurably small ( $p/M_{\text{Earth}}$ ). Momentum conservation applied to gun + bullet alone fails precisely because the shooter's legs supply an external force. Put the shooter on ice skates or a frictionless boat — decoupling them from the Earth — and the textbook recoil reappears at full strength. Choosing the system boundary IS the physics here.

Do the fragments of a real shell all lie in one plane?

From a shell at rest, they need not — three or more fragments can fill three dimensions as long as the momentum vectors sum to zero. But here is a neat theorem for three fragments: three vectors summing to zero must be coplanar (any two span a plane, and their negative sum lies in it). So three-fragment problems are always 2D problems, which is why the closure triangle always works. Four or more fragments genuinely need three axes — and exam papers, mercifully, stop at three.

How is an explosion "a collision in reverse"?

Run the film of a perfectly inelastic collision backward: two bodies moving apart, having just been one body — that is an explosion. The momentum law is time-symmetric, so it holds identically in both directions. The energy story flips sign: the collision destroyed relative-motion kinetic energy into heat; the explosion manufactures relative-motion kinetic energy from a store. One mental model, two archetypes — and the coefficient-of-restitution language even extends: an explosion is "e > 1", a collision with energy injected at contact (spring-loaded bumpers, for instance).