Conservation of Linear Momentum

In 1930 physics faced a crisis. In beta decay, a nucleus spits out an electron — and when experimenters added up the momenta and energies before and after, the books refused to balance. Some of the momentum simply vanished. Niels Bohr was prepared to abandon the conservation laws for nuclear events. Wolfgang Pauli chose the opposite bet: the law is exact, so something invisible must be carrying the missing momentum. His "desperate remedy" — a neutral, almost massless particle nobody had ever seen — was detected twenty-six years later. The neutrino is what it looks like to trust conservation of momentum more than you trust your own detectors. That is the standing of the law this article builds. Two skaters push apart, a rifle fires, a shell explodes at the top of its arc: in every case the total momentum of the system — the vector sum $\mathbf{P} = \sum_i m_i\mathbf{v}_i$ — comes out of the chaos exactly equal to what went in, each component separately. The Newtonian reason is almost embarrassingly short: internal forces come in equal-and-opposite pairs, so they cancel from the total, and only external forces can change $\mathbf{P}$ . But the short proof undersells the law. Newton's third law fails for fast-moving charges; momentum conservation does not — one just learns that fields carry momentum too. The deeper foundation, uncovered by Emmy Noether in 1918, is that momentum conservation is the direct consequence of the homogeneity of space: physics doesn't care where you stand. That is why the law survives intact into relativity and quantum mechanics, outliving the very equations it was first derived from.

Key takeaways

The total momentum $\mathbf{P} = \sum_i m_i\mathbf{v}_i$ of a system changes only in response to external forces: $\mathrm{d}\mathbf{P}/\mathrm{d}t = \mathbf{F}^{\mathrm{ext}}$ .
Internal forces — however violent — cancel in Newton-third-law pairs and can never change $\mathbf{P}$ ; with $\mathbf{F}^{\mathrm{ext}} = 0$ , momentum is conserved exactly.
It is a vector law: each Cartesian component is conserved independently — a projectile in gravity keeps its horizontal momentum even as gravity spends the vertical one.
Conservation holds in every collision or explosion of an isolated system, elastic or not — which is precisely what makes it the workhorse for inelastic processes where energy bookkeeping fails.
Equivalent statement: the center of mass of an isolated system moves in a straight line at constant velocity, deaf to all internal drama.
The deeper origin is symmetry: by Noether's theorem, momentum conservation is the homogeneity of space — which is why the law outlives Newton's third law in field theory.
Key idea: internal forces only move momentum around inside a system, never create or destroy it — so with no net external force, the total $\mathbf{P}$ is frozen, component by component.

The core insight

For any system, $\mathrm{d}\mathbf{P}/\mathrm{d}t = \mathbf{F}^{\mathrm{ext}}$ : internal forces cancel in Newton-third-law pairs and never change the total momentum, so when the net external force vanishes, $\mathbf{P} = \sum_i m_i\mathbf{v}_i$ is constant — each component separately.

Where this lives today

Momentum conservation remains physics' favorite way to see the invisible. At the LHC, detectors sum the transverse momenta of every particle a collision sprays out; the beams carry zero transverse momentum in, so the visible products must sum to zero out. When they don't, the imbalance — "missing transverse momentum" — is treated as a measurement of whatever escaped unseen. Neutrinos are logged this way millions of times a day, and the same signature, larger than neutrinos can explain, is the smoking gun that dark-matter searches at CERN watch for in 2026. The most spectacular modern application is astrophysical: gravitational waves carry momentum, and when two merging black holes radiate them asymmetrically, the merged remnant must recoil. These "gravitational kicks" — inferred from LIGO-Virgo waveforms — can exceed a thousand kilometers per second, enough to eject the black hole from its host galaxy entirely: momentum bookkeeping performed on ripples in spacetime itself. Back on the engineering floor, every satellite reaction-wheel desaturation, every ion-thruster burn, and every docking maneuver at a space station is planned as a momentum ledger, because in orbit there is nothing to push against except what you brought — and conservation of momentum is the whole rulebook.

The physical picture

Picture two skaters at rest on smooth ice, palms together, pushing off. Whatever force one feels, the other feels exactly reversed, for exactly as long — so their momenta change by equal and opposite amounts, and the pair's total stays what it was: zero. The heavier skater drifts away slowly, the lighter one quickly, in exact inverse proportion to their masses. Their combined center of mass never moves. No amount of internal pushing, kicking, or clever timing can change that; to move the total, you need the outside world.

Figure

Push-off on ice — internal forces trade momentum but the total stays zero

The right mental model is an accounting one. Momentum is a conserved currency: particles inside a system can only trade it among themselves, and every internal transaction is double-entry — the third law guarantees that what one account gains, another loses at the same instant. External forces are the only deposits or withdrawals. A rifle and its bullet share zero momentum before the trigger; afterward the bullet carries $+4\ \mathrm{kg\,m/s}$ and the rifle exactly $-4\ \mathrm{kg\,m/s}$ . The explosion — thousands of newtons, a millisecond of chemical violence — moved a great deal of currency, and created none. What gives the law its practical power is what it doesn't ask of you. It never mentions the forces' details. In a collision the internal forces are jagged, enormous, and unknowable; in an explosion they are chemistry; in a nuclear decay they are quantum. Conservation of momentum ignores all of it and connects before directly to after with algebra. And because the argument never once mentions kinetic energy, the law holds whether the collision is elastic or a sticky, energy-shredding wreck — which is exactly when you need it most.

The math: one law, two derivations

Start with the object the law protects. For $N$ particles, define the total momentum as the vector sum of the individual momenta — and note, before any dynamics, that this sum has a second identity worth its own display. Writing $M = \sum_i m_i$ for the total mass and $\mathbf{R}_{\mathrm{cm}} = \tfrac{1}{M}\sum_i m_i \mathbf{r}_i$ for the center of mass:

\mathbf{P} = \sum_{i=1}^{N} m_i \mathbf{v}_i = M\,\dot{\mathbf{R}}_{\mathrm{cm}} \qquad (1)

where: $m_i, \mathbf{v}_i$ — mass and velocity of particle $i$ ; $\dot{\mathbf{R}}_{\mathrm{cm}}$ — the center-of-mass velocity. The second equality (differentiate the definition of $\mathbf{R}_{\mathrm{cm}}$ ) says the total momentum of any system, however complicated, is that of a single imaginary particle of mass $M$ riding the center of mass. Every statement about $\mathbf{P}$ is secretly a statement about that point. The dynamics of $\mathbf{P}$ follow from summing Newton's second law over the particles, and the result is the pivot of this whole article:

\frac{\mathrm{d}\mathbf{P}}{\mathrm{d}t} = \mathbf{F}^{\mathrm{ext}} \qquad (2)

where: $\mathbf{F}^{\mathrm{ext}}$ — the vector sum of forces from outside the system only. The internal forces have vanished from the ledger — Derivation A shows the cancellation — and the payoff is immediate: if the external forces sum to zero,

\mathbf{F}^{\mathrm{ext}} = \mathbf{0} \;\Longrightarrow\; \mathbf{P} = \text{const} \qquad (3)

where: the constancy holds component by component, because (2) is a vector equation: $F^{\mathrm{ext}}_x = 0$ freezes $P_x$ even while gravity spends $P_y$ . The conserved value itself is frame-dependent — boost to a frame moving at $\mathbf{V}$ and $\mathbf{P}$ shifts by $-M\mathbf{V}$ — but its constancy in time holds in every inertial frame, and the frame where $\mathbf{P} = \mathbf{0}$ (the center-of-mass frame) is the collision analyst's favorite workbench.

Derivation AConservation from Newton's third law

We are deriving the system law (2) — total momentum responds to external forces only — and with it the conservation statement (3). The strategy is direct bookkeeping: write the second law for every particle, sum, and watch the internal forces annihilate in action–reaction pairs. The step to watch is the pairing itself: the double sum over internal forces must be regrouped into pairs before the third law can act, and the cancellation needs the strong form of the law only for the momentum result — equal and opposite is enough, collinearity is not required until angular momentum enters.

Step 1. Newton's second law for particle $i$ , with the force split into external and internal parts:

m_i \ddot{\mathbf{r}}_i = \mathbf{F}_i^{\mathrm{ext}} + \sum_{j \neq i} \mathbf{F}_{ij}

Why: each particle answers to the outside world plus every neighbor; $\mathbf{F}_{ij}$ is the force on $i$ from $j$ .

Step 2. Sum over all particles. The left side becomes $\mathrm{d}\mathbf{P}/\mathrm{d}t$ (masses constant), and the internal forces form a double sum:

\frac{\mathrm{d}\mathbf{P}}{\mathrm{d}t} = \sum_i \mathbf{F}_i^{\mathrm{ext}} + \sum_i \sum_{j \neq i} \mathbf{F}_{ij}

Step 3. Regroup the double sum into pairs and apply the third law:

\sum_i \sum_{j \neq i} \mathbf{F}_{ij} = \sum_{i<j} \left( \mathbf{F}_{ij} + \mathbf{F}_{ji} \right) = \sum_{i<j} \mathbf{0} = \mathbf{0}

Why every pair dies: the force on 1 from 2 and the force on 2 from 1 are equal and opposite at every instant — each internal transaction is double-entry, and the ledger of internal trades sums to zero identically.

Step 4. Only the external sum survives: $\mathrm{d}\mathbf{P}/\mathrm{d}t = \mathbf{F}^{\mathrm{ext}}$ , which is (2); setting $\mathbf{F}^{\mathrm{ext}} = \mathbf{0}$ gives $\mathbf{P} = \text{const}$ , which is (3). Why this is remarkable: the internal forces in an explosion are enormous and unknown, yet the conclusion needs nothing about them beyond their pairwise structure.

The Newtonian proof is short, but it rests on the third law — and the third law is not sacred. For charges in relative motion, the magnetic forces between them are not equal and opposite, yet total momentum is still conserved once the electromagnetic field's momentum is counted. Clearly the conservation law is standing on something firmer than action–reaction. That something is symmetry.

Go deeper · Derivation B · 1 minConservation from the homogeneity of space

Derivation BConservation from the homogeneity of space

The goal is to re-derive momentum conservation from a single assumption that makes no mention of forces at all: empty space is homogeneous — shifting an isolated system rigidly sideways changes nothing about its physics. The tool is the Lagrangian formulation, where the dynamics of the whole system is packaged in one function $L = \sum_i \tfrac12 m_i \dot{\mathbf{r}}_i^2 - U$ , with $U$ depending only on the particles' relative positions. We nudge every particle by the same tiny vector, demand that $L$ not care, and let the equations of motion convert that indifference into a conserved quantity. This is the simplest working example of Noether's theorem — symmetry in, conservation law out — and the reason the result outlives Newton's third law.

Step 1. Shift every particle by the same infinitesimal $\boldsymbol{\epsilon}$ : $\mathbf{r}_i \to \mathbf{r}_i + \boldsymbol{\epsilon}$ . Velocities are untouched, and $U$ depends only on differences $\mathbf{r}_i - \mathbf{r}_j$ , which cancel the shift. The change in the Lagrangian is

\delta L = \sum_i \frac{\partial L}{\partial \mathbf{r}_i} \cdot \boldsymbol{\epsilon} = 0

Why it vanishes: homogeneity of space is precisely the statement that a rigid translation is undetectable for an isolated system — for arbitrary $\boldsymbol{\epsilon}$ , the sum $\sum_i \partial L / \partial \mathbf{r}_i$ must itself be zero.

Step 2. Use the Euler–Lagrange equations, $\partial L/\partial \mathbf{r}_i = \tfrac{\mathrm{d}}{\mathrm{d}t}\,\partial L/\partial \dot{\mathbf{r}}_i$ — the Lagrangian form of each particle's equation of motion — to trade position-derivatives for time-derivatives:

\mathbf{0} = \sum_i \frac{\partial L}{\partial \mathbf{r}_i} = \frac{\mathrm{d}}{\mathrm{d}t} \sum_i \frac{\partial L}{\partial \dot{\mathbf{r}}_i}

Step 3. Identify the conserved quantity. For $L = \sum_i \tfrac12 m_i \dot{\mathbf{r}}_i^2 - U$ , the velocity-derivative is $\partial L/\partial \dot{\mathbf{r}}_i = m_i \dot{\mathbf{r}}_i$ , so

\mathbf{P} = \sum_i m_i \dot{\mathbf{r}}_i = \text{const} \qquad (4)

Why this proof is the deeper one: nothing here invoked action–reaction. Where the third law fails — electromagnetic interactions between moving charges — translation symmetry still holds for the full system of particles plus field, and conservation survives by enlarging what counts as "momentum" to include the field's share. *The same three-step template — shift the variable, demand invariance, read off the conserved quantity — applied with time-shifts yields energy conservation and with rotations yields angular momentum: one theorem, three great laws.*

Limiting cases

Case	Result	Meaning
single free particle	$m\mathbf{v} = \text{const}$	Newton's first law recovered — conservation's simplest citizen
external force along one axis	$F^{\mathrm{ext}}_x = 0 \Rightarrow P_x = \text{const}$	a projectile keeps horizontal momentum while gravity spends the vertical
explosion or breakup	$\mathbf{P}_{\mathrm{after}} = \mathbf{P}_{\mathrm{before}}$	blast forces are internal; fragments' momenta still sum to $M\mathbf{v}$
isolated system, any internal drama	$\dot{\mathbf{R}}_{\mathrm{cm}} = \text{const}$	the center of mass coasts in a straight line — momentum conservation restated
third law fails (moving charges)	$\mathbf{P}_{\mathrm{particles}} + \mathbf{P}_{\mathrm{field}} = \text{const}$	symmetry outranks action–reaction: count the field and the books rebalance
$v \to c$	$\mathbf{P} = \sum_i \gamma_i m_i \mathbf{v}_i = \text{const}$	the definition upgrades; conservation itself survives into relativity

Common trap“Momentum is conserved, so each object keeps its momentum”

Conservation is a property of the system total, never of the members. In a collision each body's momentum changes violently — that is what colliding means; only the sum stays fixed. Getting this backward produces two classic wrong answers. First, students conserve one ball's momentum through a bounce off a wall or floor: false — the wall exerts an external force on the ball, and the ball's momentum reverses; it is ball-plus-Earth whose total is conserved (the planet's share is real but hidden in a $\sim 10^{-24}\ \mathrm{m/s}$ recoil). Second, students freeze the total when an external force acts: a projectile system in gravity does not conserve $P_y$ . The discipline that prevents both errors is the same one: declare the system first, then ask what is external to it, then conserve only the components with no external force. And always add momenta as vectors — two clay balls closing head-on at equal speeds carry zero total momentum, not double.

Worked examples

1 · Rifle recoil, and where the energy hides (confidence builder)

A $4.0\ \mathrm{kg}$ rifle fires a $10\ \mathrm{g}$ bullet at $400\ \mathrm{m/s}$ . Find the recoil speed, and compare the kinetic energies of bullet and rifle.

Step 1. Declare the system: rifle + bullet. During the millisecond of firing, the explosion forces are internal and the external forces (gravity, the loose grip) deliver negligible impulse. Total momentum before: zero.

Step 2. Conserve the horizontal component with equation (3): $0 = m_b v_b + m_r v_r$ , so $v_r = -\dfrac{m_b v_b}{m_r} = -\dfrac{0.010 \times 400}{4.0} = \boxed{-1.0\ \mathrm{m/s}}$ — one meter per second backward, four hundred times slower than the bullet, in exact inverse proportion to the masses.

Step 3. Now the energy ledger: $K_b = \tfrac12 (0.010)(400)^2 = 800\ \mathrm{J}$ , while $K_r = \tfrac12 (4.0)(1.0)^2 = 2.0\ \mathrm{J}$ . The bullet takes 99.75% of the kinetic energy. Equal and opposite momenta, wildly unequal energies: since $K = p^2/2m$ and both carry the same $|p| = 4.0\ \mathrm{kg\,m/s}$ , the energy splits inversely with mass — $K_b/K_r = m_r/m_b = 400$ .

Step 4. The shooter's shoulder then absorbs $4.0\ \mathrm{kg\,m/s}$ of momentum; spread over about $0.05\ \mathrm{s}$ of recoil-pad compression, that is an average force of $4.0/0.05 = 80\ \mathrm{N}$ — brisk but comfortable, courtesy of the impulse–momentum trade.

2 · A shell explodes at the top of its arc (exam level)

A $6.0\ \mathrm{kg}$ shell moving horizontally at $24\ \mathrm{m/s}$ explodes at the apex of its trajectory into two pieces: a $2.0\ \mathrm{kg}$ fragment thrown backward at $12\ \mathrm{m/s}$ , and a $4.0\ \mathrm{kg}$ fragment. Find the second fragment's velocity, and state what the explosion did to the center of mass.

Step 1. Declare the system (both fragments) and the axis. At the apex the shell's velocity is purely horizontal. Gravity acts vertically, so during the near-instantaneous explosion the horizontal momentum component is conserved exactly, and even $P_y$ is unchanged across the blast because the explosion is internal and its duration negligible.

Step 2. Horizontal momentum before: $P_x = 6.0 \times 24 = 144\ \mathrm{kg\,m/s}$ .

Step 3. After, with the backward fragment at $v_1 = -12\ \mathrm{m/s}$ : $144 = (2.0)(-12) + (4.0)\,v_2 = -24 + 4.0\,v_2$ , so $v_2 = \dfrac{144 + 24}{4.0} = \boxed{42\ \mathrm{m/s}}$ , forward and horizontal. Sanity check: $2.0 \times (-12) + 4.0 \times 42 = -24 + 168 = 144$ ✓.

Step 4. The energy the explosion injected is measurable: $K_{\mathrm{after}} = \tfrac12(2.0)(12)^2 + \tfrac12(4.0)(42)^2 = 144 + 3528 = 3672\ \mathrm{J}$ against $K_{\mathrm{before}} = \tfrac12(6.0)(24)^2 = 1728\ \mathrm{J}$ . The blast added $1944\ \mathrm{J}$ of kinetic energy while adding zero momentum — the cleanest illustration that the two ledgers are independent.

Step 5. The center of mass never noticed. By equation (1), $\mathbf{P} = M\dot{\mathbf{R}}_{\mathrm{cm}}$ , and $\mathbf{P}$ was untouched by the internal blast — so the CM sails on at $24\ \mathrm{m/s}$ horizontally, tracing exactly the parabola the intact shell would have followed, until a fragment lands and the ground's external force enters the books.

Figure

The explosion cannot deflect the center of mass — fragments scatter, the parabola survives

Origins · 1 minEmmy Noether and how this was discovered

Emmy Noether

1882–1935 · Erlangen, Göttingen, Bryn Mawr

Noether proved the theorem that explains why momentum is conserved — and energy, and angular momentum — in a 1918 paper written to resolve a crisis in general relativity, where energy conservation seemed to misbehave. Her answer reorganized physics: every continuous symmetry of a system's dynamics begets a conservation law. Space's indifference to where you are yields momentum; to when, energy; to which direction you face, angular momentum. She did this work unpaid — Göttingen refused a woman a faculty position, and for years her lectures were advertised under David Hilbert's name, Hilbert having famously retorted that the faculty senate was "not a bathhouse." Expelled from Germany in 1933 as a Jew, she taught her final two years at Bryn Mawr College. Einstein's obituary called her the most significant creative mathematical genius produced since the higher education of women began; physicists rank the theorem itself alongside the laws it explains — it is the reason we trusted momentum conservation enough to bet the neutrino on it, and why the law still stands where Newton's third law has long since fallen.

History

1644

— Descartes proposes that the total "quantity of motion" is conserved, but reckons it unsigned — his rules fail for bodies meeting head-on.

1668

— Wallis, Wren, and Huygens, answering the Royal Society's collision challenge, balance the books by making momentum a directed quantity.

1687

— Newton's Principia grounds the conservation law in the third law: internal action-reaction pairs cannot move the total.

1918

— Noether's theorem reveals the deeper foundation — momentum conservation is the homogeneity of space itself.

1930

— Pauli, trusting the conservation laws over the beta-decay data, postulates the neutrino to carry the missing momentum; Reines and Cowan detect it in 1956.

Going further

Texts. Morin, Introduction to Classical Mechanics, Ch. 4–5 · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §1.2 · Landau & Lifshitz, Mechanics, §7–9 (the symmetry route) · Taylor, Classical Mechanics, Ch. 3 · Feynman Lectures I, Ch. 10 for the conversational version. All equations are standard results. Notable checkpoints: the elastic-collision $m_1=m_2$ limit gives velocity exchange, and the relative-velocity reversal identity holds generally; the Tsiolkovsky equation $\Delta v=u\ln(m_0/m_f)$ follows from $m\,\mathrm{d}v=-u\,\mathrm{d}m$ . No empirical constants are quoted. Worked-example numbers on this page (rifle recoil $1.0\ \mathrm{m/s}$ , energy split $800\ \mathrm{J}$ vs $2.0\ \mathrm{J}$ , shoulder force $80\ \mathrm{N}$ ; shell fragment $v_2 = 42\ \mathrm{m/s}$ with blast energy input $1944\ \mathrm{J}$ ) are illustrative arithmetic computed from the boxed formulas.

Builds on: Newton's Second Law · Newton's Third Law · Velocity and Acceleration · Vectors in Kinematics and Coordinate Systems. Leads to: Relative Motion and Galilean Transformations · Newton's First Law and Inertial Frames.

When a ball bounces off the floor, its momentum reverses. Where did the momentum go?

Into the Earth — really. The floor exerts an external force on the ball, so the ball's momentum is not conserved, and nothing requires it to be. Enlarge the system to ball-plus-Earth and the books balance perfectly: the planet absorbs the $2mv$ of the rebound. Its recoil velocity is that momentum divided by $6 \times 10^{24}\ \mathrm{kg}$ — around $10^{-24}\ \mathrm{m/s}$ , real and utterly unmeasurable. This is the general pattern with very unequal partners: conservation holds exactly, but the heavy partner hides its share in a microscopic velocity. The practical lesson is the article's discipline: declare your system before you conserve anything.

Does momentum conservation require the collision to be quick?

No — the law $\mathrm{d}\mathbf{P}/\mathrm{d}t = \mathbf{F}^{\mathrm{ext}}$ holds at every instant, fast or slow. Quickness buys you something different: permission to ignore external forces across the event. During a 1 ms collision, gravity's impulse $mg\,\Delta t$ is thousands of times smaller than the internal exchange, so "momentum just before = momentum just after" is an excellent approximation even though the system isn't isolated. For a slow interaction — two magnets drifting together over minutes while friction acts — you cannot freeze the total; you must track $\mathbf{F}^{\mathrm{ext}}$ honestly. Quick events make conservation usable, not true.

Friction destroys momentum, right? So how can momentum be conserved in a sticky collision?

Friction between the system and the outside (the road, the table) is an external force and does change $\mathbf{P}$ . But friction inside the system — the grinding, deforming, heat-generating mess where two clay balls merge — consists of internal third-law pairs, and internal pairs cannot touch the total. That is why a perfectly inelastic collision conserves momentum exactly while destroying kinetic energy wholesale: energy leaks into heat, which has no direction and no momentum ledger of its own, but every newton of internal friction on one body is matched by an equal and opposite newton on the other. Momentum has nowhere to leak.

How can a rocket accelerate in empty space with nothing to push against?

It brings along its own thing to push against: the propellant. Declare the system to be rocket-plus-exhaust and it is isolated, so its total momentum is frozen. The engine hurls exhaust backward at high relative speed; the rocket must gain exactly the momentum the exhaust carries away, each second, or the total would change. "Nothing to push on" has the physics backward — the pushing is between the rocket and its own reaction mass, an internal trade the vacuum cannot forbid. The bookkeeping, integrated over a burn, is the Tsiolkovsky rocket equation, which gets its own article in this module.

What does the Noether argument actually say, in plain words?

It says: if an experiment gives the same results here as it does one meter to the left, then there must exist a quantity that never changes — and that quantity is total momentum. "Space has no preferred location" is the symmetry; momentum conservation is its shadow in the dynamics. The power of the argument is its indifference to mechanism: it never asks what forces act or whether they obey action–reaction, only that the whole system's physics is translation-blind. That is why the law survives in electromagnetism (count the field's momentum), in relativity, and in quantum mechanics, where the same symmetry makes the momentum operator the generator of translations.

If total momentum is conserved, why does a sliding box on a floor just stop?

Because the box alone is not an isolated system — friction from the floor is an external force on it, and $\mathrm{d}\mathbf{P}/\mathrm{d}t = \mathbf{F}^{\mathrm{ext}}$ duly reduces the box's momentum. Where did it go? Into the floor, and through it the Earth: friction on the box has a third-law partner pushing the planet forward, imperceptibly. Zoom out to box-plus-Earth and the total never changed. Everyday "momentum loss" is always this: a transfer to something so massive you don't notice its share. Genuine disappearance has never been observed — which is exactly why Pauli preferred inventing the neutrino to accepting one.

Is the value of the total momentum the same for all observers?

No — and that is fine. An observer moving at velocity $\mathbf{V}$ relative to you measures $\mathbf{P}' = \mathbf{P} - M\mathbf{V}$ : a parked train has zero momentum to the platform and plenty to a passing car. What all inertial observers agree on is the law: since $M\mathbf{V}$ is constant in time, if $\mathbf{P}$ is constant for you, $\mathbf{P}'$ is constant for them. Conservation is frame-invariant even though the conserved number is not. Physicists exploit the freedom by boosting to the frame where $\mathbf{P}' = \mathbf{0}$ — the center-of-mass frame — where collisions become maximally symmetric and the algebra collapses.

Which is more fundamental — momentum conservation or Newton's third law?

Conservation, by the modern reckoning. In Newtonian point mechanics the two are nearly equivalent: the third law implies conservation (Derivation A), and conservation for all subsystems implies pairwise balanced forces. But push into electromagnetism and they part ways: the magnetic forces two moving charges exert on each other are generally not equal and opposite, yet nothing observable ever violates momentum conservation — the field itself carries momentum ( $\propto \mathbf{E} \times \mathbf{B}$ ), and the total is exact. Noether's theorem explains the asymmetry: conservation flows from translation symmetry of the complete system, a property deeper than any particular force law. The third law is a useful approximation; the conservation law is, as far as every experiment can tell, exact.