Momentum Conservation from Newton's Third Law

Derivation

Momentum Conservation from Newton's Third Law

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Statement

For an isolated system of \(N\) point particles interacting only through pairwise internal forces that obey Newton's third law, the total linear momentum \(\vec{P}=\sum_{i} m_i\vec{v}_i\) is constant in time, \(\frac{d\vec{P}}{dt}=\vec{0}\). The internal forces produce no net change in the system's momentum because they cancel in equal-and-opposite pairs, so \(\frac{d\vec{P}}{dt}\) equals the net external force alone.

Why it matters

Momentum conservation is the first genuinely universal conservation law a student meets: it holds in collisions, explosions, rocket flight, and particle decays, whether or not energy is conserved and whether or not the internal forces are known in detail. Because it follows from the third law alone, we can predict the outcome of an interaction without solving the dynamics inside it.

Deeper still, this derivation is the concrete, Newtonian face of a symmetry principle. The same law re-emerges in Lagrangian mechanics as a consequence of the homogeneity of space (Noether's theorem), and in that form it survives into relativity and quantum field theory, where Newton's third law itself does not.

Assumptions

The system is isolated: no net external force acts.If external forces are present, only the internal forces cancel; the total momentum then changes at the rate of the net external force, \(\frac{d\vec{P}}{dt}=\vec{F}^{\text{ext}}\), and conservation fails.

All internal interactions are pairwise and additive.If the force on particle \(i\) cannot be written as a sum of two-body forces from each other particle (genuine three-body forces or field self-interactions), the pair-cancellation bookkeeping no longer exhausts the internal forces.

Each internal pair obeys Newton's third law in weak form, \(\vec{F}_{ij}=-\vec{F}_{ji}\).If reaction forces are not equal and opposite, the pairwise sum leaves a residual internal force and the total momentum drifts even with no external agent.

Instantaneous action at a distance: the pair forces are defined at a single common time.If the interaction is mediated by a field with finite propagation speed, the two forces act at different retarded times and need not cancel at any instant; momentum is conserved only when the field's own momentum is included in the ledger.

The masses \(m_i\) are constant.If a particle's mass changes (variable-mass bodies, relativistic speeds), then \(m_i\vec{a}_i\) is not the whole time-derivative of its momentum and the elementary steps below must be re-derived from \(\vec{p}=m\vec{v}\) directly.

Derivation

\[ \vec{P}\equiv\sum_{i=1}^{N}\vec{p}_i=\sum_{i=1}^{N} m_i\vec{v}_i \]

Definition of total linear momentum as the vector sum of particle momenta. A

\[ \frac{d\vec{P}}{dt}=\sum_{i=1}^{N}\frac{d\vec{p}_i}{dt} \]

The time-derivative of a finite sum is the sum of derivatives; the number of particles is fixed. A

\[ \frac{d\vec{p}_i}{dt}=\vec{F}_i \]

Newton's second law in momentum form for each particle; \(\vec{F}_i\) is the total force on particle \(i\). A

\[ \vec{F}_i=\vec{F}_i^{\text{ext}}+\sum_{\substack{j=1 \\ j\neq i}}^{N}\vec{F}_{ij} \]

Resolve the force on each particle into an external part and a sum of internal forces \(\vec{F}_{ij}\) (force on \(i\) due to \(j\)); the term \(j=i\) is excluded, since a particle exerts no force on itself. B

\[ \frac{d\vec{P}}{dt}=\sum_{i=1}^{N}\vec{F}_i^{\text{ext}}+\sum_{i=1}^{N}\sum_{\substack{j=1 \\ j\neq i}}^{N}\vec{F}_{ij} \]

Substitute step 4 into steps 2–3 and separate the double sum of internal forces from the external forces. B

\[ \sum_{i}\sum_{j\neq i}\vec{F}_{ij}=\sum_{i

The ordered double sum runs over every ordered pair once; regroup so each unordered pair \(\{i,j\}\) contributes both members together. This is a pure relabelling of the same finite set of terms. C

\[ \vec{F}_{ij}+\vec{F}_{ji}=\vec{F}_{ij}+\left(-\vec{F}_{ij}\right)=\vec{0} \]

Apply Newton's third law to each pair: the reaction is equal in magnitude and opposite in direction, so every pair term vanishes identically. C

\[ \frac{d\vec{P}}{dt}=\sum_{i=1}^{N}\vec{F}_i^{\text{ext}}\equiv\vec{F}^{\text{ext}} \]

The entire internal double sum is zero, leaving only the net external force. B

\[ \vec{F}^{\text{ext}}=\vec{0}\;\;\Longrightarrow\;\;\frac{d\vec{P}}{dt}=\vec{0}\;\;\Longrightarrow\;\;\vec{P}=\text{const} \]

Impose the isolation assumption; a vanishing time-derivative means each Cartesian component of \(\vec{P}\) is constant in time. A

Result

\[ \frac{d\vec{P}}{dt}=\vec{F}^{\text{ext}}\qquad\xrightarrow{\;\vec{F}^{\text{ext}}=\vec{0}\;}\qquad \vec{P}=\sum_{i} m_i\vec{v}_i=\text{const} \]

Reading. The total momentum of an isolated system is a fixed vector for all time. Internal forces, however violent, only shuffle momentum between the particles; they can never change the total. Equivalently, the centre of mass moves in a straight line at constant velocity, since \(\vec{P}=M\vec{v}_{\text{cm}}\) with total mass \(M\) constant.

Units check. Each term \(m_i\vec{v}_i\) has SI units \(\text{kg}\cdot\text{m}\,\text{s}^{-1}\); the sum shares those units, and \(\frac{d\vec{P}}{dt}\) has \(\text{kg}\cdot\text{m}\,\text{s}^{-2}=\text{N}\), matching the force on the right of step 8. Both sides of the equation of motion carry newtons, so the relation is dimensionally consistent.

Limiting cases

Two-body system (\(N=2\)). The double sum has a single pair; \(m_1\vec{v}_1+m_2\vec{v}_2=\text{const}\) — the workhorse relation for collisions and recoil.
External force present but one axis unforced. If \(\vec{F}^{\text{ext}}\) has no component along, say, \(x\), then \(P_x\) alone is conserved even though \(\vec{P}\) is not — e.g. horizontal momentum in projectile problems under gravity.
Rigid body as a limit. Taking the internal forces stiff enough to lock all separations fixed recovers \(\vec{P}=M\vec{v}_{\text{cm}}\); the internal constraint forces still cancel pairwise.
Impulsive interactions. Over a collision so brief that external impulses \(\vec{F}^{\text{ext}}\,\Delta t\) are negligible, momentum is conserved to good approximation even for non-isolated systems.

Breaks when

Field-mediated interactions with retardation. For charges interacting electromagnetically, the force on charge 1 from charge 2 at a given instant is generally not the negative of the force on 2 from 1 (the magnetic forces between two moving charges famously violate even the weak third law). The mechanical momentum of the particles alone is not conserved; conservation is restored only by crediting momentum \(\int \epsilon_0\,\vec{E}\times\vec{B}\,dV\) to the field itself.
Genuine external forces. A block sliding on a table with friction, or a ball falling under gravity, exchanges momentum with the Earth; treating the block alone as "the system" gives \(\frac{d\vec{P}}{dt}\neq\vec{0}\). The law is not wrong — the chosen boundary excludes part of an interacting pair.
Non-inertial reference frames. In a rotating or accelerating frame, fictitious forces (centrifugal, Coriolis) act on every particle with no equal-and-opposite partner inside the system, so \(\vec{P}\) measured in that frame is not conserved.
Irreducible many-body forces. If the interaction genuinely depends on three or more particles at once (e.g. certain nuclear three-body forces), the neat pairing of step 6 no longer accounts for all internal terms and cancellation must be re-examined.

Failure modes

Adding momenta as scalars. Momentum is a vector; writing \(m_1 v_1+m_2 v_2\) without signs or components turns a head-on collision into nonsense. Always resolve along axes.
Choosing a system that cuts through an interacting pair. Selecting "the ball" instead of "ball + Earth" for a falling body, then being surprised momentum is not conserved. Draw the boundary so both members of every relevant pair are inside, or account for the external force.
Confusing conservation of momentum with conservation of kinetic energy. Momentum is conserved in every isolated interaction; kinetic energy is conserved only in elastic ones. Students routinely impose both in an inelastic collision and get contradictory equations.
Working in a non-inertial frame without adding fictitious forces. Applying \(\vec{P}=\text{const}\) inside an accelerating vehicle and ignoring the pseudo-force term.
Assuming individual momenta are conserved. Only the total is fixed; each \(\vec{p}_i\) generally changes. Writing \(\vec{p}_1=\text{const}\) for a single colliding particle is a common slip.
Mixing frames within one problem. Measuring one velocity in the lab frame and another in the centre-of-mass frame before summing.

Discussion

The heart of the derivation is a counting argument, not a dynamical one. Steps 6 and 7 never solve any equation of motion; they merely reorganise a finite sum and invoke a symmetry of the force law. This is why momentum conservation is so robust: it is indifferent to the functional form of \(\vec{F}_{ij}\), to whether the collision is elastic or inelastic, and to how complicated the trajectories are. All that matters is that internal forces come in balanced pairs.

Notice the logical dependence. The weak form of the third law (equal and opposite) is sufficient for linear-momentum conservation; the strong form (equal, opposite, and along the line joining the particles) is additionally required to conserve angular momentum. Central forces — gravity, electrostatics — obey the strong form, which is why isolated planetary systems conserve both. Linear-momentum conservation asks less of the interaction than angular-momentum conservation does.

The Newtonian route taken here is ultimately a special case of a far deeper statement. In the Lagrangian formulation, if the Lagrangian is invariant under a rigid translation of every coordinate, \(\vec{r}_i\to\vec{r}_i+\vec{\epsilon}\) — the homogeneity of space — then Noether's theorem yields a conserved quantity equal to \(\sum_i \frac{\partial L}{\partial \dot{\vec{r}}_i}=\vec{P}\). Translation invariance of \(L\) requires the potential to depend only on relative separations \(\vec{r}_i-\vec{r}_j\), which is precisely the condition \(\vec{F}_{ij}=-\nabla_i V=-(-\nabla_j V)=-\vec{F}_{ji}\) that generates equal-and-opposite pair forces. Thus "Newton's third law" and "space looks the same everywhere" are two languages for one fact, and it is the symmetry statement, not the force statement, that survives into relativistic field theory where instantaneous action at a distance is abandoned.

Common misconceptions. That the third law "explains why nothing moves" — but the paired forces act on different bodies, so they never cancel on a single object and motion is unaffected. And that a rocket "pushes against the air" — a rocket conserves momentum by expelling exhaust; the exhaust carries away backward momentum equal and opposite to the rocket's forward gain, and it works perfectly in vacuum.

Worked examples

Recoil of a rifle. A rifle of mass \(M=4.0\ \text{kg}\) fires a bullet of mass \(m=12\ \text{g}\) at muzzle speed \(u=380\ \text{m s}^{-1}\). Find the recoil speed \(V\).

Isolated system (bullet + rifle), external horizontal forces negligible during the brief firing. Set up conservation. A

\[ P_{\text{before}}=0\;\;\Longrightarrow\;\; mu-MV=0 \]

Total momentum starts at zero; after firing the bullet moves forward \((+)\) and the rifle backward \((-)\). Symbols first. A

\[ V=\frac{mu}{M} \]

Rearrange for the unknown recoil speed before substituting numbers. A

\[ V=\frac{(0.012\ \text{kg})(380\ \text{m s}^{-1})}{4.0\ \text{kg}}=\frac{4.56}{4.0}\ \text{m s}^{-1} \]

Insert values in consistent SI units (\(12\ \text{g}=0.012\ \text{kg}\)). A

\[ V\approx 1.14\ \text{m s}^{-1}\ \text{(backward)} \]

Reading. The rifle recoils at just over \(1\ \text{m s}^{-1}\) — small because it is \(330\times\) more massive than the bullet, so it carries the same momentum at \(1/330\) of the speed.

Units check. \((\text{kg})(\text{m s}^{-1})/(\text{kg})=\text{m s}^{-1}\), a speed. Momentum each way: \(mu=4.56\ \text{kg}\cdot\text{m s}^{-1}=MV\), balanced.

Perfectly inelastic collision. A \(1500\ \text{kg}\) car at \(20\ \text{m s}^{-1}\) east strikes a stationary \(1000\ \text{kg}\) car and they lock together. Find the common final velocity \(v_f\) and the kinetic energy lost.

Isolated along the road during impact (friction impulse negligible over the collision). One-dimensional. A

\[ m_1 v_1+m_2 v_2=(m_1+m_2)v_f \]

Total momentum before equals total after; the cars share one velocity because they stick. B

\[ v_f=\frac{m_1 v_1+m_2 v_2}{m_1+m_2} \]

Solve for the unknown before inserting numbers. A

\[ v_f=\frac{(1500)(20)+(1000)(0)}{1500+1000}=\frac{30000}{2500}\ \text{m s}^{-1} \]

Substitute SI values; the stationary car contributes zero momentum. A

\[ v_f=12\ \text{m s}^{-1}\ \text{east},\qquad \Delta K=-120\ \text{kJ} \]

Reading. The wreck moves off at \(12\ \text{m s}^{-1}\). Kinetic energy drops from \(300\ \text{kJ}\) to \(180\ \text{kJ}\) — \(120\ \text{kJ}\) lost to deformation — yet momentum is exactly conserved, illustrating that the two laws are independent.

Units check. \((\text{kg}\cdot\text{m s}^{-1})/(\text{kg})=\text{m s}^{-1}\). Momentum: before \(30000\ \text{kg}\cdot\text{m s}^{-1}\), after \((2500)(12)=30000\ \text{kg}\cdot\text{m s}^{-1}\). Balanced.

Problems

A \(60\ \text{kg}\) skater at rest throws a \(2.0\ \text{kg}\) ball horizontally at \(8.0\ \text{m s}^{-1}\). On frictionless ice, find the skater's recoil speed.
Solution
Isolated system, \(P_{\text{before}}=0\). So \(mu=MV\Rightarrow V=\frac{mu}{M}=\frac{(2.0)(8.0)}{60}=\frac{16}{60}=0.27\ \text{m s}^{-1}\) in the direction opposite the throw.
Two gliders on an air track: \(0.30\ \text{kg}\) moving at \(0.50\ \text{m s}^{-1}\) toward a stationary \(0.50\ \text{kg}\) glider. They couple. Find the final common speed and the fractional kinetic-energy loss.
Solution
\(v_f=\frac{(0.30)(0.50)}{0.30+0.50}=\frac{0.15}{0.80}=0.19\ \text{m s}^{-1}\). \(K_i=\tfrac12(0.30)(0.50)^2=0.0375\ \text{J}\); \(K_f=\tfrac12(0.80)(0.1875)^2=0.0141\ \text{J}\). Fraction lost \(=1-\frac{0.0141}{0.0375}=0.625\), i.e. \(62.5\%\).
A \(5.0\ \text{kg}\) shell moving east at \(30\ \text{m s}^{-1}\) explodes into two fragments. One fragment, \(2.0\ \text{kg}\), flies off at \(40\ \text{m s}^{-1}\) east. Find the velocity of the \(3.0\ \text{kg}\) fragment.
Solution
Momentum before \(=(5.0)(30)=150\ \text{kg}\cdot\text{m s}^{-1}\) east. After: \((2.0)(40)+(3.0)v=150\Rightarrow 80+3v=150\Rightarrow v=\frac{70}{3}=23.3\ \text{m s}^{-1}\) east. The explosion is internal, so momentum is conserved despite the energy release.
A \(0.045\ \text{kg}\) puck moving at \(6.0\ \text{m s}^{-1}\) collides elastically and head-on with a stationary \(0.030\ \text{kg}\) puck. Find both velocities afterward.
Solution
For a 1D elastic collision with target at rest, \(v_1'=\frac{m_1-m_2}{m_1+m_2}u_1\) and \(v_2'=\frac{2m_1}{m_1+m_2}u_1\). Here \(v_1'=\frac{0.045-0.030}{0.075}(6.0)=\frac{0.015}{0.075}(6.0)=1.2\ \text{m s}^{-1}\); \(v_2'=\frac{2(0.045)}{0.075}(6.0)=1.2\times6.0=7.2\ \text{m s}^{-1}\). Check momentum: \((0.045)(6.0)=0.27\); after \((0.045)(1.2)+(0.030)(7.2)=0.054+0.216=0.27\ \text{kg}\cdot\text{m s}^{-1}\). Balanced, and energy is conserved by construction.
In deep space a \(1200\ \text{kg}\) spacecraft ejects \(3.0\ \text{kg}\) of exhaust at \(2500\ \text{m s}^{-1}\) relative to the ship (treat the ship's mass as roughly unchanged). Estimate the change in the ship's speed.
Solution
Momentum conservation for the isolated ship+exhaust in the ship's instantaneous rest frame: \(M\,\Delta V=m\,u_{\text{ex}}\Rightarrow \Delta V=\frac{m\,u_{\text{ex}}}{M}=\frac{(3.0)(2500)}{1200}=\frac{7500}{1200}=6.25\ \text{m s}^{-1}\), directed opposite the exhaust. This is the rocket principle: it needs no medium to push against.