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For a fixed collection of two constant-mass bodies in an inertial frame, separate their mutual forces from the resultant external force on each:Adding the two Newton's second law equations and using Newton's third law givesThis is the proof of internal-force cancellation in system momentum. For more bodies, each internal force cancels with its partner once both appear in the sum. A boundary that excludes one interacting body leaves its force in the external force sum. A massive connecting spring must also be included, or its own changing momentum accounted for; its two endpoint forces are not automatically one third-law interaction pair between the carts.

Zero resultant external force gives momentum conservation; complete absence of external forces is sufficient but unnecessary. For constant total mass $M$, the centre of mass has velocity $\mathbf{V}=\mathbf{P}/M$ and obeys $Md\mathbf{V}/dt={\mathbf{F}}_{\mathrm{ext}}$. Internal interactions can move the parts apart while leaving the centre of mass moving uniformly.

Additional external forces do not invalidate the identified mutual pair. At the $4\mathrm{N}$ instant above, suppose a hand also exerts $+3\mathrm{N}$ on $A$. The mutual forces remain $-4\mathrm{N}$ and $+4\mathrm{N}$, but the net forces become $-1\mathrm{N}$ on $A$ and $+4\mathrm{N}$ on $B$. Consequently $a_A=-1m/{\mathrm{s}}^2$, $a_B=+2m/{\mathrm{s}}^2$, $dP/dt=+3\mathrm{N}$, and the centre of mass has acceleration $+1m/{\mathrm{s}}^2$. Pair equality by itself does not promise constant total momentum.