= Internal-force cancellation in system momentum
{title2=$d\mathbf P/dt=\mathbf F_{\mathrm{ext}}$}
For a fixed collection of two constant-<mass> bodies in an <inertial frame>, separate their mutual <forces> from the resultant <external force> on each:
$$
\frac{d\mathbf p_A}{dt}=\mathbf F_{B\mathrm{\ on\ }A}+\mathbf F_{A,\mathrm{ext}},\qquad
\frac{d\mathbf p_B}{dt}=\mathbf F_{A\mathrm{\ on\ }B}+\mathbf F_{B,\mathrm{ext}}.
$$
Adding the two <Newton's second law> equations and using <Newton's third law> gives
$$
\mathbf P=\mathbf p_A+\mathbf p_B,\qquad
\frac{d\mathbf P}{dt}=\mathbf F_{A,\mathrm{ext}}+\mathbf F_{B,\mathrm{ext}}.
$$
This 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 $M\,d\mathbf V/dt=\mathbf F_{\mathrm{ext}}$. 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 $a_A=-1\,\mathrm{m/s^2}$, $a_B=+2\,\mathrm{m/s^2}$, $dP/dt=+3\,\mathrm N$, and the <centre of mass> has <acceleration> $+1\,\mathrm{m/s^2}$. Pair equality by itself does not promise constant total <momentum>.
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