= Solution
Put $M=m_1+m_2$, define the <center of mass> $\mathbf R=(m_1\mathbf x_1+m_2\mathbf x_2)/M$, and let $\mathbf r=\mathbf x_1-\mathbf x_2$ be relative <position>. Adding the two <Newton's second law> equations gives
$$
M\ddot{\mathbf R}=\mathbf F_1+\mathbf F_2=0,
\qquad
\boxed{\mathbf R(t)=\mathbf R(0)+t\dot{\mathbf R}(0).}
$$
The <center of mass> has constant <velocity>, by conservation of total <momentum>.
Subtracting the particle <accelerations> instead gives
$$
\ddot{\mathbf r}=\frac{\mathbf f(\mathbf r)}{m_1}+\frac{\mathbf f(\mathbf r)}{m_2}
=\frac{\mathbf f(\mathbf r)}\mu,
\qquad
\boxed{\mu\ddot{\mathbf r}=\mathbf f(\mathbf r),\quad
\mu=\frac{m_1m_2}{m_1+m_2}.}
$$
Thus relative motion is a one-particle problem with the <reduced mass> $\mu$, while <center-of-mass motion> is uniform. The same <reduced mass> appears in the <kinetic energy> decomposition $T=\tfrac12M|\dot{\mathbf R}|^2+\tfrac12\mu|\dot{\mathbf r}|^2$.
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