= Solution
Away from a collision, let $r=q_1-q_2$ and $D=|r|>0$. The <Lagrangian> is
$$
\boxed{L=\frac12m_1|\dot q_1|^2+\frac12m_2|\dot q_2|^2-V(D).}
$$
Differentiating $D^2=r\cdot r$ with respect to $x_1$ and $x_2$ gives $2D\partial_{x_1}D=2(x_1-x_2)$ and $2D\partial_{x_2}D=-2(x_1-x_2)$. Therefore
$$
\boxed{\partial_{x_1}D=-\partial_{x_2}D=\frac{x_1-x_2}{D}.}
$$
The other coordinates give $\nabla_{q_1}D=r/D$ and $\nabla_{q_2}D=-r/D$. Substituting these into the <Euler-Lagrange equations> yields equal and opposite central forces:
$$
\boxed{m_1\ddot q_1=-\frac{V'(D)}D r,\qquad
m_2\ddot q_2=\frac{V'(D)}D r.}
$$
Adding the equations shows that the <derivative> of the total <linear momentum> is zero, so $P=m_1\dot q_1+m_2\dot q_2$ is constant. For the total <angular momentum>, the terms $\dot q_i\times\dot q_i$ vanish, giving
$$
\begin{aligned}
\frac d{dt}(m_1q_1\times\dot q_1+m_2q_2\times\dot q_2)
&=q_1\times\left(-\frac{V'}D r\right)+q_2\times\left(\frac{V'}D r\right)\\
&=-\frac{V'}D(q_1-q_2)\times r=0.
\end{aligned}
$$
Thus both stated vectors are <conserved quantities>.
\b[The total energy is also constant.] Define $E=\tfrac12m_1|\dot q_1|^2+\tfrac12m_2|\dot q_2|^2+V(D)$. Since $\dot D=r\cdot(\dot q_1-\dot q_2)/D$,
$$
\begin{aligned}
\dot E&=m_1\dot q_1\cdot\ddot q_1+m_2\dot q_2\cdot\ddot q_2+V'\dot D\\
&=-\frac{V'}D r\cdot(\dot q_1-\dot q_2)+V'\frac{r\cdot(\dot q_1-\dot q_2)}D=0.
\end{aligned}
$$
Hence $\boxed{E=T+V(D)\text{ is conserved}}$. The explicit cancellation establishes energy conservation for an arbitrary smooth separation-dependent potential, throughout any collision-free trajectory.
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