 Solution

Away from a collision, let $r=q_1-q_2$ and $D=|r|>0$. The Lagrangian isDifferentiating $D^2=r\cdot r$ with respect to $x_1$ and $x_2$ gives $2D{\partial }_{x_1}D=2(x_1-x_2)$ and $2D{\partial }_{x_2}D=-2(x_1-x_2)$. ThereforeThe 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: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, givingThus both stated vectors are conserved quantities.

The total energy is also constant. Define $E=\frac{1}{2}{m}_1|{\dot{q}}_1{|}^2+\frac{1}{2}{m}_2|{\dot{q}}_2{|}^2+V(D)$. Since $\dot{D}=r\cdot ({\dot{q}}_1-{\dot{q}}_2)/D$,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.