Haven't got time for conservation

For simplicity, consider the potential $V(r_1,r_2)$ for two particles in 1 dimension.

Newton's second law state that the time derivative of momentum is given by minus the space derivative of the potential energy. Thus, the rate of change of each particle's momentum is given by

$\begin{aligned} \frac{dp_1}{dt} &= -\frac{\partial V}{\partial r_1} \\ \frac{dp_2}{dt} &= -\frac{\partial V}{\partial r_2} \end{aligned}$

The energy of the system is given by

$E = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + V(r_1,r_2)$

The time derivative of the energy is thus given by

$\begin{aligned} \dot{E} &= m^{-1}\left(p_1\dot{p}_1 + p_2\dot{p}_2\right) + \frac{dV}{\partial t} \\ &= - \sum_{i\in \{1,2\}} \frac{\partial V}{\partial r_i}\frac{\partial r_i}{\partial t} + \frac{dV}{\partial t} \end{aligned}$

However, since $V$ has no explicit time dependence, we have

$\frac{dV}{dt} = \sum \frac{\partial V}{\partial r_i}\frac{\partial r_i}{\partial t}$

Thus, the rate of change in energy is given by

$\begin{aligned} \dot{E} &= -\sum_{i\in \{1,2\}} \frac{\partial V}{\partial r_i}\frac{\partial r_i}{\partial t} + \sum _{i\in \{1,2\}} \frac{\partial V}{\partial r_i}\frac{\partial r_i}{\partial t} \\ &= 0 \end{aligned}$

So, the rate of change of the total energy is equal to zero, and energy is conserved in the system.