Masses on Rails

Let the particle numbered as 1 have coordinates $(0,y)$ and that numbered as 2 have coordinates $(x,0)$ . At any given instant of time the total energy of the system is:

$E = T + V$

Where $T$ is the total kinetic energy and $V$ is the total potential energy. The expressions for energies are:

$E = T + V = \left(\frac{m\dot{x}^2}{2} + \frac{m\dot{y}^2}{2}\right) + \left(mgy + \frac{1}{2}k\left(\sqrt{x^2 + y^2} - L_o\right)^2\right)$

Now, since there are no dissipative elements, the derivative of the total energy is zero. DIfferentiating and rearranging gives:

$\frac{dE}{dt} = 0 = \left( mg + m\ddot{y} + \frac{k\left(\sqrt{x^2 + y^2} - L_o\right)y}{\sqrt{x^2 + y^2}}\right)\dot{y} + \left(m\ddot{x} + \frac{k\left(\sqrt{x^2 + y^2} - L_o\right)x}{\sqrt{x^2 + y^2}}\right)\dot{x}$

Now, the only way the derivative of the energy is zero is if the expressions in the brakets are zero. If $\dot{x}$ and $\dot{y}$ are each zero, that means there would be no motion which is not possible. Hence, two equations of motion are obtained which are:

$mg + m\ddot{y} + \frac{k\left(\sqrt{x^2 + y^2} - L_o\right)y}{\sqrt{x^2 + y^2}} = 0$

$m\ddot{x} + \frac{k\left(\sqrt{x^2 + y^2} - L_o\right)x}{\sqrt{x^2 + y^2}} = 0$

Initial conditions are:

$x(0) = 1; y(0) = 1; \dot{x}(0) = 0; \dot{y}(0) = 0$

Using these initial conditions, the system of equations is solved numerically over a period of time until $x$ becomes zero. At that instant, the corresponding $y$ coordinate of the other particle is the required answer. The value is: $\boxed{y_{answer} \approx -4.32}$ . I have left out the details of the numerics here and have focussed on arriving at the equations of motion. Of course, Lagrangian mechanics is an alternate approach.