A massless elastic cord (that obeys Hooke's Law) will break if the tension in the cord exceeds \(T_{max}\). One end of the cord is attached to a fixed point, the other is attached to an object of mass \(3m\). If a second, smaller object of mass \(m\) moving at an initial speed \(v_0\) strikes the larger mass and the two stick together, the cord will stretch and break, but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one-dimensional collision, the cord will still break, and the larger mass will move off with a final speed of \(v_f\). All motion occurs on a horizontal, frictionless surface.

Find the ratio of the total kinetic energy of the system of two masses after the perfectly elastic collision and the cord has broken to the initial kinetic energy of the smaller mass prior to the collision. The original answer is in \(\frac{k}{m}\) format, please give the value of k + m.

Let's think step by step and output the final answer within \boxed{}.
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