Energy of single particle is equal to energy of multiparticle system it consists of?

The solution to a problem of atom fission led me to this question. In this problem the mass of original atom nucleus, masses of two child atom nuclei as well as zero kinetic energy of first atom nucleus were given. The problem asked to find combined kinetic energy of two child nuclei when they are far apart.

The solution was provided by following formula: $$M_{parentnucleus} c^2 + K_{parent} = 2 M_{childnuclues} c^2 + K_{combined} + U$$

As $K_{parent} = 0$ and potential energy of system $U = 0$ as nuclei are far apart.

However, I doubt that single particle energy equals to energy of system of particles it consisted of.

Let's imagine simple example of system that consists of two protons with speeds much less than speed of light.

If we consider this as system consisting of two particles, then $$E_{sys} = m c^2 + m c^2 + \frac{m v_1^2}{2} + \frac{m v_2^2}{2} + U$$ On the other hand, if we consider this system as single particle with velocity $\vec{v} = \frac{\vec{v_1} + \vec{v_2}}{2}$, then $$E_{sys} = 2m c^2 + \frac{2mv^2}{2} = 2m c^2 + \frac{m v_1^2}{2} + \frac{m v_2^2}{2}- \frac{m|\vec{v_1} - \vec{v_2}|^2}{4}$$

Evidently, $U ≠ -\frac{m|\vec{v_1} - \vec{v_2}|^2}{4}$ always. I may fix the distance between protons and thus fix $U$ value and pick such velocities that terms won't be equal.

So my question: is it correct to use that equation for solution to atom fission problem? If yes, why? Aren't my reasoning and contradiction shown correct?

• The original equation you wrote is correct as guarenteed by conservation of energy. When you consider a two particle system as a single particle with a CoM motion you can't naively add the masses the way you are doing (there is additionally internal energy as your particle is not a point particle). – jacob1729 May 24 '18 at 15:30
• @jacob1729 what is CoM motion? – Alex Velickiy May 29 '18 at 16:50