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This is a general question regarding the motion of fragments of a single particle at rest, with no external force acting on it.

Say I have a (point like) body of mass $m$, situated at the origin of my coordinate system. Due to some internal forces, if the body happens to split into two symmetrical fragments, each of mass $m/2$. We know that it will move in a straight line due to conservation of linear momentum, with equal but opposite velocities.

Now consider the same mass $m$, but let's say it splits into 3 symmetrical fragments, each of mass $m/3$. Can we say that the fragments will certainly move in the direction of the vertices of an equilateral triangle, assuming the origin to be the centroid of the triangle, with equal speeds?

Further, consider the same body, $m$ splits into 4 symmetrical fragments, each of mass $m/4$. Can we say that the fragments will certainly move in the shape of a square, with increasing side length? If so, why not in the shape of a regular tetrahedron? Even that is symmetrical (regarding the conservation of linear momentum) for the particles.

In a nutshell, is it even possible to accurately predict the motion of the fragments of an ideal body given that it splits into equal masses?

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In general, you would need very precise, and often unattainable, information about the internal structure of the fragmented body, the exact nature of the stresses that caused it to fracture, and a careful analysis of the evolution of the body subject to these forces as it started to break to predict the specific motion of the fragments. This is, of course, not possible in many cases. I answered a question here that addressed the same idea for a shattering object like a vase on a floor. A further consideration is whether there were any sorts of dissipative losses to complicate things even further. Friction, air resistance, the creation of a sound wave when an object breaks up, etc. can all make this already complicated problem more unwieldy. In only the most ideal situation will you see the motions that you described: I.e. the pieces are all perfectly uniform in mass, broke according to perfectly symmetrical internal forces, and therefore the pieces scatter in a completely symmetrical outward fashion. This is a really atypical case, even though we do see a sort of really vague symmetry when some objects break apart. But, as your intuition suggests, there is generally no realistic way to predict the outcome of an object breaking up unless you happen to have exquisitely detailed information about the object and its internal forces that are typically not attainable.

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is it even possible to accurately predict the motion of the fragments of an ideal body given that it splits into equal masses?

No, at least, not if all we are given is that the system conserves momentum.

If momentum conservation is the only consideration that only implies that momenta of the N fragments forms a closed N-sided figure when laid “tip to tail”.

For N=2 they are equal and opposite.

For N=3 they need not be equal. As long as the momenta form a closed triangle, momentum is conserved. Any triangle will do, it doesn’t need to be equilateral.

For N=4 and higher the momenta need not even be coplanar.

In each case N-1 of the vectors can have any random values, and then the remaining 1 is determined.

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In a nutshell, is it even possible to accurately predict the motion of the fragments of an ideal body given that it splits into equal masses?

Why wouldn't it?

For conservation of momentum, if the center of mass (COM) of the body is initially at rest at a specified location in a specific coordinate system, and there are no external forces acting upon the body, then the COM of all the fragments of the body will also be at rest at the same location of the coordinate system. Therefore, the scenarios you describe can theoretically be predicted, the low probability of the scenarios actually occurring (all fragments being of identical mass) not withstanding.

As far as the associated total kinetic energy of the fragments is concerned, that energy can be variable as it depends on the amount and type of potential energy (elastic, chemical, etc.) that was converted to kinetic energy.

Hope this helps.

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If we assume that all the fragments have the energy, then for three fragments, we have that the momentum vectors have equal lengths, and their sum is zero. The sum of vectors is geometrically represented by laying them tail to head, so them adding to zero is represented by the head of the last one lining up with the tail of the first, i.e. a closed loop. A closed loop of three equal-length line segments is an equilateral triangle. So for three fragments, we do indeed have an equilateral triangle.

For four fragments, we have a quadrilateral with equal length sides. That's not necessarily a square; rather, it's a rhombus. However, it's not guaranteed to be in a single plane, so more generally it's a "bent" rhombus. One of these bent rhombuses does indeed correspond to a tetrahedron (remember, the rhombus is what results when finding the sum by laying the vectors tail to head; the motion of the fragments in physical space is dispersion from a central point, so the vectors are tail to tail). The more fragments you add, the more complicated, and more varied, the possibilities get, but all of them involve a closed loop when the vectors are added.

Again, however, the equilateral triangle results from the assumption of equal energy (that the vectors add to zero, however, does not). Without that assumption, a wide variety of possibilities exist; it's not possible just from conservation of momentum and energy to determine which will be realized.

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Yes, it is possible to predict the movement of fragments that break apart in parts of equal mass. Consider that the energy must be evenly divided among the bodies, which results in all bodies having the same velocity. Thus, we can associate a scattering angle with this and determine which regular shape we can observe.

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    $\begingroup$ “Consider that the energy must be evenly divided among the bodies,”. Why would that necessarily be so? $\endgroup$
    – Bob D
    Commented 2 days ago

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