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I recall in Highschool physics receiving a problem of the form "assume 2 spherical bodies" $P_1, P_2$ enter a collision where $P_1$ has momentum $m_1 V_1$ and $P_2$ has momentum $m_2 V_2$, both kinetic energy and momentum are conserved and neither changes any medium except the motion of these bodies.

The solution I though was very simple, we have that

$$ m_1 V_1 +m_2 V^2 = m_1 V_1 ' + m_2 V_2 ' $$

$$ m_1 ||V_1||^2 + m_2 ||V_2||^2 = m_1 ||V_1'||^2 + m_2 ||V_2'||^2$$

But now I realized that if $V_i, V_i' \in \mathbb{R}^n$ where $n > 1$ this system may have multiple solutions of $V_1', V_2'$.

For example if a very large body slams into a very small body both headed directly towards each other, you can either completely transfer momentums and kinetic energy, or partially transfer (some average) of momentum and kinetic energy to each body, (now assuming rigid they cannot have a direction vector towards each other) so they move in opposite directions.

But why is complete transfer so much more common in my intuition and textbooks than say 50% of total energy/momentum entering the little body and the other 50% to the big, or even weirder things like $e^{\pi}$%. From a conservation viewpoint these could all be fair game I would think, yet they don't happen.

What law is deciding this?

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  • $\begingroup$ In higher dimensions, you need to know how the bodies are approaching each other, and also know the full expression for the force $F(r)$ between them. Once you know that, you know everything. (Also, I think your examples of different possible collision results aren't actually possible. There's a much easier way to see that multiple results are possible: in 2D and higher, the particles might just miss each other, transferring 0% of their energy.) $\endgroup$
    – knzhou
    Commented Mar 28, 2016 at 5:00
  • $\begingroup$ So that is to the say "two rigid bodies collide at some angle" is meaningless, since alone that doesn't tell you how the forces act. $\endgroup$
    – Sidharth Ghoshal
    Commented Mar 28, 2016 at 5:01
  • $\begingroup$ Well, when we say 'rigid body' we often assume the force is zero when they're not touching and infinite when they do, so there's a well-defined solution. But for two bodies in general, yes. $\endgroup$
    – knzhou
    Commented Mar 28, 2016 at 5:03

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You are right that the system of equations has multiple solutions. None of these solutions is better or worse than another. Given masses and initial velocities it is NOT possible to find out the velocities after collision. Just imagine the collision of two billiard balls: how they move depends not only on initial velocities, but also on position of the balls at the moment of collision (the collision may be central not).

Complete transfer of energy is NOT common. F.e. if the initial velocity of one of the bodies is zero, the complete transfer of energy is only possible if both bodies have equal masses and the collision is central. Only in this case the first (initially moving) body would stop and hence it's total momentum and energy would be transferred to the other body.

You say: "For example if a very large body slams into a very small body both headed directly towards each other, you can either completely transfer momentums and kinetic energy,..."

This can happen only in a system of reference where total momentum of these two bodies is zero (both bodies move towards each other and bounce back as if they collide with a wall). But you can look at this collision from some other point of reference moving with a very large speed. In this system of reference both bodies move with a very large speed before and after the collision, and majority of energy belongs to the large body. No "complete transfer of energy" in this case (in this system of reference).

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  • $\begingroup$ oops. In the situation you described only momentum is fully transferred, not kinetic energy. It's easy to prove that both two bodies can exchange both momentum and energy only if they have equal masses. But my point remains unchanged: complete transfer of momentum/energy is not common at all. $\endgroup$
    – lesnik
    Commented Mar 28, 2016 at 10:21

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