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There is a high probability, I think, that this question is a duplicate of some other question ... but to may knowledge, it hasn't been posed in this exact manner:

Assume we have 2 points, $P_1$ and $P_2$, of mass $m_1$ and $m_2$ in a world coordinate system $(O, \vec{i}_0, \vec{j}_0, \vec{k}_0)$. The point $P_1$ is moving with the constant velocity $\begin{bmatrix} v_{x1i}\\ v_{y1i}\\ v_{z1i}\end{bmatrix}$ while the point $P_2$ is stationary. The point $P_1$ undergoes a perfectly elastic collision with $P_2$. How will these two points move after the collision?

My attempt

This problem is about the conservation of linear momentum: therefore the momentum of the system formed by these two points remains constant. Before the collision the momentum of the system is: $$\vec{P}_{init} = m_1\cdot \begin{bmatrix} v_{x1i}\\ v_{y1i}\\ v_{z1i}\end{bmatrix} + m_2 \cdot \begin{bmatrix} 0\\0\\0\end{bmatrix}$$

After the collision, the linear momentum of the system is: $$ \vec{P}_{fin} = m_1\cdot \begin{bmatrix} v_{x1f}\\ v_{y1f}\\ v_{z1f}\end{bmatrix} + m_2 \cdot \begin{bmatrix} v_{x2f}\\v_{y2f}\\v_{z2f}\end{bmatrix}$$ The unknows are $v_{x1f}, v_{y1f}, v_{z1f}, v_{x2f}, v_{y2f}, v_{z2f}$. But we have only three equations $\vec{P}_{init} = \vec{P}_{fin}$ and six unknowns ... One can also use the law of conservation of energy to obtain another equation: $$ \frac{m_1}{2} \cdot \left(v_{x1i}^2 + v_{y1i}^2 + v_{z1i}^2\right) = \frac{m_1}{2}\cdot \left( v_{x1f}^2+v_{y1f}^2+v_{z1f}^2\right) + \frac{m_2}{2}\cdot \left( v_{x2f}^2+v_{y2f}^2+v_{z2f}^2\right)$$ but there are still only four equations and six variables ...

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    $\begingroup$ Largely a duplicate of this: physics.stackexchange.com/q/453393 I think $\endgroup$ – jacob1729 Jan 21 '19 at 13:45
  • $\begingroup$ If you know the exact direction in which the balls will move after the collision, assume that to be the x-axis. $\endgroup$ – Harshit Joshi Jan 21 '19 at 15:31
  • $\begingroup$ Your problem will then be reduced to solving two variables with two equations. You can then write it using the original coordinate system. $\endgroup$ – Harshit Joshi Jan 21 '19 at 15:41
  • $\begingroup$ However if you really want to solve using this method, you can get some equations by using the fact that the velocities of both the bodies will be along the line of contact. All the components perpendicular to this will be zero. $\endgroup$ – Harshit Joshi Jan 21 '19 at 15:45
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Because there is an infinite number of solutions. Even if you assume conservation of the energy, a given collision that results in final components of the momentum outside the line of initial motion, will be degenerated by a rotation around that axis. The degeneracy is double (six variables, four equations), because you have degeneracies across each of the two axes perpendicular to the orignal trajectory of motion.

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