# Particular direction in a collision of particles: [closed]

I have two particles, each of mass $m$. They move on the $x$-axis with initial velocities that are equal in magnitude but opposite in direction ($v$ and $-v$). They collide in the origin of the coordinates and, assuming that the collision is perfectly elastic, I show that after the collision the final velocities have the same magnitude as the initial velocities. In particular, if $\alpha$ and $\beta$ are the angles that the velocity vector forms with the $x$-axis after the collision,I have that $\alpha+\beta=180°$. With what condition could I get that $\alpha=90°$ and $\beta=90°$ such that the components of final velocities are only on the $y$-axis?

• Do you mean to make one of those final angles negative? If they are both heading in the same upward direction after collision then momentum is not conserved. Aug 3, 2018 at 10:43
• No i may have explained myself wrong. I would that a particle to go in the positive direction of the axis $y$ and the other in the negative direction. In this way the momentum before the collision is $p_{i,x}=mv-mv=0$, $p_{i,y}=0$ and after the collision $p_{x,f}=0$ and $p_{f,y}=mv-mv=0$. Aug 3, 2018 at 11:13
• What type of conditions are you looking for specifically? The objects would just need to interact in such a way so that the internal force between them has both horizontal and vertical components such that the objects stop moving horizontally and start moving vertically. Aug 3, 2018 at 12:34
• I am not looking for a particular condition, the book says that " the distance between their parallel approach lines has been arranged in such a way that after the collision they move along the $y$- axis at the same and opposite final velocities. I was trying to figure out if he had put himself in a particular case or could be justified in some way, because in general the final velocities have the same magnitute and opposite direction but this isn't only the axis $y$ direction. Aug 3, 2018 at 13:14
• I'm trying to figure out just what your question is. This scattering event is quite possible, so do you need to figure out what the impact parameter is, or what? Aug 3, 2018 at 21:53

## 1 Answer

The difficulty is that this problem is indeterminate for point particles, which have no size or structure.

There are 4 unknown variables after the collision : 2 speeds and 2 directions. The (scalar) conservation of kinetic energy provides 1 constraint, and the (vector) conservation of linear momentum provides 2 constraints. There are 4 unknowns but only 3 constraints - not enough constraints to solve the problem uniquely. Any angle $\alpha$ is possible.

The difficulty can be solved if the particles have finite size and structure - eg 2 smooth rigid circles colliding in 2D. The impulse between these circles and the change in momentum for each lies along the line joining their centres at the point of contact. The direction of this line of impulse gives a 4th constraint. In the case of point particles it is impossible to decide what direction this line points in, because the point particles are their own centres, and these points coincide at impact, so the line joining their centres cannot be defined.

• The book says that " the distance between their parallel approach lines has been arranged in such a way that after the collision they move along the y- axis at the same and opposite final velocities. Is it a particular case or what? I do not understand how to justify this sentence. Aug 3, 2018 at 13:31
• Yes $\alpha=90^{\circ}$ is a particular case, because $\alpha$ could have any value. The distance you refer to is distance $b$ in the diagram. The angle of impact should be $\theta=45^{\circ}$ in this diagram in order to have $\alpha=\beta=90^{\circ}$. Note that $b=2R\sin\theta=R\sqrt2$. Aug 3, 2018 at 14:26
• Excuse me, just one last question. What is the link between $\alpha$ or $\beta$ with $\theta$? Aug 3, 2018 at 16:14
• If the collision is elastic then $\alpha=2\theta$. $\theta$ is the "angle of incidence" at which the circles collide. $\alpha$ is the angle of deviation/deflection. Aug 3, 2018 at 16:30
• Thanks a lot. Where can I find the proof of this result? Aug 3, 2018 at 17:08