Elastic Collisions angle 90 degrees

[yitriana]

Why is the angle between products in non head on, perfectly elastic collisions always 90 degrees?

 

[wavingerwin]

Very good question!
I unfortunately do not know why

I have tried proving it with principles of conservation of momentum, conservation of energy, impulse and forces, but still cannot get any luck.

My guess is that on non head-on collision involving spherical objects, (and as long as one of the object is initially stationary) the force that the moving object (1) exerts on the stationary one (2) is directed towards the center of (2).

The reaction force that (2) exerts on (1) is equal in magnitude with the action force and thus is proportional to the initial momentum of (1). (impulse principle)

With this, collision is such that the reaction force will always cause the initial momentum of (1) to change direction making a resultant that has 90 degrees angle with the direction of (2). This may be able to be proven by solving energy conservation (just a guess)

i attached a picture to demonstrate my idea
Hope it is right and helps

 

[A.T.]

Why is the angle between products in non head on, perfectly elastic collisions always 90 degrees?

I think this is only true for two equal masses. You can get there like this:

- For a 1D elastic collision of two equal masses A and B with B initially at rest, you can show (via momentum and kin.energy conservation) that the momentum of A will be completely transferred to B.

- For 2D/3D the above still holds along the collision's normal direction, along which B moves after the collision. Since the momentum along that direction is completely removed from A's momentum, A is left only with the momentum perpendicular to it.

- Obviously: If this holds true in the frame of reference where B is initially at rest, it has to be true in every frame, because angles are preserved under frame transformations.

 

[Bob S]

Why is the angle between products in non head on, perfectly elastic collisions always 90 degrees?

The recoils of the two products are always 90 degrees apart if both incident masses are the same. It is proved somewhere in Goldstein's book on Classical Mechanics. Both masses have to be equal, and one has to be stationary before the collision.

 

[Nanyang]

Let me try to 'show' it:

Consider two particles of same mass and one of them approach the other (not necessarily head on) with one of them originally at rest.

The total momentum before and after the collision will thus be:

p₁ = p'₁ + p'₂

The sum of the kinetic energy of the center of masses of the particles is also conserved for elastic collisions. The relation p²/2m = T can be used to express the conservation of energy for the collision:

p₁²/2m = p'₁²/2m + p'₂²/2m

Thus p₁² = p'₁² + p'₂²

The above equation implies that p'₁ and p'₂ must be perpendicular to each other so that the sum of their squares give another square p₁² because of the Pythagoras Theorem, provided that initially they are not head on. If it is a head on collision, then the momentum of the approaching particle will be zero after the collision since there cannot be any momentum in the direction parallel to the direction of the momentum of the particle originally at rest. Otherwise it would not be head on in the first place.

This moving off perpendicularly thing seems to only be possible from the frame of reference where originally one of the particles are at rest. I'm not sure about angles being preserved under a frame transformation... because if I consider relative to K' the velocities are perpendicular, but relative to K I have to add the relative velocity between K' and K and this would imply that relative to K the velocities are not perpendicular. I don't know if I'm correct on this point.

 

[DrGreg]

This moving off perpendicularly thing seems to only be possible from the frame of reference where originally one of the particles are at rest. I'm not sure about angles being preserved under a frame transformation... because if I consider relative to K' the velocities are perpendicular, but relative to K I have to add the relative velocity between K' and K and this would imply that relative to K the velocities are not perpendicular. I don't know if I'm correct on this point.

You are correct. Just imagine what the velocities would be relative to an observer who is flying past at a speed, in the original frame, that is 100 times greater than either of the particles' speeds.

Therefore...

If this holds true in the frame of reference where B is initially at rest, it has to be true in every frame, because angles are preserved under frame transformations.

...not true.

 

[A.T.]

Just imagine what the velocities would be relative to an observer who is flying past at a speed, in the original frame, that is 100 times greater than either of the particles' speeds.

Yeah you're right, angles between velocities are of course frame depended. What was I thinking. So it's 90° only when one mass is at rest. Still useful in billiard.

 

[Bob S]

Yeah you're right, angles between velocities are of course frame depended. What was I thinking. So it's 90° only when one mass is at rest. Still useful in billiard.

I used to believe that billiad ball collisions were ideal two-body collisions. But 2/7 of the total kinetic energy of a billiard ball is rotational, not linear (translational) kinetic energy, because their moment of inertia is 2/5 mR².