Hi,
I Have set up 4 collisions between two Ovals (AA,BB,CC,DD) and I would like to know if the outcome will be identical for all 4 of them.
First of, when looking at this orange collision that is going to take place, one can draw straight vertical lines between the two centers ...
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… and if you fly along with the lower oval, and draw it out; than it looks as if they are approaching each other in a straight line:
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Next, look at the 4 collisions (AA,BB,CC,DD) in the image below, than you can draw 4 times the same visual thing as the image above, when flying each time along with the lower Oval ABDC, be it for collision AA, BB, CC or DD.
The right frame of reference to calculate collisions in is center of mass frame ... and when looking at this from a 'set' view from the top, than the outcome would be 4 x times different, but is it also 4 times different? ... cause they are all coming from different directions, and are differently orientated before they collide, and inertia-wise ... so my guess would be that the out comes are in fact also really different each time, although they approach each other visually in the same way.
![]()
A friend gave me this explanation (see quote below), but it doesn't seem to be right ... so that's why I'm posting it here, just to be sure.
Looking forward to hear your opinion(s), thanks!
m.
I Have set up 4 collisions between two Ovals (AA,BB,CC,DD) and I would like to know if the outcome will be identical for all 4 of them.
First of, when looking at this orange collision that is going to take place, one can draw straight vertical lines between the two centers ...
… and if you fly along with the lower oval, and draw it out; than it looks as if they are approaching each other in a straight line:
Next, look at the 4 collisions (AA,BB,CC,DD) in the image below, than you can draw 4 times the same visual thing as the image above, when flying each time along with the lower Oval ABDC, be it for collision AA, BB, CC or DD.
The right frame of reference to calculate collisions in is center of mass frame ... and when looking at this from a 'set' view from the top, than the outcome would be 4 x times different, but is it also 4 times different? ... cause they are all coming from different directions, and are differently orientated before they collide, and inertia-wise ... so my guess would be that the out comes are in fact also really different each time, although they approach each other visually in the same way.
A friend gave me this explanation (see quote below), but it doesn't seem to be right ... so that's why I'm posting it here, just to be sure.
Looking forward to hear your opinion(s), thanks!
m.
Quote:
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ABCD are all physically the same collision. In each case, you can transform to a reference frame in which they approach not along the black arrows, but along the orange ... yellow arrows. This transformation is precisely just subtracting out the average velocity. The outcomes must be the same physically, because the situation is the same physically. The way the particles behave cannot depend on the camera you choose to observe them with. I should add, the thing that makes ABCD _appear_ to be different collisions is that you not only subtract their average velocity before calculating the collision, but you add it back on afterwards. So A appears to bounce in a primarily right-wards direction, because the average velocity in the case of A is primarily pointing rightwards, and so on. |