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multiple collision : balls

Started by Agemaniac Jul 20, 2003 at 2:51 AM 1 reply 400+ views

Original Post

Jul 20, 2003 at 2:51 AM

ok, imagine the following situation...the balls going to collide. ball 1 is in the position (20,0,0). ball 2 is in (-20,0,0) and ball 3 is in (0,0,0). if ball 1 and 2 is going to the origin velcity1 = (-10,0,0) and velocity2 = (10,0,0) how should i calculate the impulse? first compute the impulse of the the collision 1-3 and than calculate the impulse of the collision 2-3 and sum both??? if i do it i think that the ball 3 should stay in the (0,0,0) and the other two stop in the point of collison....... note: i''m assuming 3 identical balls...so a collision between two balls should change the velocities between then...if i handle the collision 1-3..the result should be velocity1 = (0,0,0)..velocity3(-10,0,0)..but i have to compute impulse from 2-3 to aply in 3 as well..ok..them in 2-3 velocity3 become (0,0,0) and velocity2 (0,0,0)..so i think it is wrong.........what i''m doing wrong?? can someone explain me or give some resources on the web when i can read about it? thanks...

Atheist

Jul 20, 2003 at 7:32 AM

a)
Correct solution (assuming a perfectly elastic collision):

before:
2 ------> 3 <------- 1

after hit:
<----- 2 3 1 ------>

in words: 1 and 2 bounce of (v1=(-v1x,0,0), v2=(-v2x,0,0)), 3 remains at rest in (0,0,0).

b) "impulse" is called momentum. Those people from old europe don´t seem to realize what´s going on in the world

c) momentum is calculated by p = m*v, where m is the mass of your sphere (but I don´t think that really was your question).

d) if both collisions (C1: 2->3, C2: 3<-1) take place the same time I think you can still calculate them as if they didn´t (at least you can do that as long as you have perfectly elastic colisions). Take C1 first: 2 transfers it´s momentum to 3. 2 is at rest, 3 moves towards 1. 3 hits 1, both of them bounce away, so v1 = (-v1x, 0, 0). 3 bounces back towards 2 which is at rest. They hit and 2 takes over the momentum from 3. Hence 3 is at rest and v2=(-v2x,0,0)...
This should also work if |v1x| <> |v2x| (balls 1 and 2 exchange their momentum,, then) but you might run into trouble when having unelastic collisions.
NOTE: This might not work with every way of detecting collisions (don´t know what the standard way of detecting collisions I just know how I´d do it).

Waverider

108

Waverider

Jul 20, 2003 at 7:40 AM

I did my billiards demo the same way. Treat each collision separately. Handle the first one, then the next, etc.

My algorithm does have a problem with billiards that are absolutely flush against each other for some reason, but in theory, it should work.

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