Collision with more than two objects at once in code

[Luke_M]

Balls A,B,C,D are traveling at V=1 to the right next to each other like a train they hit ball E whom is stationary. At the exact instant ball F is traveling down and hits A, F is traveling at V=1 but down

Here is a diagram this is birds eye view:

*=collision
both collisions have happened at the same time
|
|
V
F
*
-------->ABCD*E



A: Vx=1 m=10
B: Vx=1 m=10
C: Vx=1 m=20
D: Vx=1 m=10
E: Vx=0 m=60
F: Vy=1 m=10

will object E receive any Vy ?
why not ?
what are the new V's ?
I need to know how to work this out as I am making a physics model in my game.

If you are a programmer maybe you can also answer this question:
In the real world collisions can happen all at once like in my above example. How can I do this in code ? I can only do them sequentially with for loops. is this normal practice in constructing a physics engine. Because if I do the collision between F and A after the other collisions A's new V will incorrectly not be included in the calculation.
Thanks in advance.

 

[CSBME]

For programming I think that you need to take a look at a concept called "threading" not all languages support it, but Java does and I believe C++ does as well.

Here is a great place to start:
http://java.sun.com/docs/books/tutorial/essential/concurrency/

Plus why are you using loops, can't you just use a couple of equations to model this and just put those into your code?

 

[AlephZero]

IN the real world collisions don't happen "all at once". They take a finite amount of time, and the effects propagate through the bodies at the speed of sound, so if one side of a ball collides with something the other side of the ball doesn't "know" anything about that till a finite amount of time later. The typical speed of sound in a solid is a few km/sec, or a few mm per microsecond.

The easy way to program this is to pretend the balls are all a small distance apart. So in your example first, F collides with A and D collides with E (and those two collisions don't interact with each other).

Next, because D has slowed down, C collides with D, and so on.