# Collisions in series

1. Mar 18, 2006

### MelissaSweet

Hi guys!

I was hoping that some of you smart guys and gals might be able to help me.

I'm looking for the simplest way to describe a series of collisions. Basically, an object colliding with another object, then a time delay and then another object colliding with the first two and so on.

Is there an easy way to calculate the effects, or must each function be performed separately?

Thanks!

Melissa

2. Mar 19, 2006

### vanesch

Staff Emeritus
That description is too vague to allow us to help you: is this, say, for a computer game, or any other "rough" simulation ? Or is it for a calculation that needs some accuracy in its results ? What are the objects involved ? Neutrons ? Ping pong balls ? Cars ? Fish ? Customers ? Birds (with or without bird flu ?) ? Are there 5 of them, or 10000, or 10^20 ?

3. Mar 19, 2006

### MelissaSweet

Oh, okay sweetie.

I'm looking at a 3-body collision. Body x is in a stationary frame relative to the observer. Body y is moving to intercept body x and body y2 will intercept xy shortly thereafter in an exactly opposite trajectory to y, stopping xyy2 relative to the observer.

How do you figure the distance x traveled?

Melissa

4. Mar 20, 2006

### vanesch

Staff Emeritus
So, x and y are undergoing an inelastic collision ? (meaning, x and y get "glued together" on their interaction, right)

Can we approximate the bodies as mass points, or do we have to take their spatial extension into account ?

I will presume that we can do with mass points. In that case, the thing that prevails is momentum conservation.

Body x, in the observer's frame, has no velocity and hence no momentum. Body y has a certain mass and velocity, and hence a certain momentum.
The total momentum of this system (xy) is then simply the momentum of y, which should be, after collision, equal to the momentum of the entire system xy (with mass = mass of x plus mass of y).
This allows you to calculate the velocity of the xy system.

Next, you do the same with the collision of (xy) and y2. Conservation of momentum will allow you to say what will be the final momentum. Now, if y and y2 have equal masses, and have opposite velocities, you can already see that the final momentum will be 0.
However, to know exactly the *trajectory* of x (which will start moving upon the first impact, and stop at the second impact), you'll have to say exactly when or where the first and the second y will hit. This is in the details of the description of your problem.

cheers,
Patrick.

5. Mar 20, 2006

### MelissaSweet

Hiya Patrick,

I like that name (being of Irish descent myself). In fact I found a four-leaf clover in my yard just yesterday so I'm really feeling the luck o' the Irish is with me!

Anyway, yes. You can presume the problem in point masses (but it's not pointless!) and inelastic collisions.

y and y2 are intercepting x in equal and opposite trajectories, but y2 is initially farther away from x than y is.

I know the trajectory of x will be in the same trajectory as y, but what I need is a formula to determine the distance x travels in this trajectory relative to the observer. I believe this formula must incorporate a function of time, but I don't know how to do it.

f=ma just tells me that overall the system is stationary to the observer, but obviously x must move some distance. How do I determine that distance?

Thanks!

Melissa

Last edited: Mar 20, 2006
6. Mar 21, 2006

### heman

I just got to ask one thing ...why do we always have to conserve momentum during collisions...isn't there any else way out to solve them..I actually solved such questions many years before,,so may be point is missing somewhere.

7. Mar 21, 2006

### vanesch

Staff Emeritus
Well, if you know WHEN y hits x and you know where y2 is AT THAT POINT, you now have x (in the couple xy) moving from that position onwards with the velocity you can calculate, and y2 moving towards xy with its initial velocity.

That's a problem of the kind: train A leaves station D at noon, towards station E, and has speed v1 ; train B leaves station E, also at noon, towards station D, with speed v2. At what moment, and where, does train A cross train B on the track between station D and station E...

8. Mar 21, 2006

### MelissaSweet

Patrick,

Okay, that's as I thought. There isn't an easier way. Each calculation is a separate function.

Thanks

Melissa