# Momentum, why can we change the system?

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1. Feb 22, 2015

### x86

Here is our momentum equation for when two separate masses m1 and m2 collide with each other and couple together

m1v1 + m2v2 = (m1+m2)v3

In the LHS of our equation, we have two systems. m1 and m2.

But in the RHS of our equation, we use a totally different system. m3, the combination of m1 and m2.

How is this possible? I am confused

2. Feb 22, 2015

### Bandersnatch

A 'system' here is a collection of all the masses. As long you never add or deduct any mass from the total, you've got the same system.

For example, m1 could very well be treated as two masses stuck together: m1' and m1'', where m1'+m1''=m1. You could do that for all the atoms making up m1 and m2. It's still the same system.

3. Feb 22, 2015

### Staff: Mentor

I wouldn't interpret it that way. I would interpret the system as being the same on both sides. On both sides the system consists of the combination of m1 and m2. On the left hand the two parts of the system are moving at different speeds, but that doesn't mean they are not part of the same system.

4. Feb 22, 2015

### x86

I see. Am I able to change the system though for multi part questions? Like suppose system 1 is a boy and a green cart. A boy jumps from the cart. Then I find the boys velocity midair. Can I then pick a different system to use, say system 2 with a boy and a red cart; and use the boys velocity I found from part 1 to solve the momentum equations for system 2?

5. Feb 22, 2015

### Staff: Mentor

Sure. What constitutes the "system" is arbitrary, you can choose whatever is most convenient and you can change your system's boundaries as needed.

However, you do need to be careful. The boy+green cart and the boy+red cart are different systems, so you need to be careful and not assign one system the other system's momentum etc., and make sure to keep track of which forces are internal or external for which system.

6. Feb 22, 2015

### x86

Okay, thanks for the tip (: