# Is momentum conserved in all inertial frames of reference?

1. Feb 1, 2006

### Vereinsamt

hi,

in the case of elastic collision of two balls in constant speeds, if we took the origin in the center of of one of the balls, so the speed of the origin ball will be always zero, and the second ball will has the speed of v befor the collision and -v after the collision.

so how the conservation of momentum would be?

2. Feb 1, 2006

### Hurkyl

Staff Emeritus
No inertial frame of reference can have any of the balls perpetually at the (spatial) origin.

3. Feb 1, 2006

### Staff: Mentor

But that would not be an inertial frame--the ball accelerates! (To rephrase what Hurkyl already explained.)

4. Feb 5, 2006

### Vereinsamt

thank u Hurkyl and Doc Al. and sorry for leaving the thread last few days

Actully it's somehow working. I'll tell u the story from the begining:

usually in two balls elastic collision problems we solve the equations of consevation of momentum and kinetic energy togather an then we find this equation:

v+v'=u+u' (1)

v for the first ball and u for the second and the prime is for after collision velocities.

I found that this result can be derrived by only by taking on of the balls as an origin of coordinates, and using the fact that v=-v' in this frame of reference then we will get the equation (1).

what I want to know where I got v=-v' from? what consevation is this? or is it newton's third law?

and why taking one of the balls as an origin is giving a right result?

Last edited: Feb 5, 2006
5. Feb 5, 2006

### Staff: Mentor

This is perfectly correct and is derived, as you seem to realize, by combining conservation of energy with conservation of momentum.

No. Equation (1) is true for any inertial frame. Furthermore, as Hurkyl and I have pointed out, the ball's frame of reference is not an inertial frame; so you can't just take a ball as being fixed at the origin of an inertial frame. (See the exception below.)

By setting one of the balls as being at the origin of an inertial frame you are essentially assuming that its mass is much greater than the other ball. If the second ball has a mass much greater than that of the first ball (say a ping pong ball colliding with a bowling ball), then it's true that v=-v', but that's a special case. (In that case you can take the bowling ball as being approximately in an inertial frame throughout the collision.)

It doesn't, except in the special case noted above.

What you need to do is learn how your equation (1) is really derived. Start with conservation of momentum:
$$m_1u + m_2v = m_1u' + m_2v' \mbox{ [a]}$$

Which can be rewritten as:
$$m_1(u - u') = m_2(v' - v) \mbox{ [a']}$$

Now take conservation of energy:
$$m_1u^2 + m_2v^2 = m_1u'^2 + m_2v'^2 \mbox{ }$$

Which can be rewritten as:
$$m_1(u^2 - u'^2) = m_2(v'^2 - v^2) \mbox{ [b']}$$

I'll leave the last step to you: combine equations a' and b'.

Last edited: Feb 5, 2006
6. Feb 9, 2006

### Vereinsamt

sorry again

Why do you I am that weak? it is very easy
actully derivations is my favorite parts in math and physics.

ok so look what I've got:
$$v_r_m_2=v-u$$ (2) ($$rm_2$$ refers that the speed is relative to the other mass or ball)
$$v'_r_m_2=v'-u'$$ (3)
$$v_r_m_2=-v'_r_m_2$$ (4)
$$v-u=-(v'-u')=u'-v'$$ (5)

and (5) is the same as (1) which we know that is right.

:uhh:

Last edited: Feb 9, 2006
7. Feb 9, 2006

### Staff: Mentor

Ah... I think I see what you've been trying to say. You weren't taking one of the balls as the origin of an inertial frame (which it is not), you were describing things in terms of relative velocity! Nothing wrong with that. (But the fact that the relative velocity is reversed in an elastic collision is what you are trying to prove--you can't just assume it's true!)

Realize that your equation (4) is just another way of saying (5); they are equivalent. What you need to do is prove that it's true.

8. Feb 10, 2006

### Vereinsamt

I think its the same think to say that the speed is relative to the other ball or to say the speed is in the ball's frame of reference.
ok lets say that its not an inertial frame of reference my questions is why it is giving a right result?

by the way, can we think of it like this?: the frame is inertial before the collision and after the collision but not during the collision and we dont care about "during the collision" case because conservation of momentum is just about "before" and "after" cases

i think the proof is just by changing the order of the pervious equations like this:
[a]
[a']

[b']
(1)
(2)
(3)
and then (4) QED.

I hope to hear your opinions

9. Feb 10, 2006

### Staff: Mentor

In that case the frame is not an inertial reference frame for the purposes of analyzing the collision. You are using one inertial reference frame before the collision, and a different inertial reference frame after the collision. You must use the same inertial reference frame before and after the collision.

10. Feb 10, 2006

### Staff: Mentor

The reason why it works is because while you think you are using the reference frame of one ball, you are actually using a standard inertial frame in which the balls have speeds of $u$ & $v$ before and $u'$ & $v'$ after the collision. (This is often called the "laboratory frame".) Your expression $v_{rel} = - v'_{rel}$ is written in terms of relative velocity, but it's not from the frame of one ball. (And this expression is equivalent to what you are trying to prove--so your reasoning is circular!)

If you really used the frame of ball one, then its momentum and energy would be zero. The total momentum before the collision in that non-inertial frame would be: $m_2 (v - u)$; and the total momentum after the collision would be: $m_2 (v' - u')$. Momentum would not be conserved. (Assuming it were conserved in that frame would lead to the incorrect equation $v - u = v' - u'$.)

jtbell explained this one; mark his words.

Realize that (2) & (3) are just definitions and that (4) and (1) are equivalent.

The real proof is to combine [a'] & [b'] (which are true due to conservation of momentum and energy) to derive (1). Do this.