Conservation of Momentum in an Atom Interaction

AI Thread Summary
The discussion focuses on the conservation of momentum in a system of two equal mass atoms interacting in opposite directions. It is established that both linear and angular momentum are conserved due to the absence of external forces. The center of mass remains stationary with a velocity of zero throughout the interaction. The velocity of the second atom after the interaction is determined to be the negative of the first atom's velocity, and its trajectory is described as a straight line. The relationship between impact parameters and velocities is explored, leading to the conclusion that energy conservation implies V0 must equal V1.
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Homework Statement


Two atoms of equal mass m, that move with the same speed but opposite direction, interact when they're in some region R of space, as in fig.1. After the interaction, one of the atoms moves with velocity ## \vec{V1} ## as in fig.2.

a) Are the linear and angular momentum of the system conserved?
b) Calculate the velocity of the center of mass, before, during and after the interaction.
c) Find the position of the center of mass, before, during and after the interaction.
d) What is the velocity of the other atom after the interaction ?
e) Find the trajectory of the other atom after the interaction.
f) Compare ## \vec{V1} ## with ## \vec{V0} ## for different values of the impact parameter a, that is, in the cases where a>b, a=b, a<b.

Homework Equations



## \vec{p} \equiv m\vec{v} ##

## \vec{L} = \vec{r} X \vec{p} ##

The Attempt at a Solution



a) I thought that since there are no external forces acting on the system, both angular and linear momentums are conserved.
c) Since both atoms have the same mass and speed but opposite directions, and are at the same distance from the center O of the coordinate system, the center of mass stays there before, during and after the interaction.
b) From the above, the velocity of the center of mass equals cero at all times.
d) Since momentum must be conserved, the velocity of the other atom must be ## \vec{V} = - V1 \hat{j} ##
e) The trajectory of the other atom should be a straight line at X=b, for Y<0. Is this a correct way to put it?
f) This I don't know quite well how to put it, it seems to me that for a=b , V=V1 from simmetry, but I don't know in the other two cases.

Can anyone give me a hint? Thanks in advance.
 
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Use conservation of angular momentum.
 
Ok, using your hint, I thought that since a is the only component of the distance ## \vec{r} ## that's involved in the cross product, I get for the first case:

## \vec{L} = 2a m V0 ##

and using the same reasoning in the second case I get

## \vec{L}= 2bmV1 ##

Since they should be equal, I finally get a/b = V1/V0, from where I can relate the impact parameters with the velocities. Is this correct?.
 
Looks good. This problem seems a little strange to me since I don't see how you can get anything other than V0 = V1 if energy is to be conserved.
 
Thanks!
 
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