
#1
Apr813, 12:03 AM

P: 783

Suppose that a mass M1 is moving with speed V1 and collides with mass M2 which is initially at rest. After the elastic collision they make, both momentum and kinetic energy are conserved.
[tex] m_{1}v_{1f} + m_{2}v_{2f} = m_{1}v_{1i} [/tex] [tex] \frac{1}{2}m_{1}v_{1i}^{2}= \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2} [/tex] Derive the following equations: [tex] v_{1f} = \frac{m_{1}m_{2}}{m_{1}+m_{2}}v_{1i} [/tex] [tex] v_{2f} = \frac{2m_{1}}{m_{1}+m_{2}}v_{1i} [/tex] Resnick & Halliday give a fairly staightfoward proof. But in the proof it fails to recognize the fact that the v values in the momentum conservation are vectors, whereas those in the energy conservation are scalars. So the proof is not rigorous. I was curious how one would prove this rigorously, (preferably without casework), given this remark. Thanks! BiP 



#2
Apr813, 04:48 AM

Mentor
P: 10,813

Those equations are true in the 1dimensional case only, where you can treat the velocity as scalar.
For two dimensions, you get an additional degree of freedom in the collision, and there are no equivalent fixed equations for the velocity. 



#3
Apr813, 08:18 AM

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P: 11,239

Some 2D collisions can be solved, but you need more information than just the masses and initial (vector) velocities. For example, consider two spheres (or rather, circular pucks sliding on a frictionless surface), with one at rest initially. In addition to the initial velocity of the other puck, you need to specify the impact parameter: the transverse distance between the path of the moving puck's center, and the the center of the stationary puck. This specifies whether the collision is headon, slightly off center, lightly glancing, etc.
Halliday/Resnick don't discuss this, but you can probably find it in a higherlevel classical mechanics book, or maybe with a suitable Google search. 



#4
Apr813, 10:02 AM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,351

Elastic collisions proof
You didn't say exactly what part of the proof you think is not rigorous, but I suppose one objection to it is this: if two particles moving northsouth collide with each other, it is an assumption (with no justification) that there they have no velocity components in the eastwest direction after collision.
It would be possible to justify that by a symmetry argument. Suppose after the collision particle 1 moves east and particle 2 moves west. In that case there would be another solution where particle 1 moves west and particle 2 moves east. Since Newtonian mechanics is assumed to be deterministic, there is no reason to choose one of these solutions rather than the other one, therefore the eastwest velocity components must be zero. Note that as the previous answers said or implied, this result is only true for point particles, not for finite sized objects which can have rotational kinetic energy and angular momentum. 



#5
Apr813, 11:37 AM

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P: 11,239

I've done this derivation only for the frictionless case. 



#6
Apr813, 03:24 PM

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P: 10,813

If you want to go into detail, you can also consider the shape and orientation of the objects  if they are not round disks, the shapes are relevant, too.



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