What is the rigorous proof for elastic collisions?

[Bipolarity]

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.

$$m_{1}v_{1f} + m_{2}v_{2f} = m_{1}v_{1i}$$

$$\frac{1}{2}m_{1}||v_{1i}||^{2}= \frac{1}{2}m_{1}||v_{1f}||^{2} + \frac{1}{2}m_{2}||v_{2f}||^{2}$$

Derive the following equations:

$$v_{1f} = \frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1i}$$
$$v_{2f} = \frac{2m_{1}}{m_{1}+m_{2}}v_{1i}$$

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

 

[mfb]

Those equations are true in the 1-dimensional 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.

 

[jtbell]

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 head-on, slightly off center, lightly glancing, etc.

Halliday/Resnick don't discuss this, but you can probably find it in a higher-level classical mechanics book, or maybe with a suitable Google search.

 

[AlephZero]

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 north-south collide with each other, it is an assumption (with no justification) that there they have no velocity components in the east-west 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 east-west 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.

 

[jtbell]

you need to specify the impact parameter

I just remembered that it also makes a difference whether the pucks collide with or without friction against each other. If there is (sliding) friction between them when they come in contact, that generally causes each of the pucks to rotate around their centers of mass after the collision. This gives them rotational kinetic energy which has to be accounted for when applying conservation of energy.

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

 

[mfb]

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.