Relation between Work-Energy Theorem and Collisions I am stuck on this idea relating the Work-Energy theorem and collisions. The derivation for the Work-Energy theorem is entirely mathematical, and simply involves
Net work on a system = $$W_{net} = \int_A^B \vec{F_{net}}\cdot d\vec{r} =\ldots = \frac{m}{2}\int_A^B\left(\frac{d}{dt} v^2\right) dt = \Delta K_{sys}$$
My interpretation is that, if this is an equality, then the change in the kinetic energy must be due to the work of some external force on our system. But if we consider a collision involving two balls with no net external forces, by definition (and experimentation and the like), $\Delta K_{sys} = 0 $ for elastic collisions, $\Delta K_{sys} < 0$ for inelastic collisions, and $\Delta K_{sys} > 0$ for superelastic collisions. Textbooks online and Kleppner and Kolenkow attribute the change in/constancy of kinetic energy of the system to interaction forces during the collision and the work they do. But how are these interaction forces between balls of any relevance if they are internal to the system? How can the work of these internal forces change the system's kinetic energy if such a change can only be attributed to the work of external forces?
Similarly, even if we were to try to compute in some way the relevance of these interaction forces, how does them being conservative or non-conservative affect the velocities/kinetic energies of the balls? As far as I know, the conservativity of a force only relates to the work it does being path-independent or equivalently, that it is the gradient of some potential function, which is irrelevant to kinetic energies.
For further confusion, KK differentiates between elastic collisions having conservative interaction forces and inelastic collisions having non-conservative interaction forces, but doesn't say which for superelastic collisions. The example KK uses for superelastic collisions is the collision of two cocked mouse traps which release the energy stored in the springs. But the spring forces there should be conservative. In another example I found online, the instructor gave an example of a superelastic collision where the balls had some explosive chemical that ignited during the collision and increased the kinetic energies post-collision. I am not sure about the conservativity of the forces involved in the chemical explosion, but I would guess they are not conservative. Is there a defined category of interaction forces involved in superelastic collisions?
Thanks!
 A: As has been indicated, you have asked multiple questions, whereas you should focus on one question per post. However, I think a major part of the problem you are having is due to the mixing of two different but related concepts:
(1) The change in kinetic energy of a particle or rigid object due to net work done on the particle or object is net work done by all forces, not just external forces, per the work energy theorem
(2) The change in total mechanical energy (KE + PE) of a system, due to net external work done on a system.
Although you have referred to the work energy theorem as the the net work done on a "system", you will find that the most most common statement of the theorem is the "net work done on a particle or object equals its change in kinetic energy". In my view, the use of the term "system" in connection with the work energy theorem, as is sometimes done, is problematic because a "system" can undergo a change potential energy (PE) as well as kinetic energy (KE) as a result of net work on the system, whereas an particle or object alone cannot undergo a change in PE. And that's because PE is a system property and not a property of an object alone.

$\Delta K_{sys} < 0$ for inelastic collisions

For inelastic collisions conservation of mechanical KE does not apply. You can use the WET to analyze collisions with respect to the average impact force during a collision. An example is given here: http://hyperphysics.phy-astr.gsu.edu/hbase/carcr.html#cc1

$\Delta K_{sys} > 0$ for superelastic collisions

Here the increase in KE has to do with conservation of mechanical energy of an isolated system (concept 2 above), which can be stated as, in the absence of external work,
$$\Delta KE+\Delta PE=0$$
So for the "superelastic" collision, the increase in KE simply equals the decrease in PE stored in the system (elastic potential energy in the case of the example of sprung mousetraps, or stored chemical potential energy in the case of explosives, though chemical potential energy is not "mechanical" energy)

How can the work of these internal forces change the system's kinetic
energy if such a change can only be attributed to the work of external
forces?

Because the change in the KE of a system is not only attributable to external forces. The change in KE plus the change in PE of a system is.  Internal forces can convert KE to PE or PE to KE such that for the system $\Delta KE+\Delta PE=0$.

As far as I know, the conservativity of a force only relates to the
work it does being path-independent or equivalently, that it is the
gradient of some potential function, which is irrelevant to kinetic
energies.

Conservative forces are not irrelevant to kinetic energies. A change in PE equals the negative of the work done by a conservative force. For an isolated system, not subjected to dissipative forces, an increase/decrease in PE must be accompanied by a decrease/increase in KE of an equal amount.
Hope this helps.
A: First word of caution, you need to be careful about which kinetic energy you are talking about. The theorem you quoted can be applied to a point, but when you apply it to a system, the $\vec v$ is the speed of the center of mass and $m$ the total mass, so your $K_{sys}$ is not the sum of the kinetic energies of individual particles, which I will write $K_{tot}$, but rather the kinetic energy of the overall translation movement, so you should rather write it $K_G$. Actually, due to Huygen's theorem, $K_{tot} \geq K_G$ with the difference being the kinetic energy of the system in the frame moving at speed $\vec v_G$ compared to your reference frame.
If you're interested in $K_G$, your concern still stands. As you pointed out, the fact that the interaction forces are conservative is not necessary for $K_G$ to remain constant. The relevant condition is Newton's 3rd law, that is every force is matched with an equal and opposite force. Since this is the case for conservative forces, the conservative criterion makes it a sufficient condition for total kinetic energy conservation (but, I will reiterate, not necessary).
For example, the usual example of interaction forces violating Newton's 3rd law would be the full electromagnetic one (including relativistic corrections). In this case, the system will lose some $K_G$, and it turns out that it goes to the EM field.
However, if you're interested in $K_{tot}$, the conservative nature of the interaction is extremely relevant as $K_{tot}$ can vary even if Newton's 3rd law holds. This should resolve your confusion surrounding the classification of elastic, non-elastic and super-elastic collisions. In particular, the final examples of super-elastic show that you have an increase of $K_{tot}$, but constant $K_G$.
Finally, a final note on the conservative nature of forces. Conservation of energy is always true, so every force you'll meet in nature is conservative. However, if your model does not incorporate the full picture, they can appear to be non-conservative. For example, the drag force in fluid mechanics is non-conservative when only considering the object moving through the fluid but becomes conservative if you take into account the full energy of the fluid (and its internal energy if it is viscous). This is why in your final example of superelastic collisions, while the underlying forces are conservative, from your interacting particles perspective they are non-conservative.
Hope this helps and tell me if you need more clarification.
