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We see that our calculations always gives a loss of Kinetic energy in inelastic collisions while the net momentum of the system remains the same. Where does the lost energy go while there is no trace of friction or air drag in our calculation?

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In a classical frame, the total energy of a system $E$ is the sum kinetic $E_K$, potential $E_P$, and internal $U$. Any event rebalances the energies. In a frame with no potential energy, the loss of $E_K$ goes to $U$.

For example, a mass of water $m$ starts at the top of a water fall with no (vertical) $E_K$, a potential $E_P = m g \Delta h$, and a certain internal energy $U$ as indicated primarily by its temperature. At the bottom of the water fall, just before the falling water $m$ hits the stagnant water at the bottom, $m$ has translated $E_p$ to $E_K$ (assuming that the fall is essentially an isothermal process so that $\Delta U$ is zero). As $m$ now stops moving vertically, it translates $E_K$ to $U$. This causes an increase in the temperature of the water.

The above example is drawn from a common problem in engineering thermodynamic textbooks.

By further reference, an inelastic collision does not directly imply that we must consider friction. Picture two spheres at the same $mv$ and $E_K$ that collide, stick, and stay in one place. The collision is entirely inelastic. Friction at the macroscopic level does not need to be invoked to explain this event. It can be explained entirely by recognizing a permanent deformation for the spheres themselves.

Friction, when it does occur, is a source of irreversibility in a process. Friction is translated typically to be a heat loss from the system to the surroundings.

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  • $\begingroup$ Maybe one can stress that there indeed is friction in an inelastic collision, contrary to what's claimed in the question: The friction is internal to the deforming bodies of fluids, and dissipated the kinetic energy into heat. $\endgroup$
    – Toffomat
    Commented Jul 27, 2020 at 12:57
  • $\begingroup$ @Toffomat Done. Friction is not required as a defining way to explain the inelastic collision between two bodies. $\endgroup$ Commented Jul 27, 2020 at 23:23
  • $\begingroup$ I guess it's more about words than content, but I'd call the processes that transform kinetic into internal energy "friction". Also, friction doesn#t really casue heat loss, but rather loss of kinetic energy to heat. $\endgroup$
    – Toffomat
    Commented Jul 28, 2020 at 12:24
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    $\begingroup$ @Toffomat Words and their definitions absolutely matter. Friction is NOT a generic substitution for "the process that transforms kinetic energy into internal energy." Friction is the resistance between two surfaces that rub against each other. That is it. $\endgroup$ Commented Jul 29, 2020 at 0:40
  • $\begingroup$ Well, I would certainly attribute the heating of an inelastically deformed solid to "internal friction", but I guess we can leave it at that. $\endgroup$
    – Toffomat
    Commented Jul 29, 2020 at 8:31
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The other answers have sufficiently answered you query.Still,I am giving a different rather philosophical approach.

Well the lost energy,first of all, must go somewhere.Now there are tons of ways to do this.Sound energy , temperature increase , friction , air drag or any other way are all possible answers depending on environmental factors(say).

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According to work energy theorem change in Kinetic energy is equal to work done by all forces.

Since there is no external force on the bodies other than friction, air drag,sound or the force that causes deformation, the change in Kinetic energy is equal to work done by all the mentioned forces.

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The lost energy is stored in the form of potential energy which exists in the bodies after collision because of the deformation of balls on molecular level.Hence, in perfectly elastic collisions where there is no sort of deformation after collision; you can conserve kinetic energy in it.

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