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During inelastic collision the kinetic energy does not remain constant. In many texts, I have seen it is because there is loss in energy from the system. If there is loss in energy, then how can mass and thereby momentum remain constant?

I think kinetic energy changes because it calculates the energy required when the body is accelerated, and during inelastic collision the body doesn't retard or accelerate constantly to zero velocity and then reverse direction. Consider it is a linear collision. Is my logic correct?

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    $\begingroup$ No, you cannot "lose energy." In an inelastic collision, kinetic energy is dispersed into tons of molecular-level bits of kinetic energy, some of which may well be converted to potential energy if bonds or material structures change. $\endgroup$ – Carl Witthoft Jan 2 '16 at 20:08
  • $\begingroup$ I will edit the question because you are mistaking my question. $\endgroup$ – N.S.JOHN Jan 3 '16 at 7:26
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When we describe an inelastic collision of macroscopic objects we normally say the missing energy ends up as heat. That is, the temperature of the colliding objects is higher after the collision than before it.

However this is a simplified description because heat is just the kinetic energy of the atoms and molecules making up the objects. Suppose you take a (very) simplified system of two dumbbells colliding:

Collision

If we treated the dumbbells as point objects with some mass $M$ then the initial energy would be:

$$ E_i = \tfrac{1}{2}M_1v_{1i}^2 + \tfrac{1}{2}M_2v_{2i}^2 $$

After the collision we have to count the rotational energy $\tfrac{1}{2}I\omega^2$ as well, so the final energy is:

$$\begin{align} E_f &= \tfrac{1}{2}M_1v_{1f}^2 + \tfrac{1}{2}M_2v_{2f}^2 \\ &+ \tfrac{1}{2}I_1\omega_1^2 + \tfrac{1}{2}I_2\omega_2^2 \end{align}$$

And of course conservation of energy tells that the initial and final energies must e equal $E_i = E_f$. But suppose we ignored the rotational kinetic energy and argued that it was just internal energy, i.e. heat, and shouldn't be counted. In that case we'd have:

$$ E_f = \tfrac{1}{2}M_1v_{1f}^2 + \tfrac{1}{2}M_2v_{2f}^2 $$

and we would conclude that $E_f \lt E_i$ so we had an inelastic collision.

Real objects are made up of huge numbers of atoms and molecules, not just two, but the same argument applies. We ignore all the internal motions and just describe them as an increase in temperature. That's why the collision looks inelastic.

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  • $\begingroup$ @JR I knew this aspect which I have seen earlier in some texts. I will make clearer which aspect I am having a doubt in. $\endgroup$ – N.S.JOHN Jan 2 '16 at 11:29
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during inelastic collision the body doesn't retard or accelerate constantly to zero velocity and then reverse direction. Consider it is a linear collision. Is my logic correct?

I think the term "inelastic" signifies the idea that in such interactions the bodies involved carry such 'events' which are energy

dissipation processes and thus the energy before and energy after the collision can not be equated.

Its not a priory condition that bodies involved will have retarding forces and the direction of motion will be reversed.

Therefore the above statement seems not to be correct.

There can be situations in which inter molecular forces and the structure of the colliding bodies get involved in 'inelastic deformations' which leads to non conservation of energy.

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