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Let's assume that we have an isolated system, if two object collide they will both exert equal and opposite forces on each other, they will both exert these forces for the same distance, hence they both did the same amount of work on each other. If $\vec{F}_{net} \cdot \mathrm{d}\vec{x}$ is equal to the change in kinetic energy, therefore both objects will experience the same net force (referring to the fact that this is an isolated system) for the same distance, now with this in mind how is it possible for an inelastic collision to occur if one object gains the same amount of kinetic energy as the other loses that exact same amount?

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  • $\begingroup$ It doesn't. Some energy is lost to e.g. deforming the material or heat or something else which makes this inelastic $\endgroup$ – Steeven Apr 27 '17 at 10:33
  • $\begingroup$ But would that not go against the mathematics which I have presented ? $\endgroup$ – LM26 Apr 27 '17 at 10:36
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    $\begingroup$ Duplicate physics.stackexchange.com/q/288835 and physics.stackexchange.com/q/93739 and . . . . . . $\endgroup$ – Farcher Apr 27 '17 at 10:38
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    $\begingroup$ You are mixing up "the real world" with an idealized system consisting of two perfectly rigid bodies. In the real world, the collision can transform some of the mechanical (kinetic) energy into other forms like heat, electromagnetic radiation, etc. If the bodies can deform (and no real-world body is perfectly rigid) some of the KE can end up as internal vibration in the body, not as "movement of its center of mass". In fact the definition of "inelastic collision" is simply that "mechanical energy is not conserved." $\endgroup$ – alephzero Apr 27 '17 at 13:16
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The law of conservation of energy is about the total energy in the isolated system, not just the kinetic energy.

It is the total kinetic +potential+ radiative energy that is conserved.

For example, one of the balls hitting the other may get stuck on a high shelf. One has to include the gravitational potential energy it acquired when it reached the shelf, in addition to the adhesion energies of molecules that kept it there, or transfer to vibrations of the shelf and all the other energy forms in the discussions above ,plus any loss of energy in radiation due to triboelectric effects.

If one goes in systems where special relativity has to be used, part of the energy can turn into mass.

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  • $\begingroup$ A much better answer than mine. $\endgroup$ – Apoorv Khurasia Apr 27 '17 at 11:01
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It is true that they exert equal and opposite forces on each other but the work done by them on each other may not be converted to Kinetic Energy. In inelastic collisions, some (or all) energy is ultimately lost as heat, sound, or other forms (e.g., in making permanent deformities on one or both of them).

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If $\vec{F}_{net} \cdot \mathrm{d}\vec{x}$ is equal to the change in kinetic energy, therefore both objects will experience the same net force (referring to the fact that this is an isolated system) for the same distance, ... how is it possible for an inelastic collision to occur?

The answer is that the distance is not the same; nothing requires that the objects move the same distance in the collision.

To see why this is the case, suppose that two objects of unequal masses $m_1 < m_2$ collide with each other; suppose further that $m_1$ is moving with speed $+v_0$ initially, while $m_2$ is at rest. For simplicity, let's also assume that the magnitude of the force $F$ between the objects is constant over some short period of time $\Delta t$. The acceleration of $m_1$ during this time is $-F/m_1$; so the distance it travels during the collision will be $$ \Delta x_1 = v_0 \Delta t - \frac{F}{2 m_1} (\Delta t)^2. $$ The acceleration of $m_2$ is $+F/m_2$, and so the distance it travels will be $$ \Delta x_2 = \frac{F}{2 m_2} (\Delta t)^2, $$ There is no particular reason that $\Delta x_1$ and $\Delta x_2$ need to be the same; and so although the force acting on each object is the same over the time period of the collision, the work done on them is not.

Conservative forces evade this calculation by necessarily being time-dependent. In any collision, the rate at which mechanical work is being done on each object (i.e., the mechanical power imparted) is not necessarily the same at any given time. However, conservative forces have the ability (by definition) to balance things out, so that the time integrals of the power imparted in each case end up being equal.

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  • $\begingroup$ How can they not move the same distance during a collision , this is not matching up with my intuition , during the period over which they exert forces on each other , then both can be considered as one body , since the only time they exert forces on each other is if they are in physical contact ( interacting with one another ), then how can one body move two distances . Your math makes sense , it is just that I can't picture the situation . $\endgroup$ – LM26 Apr 27 '17 at 19:23
  • $\begingroup$ @LM26: You're assuming that the bodies are perfectly rigid. The problem is that bodies generally deform in a collision, and so different parts of the "composite body" you're envisioning will be moving at different speeds (and so move different distances in the same interval of time.) $\endgroup$ – Michael Seifert Apr 27 '17 at 19:53
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Some of the energy is transformed into thermal energy.

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  • $\begingroup$ ... or others, such as acoustic energy etc. $\endgroup$ – ZeroTheHero May 2 '17 at 23:45
  • $\begingroup$ all energy "lost" during an inelastic collision is transformed into thermal energy after reaching a steady mechanical state $\endgroup$ – numberspie May 4 '17 at 0:37

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