Kinetic energy conservation in collision.
In any type of collision type of collision if we take the system to comprised of both bodies then the net external force is zero. So the work done by external force is also zero. By the relation $W=K_f-K_i$ as $W$ is zero hence kinetic energy is conserved. But this not the case.
Can you tell me what is the flaw in my argument? Some people say that the argument has neglected the fact that some kinetic energy is transformed into heat. But heat is created due to the interactions between bodies during collision which is thus an internal force and thus shouldn't affect kinetic energy.
Answer
The equation $W_\text{net,ext}=K_f-K_i$ is only correct when the only form of energy being transformed is kinetic. If you have other forms of energy that change value, this equation won't work.
If you want to look at the system of the two colliding objects, you are correct that $W=0$ (though for a slightly different reason than what you stated; the net force is zero, but this doesn't mean the net work by external forces is zero.) A more encompassing equation than the one you're using is
$$W_\text{net,ext}=\Delta E_\text{tot}=\Delta K + \Delta E_\text{thermal} + \Delta U_\text{potential}+\cdots.$$
So, since $W=0$, the decrease in kinetic energy is accompanied by an increase in other forms of energy, such as thermal energy, acoustic, etc. (not sure what the "etc." is actually).
To more directly address your concern, internal forces absolutely can and do affect kinetic energy, even if the work done by external forces is zero. But just knowing that $W_\text{net,ext}=0$ doesn't tell you how the internal energies transform; only that the total sum is constant.
(Above I assumed that heat $Q$ added to the system is zero.)
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