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Why does conservation of energy break if the collisions are not elastic. And why conservation of momentum works in both cases?

If this is the case, how do we derive conservation of energy then? Because we don't make such assumptions before deriving it.(Atleast my textbook doesn't do so)

PS: Although, there are answers for the first part, no one explains the second part.

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  • $\begingroup$ What textbook is that? And what assumption is not made? $\endgroup$
    – nasu
    Feb 3, 2023 at 16:33

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It is not "conservation of energy" that breaks during non-ellastic collisions. It is just that the kinetic energy is not conserved. There is no such thing as conservation of kinetic energy so I doubt that your textbook is deriving conservation of kinetic energy. In general, in any process the energy is converted between different forms (kinetic, potential, thermal, electromagnetic radiation etc). Only the total energy is conserved, if the system is isolated from the outside word. During a collision some of the kinetic energy is converted into potential energy and some other forms (sound, thermal). Only in very special (ideal actually) cases the only conversion is from kinetic to potential only and then all this potential energy is converted back into kinetic energy. We call these special cases elastic collisions. So, you may say that the kinetic energy by itself is not conserved because it is not the only form of energy and it can be converted to other forms. Momentum does not have this "option". There is only one momentum. Technically, the difference comes from the fact that internal forces in a system cannot change the momentum but can change the kinetic energy. This is what happens in collision: the internal forces between the two objects cannot change the total momentum but can change the kinetic energy. Both these behaviours result from Newton's laws applied to a system, with no other assumptions.

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assume two mass collied

the conservation of the linear momentum is the conservation of the center of mass velocity bevor and after the collision , no external forces are applied.

the equation

$$\frac{m_1\,v_1+m_2\,v_2}{M}=\frac{m_1\,u_1+m_2\,u_2}{M}\quad\Rightarrow\\ m_1(v_1-u_1)+m_2(v_2-u_2)=0\quad \text{conserved !}$$

the total energy is

$$2\,T=m_1\,(v_1^2-u_1^2)+m_2\,(v_2^2-u_2^2)= m_1\,(v_1+u_1)\,(v_1-u_1)+m_2\,(v_2+u_2)\,(v_2-u_2)$$

and with the momentum conservation you obtain

$$2\,T=m_2\,(v_2-u_2)\,\underbrace{(v_2-v_1+(u_2-u_1))}_{=0}$$

thus the total energy is also conserved

in case of elastic collision

$$2\,T=m_2\,(v_2-u_2)\,\underbrace{(v_2-v_1+\epsilon\,(u_2-u_1))}_{\ne 0}$$

where $~\epsilon~,(0\le \epsilon\le 1) ~$ is per definition

$$\epsilon=-\frac{v_2-v1}{u_2-u1}$$

so only for inelastic case $~\epsilon=1~$ the total energy is conserved

  • $~u_i~$ velocities bevor the collision
  • $~v_i~$ velocities after the collision
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For your first question: technically speaking, conservation of energy always holds - it is a fundamental law of the universe. When your textbook says that it doesn't hold in a certain case, they are really making an approximation in which they try to account for the effects of the environment on the collision. So in the case of an inelastic collision, some of the energy is "carried away" in the form of sound waves, heat, and other effects that happen when the two bodies crash into each other. This happens to a lesser extent for elastic collisions too, but it is considered fairly negligible. And as to your second question: once again, conservation of momentum is not something that we can turn on or off in certain cases, but a fundamental law that we can always apply. The difference is that while energy can be "carried away" by the air or the environment, momentum really can't be - the only way that momentum could be removed from the system of the two bodies is if the air were to carry away in the form of wind, but this effect probably wouldn't be appreciable.

I'm guessing this isn't quite as satisfying as the answer you were looking for, but this kind of approximation is common in a lot of physics equation. Just keep in mind that both CoE and CoM hold in all cases when applied to the entire system (that is, the bodies + the air molecules around them + any other variables we need to account for). But if we are just interested in the motion of the two bodies, sometimes they will appear to violate one of these equations because of their interactions with the environment. A similar situation comes up in quantum mechanics when the degree of quantum coherence in a system can be "carried away" by its interactions with the environment.

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