I bow to the immortal law of conservation of momentum but consider my argument below and see if you can disprove it:

Example 1: Momentum cannot be conserved during an inelastic collision. When two bodies undergoing inelastic collision, they loose kinetic energy which might go into heat, light etc. Consider the situation in which the kinetic energy goes into heat energy, speeding up the molecules of air, consequently increasing its temperature. (Thermodynamics tells us that during transfer of energy (heat), temperature rises because of increase in the movement of molecules.) This means molecules are moving faster than they were before the collision. So they must have been transferred some momentum from the two bodies.

Putting the whole story together: " In inelastic collision, momentum in not conserved since some energy is dissipated and so should the momentum be not the same all the time."

Due to the ambiguity in the above example, I have presented another one much better than the previous one:

**Example 2: Consider an isolated system consisting of two bodies. They collide with each other inelastically and some kinetic energy is turned into vibrational energy of the atoms (of one of the bodies), causing a heating effect. Now in order for the vibrational energy to increase some momentum must have to be either gained or lost to those atoms. Isn't it? So, how should the momentum of two bodies be conserved without taking into account the momentum lost or gained to atoms of the two bodies?**

I need to rule out this argument. Where am I wrong?

(Note: I am not asking you to prove that momentum remains conserved, I fully agree with it but rather to disprove my above argument.)

Thanks :)

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TopNewestIn physics, systems are very important. Without knowing what your system is, applying any conservation laws makes no sense.

Let's consider a simple situation: two rigid bodies colliding in vacuum, elastically. Kinetic energy is conserved. Bodies have same total kinetic energy before and after the collision. Momentum is conserved. Bodies have same total momentum before and after the collision. It's an ideal collision: absolutely no energy is dissipated.

In this system, we had two bodies. That's it. Nothing else. No air molecules, no friction. Even the bodies we are considering are rigid: their particles do not vibrate or rotate. They're just sitting there, moving along with their centre of mass. Here, there is no question about the validity of any conservation laws. Where that becomes a problem is at the atomic level.

A slightly more complex situation: two rigid bodies colliding in the air (neglect gravity). Here, air molecules are present and

interactingwith the two bodies we took as our system the last time. If we consider (and mind you, this is an arbitrary choice) the air molecules to beoutsidethe system, then momentum isn't conserved for the two-body system. Somehow, when you sum the initial and final momenta, they don't tally. When you sum the initial and final kinetic energies, they don't tally either. So you give up physics and start selling coconuts in Hawaii. What do we do now? Where has the momentum gone? Where has the energy gone? Should you sell beverages instead?Now, listen carefully:

If you'd wanted to study conservation of momentum/energy, you should have taken the entire region to be your system: the two bodies, air particle 1, air particle 2, air particle 3, coconut 475, etc (assuming, of course, that these entities interact with the colliding objects). And then, when the two bodies collide, you sum up the individual momenta for body 1, body 2, air particle 1, 2, 3, 4 and the coconut (every single object/particle interacting with the colliding objects). And voila, momentum is conserved.

Do the same thing for kinetic energy. But the thing is, here, you can't actually see the kinetic energy of the particles: the exotic dance of rotation, translation and vibration. If you had really good eyes, maybe you could, but if you don't, you calculate their kinetic energies differently: by heat. The macroscopic effect of microscopic motion is heat. If you sum up the kinetic energies of the bodies (easy, because they're rigid) and the kinetic energies of the particles 1, 2, 3, 4 and the coconut (by using heat energy, measured as temperature), that quantity will be conserved. In one of the initial chapters of Feynman's Lectures, Richard Feynman talks about energy and Dennis the Menace (wonderful cartoon!). Check that out.

If you zoom down at the atomic scale and look at the momenta of the individual particles as well, you'd get a conservation. It's just that you can't

actuallysee the particles to calculate their momenta. If you could calculate them (maybe not in the classical way, but some quantum mechanical way; I have no clue how one would go about doing that), then you would definitely see momentum conserved.DISCLAIMER: This is according to the traditional view of mechanics I've heard many times, and may not be entirely accurate in light of ground-breaking research. Nonetheless, it would be a good thing to read it in any case. Also, coconut fluid can be used as an IV drip. Hence, carry a coconut with you wherever you go. Warning: that is not professional medical advice. – Raj Magesh · 3 years, 9 months ago

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First, momentum isn't used up. That doesn't even make sense since it is a vector quantity not scalar. Second, the molecules aren't part of your initial system so it makes no sense to talk about them. Third, even if you do consider the molecules as part of your system, the random directions of motion will cause their momentum vectors to effectively cancel out. – Yash Farooqui · 3 years, 9 months ago

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– A L · 3 years, 9 months ago

This has the least waffle of all answers here, +1. It may be worth noting that the momentum vectors cancel because of Newton's 3rd law, though.Log in to reply

– Lokesh Sharma · 3 years, 9 months ago

Why doesn't it makes sense to consider molecules those aren't in the system?Log in to reply

– A L · 3 years, 9 months ago

The total momentum of an isolated system is constant, the momentum of a system interacting with something else is not. Example: an object in a gravitational field does not have constant momentum as its velocity is not constant. However, if you include in the system the momentum of the massive body inducing the gravitational field, the system's momentum will be constant as that massive body will be drawn slightly towards the smaller body, equalling the momentum change of the smaller body.Log in to reply

– Innocent Zaman · 3 years, 9 months ago

BrillienTLog in to reply

What is momentum?it is not the the sum of all momenta of individual particles due to temperature.when temperature is raised ,particles momenta will increase in all possible direction .so net change in momenta is 0.so momenta inclement of particles due to temperature increase is zero.because it will increase momenta in all possible direction. – Hare Krishna Physics · 3 years, 9 months ago

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The molecules are part of the body, the momentum of the body is just the sum of the momentum of its particles. In a hot body, the momenta of molecules a higher, but they still a summed to the same overall momentum, corresponding to the motion of the body as a whole. – Aleksey Korobenko · 3 years, 9 months ago

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The heat probably moves in pretty random directions, so, since momentum is a vector quantity, the sum of the momenta of the particles due to heat should be pretty negligible. – Benjamin Chen · 3 years, 9 months ago

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– Lokesh Sharma · 3 years, 9 months ago

This seems plausible but I just wish that there be better reason as this could lead to inconsistent result sometimes. Think of a gun releasing heat opposite to the direction of travelling bullet instead of releasing heat in all directions. If we calculate speed of bullet using conservation of momentum of the two bodies that is gun and bullet and neglecting the effect of heat released at the back, we would get wrong results. Further if you are correct it would imply momentum is not conserved in inelastic conditions.Log in to reply

You can't take a thermodynamic objection to conservation of momentum (one that as I pointed out makes no sense since it's unclear what it would mean for momentum to be "used up") and then complain that it fails in situations where thermodynamics doesn't work. It's just bad logic.

It's true that conservation of momentum doesn't ALWAYS hold (despite it being postulated in Newton's third law) but in this simple case it does. And if you extend your conception of a field to having momentum, the law of conservation of momentum can be saved even in those circumstances. – Yash Farooqui · 3 years, 9 months ago

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– Yash Farooqui · 3 years, 9 months ago

Seriously what do you mean by "so should the momentum be dissipated"? Does the magnitude decrease? I honestly have no idea what that means for a vector quantity.Log in to reply

– Lokesh Sharma · 3 years, 9 months ago

I got your point. You are right. The increase in temperature is result of increase in kinetic energy of molecules not momentum which I was assuming throughout. Momentum is a vector quantity though I treated it like a scalar. Thanks to you, the argument is resolved now.Log in to reply

– Hamdan Ridwan · 3 years, 9 months ago

What he is trying to say is, I guess, Kinetic energy is getting dissipated.Log in to reply

sorry my horrible english, I am Brazilian, but, the moment is only conserved in isolated systems ,however, the system above is not isolated since there is particles, so the moment is transfered to to them. – Israel Oliveira · 3 years, 9 months ago

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– Lokesh Sharma · 3 years, 9 months ago

I think you are right but please look at the other example at the top I just gave in which I have taken isolated system but there is still a problem.Log in to reply

– Paramjit Singh · 3 years, 9 months ago

That is the actual answer!!!Log in to reply

Momentum is a vector quantity. Since the motion of air molecules or the particles of the body is random their net momentum is zero so momentum of the system is still conserved. – Rohit Shah · 2 years, 10 months ago

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The atoms in the two bodies would vibrate both forwards and backwards in a uniform manner, which would result in a net momentum gain of 0. – Sam Iddles · 3 years, 9 months ago

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agree with aloshya l – Kaustubh Deshpande · 3 years, 9 months ago

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The answer is simple. You are talking about an inelastic collision, which by definition are those collisions in which kinetic energy of the system is not conserved. But you should also know that they are called ''inelastic" because that is what is the colliding bodies are: they do not come back to their original shape after the collision occurs, i.e. they are not elastic. Hence, ideally, there is no vibration in those bodies. (Vibration occurs when a body tries to regain its original shape after a small deformation.) What happens is that ALL the KINETIC ENERGY is used up to change the shape of the colliding bodies. The bodies remain in the new deformed shape.. But in an elastic collision, the bodies come back to the original shape even after a small deformation. – Hamdan Ridwan · 3 years, 9 months ago

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momentum is only conserved in a closed system with no external forces.. if there's air right next to your gun, then thats not quite a closed system – Jord W · 3 years, 9 months ago

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– Lokesh Sharma · 3 years, 9 months ago

Please take another look at the other example I just gave above in which system is isolated though still not clear if momentum is conserved in it or not.Log in to reply

The overall momentum of every molecule remains the same since heat, too, is actually the motion of particles. What do you mean by energy is dissipated? Even though it is dissipated total energy in a closed system does remain the same. – Yash Talekar · 3 years, 10 months ago

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– Lokesh Sharma · 3 years, 9 months ago

That's what I am saying. The overall momentum remains conserved but if you choose the two masses colliding as your system then momentum is lost to heat up the air molecules. The air molecules and rest of the world are not in our system. If you agree with this fact then that would imply momentum is not conserved in inelastic collision which is not the case. To find out the speed with which a bullet is shot from a gun we take into account the conservation of momentum of bullet and gun. How is this possible if momentum is being also lost to air molecules?Log in to reply