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I had the impression that conservation of momentum is valid only when we consider no loss of kinetic energy into heat etc. My tutor says that even if we have loss of kinetic energy as heat in a collision, we still have conservation of momentum. Is this true?

EDIT: Alright Bill, but tell me something please: We have two cars that collide and stay together. Or two balls of plasteline or mud or whatever. Obviously all the momentum of the system is lost. How do you say that always momentum is conserved? For this momentum conservation theorem to work, do we ASSUME that kinetic energy is conserved? Does it ONLY work for perfect elastic collisions? Please answer my specific question, because so far all the answers are about everything apart from what I am specifically asking! My specific question (in addition to the above) is: If in a collision there is a coefficient of restitution BELOW 1, doesn't that mean that the collision is INELASTIC? YES OR NO! And if that means that the collision IS INELASTIC, is it correct to use AT THE SAME TIME momentum conservation equation? YES OR NO!

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  • $\begingroup$ So you're thinking of equal-mass cars colliding at equal velocity and opposite directions? That same situation where the total momentum (vector sum of the individual momenta) is zero? How exactly is this a contradiction? $\endgroup$ – Emilio Pisanty Feb 3 '16 at 18:05
  • $\begingroup$ They dont have to be the same mass. It's just that their whole kinetic energy was converted into structural deformations. Is this IMPOSSIBLE to happen? $\endgroup$ – ergon Feb 3 '16 at 20:35
  • $\begingroup$ If they don't have the same mass, it is impossible for them to come to a complete stop, and the final momentum will be nonzero. $\endgroup$ – Emilio Pisanty Feb 3 '16 at 20:58
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    $\begingroup$ If you want to be taken seriously, then a slightly more respectful tone would not come amiss, and if you want us to address your (superficial) analysis then you should provide the analysis instead of unproven claims. However much force is used to deform the objects, the net force of one object on the other must still be the same, and equal to the reaction force on the other object. Hence if the masses are different the accelerations will be asymmetric and the joined bodies will not remain at rest. To be clear: there is no part of the forces which does not cause acceleration of the COM. $\endgroup$ – Emilio Pisanty Feb 4 '16 at 0:38
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    $\begingroup$ That analysis is too big for a comment, but if you rephrase the question (or post a new one) to make it clear that this is the issue, I'll be happy to post an answer. (I won't shoot a moving target, though: provide a clear question, and don't change it afterwards.) The short of it is that for the front of the car to get crumpled, it needs to be squeezed by the back, i.e. the back provides a forwards force on the front. By Newton's third law, the back exerts exactly that much force on the back. $\endgroup$ – Emilio Pisanty Feb 4 '16 at 16:22
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In a collision momentum is conserved if there are no external forces. Energy is also conserved but in the examples kinetic energy is an important parameter. Elastic collisions are ones in which kinetic energy is conserved. In non-elastic or inelastic collisions kinetic energy is not conserved and some kinetic energy can be converted into heat, sound and in doing work to permanently deform the colliding objects.

You can also have super-elastic collisions where the kinetic energy actually increases (and momentum is conserved). An example is a bullet being shot out of a rifle where the kinetic energy comes from the chemical energy of the propellant.

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  • $\begingroup$ You don't answer my question. In non elastic collisions where we have kinetic energy loss into heat etc, do we have momentum conservation???? $\endgroup$ – ergon Feb 2 '16 at 15:57
  • $\begingroup$ Sorry I thought I had answered your question. If there are no external forces momentum is conserved whether or not kinetic energy is conserved. $\endgroup$ – Farcher Feb 2 '16 at 16:01
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Momentum, energy, angular momentum, and charge are conserved locally, globally, and universally. One must remember that conservation locally (within a defined system) does not mean constancy. Constancy occurs only when the system is closed/isolated from the rest of the universe.

Conservation means that these quantities cannot spontaneously change. Let's consider momentum: the momentum of a system at a later time must equal the momentum at an earlier time plus the sum of the impulses applied to a system. The impluses in this sum could be adding or removing momentum from the system, but never creating nor destroying momentum: $$\vec{p}_{later}=\vec{p}_{before}+\Sigma\vec{J}_{during}.$$

For an isolated collision, without outside influence, $\vec{J}_{during}=0$, and $\vec{p}_{later}=\vec{p}_{before}$.

For the energy: $E_{later}=E_{before}+W+Q+\mathrm{radiation}$

For angular momentum: $\vec{L}_{later}=\vec{L}_{before}+\Sigma\vec{\Gamma}_{outside}$ ($\Gamma$ is torque on system)

For charge: $ Q_{later}=Q_{before}+\int I\;\mathrm{d}t$

In the case of kinetic energy, it is not universally conserved. It can appear and disappear as energy is transformed to different manifestations:heat internal energy, gravitational, electromagnetic, nuclear, all of which are energy. The total energy is conserved in a system (not necessarily constant), with the transfer agent being work/radiation/heat. The elastic collision is defined to be one in which the kinetic energy of the system remains constant.

Note that if you define a single object as the system of interest, neither the momentum nor the kinetic energy will remain constant during a collision with another object or while it falls in a gravitational field, but the momentum will be conserved (the object is subjected to impluses) and the energy of the object is conserved (outside forces do work).

Bottom line: Define a system, look for transfers of momentum (impulse), energy (work, etc), angular momentum (torque), and charge (current) into or out of the system. Then see if any of those conserved properties are also constant for your situation.

EDIT - Response to OP specific questions:

My specific question (in addition to the above) is: If in a collision there is a coefficient of restitution BELOW 1, doesn't that mean that the collision is INELASTIC? YES OR NO!

Yes. One may also call it partially elastic. If the coefficient of restitution is zero (0), the collision is completely inelastic.

And if that means that the collision IS INELASTIC, is it correct to use AT THE SAME TIME momentum conservation equation? YES OR NO!

Yes. Momentum is conserved in all collisions and explosions. And sometimes it might even be constant for short periods of time.

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  • $\begingroup$ Isn't part of the impulse used to create plastic deformation and other things that are not fully reversed by a reaction? $\endgroup$ – ergon Feb 2 '16 at 17:12
  • $\begingroup$ Impulse changes momentum (by definition). You can't partition it and say "some of the impulse doesn't change momentum." Work creates deformation. Both of them are related to force. Impulse is the integral of force w.r.t. time. Work is the integral of force w.r.t. distance. Work and impulse are two differing effects of force. $\endgroup$ – Bill N Feb 2 '16 at 21:06
  • $\begingroup$ We have two balls of plasteline colliding and sticking. Obviously all the momentum of the system is lost. How can momentum is always conserved? For this momentum conservation theorem to work, do we ASSUME that kinetic energy is conserved? Does it ONLY work for perfect elastic collisions? My specific question (in addition to the above) is: If in a collision there is a coefficient of restitution BELOW 1, doesn't that mean that the collision is INELASTIC? YES OR NO! And if that means that the collision IS INELASTIC, is it correct to use AT THE SAME TIME momentum conservation equation? YES OR NO! $\endgroup$ – ergon Feb 3 '16 at 20:23
  • $\begingroup$ No, the momemtum of the system is not lost. Momentum is a vector quantity, so direction matters. If your plastic balls have equal but opposite momenta, the net momentum is zero, so if they collide and stick, the result will stop. Kinetic energy is NOT conserved in that case. Conservation of momentum works for all interactions. Sometimes it remains constant. If a system is not under the influence of outside forces, the momentum will be constant. $\endgroup$ – Bill N Feb 5 '16 at 16:17
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I had the impression that conservation of momentum is valid only when we consider no loss of kinetic energy into heat etc.

Nope.

My tutor says that even if we have loss of kinetic energy as heat in a collision, we still have conservation of momentum. Is this true?

Yup.

EDIT: Alright Bill, but tell me something please: We have two cars that collide and stay together.

P = mv

But that P is all of the mv's in the system. So...

Lets say, to keep the math simple, that both cars mass Mc and are travelling at speed Vc towards each other. So then in that case, before the collision you have:

P = (McVc) + (MC(-Vc)) = 0

and after, it's:

P = (Mc0) + (Mc0) = 0

Momentum is conserved, it was zero before and zero after.

Wait, you cry, where did that negative sign come from? Well, what's the definition of velocity, as opposed to speed?

Mystery solved!

As always, it's in the definitions where you have to be careful. Feel free to apply it to your other examples - squishy balls and such. They'll all be the same.

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