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It is possible to conserve momentum without conserving kinetic energy, as in inelastic collisions. Is it possible to conserve the total kinetic energy of a system, but not its momentum? How?

To clarify, I am not necessarily talking about an isolated system. Is there any scenario which we could devise in which momentum is not conserved but kinetic energy is?

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  • $\begingroup$ Did you want angular momentum not conserved, too? $\endgroup$ – JdeBP Dec 18 '19 at 14:19
  • $\begingroup$ Would a moving charge in a magnetic field work for you as an example? $\endgroup$ – Nathaniel Dec 18 '19 at 15:18
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    $\begingroup$ The trick here is to note that momentum is a vector while KE is a scalar. By requiring KE to be constant, you're pinning the magnitude of the momentum vector, but you still have a degree of freedom in the vector's direction. $\endgroup$ – Dancrumb Dec 18 '19 at 21:44
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    $\begingroup$ It seems you mean a closed system. If so, consider editing to clarify that $\endgroup$ – Ben Dec 19 '19 at 10:36
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    $\begingroup$ Bounce (in an inelastic collision) the object into another of equal mass and velocity moving in the opposite direction. $\endgroup$ – Hot Licks Dec 20 '19 at 1:43
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In order for momentum to be conserved, it must be the case that $$\mathbf F_\text{net}=\frac{\text d\mathbf p}{\text dt}=0$$

In order for kinetic energy to be conserved, it must be the case that $$\text dK=\text dW_\text{net}=\mathbf F_\text{net}\cdot\text d\mathbf x=0$$ at all instants in time.

So, is there a case where the net work done on an object is $0$, yet there is still a net force acting on the object? The answer is yes! We just need $\mathbf F_\text{net}\neq0$ to be perpendicular to the path of the object at all times. A simple example of this is an object undergoing uniform circular motion. The object's kinetic energy is not changing (as its speed remains constant), yet the momentum is constantly changing due to the non-zero net force.

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    $\begingroup$ I was coming to discuss the Earth-moon system (or any orbitting cluster) as an example of this cases. $\endgroup$ – dmckee --- ex-moderator kitten Dec 17 '19 at 19:31
  • $\begingroup$ Well, but an isolated object can't be undergoing uniform circular motion - it can do only because it's attracted by something else, but it's equally attracking that something, changing its momentum. If it's in circular motion because it's orbiting something, then the momentum of the total system (e.g. earth-moon) would be conserved. So it's pretty much a tautology "momentum of a system can be changing if it's caused by an interaction with another system changing its momentum in exact opposite way". $\endgroup$ – Peteris Dec 18 '19 at 11:59
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    $\begingroup$ @Peteris couldn't a single rotationally symmetric object spinning about its center of mass be undergoing uniform circular motion? $\endgroup$ – UuDdLrLrSs Dec 18 '19 at 12:13
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    $\begingroup$ @Peteris I never said anything about a single isolated object. Furthermore, what you define as "the system" is subjective. I have defined my example system as an object experiencing an external force causing the object to undergo uniform circular motion. You can always widen your scope to include enough to say everything is conserved, but that doesn't mean everything is conserved for all systems. $\endgroup$ – BioPhysicist Dec 18 '19 at 13:17
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    $\begingroup$ @DaveInCaz At that point the system is all of the particles that make up that body. The total momentum would then be constant as no net force is acting on the spinning body. $\endgroup$ – BioPhysicist Dec 18 '19 at 21:01
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Suppose that energy is conserved in one frame of reference, and you want it to be conserved in all other frames as well. Conservation of momentum is exactly the condition you need in order to make this happen in all frames.

To see this, consider what happens when you change to a different frame of reference, $v\rightarrow v+u$. Then all kinetic energies transform according to $K\rightarrow K+muv+\text{const.}$ (Potential energies don't change under this transformation.)

Let's say we write your question as a conjecture: --

If energy is conserved and total KE is conserved, then momentum is conserved.

Then your conjecture can actually be strengthened to read: --

If energy is conserved, then momentum is conserved.

(This is implicitly assuming that we want all frames of reference to be valid.)

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    $\begingroup$ I thought the OP was asking about conserving KE and not conserving momentum. $\endgroup$ – BioPhysicist Dec 17 '19 at 19:28
  • $\begingroup$ Ah, this is nice. The example that @Aaron and I were thinking of has an implicit choice of frame: one where the motion is perpendicular to the force. $\endgroup$ – dmckee --- ex-moderator kitten Dec 17 '19 at 19:33
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    $\begingroup$ @AaronStevens: My phrasing of the conjecture is just the logical negation of what they're asking about. $\endgroup$ – user4552 Dec 17 '19 at 21:08
  • $\begingroup$ I guess I wasn't viewing their conjecture as an "if-then statement", but yes, now I see the equivalence. $\endgroup$ – BioPhysicist Dec 17 '19 at 21:42
  • $\begingroup$ Isn't the example of the Force being perpendicular to the velocity in 3D Coordinates Contradicting your statement (which was just derived in 1D)? - Ok nevermind :) - I just figured out that this example conserves kinetic Energy only in a special class of frames of reference. $\endgroup$ – Quantumwhisp Dec 18 '19 at 6:14
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Is there any scenario which we could devise in which momentum is not conserved but kinetic energy is?

When a ball bounces off the ground or a wall. The momentum is flipped but the kinetic energy stays about the same.

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    $\begingroup$ If we look in detail the KE is restored, rather than preserved - in mid bounce the KE is zero and the energy is stored as elastic potential in the ball and/or surface. $\endgroup$ – bdsl Dec 18 '19 at 11:09
  • $\begingroup$ You've neglected the effect of the ball on the ground. Try doing the math where you track the momentum of the Earth as well as the ball before and after this interaction; do you still conclude that the momentum of the whole system has "flipped"? Or is it rather the case that momentum is conserved for the whole system, and we have just transferred some momentum from the ball to the Earth? $\endgroup$ – Eric Lippert Dec 19 '19 at 1:14
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    $\begingroup$ @EricLippert Of course, but the OP asked for a system where momentum was conserved and energy wasn't. So I chose the system of the ball, which is a common and good choice. If you insist on zooming out all the way, then in classical mechanics both momentum and energy are always conserved -- which makes the question rather pointless. $\endgroup$ – knzhou Dec 19 '19 at 1:17
  • $\begingroup$ @knzhou at least when I learned, interactions which cross system boundaries invalidate conservation laws. So taking the ball as a system and introducing an interaction external to the system which changes the momentum of the system voids the question. You may as well have other things crossing the boundary, say going from one flying duck to four ducks flying at half the speed. $\endgroup$ – Pete Kirkham Dec 19 '19 at 13:52
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If non-isolated system are of interest, then what you’re looking for is an external force that does no work.

  • The central force in a circular orbit: the satellites energy in unchanged, but its momentum is continuously changing.

  • An electron moving across a constant magnetic field: ditto

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Yes, it is possible to conserve the total kinetic energy of the system but not the momentum.

Let me give you an intuitive explanation.

  1. Suppose we have two charges, in which one is fixed and other is free to move and then they are released at some distance. In this case since one charge is fixed so the net external force on the system is not zero, so momentum of the system will not remain conserved, but kinetic energy + potential energy of the system will remain conserved, as there are no dissipative non conservative forces involved in the system.

  2. Suppose a particle is tied to a point with the help of a string and is performing uniform circular motion on the horizontal surface. In this case the kinetic energy of the system will remain same but not the momentum.

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