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[This question is inspired by an astute observation from a student of mine.]

When we discuss conservation of momentum, students often ask, "When is momentum conserved?" And the lazy, mechanical response is often, "Momentum is always conserved."

A thoughtful student might then reply, "But what about falling objects? If I drop my pen, it accelerates towards the ground and gains momentum. Clearly, momentum is not always conserved."

This is a fair criticism. But one with an easy answer: "You aren't taking into account the momentum of the earth," we say, "As the earth pulls down on the pen, the pen (by Newton's Third Law) pulls up on the earth, accelerating it. Therefore, the downward momentum of the pen is cancelled out by the upward momentum of the earth.

"The more correct statement," we conclude, "is that momentum is always conserved in a closed system - namely, in a system containing all of the equal-and-opposite force pairs. None of the forces are allowed to cross the boundary of the system (like the weight of the pen does when you watch it fall without considering the earth)."

"Alright," the student concedes, "I'll buy that. But even when we neglect gravity, why doesn't momentum always appear to be conserved?"

"How do you mean?"

"I mean this: imagine floating in empty space - just you and a baseball. You look around and notice that you have zero momentum. And why shouldn't you? After all, you are in your own center of mass reference frame (where momentum is always zero).

"But then you throw the baseball," the student continues, "and it begins to move away from you with some momentum. However, from your point of view, you still are not moving. The system of you and the baseball is closed (there are no external forces), so momentum should be conserved. Yet before you threw the ball, momentum was zero, and afterwards, it is non-zero.

"What happened? Why was the momentum of this closed system not conserved?"

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  • $\begingroup$ My only comment is that it took me a couple of passes to correctly parse "in a participant's frame of reference" and thereby make sense of the question (and see the answer). But I'm not sure how to improve the situation and retail the proper brevity in the title. $\endgroup$ – dmckee Oct 17 '17 at 2:33
  • $\begingroup$ @dmckee Yeah, it was the best I could come up with. Maybe "in the frame of reference of one of the colliding objects"? It's certainly clearer that way. I'm happy for it to be edited to improve clarity. $\endgroup$ – Geoffrey Oct 17 '17 at 2:38
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Awakening from from our stupor, we stroke our collective chin for a moment before realizing the answer.

"You're not in an inertial reference frame," we reply. "Remember when we first started discussing Newton's Laws, and we talked about inertia? It's basically the tendency of an object to not change its current state of motion. Well, I may have neglected to mention that Newton's Laws (particularly, #2: $F=ma$) are only true in what are called 'inertial reference frames.'

"A reference frame is an observer's perspective, his or her vantage point. And an inertial reference frame is a special kind of perspective in which the observer's motion is not changing (he has succumbed to his inertia, if you like). The reference frame that you chose in this problem starts out as inertial (before throwing the ball) and is inertial at the end (after throwing the ball), but in the middle (while throwing the ball) it is not inertial because your (read: the observer's) motion is changing.

"Broadly, Newton's Laws don't apply in non-inertial (or accelerating) reference frames, so you can't use them to make sensible deductions without being very, very careful. So the truly final and correct rule about Conservation of Momentum is this:

"Momentum is always conserved in closed systems from the perspective of inertial observers."

And I hope I haven't missed anything this time!

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  • $\begingroup$ Momentum is always conserved because it is exchanged between objects. The momentum of a closed system is constant, which is a special case of conservation. If make a system small enough (e.g., a single object) the momentum will change during any collision, but it is still conserved because true conservation involves a continuity equation. For momentum it is $\vec{p}_{final}=\vec{p}_{initial}+\int \vec{F} dt$. The impulse integral is the exchange of momentum with an outside system. Momentum is never created or destroyed. It is transferred. ALWAYS conserved. $\endgroup$ – Bill N Oct 17 '17 at 3:00
  • $\begingroup$ @BillN Granted. If you choose a large enough system such that no forces are ever external to that system, then momentum is always conserved. That said, "conservation" (as an English word) general means something like "preserving what is there." I think it is disingenuous to argue that calculating an impulse integral to determine the change in momentum due to external forces constitutes "conservation." By its very nature, a conserved quantity is constant. It's not a special case; it's the only case. Also, I wouldn't recommend trying to split hairs like that with non-major undergrads. $\endgroup$ – Geoffrey Oct 17 '17 at 3:08
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    $\begingroup$ Geoffrey, some authors of introductory texts have been making a distinction between conserved (a global absolute) being constant in a particular system (a local property specific to the system selected and what interaction it is currently experiencing). When I was taught we used "conserve" for both cases and had to distinguish between the global and local sense explicitly. But @Bill is older than me, so this may be a thing that has been slow to catch on, or it may have fallen by the wayside only to be re-introduced. $\endgroup$ – dmckee Oct 17 '17 at 5:48
  • $\begingroup$ @dmckee Hmm, I see. I had no idea that this was a pedagogical issue. Thanks for the context. $\endgroup$ – Geoffrey Oct 17 '17 at 5:55
  • $\begingroup$ Many (most?) axiomatizations of classical mechanics define an inertial frame to be a reference frame in which Newton's Second Law holds—or, equivalently, momentum is conserved. This makes this answer tautological. (Not that it's in any way false; it's the answer I would have given myself.) $\endgroup$ – Michael Seifert Oct 17 '17 at 13:46
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But then you throw the baseball," the student continues, "and it begins to move away from you with some momentum.

yes, and and you are moving away from it with equal and opposite momentum. You are in the center of mass system of "ball +you"

However, from your point of view, you still are not moving

You are mixing two systems. You start with a center of mass (CM) of "you+ball" at rest, and make a sudden transformation from that to your at rest system.

The system of you and the baseball is closed (there are no external forces), so momentum should be conserved. Yet before you threw the ball, momentum was zero, and afterwards, it is non-zero.

The transformation to your CM will be d(p) dependent ,from p the equal and opposite momentum to the ball in the overall cm, to zero , in your cm, and thus it is not a transformation between two inertial frames.

The apparent non conservation of momentum comes from mixing the two frames of CM.

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