During a collision, why is momentum not conserved in a participant's frame of reference? [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?"
 A: 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! 
A: 
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.
