Total Momentum in the Center of Mass Reference Frame For solving problems involving elastic collisions it is useful to use the center of mass reference frame as to avoid quadratic equations. However, I am confused about why the total momentum in the center of mass reference frame is equal to zero. I understand that the change in momentum would be zero, but why is the total momentum equal to zero in this frame?
 A: Just call it the center of momentum frame instead and it will seem less mysterious.
If you take the energy-momentum four-vector for each part of the system and add them up, you get a total energy-momentum four-vector. It will be timelike. And in the frame of that vector you get zero total momentum.
It's a useful frame. And calling it the center of mass frame just confuses the issue. Technically the center of mass frame is the center of momentum frame with a specific choice of origin, but an origin hardly matters for anything.
A: 
However, I am confused about why the total momentum in the center of mass reference frame is equal to zero.

The center of mass frame of a system of particles is defined as having two key qualities. Less important is that the center of mass at some time $t_0$ is located at the origin. Much more importantly, the time derivative of the center of mass at that time is the zero vector. It is always possible to define such a frame in both classical mechanics and in quantum mechanics. (Note well: This is not so well definable in general relativity, at least on the scale of the universe. I'll leave that aside.)
If no external forces act on a system of particles, and if conservation of linear momentum holds true, then by definition, the center of mass of that system of particles moves at a constant velocity. It is extremely convenient to choose a frame in which that constant velocity is zero. The center of mass frame, by definition, satisfies those constraints (assuming those two very big ifs).
A: It's best to consult the defining equations. The velocity of the center of mass (in any frame) is defined as
$v_{cm} = \frac{1}{M}(\sum_{i=1}^{n} m_iv_i$)
Therefore $Mv_{cm} = \sum_{i=1}^{n} m_iv_i = P_{tot}$.
The center-of-mass frame is defined as the frame in which the center of mass is at the origin. Therefore, in this frame, $v_{cm} = 0$, by definition. But by the above expression, that means that, in this frame, $P_{tot} = 0$, again, by definition.
So the fact that the momentum is zero in the center-of-mass frame is not some interesting fact that needs explanation. It is just an artifact of how we define this particular inertial reference frame. Albeit often a convenient one.
