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I was watching one of Walter Lewin's lectures, in which he mentions that the total momentum of a system in the center of mass frame is always 0. I don't quite understand this. Here is an example which I'm trying to make sense of. Say we have two objects A and B that have the same mass. B is to the right of A and is at rest. A travels to the right at 2 m/s until it collides with B in a completely elastic collision. Before the collision, since A moves to the right with a speed of 2, so does the center of mass (according to my understanding, which may be incorrect). So A is stationary relative to the COM. B, on the other hand, moves to the left with a speed of 2 relative to the COM. So how is the total momentum even 0 in the COM frame before the collision ever happens?

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    $\begingroup$ Are you sure that the center of mass initially moves at 2m/s? $\endgroup$ – Javier Jul 22 '19 at 14:14
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    $\begingroup$ if the COM travels at 2 m/s as A travels at 2 m/s, when A hits B, is the COM on the other side of B? This doesn't seem right. $\endgroup$ – Vangi Jul 22 '19 at 14:18
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Before the collision, since A moves to right with a speed of 2, so does the center of mass (according to my understanding, which may be incorrect).

Your understanding is indeed incorrect. The velocity of the center of mass is the average velocity of all objects in the system weighted by their masses. For this system where object $A$ moves at a velocity of $\mathbf{v}_A = (2\,\mathrm{m/s})\hat{\mathbf{x}}$ and object $B$ is at rest, it would be $$ \mathbf{v}_{CM} = \frac{m_A \mathbf{v}_A + m_B \mathbf{v}_B}{m_A + m_B} = \frac{m_A}{m_A+m_B}\left( 2 \,\mathrm{m/s}\right)\hat{\mathbf{x}}, $$ which is only equal to $(2\,\mathrm{m/s})\hat{\mathbf{x}}$ if $m_B = 0$.

Moving to the center of mass frame, the velocity of object $B$ would be $\mathbf{v}_B' = -\mathbf{v}_{CM}$ and the velocity of object $A$ would be $\mathbf{v}_A' = \mathbf{v}_A - \mathbf{v}_{CM}$. Calculating the total momentum in this frame gives $$ m_A \mathbf{v}_A' + m_B\mathbf{v}_B' = m_A(\mathbf{v}_A - \mathbf{v}_{CM}) - m_B\mathbf{v}_{CM} = m_A \mathbf{v}_A - (m_A+m_B)\frac{m_A\mathbf{v}_A}{m_A + m_B} = 0, $$ as expected.

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An inertial reference frame is any which the combined center of mass moves with a constant speed. It makes no difference to the system what the speed of the center of mass is. You can always define a co-moving reference frame which moves together with the center of mass, and thus in that reference frame, the combined momentum is zero.

It is easy to show that in calculating momentum the velocity vector of the center of mass $\boldsymbol{v}_{CM}$ is grouped outside of the expression.

$$ \boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i = \left( \sum_i m_i \right) \boldsymbol{v}_{\rm CM} =m \, v_{\rm CM} $$

So in the reference frame where $\boldsymbol{v}_{\rm CM}=0$, by definition momentum, will also be $\boldsymbol{p}=0$.

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For a system of particles of $n$ particles, $$\vec{P_\text{system}} = \sum_{i=1}^{n}{m_i v_i}$$ and from this expression we can see that $\vec{P_{\text{CM}}} = \vec{P_{\text{system}}}$, and since in CM frame $\vec{P_{\text{CM}}} = 0$, it follows that $\vec{P_{\text{system}}}$ is also equal to $0$. Hence the total linear momentum in the CM frame is always zero.

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