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The COM frame is defined as the coordinate system in which the centre of mass of the system is at rest. In the case of a 2-body system the center of mass coordinate is: $$\vec{R_{CM}} = \frac{M_A \vec{R_A} + M_B \vec{R_B}}{M_A + M_B}$$

And the velocity of the COM is:

$$\vec{U_{CM}} = \frac{M_A \vec{U_A} + M_B \vec{U_B}}{M_A + M_B}$$

Now my book affirms that momentum of the projectile particle $A$ ($B$ is at rest in laboratory frame) is related by the following: $$\vec{p_A} = \vec{q_A} - M_A \vec{V_{CM}}$$

where $\vec{q_A}$ is the momentum of particle $A$ in the laboratory frame and $\vec{V_{CM}}$ is the velocity of the COM (in the laboratory frame). How does this result follow?

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    $\begingroup$ $p_A=m_A(V_A-V_{CM})$ $\endgroup$ – Wolphram jonny Jan 17 at 19:07
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The derivation was quite easy, I just got a bit confused. I just had to use the transformation between velocities: $\vec{U} = (\vec{U})_L - \vec{V_{CM}}$ $$\vec{p_A} = M_A (\vec{U_A}) = M_A ((\vec{U_A})_L - \vec{V_{CM}}) = \vec{q_{A}} - M_A \vec{V_{CM}}$$

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