# Momentum of projectile in the centre of mass frame in a two body system

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?

• $p_A=m_A(V_A-V_{CM})$ – Wolphram jonny Jan 17 at 19:07

The derivation was quite easy, I just got a bit confused. I just had to use the transformation between velocities: $$\vec{U} = (\vec{U})_L - \vec{V_{CM}}$$ $$\vec{p_A} = M_A (\vec{U_A}) = M_A ((\vec{U_A})_L - \vec{V_{CM}}) = \vec{q_{A}} - M_A \vec{V_{CM}}$$