Center of Mass and Systems of Particles in Galilean relativity

Question

Consider a reference frame in which two particles move with constant velocities $\vec{v}_1 = v_1 \hat{i}$ and $\vec{v}_2 = -v_2 \hat{i}$. Their center of mass would be the vector $\vec{R} = \frac{(m_1v_1t - m_2v_2t)}{m_1 + m_2}\hat{i}$. Now, if we consider the same situation from the reference frame of particle $1$, we find that, since particle $1$ is stationary and particle $2$ moves with velocity $-(v_1+v_2)\hat{i}$, the center of mass in that frame is given by $\vec{R'} = \frac{- m_2(v_1+v_2)t}{m_1 + m_2}\hat{i}$. But if we apply the Galilean transformation $x'=x-vt$, where $v=v_1$, we don't get that same $\vec{R'}$. What does that say about the center of mass as we move from one frame to another? And what does it say about the force on a system of particles, which is normally given by the acceleration of their center of mass. If the $COM$s are different, does that mean the forces are different too? I haven't calculated the force explicitly, but I'm more bothered about the transformation for the center of mass vector.

Answer 1

Your expressions for the COM supposes that $R = 0$ for $t = 0$, and also $R' = 0$, when $t = 0$. It is of course not possible.

The most general expression for the position of each particle is $r_0 + vt$, and not only $vt$