Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass: Show $\text{p}=\text{p}'_\text{i}+\text{m}_\text{i}\text{V}$ where $p_i$ is the momentum of the ith particle (of mass $m_i$ ) and $p′_i = m_iv′_i.$ Note: $v′_i$ is the velocity of the ith particle relative to the centre of mass. Also, prove using the definition of the centre of mass.
Exercise
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Take a system of i moving particles. Mass of the $i^{th}$ particle = $m_i$ Velocity of the $i^{th}$ particle = $v_i$ Hence, momentum of the $i^{th}$ particle, $\text{p}_\text{i}=\text{m}_\text{i}\text{v}_{\text{i}}$ Velocity of the centre of mass = V The velocity of the $i^{th}$ particle with respect to the centre of mass of the system is given as, $\text{v}'_{\text{i}}=\text{v}_\text{i}-\text{V}\ ...(\text{i})$ Multiplying $m_i$ throughout equation (i), we get, $\text{m}_\text{i}\text{v}'_\text{i}=\text{m}_\text{i}\text{v}_\text{i}-\text{m}_\text{i}\text{V}$
$\text{p}'_\text{i}-\text{m}_\text{i}\text{V}$ Where, $\text{p}'_\text{i}=\text{m}_\text{i}\text{v}_\text{i}'=$ Momentum of the $i^{th}$^ particle with respect to the centre of mass of the system, $\therefore\text{ p}_\text{i}=\text{p}'_\text{i}+\text{m}_\text{i}\text{V}$ We have the relation, $\text{p}'_\text{i}=\text{m}_\text{i}\text{v}_\text{i}'$ Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get $\sum_\limits\text{i}\text{p}'_\text{i}=\sum_\limits\text{i}\text{m}_\text{i}\text{v}'_\text{i}=\sum\limits_\text{i}\frac{\text{dr}'_\text{i}}{\text{dt}}$ Where, $\text{r}'_\text{i}=$ Position vector of $i^{th}$ particle with respect to the centre of mass. $\text{v}'_\text{i}=\frac{\text{dr}'_\text{i}}{\text{dt}}$ As per the definition of the centre of mass, we have, $\sum\limits_\text{i}\text{m}_\text{i}\text{r}'_\text{i}=0$
$\therefore\ \sum\limits_\text{i}\text{m}_\text{i}\frac{\text{dr}'_\text{i}}{\text{dt}}=0$
$\sum\limits_\text{i}\text{p}'_\text{i}=0$
$\text{p}'_\text{i}-\text{m}_\text{i}\text{V}$ Where, $\text{p}'_\text{i}=\text{m}_\text{i}\text{v}_\text{i}'=$ Momentum of the $i^{th}$^ particle with respect to the centre of mass of the system, $\therefore\text{ p}_\text{i}=\text{p}'_\text{i}+\text{m}_\text{i}\text{V}$ We have the relation, $\text{p}'_\text{i}=\text{m}_\text{i}\text{v}_\text{i}'$ Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get $\sum_\limits\text{i}\text{p}'_\text{i}=\sum_\limits\text{i}\text{m}_\text{i}\text{v}'_\text{i}=\sum\limits_\text{i}\frac{\text{dr}'_\text{i}}{\text{dt}}$ Where, $\text{r}'_\text{i}=$ Position vector of $i^{th}$ particle with respect to the centre of mass. $\text{v}'_\text{i}=\frac{\text{dr}'_\text{i}}{\text{dt}}$ As per the definition of the centre of mass, we have, $\sum\limits_\text{i}\text{m}_\text{i}\text{r}'_\text{i}=0$
$\therefore\ \sum\limits_\text{i}\text{m}_\text{i}\frac{\text{dr}'_\text{i}}{\text{dt}}=0$
$\sum\limits_\text{i}\text{p}'_\text{i}=0$
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