# Q7.33     Separation of Motion of a system of particles into motion of the centre of mass and motion about the               centre of mass :             (a) Show $P=P_{i}^{^{'}}+m_{i}V$              where $P_{i}$ is the momentum of the ith particle (of mass $m_{i}$)and  $P_{t}^{'}=m_{t}V_{t}$ Note$V_{t}^{'}$  is the velocity            of the ith particle relative to the centre of mass.Also, prove using the definition of the centre of mass$\sum P_{t}'=0$

Answers (1)

The momentum of ith particle is given by :             $p_i\ =\ m_iv_i$

The velocity of the centre of mass is V.

Then the velocity of ith particle with respect to the center of mass will be :     $v_i'\ =\ v_i\ -\ v$

Now multiply the mass of the particle to both the sides, we get :

$mv_i'\ =\ mv_i\ -\ mv$

or                                                             $p_i'\ =\ p_i\ -\ p$                                          (Here p'i is the momentum of ith particle with respect to center of mass.)

or                                                             $p_i\ =\ p_i'\ +\ p$

Now consider   $p_i'$  :

$\sum p_i'\ =\ \sum m_iv_i'\ =\ \sum m_i\frac{dr_i}{dt}$

But as per the definition of centre of mass, we know that :

$\sum m_ir_i'\ =\ 0$

Thus                                                             $\sum p_i\ =\ 0$

Exams
Articles
Questions