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} NoteV_{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

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