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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)

best_answer

The momentum of i_th 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 $\sum p_i\ =\ 0$

Posted by

Devendra Khairwa

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