Question

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Answer 1

Consider the $\sf{n}$ particles having masses $\sf{m_i}$ each and moving with velocities $\bf{v_i}$ each, where $\sf{i=1,\ 2,\ 3,\,\dots\,,\ n.}$

Then the net momentum:-

$\sf{\longrightarrow\mathbf{p_1}=m_1\mathbf{v_1}+m_2\mathbf{v_2}+m_3\mathbf{v_3}+\,\dots\,+m_n\mathbf{v_n}}$

or,

$\displaystyle\sf{\longrightarrow \mathbf{p_1}=\sum_{i=1}^nm_i\mathbf{v_i}\quad\quad\dots(1)}$

The velocity of the center of mass of these particles,

$\longrightarrow\mathbf{\bar v}=\dfrac{\displaystyle\mathsf{\sum_{i=1}^nm_i}\mathbf{v_i}}{\displaystyle\mathsf{\sum_{i=1}^nm_i}}$

$\longrightarrow\mathbf{\bar v}\displaystyle\mathsf{\sum_{i=1}^nm_i}=\displaystyle\mathsf{\sum_{i=1}^nm_i}\mathbf{v_i}\mathsf{\quad\quad\dots(2)}$

Now consider the single particle whose mass is the sum of masses of those $\sf{n}$ particles.

$\displaystyle\sf{\longrightarrow M=\sum_{i=1}^nm_i}$

and this particle is moving with the velocity of the center of mass of the particles.

$\bf{\longrightarrow V=\bar v}$

Then, momentum of this single particle,

$\sf{\longrightarrow \mathbf{p_2}=M\mathbf{V}}$

$\displaystyle\sf{\longrightarrow \mathbf{p_2}=\mathbf{\bar v}\sum_{i=1}^nm_i}$

From (2),

$\displaystyle\sf{\longrightarrow\mathbf{p_2}=\sum_{i=1}^nm_i\mathbf{v_i}}$

and from (1), we get,

$\displaystyle\bf{\longrightarrow\underline{\underline{p_1=p_2}}}$

Hence the Proof!

Answer 2

☘️ See the attachment picture for Explanation.

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