Show that the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of the centre of mass.

(N/A) Let a system consist of $n$ particles with masses $m_1, m_2, \dots, m_n$ moving with velocities $\vec{v}_1, \vec{v}_2, \dots, \vec{v}_n$ respectively.
The total linear momentum $\vec{P}$ of the system is the vector sum of the individual momenta of all particles:
$\vec{P} = \vec{p}_1 + \vec{p}_2 + \dots + \vec{p}_n = m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n$ --- $(1)$
The velocity of the centre of mass $\vec{V}$ is defined as:
$\vec{V} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n}{m_1 + m_2 + \dots + m_n}$
Let the total mass of the system be $M = \sum_{i=1}^{n} m_i$. Then:
$\vec{V} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n}{M}$
Rearranging this,we get:
$M \vec{V} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n$ --- $(2)$
Comparing equation $(1)$ and $(2)$,we find:
$\vec{P} = M \vec{V}$
Thus,the total momentum of the system is equal to the product of the total mass of the system and the velocity of the centre of mass.