Let $m_1, m_2, m_3, \ldots, m_{\mathrm{n}}$ be the masses of $n$ particles constituting a system. Assume that these particles are distributed in three-dimensional space and the coordinates of these particles are $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right),\left(x_3, y_3, z_3\right), \ldots \ldots\left(x_n, y_n, z_n\right)$, respectively. The location of centre of mass of the given system is mass-based average position and can be written as follows:

$$ X_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3+\ldots+m_{\mathrm{n}} x_{\mathrm{n}}}{m_1+m_2+m_3+\ldots . .+m_{\mathrm{n}}}=\frac{\sum\limits_{i=1}^n m_{\mathrm{i}} x_{\mathrm{i}}}{\sum\limits_{i=1}^n m_{\mathrm{i}}} $$

$Y_{\mathrm{cm}}=\frac{m_1 y_1+m_2 y_2+m_3 y_3+\ldots+m_{\mathrm{n}} y_{\mathrm{n}}}{m_1+m_2+m_3+\ldots . .+m_{\mathrm{n}}}=\frac{\sum\limits_{i=1}^n m_{\mathrm{i}} y_{\mathrm{i}}}{\sum\limits_{i=1}^n m_{\mathrm{i}}}$

$Z_{\mathrm{cm}}=\frac{m_1 z_1+m_2 z_2+m_3 z_3+\ldots+m_{\mathrm{n}} z_{\mathrm{n}}}{m_1+m_2+m_3+\ldots . .+m_{\mathrm{n}}}=\frac{\sum\limits_{i=1}^n m_{\mathrm{i}} z_{\mathrm{i}}}{\sum\limits_{i=1}^n m_{\mathrm{i}}}$

X-coordinate of the centre of mass can be written as follows :

$$ X_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}=\frac{m_1(0)+m_2(d)}{m_1+m_2}=\frac{m_2 d}{m_1+m_2} $$

Note :

1. We know that the centre of mass of a two-particle system lies on the line joining the particles.

2. The distance of the centre of mass from particle $m_1$ is $\frac{m_2 d}{m_1+m_2}$ and so the distance of centre of mass from $m_2$ will be $d-\frac{m_2 d}{m_1+m_2}=\frac{m_1 d}{m_1+m_2}$

For continuous mass distribution the center of mass can be located by replacing summation sign with an integral sign. Proper limits for the integral are chosen according to the situation

$x_{c m}=\frac{\int x d m}{\int d m}, $

$y_{c m}=\frac{\int y d m}{\int d m}, $

$z_{c m}=\frac{\int z d m}{\int d m}$

Note: If an object has symmetric mass distribution about $\mathrm{x}$ axis then $\mathrm{y}$ coordinate of COM is zero and vice-versa.

$$ x_{com}=\frac{\int_0^L x d m}{\int d m}=\frac{\int_0^L(x)\left(\frac{M}{L} d x\right)}{M}=\frac{1}{L} \int_0^L x d x=\frac{L}{2} $$

The $y$-coordinate of COM is $y$_com $=\frac{\int y d m}{\int d m}=0$.

Similarly, $z_{c o m}=0$

i.e., the coordinates of $\mathrm{COM}$ of the rod are $\left(\frac{\mathrm{L}}{2}, 0,0\right)$, i.e. it lies at the center of the rod.

$\mathrm{y}_{\mathrm{cm}}=\frac{2 \mathrm{R}}{\pi}$

$\mathrm{y}_{\mathrm{cm}}=\frac{4 \mathrm{R}}{3 \pi}$

$\mathrm{y}_{\mathrm{cm}}=\frac{3 \mathrm{R}}{8}$

$y_{\mathrm{cm}}=\frac{\mathrm{R}}{2}$

(i) $\quad \vec{r}_{\text {COM }}=\frac{m_1 \vec{r}_1-m_2 \vec{r}_2}{m_1-m_2}$ or $\vec{r}_{\text {COM }}=\frac{A_1 \vec{r}_1-A_2 \vec{r}_2}{A_1-A_2}$

(ii) $\mathrm{x}_{\operatorname{com}}=\frac{\mathrm{m}_1 \mathrm{x}_1-\mathrm{m}_2 \mathrm{x}_2}{\mathrm{~m}_1-\mathrm{m}_2}$ or $\mathrm{x}_{\operatorname{com}}=\frac{\mathrm{A}_1 \mathrm{x}_1-\mathrm{A}_2 \mathrm{x}_2}{\mathrm{~A}_1-\mathrm{A}_2}$

$y_{\mathrm{COM}}=\frac{m_1 y_1-m_2 y_2}{m_1-m_2}$ or $y_{C O M}=\frac{A_1 y_1-A_2 y_2}{A_1-A_2}$ and

$$ \mathrm{z}_{\mathrm{COM}}=\frac{\mathrm{m}_1 \mathrm{z}_1-\mathrm{m}_2 \mathrm{z}_2}{\mathrm{~m}_1-\mathrm{m}_2} \text { or } \mathrm{z}_{\mathrm{COM}}=\frac{\mathrm{A}_1 \mathrm{z}_1-\mathrm{A}_2 \mathrm{z}_2}{\mathrm{~A}_1-\mathrm{A}_2} $$

Here, $m_1, A_1, \vec{r}_1, x_1, y_1$ and $z_1$ are the values for the whole mass while $m_2, A_2, \vec{r}_2, {x}_2, y_2$ and $z_2$ are the values for the mass which has been removed. Let us see two examples in support of the above theory.

If ${\mathbf{r}}_1,{\mathbf{r}}_2,\dots {\mathbf{r}}_n$ are the position vectors of masses $m_1,m_2,\dots$ $m_n$, respectively, thle position vector of the centre of mass of the system of particles is given by

${\mathbf{r}}_{\mathrm{CM}}=\frac{m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2+\cdots +m_n{\mathbf{r}}_n}{m_1+m_2+\cdots +m_n}$

The velocity of the centre of mass is given by

$\begin{array}{rl}{\mathbf{v}}_{\mathrm{CM}}& =\frac{d{\mathbf{r}}_{CM}}{dt}\\ & \\ & =\frac{m_1\frac{d{\mathbf{r}}_1}{dt}+m_2\frac{d{\mathbf{r}}_2}{dt}+\dots +m_n\frac{d{\mathbf{r}}_n}{dt}}{m_1+m_2+\dots +m_n}\\ & \\ {\mathbf{v}}_{\mathrm{CM}}& =\frac{m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2+\dots +m_n{\mathbf{v}}_n}{m_1+m_2+\dots +m_n}\end{array}$

The acceleration of the centre of mass is given by

$\begin{array}{rl}{a}_{\mathrm{CM}}& =\frac{d{\mathbf{v}}_{\mathrm{CM}}}{dt}\\ & \\ & =\frac{m_1\frac{d{\mathbf{v}}_1}{dt}+m_2\frac{d{\mathbf{v}}_2}{dt}+\cdots +m_n\frac{d{\mathbf{v}}_n}{dt}}{m_1+m_2+\cdots +m_n}\end{array}$

$\Rightarrow {\mathbf{a}}_{\mathrm{CM}}=\frac{m_1{\mathbf{a}}_1+m_2{\mathbf{a}}_2+\cdots +m_n{\mathbf{a}}_n}{m_1+m_2+\cdots +m_n}$

${\mathbf{a}}_{\mathrm{CM}}=\frac{{\mathbf{F}}_1+{\mathbf{F}}_2+\cdots +{\mathbf{F}}_n}{m_1+m_2+\cdots +m_n}=\frac{\sum {\mathbf{F}}_{\mathrm{ext}}}{M}$

If $\sum {\mathrm{F}}_{\mathrm{ext}}=0$, then ${\mathrm{a}}_{\mathrm{CM}}=0$, i.e ${\mathrm{v}}_{\mathrm{CM}}=$ constant. Hence if no net external force acts on a system, its centre of mass will remain at rest or will move with a constant velocity.

The linear momentum $\overrightarrow{(p)}$ of an object of mass $(m)$ moving with velocity $(\vec{v})$ is given as :

$\vec{p}=m \vec{v}$

Let us consider a case of $n$-particle system and total linear momentum of the system can be written as follows :

$$ \vec{P}=m_1 \vec{v}_1+m_2 \vec{v}_2+\ldots .+m_{\mathrm{n}} \vec{v}_{\mathrm{n}} $$

If $\vec{V}_{\mathrm{cm}}$ is the velocity of centre of mass of the system and $M$ is the total mass of the system then

$$ M \vec{V}_{\mathrm{cm}}=m_1 \vec{v}_1+m_2 \vec{v}_2+\ldots . .+m_{\mathrm{n}} \vec{v}_{\mathrm{n}} $$

Hence, the total linear momentum of the system can be written as follows :

$$ \vec{P}=M \vec{V}_{\mathrm{cm}} $$

On differentiating above equation we get the following relation :

$$ \begin{aligned} & \frac{d \vec{P}}{d t}=M \frac{d \vec{V}_{\mathrm{cm}}}{d t} \\\\ \Rightarrow \quad & \frac{d \vec{P}}{d t}=M \vec{a}_{\mathrm{cm}} \end{aligned} $$

We know that $\vec{F}_{\text {ext }}=M \vec{a}_{\mathrm{cm}}$;

hence above equation can be written as follows :

$$ \frac{d \vec{P}}{d t}=\vec{F}_{\text {ext }} $$

If $\vec{F}_{\text {ext }}=0 \Rightarrow $ $ \frac{d \vec{P}}{d t}=0 \Rightarrow \vec{P}=$ constant.

Thus, if no net external force acts on a system, the total linear momentum of the system remains constant; the total linear momentum being the vector sum of linear momentum of individual particles, i.e.

$$ \vec{P}=\vec{P}_1+\vec{P}_2+\cdots+\vec{P}_n $$

Impulse of force is defined as average force multiplied with time interval for which it acts on an object.

$\vec{I}=\vec{F} \Delta t=\vec{p}_2-\vec{p}_1$

Before the gun is fired, both the gun and the bullet are at rest. Therefore, the total momentum of the gun-bullet system is zero. After the gun is fired, the bullet moves forward and the gun recoils backwards. Let $m_{b}$ and $m_{g}$ be the masses of the bullet and the gun. If $\mathbf{v}_{b}$ and $\mathbf{v}_{g}$ are their respective velocities after firing, the total momentum of the gun-bullet system after firing is $\left(m_{b} \mathbf{v}_{b}+m_{g} \mathbf{v}_{g}\right)$. From the law of conservation of momentum, the total momentum after and before the gun is fired must be the same, i.e.

$$ \begin{aligned} &m_{b} \mathbf{v}_{b}+m_{g} \mathbf{v}_{g}=0 \\\\ &\text { or } \quad \mathbf{v}_{g}=-\frac{m_{b} \mathbf{v}_{b}}{m_{g}} \\\\ &\text { The negative sign indicates that the gun recoils in a } \\ &\text { direction opposite to that of the bullet. In terms of mag- } \\ &\text { nitudes, we have } \end{aligned} $$

$$ v_{g}=\frac{m_{b} v_{b}}{m_{g}} $$

$$ e=-\frac{\text { velocity of separation along line of impact }}{\text { velocity of approach along line of impact }}=\frac{v_2-v_1}{u_1-u_2} $$

Value of $e$ is 1 for elastic collision, 0 for perfectly inelastic collision and $0