Question

State and Prove the principle of conservation of linear momentum.​

Answer 1

Law of Conservation of Linear Momentum

The law states that, if no external force is acting on a system of particles, then the total linear momentum of the system remains as a constant.

Proof based on Newton's Second Law of Motion:

Consider a system of masses $m_1,\ m_2,\ m_3,\ \dots,\ m_n$ moving with velocities $\bf{v_1},\ \bf{v_2},\ \bf{v_3},\ \dots,\ \bf{v_n}$ respectively. The velocity of each mass is same but they're separated to show individuality.

Hence the total linear momentum of the system is given by $\bf{p}=\displaystyle\sum_{i=1}^nm_i\bf{v_i}.$

Well, there's no need to find this!

What Newton's Second Law of Motion states is that the external applied force acting on a system is equal to the rate of change of linear momentum. So we have,

$\mathbf{F}\propto\dfrac{d\mathbf{p}}{dt}$

But what Law Of Conservation of Linear Momentum says is that the law holds true if and only if no external force is acting on it. Hence, we have,

$\mathbf{F}=0\\\\\\\dfrac{d\bf{p}}{dt}=0$

And it's true that the derivative of a constant is 0, or we say the integral of 0 is a constant. Hence,

$\bf{p}\ \text{is a constant.}$

Proof based on Newton's Third Law of Motion:

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Consider two masses $m_1\quad\&\quad m_2$ moving in same direction with initial velocities $\mathbf{u_1}\quad\&\quad \mathbf{u_2}$ respectively, considered $\mathbf{u_1}>\mathbf{u_2}$ so that there's a chance for both to be collided. After collision let the velocities be $\mathbf{v_1}\quad\&\quad \mathbf{v_2}$ respectively.

We have $\mathbf{F}=m\mathbf{a}=m\left(\dfrac{\mathbf{v}-\mathbf{u}}{t}\right)$

During collision,

$\mathbf{F_{21}}=m_2\left(\dfrac{\mathbf{v_2-u_2}}{t}\right)$

And,

$\mathbf{F_{12}}=m_1\left(\dfrac{\mathbf{v_1-u_1}}{t}\right)$

where $\mathbf{F_{AB}}$ indicates the force exerted on A by B.

What Newton's Third Law of Motion states that there's an equal and opposite reaction to every action. So if we consider $\bf{F_{21}}$ as action and $\bf{F_{12}}$ reaction, both have same magnitude but they're opposite in directions, i.e., one is negative to the other. So,

$\mathbf{F_{21}}=-\mathbf{F_{12}}$

The converse is also possible.

Well,

$\begin{aligned}&\mathbf{F_{21}}=-\mathbf{F_{12}}\\\\\implies\ \ &m_2\left(\dfrac{\mathbf{v_2-u_2}}{t}\right)=-m_1\left(\dfrac{\mathbf{v_1-u_1}}{t}\right)\\\\\implies\ \ &m_2\left(\mathbf{v_2-u_2}}\right)=-m_1\left(\mathbf{v_1-u_1}}\right)\\\\\implies\ \ &m_2\mathbf{v_2}-m_2\mathbf{u_2}=-m_1\mathbf{v_1}+m_1\mathbf{u_1}\\\\\implies\ \ &\boxed{m_1\mathbf{v_1}+m_2\mathbf{v_2}=m_1\mathbf{u_1}+m_2\mathbf{u_2}}\end{aligned}$

Here we get that, total momentum after collision = total momentum before collision. Or we can say that the total momentum remains unchanged or constant. But remember, this is only true if and only if no external force is acting on the system.

Answer 2

$\huge{\underline{\bigstar{\mathtt{Answer!!}}}}$

The principle of conservation of momentum states that if two objects collide, then the total momentum before and after the collision will be the same if there is no external force acting on the colliding objects.