Conservation's Law of linear momentum and its Derivation

Derivation of Conservation's law of Linear momentum from Newton's Second Law and Statement: Let us consider, A particle that has mass $m$ and is moving with velocity $v$ then According to Newton's second law the applied force on a particle is

Derivation From General Form Derivation from Differential Form
$F=ma$

$F=m \frac{\Delta v}{\Delta t}$

$F= \frac{\Delta (mv)}{\Delta t}$

$F= \frac{\Delta P}{\Delta t} \qquad \left( \because P=mv \right)$

If the applied force on a body is zero then

$\frac{\Delta P}{\Delta t}=0$

$\Delta P=0$

Here $\Delta P$ is change in momentum
of the particle .i.e

$P_{2}-P_{1}=0$

$P_{2}=P_{1}$

$P=Constant$
$F=ma$

$F=m \frac{dv}{dt}$

$F= \frac{dmv}{dt}$

$F= \frac{dP}{dt} \qquad \left( \because P=mv \right)$

If the applied force on a body is zero then

$\frac{dP}{dt}=0$

$dP=0$

On integrating the above equation

$P=constant$
From the above equation, we conclude the statement i.e.
When no external force is applied to a particle, the total momentum of a particle is always conserved.

Derivation of Conservation's law of Linear momentum from Newton's Third Law and Statement: Let us consider, The two-particle that have mass $m_{1}$ and $m_{2}$ are moving towards each other. If
Conservation's Law of Linear Momentum From Newton’s Third Law
The velocity of the particles $m_{1}$ before collision is = $u_{1}$

The velocity of the particles $m_{2}$ before collision is = $u_{2}$

The velocity of the particles $m_{1}$ after collision is = $v_{1}$

The velocity of the particles $m_{2}$ after the collision is = $v_{2}$

When these particles collide to each other than according to Newton's third law

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

$m_{1}a_{1}=m_{2}a_{2}$

$m_{1}\frac{\Delta V_{1}}{\Delta t}=m_{2} \frac{\Delta V_{2}}{\Delta t} \quad (1) \qquad \left(\because a=\frac{\Delta V} {\Delta t} \right)$

Where
$\Delta V_{1}$= Change in Velocity of mass $m_{1}$ i.e $\left( v_{1}-u_{1} \right)$
$\Delta V_{2}$= Change in Velocity of mass $m_{2}$ i.e $\left( v_{2}-u_{2} \right)$

So from equation $(1)$

$m_{1}\frac{\left( v_{1}-u_{1} \right)}{\Delta t}=m_{2} \frac{\left( v_{2}-u_{2} \right)}{\Delta t}$

$m_{1}\left( v_{1}-u_{1} \right)=m_{2} \left( v_{2}-u_{2} \right)$

$m_{1} v_{1}- m_{1} u_{1}=m_{2}v_{2}- m_{2} u_{2}$

$m_{1} u_{1}+ m_{2} u_{2}=m_{1}v_{1} + m_{2} v_{2}$

From the above equation, we conclude the statement i.e.

When two particle collide two each other. If the total momentum of the particles before the collision and total momentum of the particles after is collision is same then it is known as conservation law of linear momentum.

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