Conservation of Momentum: Key Concepts and Examples

Conservation of linear momentum is a key principle in physics, stating that the total linear momentum of a closed system remains unchanged when no external forces are present. During interactions, such as collisions or explosions, the total momentum before the event equals the total momentum afterward, as long as external influences are absent. This principle is essential for comprehending various phenomena, ranging from everyday collisions to intricate systems in particle physics.

1.0Statement of Conservation of Linear Momentum

• When a system of interacting particles experiences no extrinsic forces, the total linear momentum of the system remains conserved. This total linear momentum is calculated as the vector sum of the individual linear momenta of all the particles within the system.
• According to Newton’s Second law, if net external force on the system is zero, then linear momentum of the system remains conserved. According to Newton's Second Law.

${\vec{F}}_{ext}=\frac{\overrightarrow{dp}}{dt}$

If ${\vec{F}}_{\text{ext }}=0\Rightarrow \frac{\overrightarrow{dp}}{dt}=0\Rightarrow \vec{p}=\text{ constant }\Rightarrow {\vec{p}}_{\text{initial }}={\vec{p}}_{\text{final }}$

For two particle system ${\vec{p}}_1+{\vec{p}}_2=\text{ constant }$

$\Delta {\vec{p}}_1+\Delta {\vec{p}}_2=\text{ constant }$

$\Delta {\vec{p}}_1=-\Delta {\vec{p}}_2$

$\text{ Change in momentum of }{1}^{\text{st }}\text{ particle }=\text{ Change in momentum of }{2}^{\text{nd }}\text{ particle }$

2.0Derivation of Conservation of Linear Momentum

• In an isolated system it consist of n particles ,having masses $m_1,m_2,m_3,\dots \dots ...m_n$

And are moving with velocities $v_1,v_2,v_3,\dots \dots .v_n$ respectively.

• Total linear momentum of the system is,

$\vec{p}=m_1{\vec{v}}_1+m_2{\vec{v}}_2+m_3{\vec{v}}_3+\dots \dots \dots \dots m_n{\vec{v}}_n$

$\vec{p}={\vec{p}}_1+{\vec{p}}_2+{\vec{p}}_3+\dots ...{\vec{p}}_n$

$\vec{F}=\frac{\overrightarrow{dp}}{dt}\Rightarrow \text{ External Force acting on the system }$

For an isolated system $\vec{F}=0or\frac{\overrightarrow{dp}}{dt}=0$

$\vec{p}=\text{ Constant }$

${\vec{p}}_1+{\vec{p}}_2+{\vec{p}}_3+\dots \dots ..{\vec{p}}_n=\text{ Constant }$

$\Rightarrow$ In the absence of any extrinsic force, the total linear momentum of the system is constant.

• Considering two bodies X and Y having mass $m_1$ and $m_2$
• respectively moving in same direction along a straight line with velocities $u_1$ and $u_2$ and they collide for time $\Delta t.\left(u_1>u_2\right)$. After collision their velocities $v_1$ and  $v_2$.
• During collision Body X and Y exert forces on each other, From NLM 3 we can write as,

${\vec{F}}_{XY}=-{\vec{F}}_{YX}$

• Impulse of ${\vec{F}}_{XY}={\vec{F}}_{XY}\Delta t$ =change in momentum of $X=m_1{\vec{v}}_1-m_1{\vec{u}}_1$ Impulse of${\vec{F}}_{YX}={\vec{F}}_{XY}\Delta t$ =change in momentum of $\mathrm{Y}=m_2{\vec{v}}_2-m_2{\vec{u}}_2$

${\vec{F}}_{XY}.\Delta t=-{\vec{F}}_{XY}\Delta t$

$m_1{\vec{v}}_1-m_1{\vec{u}}_1=-\left(m_2{\vec{v}}_2-m_2{\vec{u}}_2\right)$

$m_1{\vec{u}}_1+m_2{\vec{u}}_2=m_1{\vec{v}}_1+m_2{\vec{v}}_2$

• Total linear momentum before collision=Total linear momentum after collision

3.0Examples of Conservation of Linear Momentum

1. Recoil of a Gun

Total momentum before firing =Total momentum after firing

$0=m\vec{v}+M\vec{V}$

$M\vec{V}=-m\vec{v}$

$\vec{V}=-\frac{m}{M}\vec{v}$

$\Rightarrow$ Negative sign shows that $\vec{V}$ and $\vec{v}$ are in opposite directions, the gun recoils in the backward direction.

2. Explosion of a bomb - Before an explosion, a stationary bomb has zero total momentum. Upon detonation, internal energy causes it to shatter, propelling fragments in various directions. The total momentum of these fragments must also sum to zero; for instance, if one fragment moves east, another must move west with an equal momentum. 

4.0Sample Questions on Conservation of Linear Momentum

Q-1.How can Newton's Third Law of Motion be derived from the law of conservation of momentum?

Solution: According to the law of conservation of linear momentum, the net change in linear momentum in the absence of external force is zero.

$\Delta {\vec{p}}_1+\Delta {\vec{p}}_2=0$

$\Delta {\vec{p}}_2=-\Delta {\vec{p}}_1$

$\frac{\Delta p_2}{\Delta t}=\frac{\Delta p_1}{\Delta t}$

Rate of change of momentum $m_2$ = -Rate of change of momentum $m_1$

$\text{ Force on }{m}_2=-\text{ Force on }{m}_1$

$\text{ Action }=-\text{ Reaction }$

This proves Newton's Third Law of Motion

Q-2. Two cars with masses 5 kg and 10 kg respectively are moving towards each other and collide and come to rest. Initially a car retaining the mass 10 kg moves towards the east with a velocity of 5m/s . Find initial velocity of the car with mass 5 kg with respect to ground.

Solution:

$p_{\text{final }}=0$, as the cars comes to rest after collision

$p_{\text{initial }}=p_{\text{final }}$

$m_1{\vec{u}}_1+m_2{\vec{u}}_2=0$

$10\times 5+5\times {\vec{u}}_2=0$

$50=-5u_2$

$u_2=-10\mathrm{m}/\mathrm{s}$

Q-3. A gun fires n bullets in 1 second, with each bullet having a mass m and a speed v. What is the average recoil force of the gun?

Solution:

Change in momentum of each bullet $(\Delta p=mv)$

Average time of firing of each bullet = $\frac{1}{n}S$

Average force of recoil = $F=\frac{\Delta p}{\Delta t}=\frac{mv}{\frac{1}{n}}=mnv$

Q-4. A body with mass M is moving at velocity V when it explodes into two equal parts. If one part comes to rest while the other moves with velocity v, what is the value of v?

Solution: Applying Conservation of linear momentum,

$MV=\frac{M}{2}\times 0+\frac{M}{2}\times v$

$\Rightarrow v=2V$