Linear Momentum of a System of Particles: Definition, Formula, Application, Solved Prolems & Notes

The linear momentum of a system of particle is the vector sum of the linear momentum of all the individual particles in the system. 

Suppose a system consists of n particles with masses  $m_1,m_2,m_3,...,m_n$ and velocities ${\bar{v}}_1,{\bar{v}}_2,{\bar{v}}_3,...,{\bar{v}}_i$ , then the total linear momentum $\bar{P}$ of the system is expressed as: $\vec{P}=\underset{i=1}{\overset{n}{\sum }} m_i{\vec{v}}_i$

Here:

$\stackrel{\to }{p_i}=m_i\stackrel{\to }{v_i}$ is the momentum of $i^{\text{th}}$ particle.

$\stackrel{\to }{P}$ denotes the total momentum of the system.

Important Links:

NCERT Class 12 notes	CBSE Class 12 Chemistry NCERT notes
Class 12 Maths NCERT notes	Physics Class 12 NCERT notes

Key Points: 

1. The momentum has both magnitude and direction (Vector Quantity).
2. The total momentum of a system is expressed as the sum of the momenta of individual particles.
3. Momentum always depends on the frame of reference (usually inertial). 

The centre of mass helps to decide the overall motion of a system of particles. The average position of all the masses in a system, weighted by their respective masses, is the center of mass. The motion of the center of mass plays a key role in understanding how the system behaves as a whole in terms of linear momentum a system. 

Also Read: NCERT Solutions

Center of Mass:

The position vector of the center of mass ${\bar{R}}_{cm}$ for a system of n particles with masses $m_1,m_2,m_3,...,m_n$ located at a position ${\bar{r}}_1,{\bar{r}}_2,...,{\bar{r}}_n$   is given as 

${\bar{R}}_{cm}=\frac{1}{M}\underset{i=1}{\overset{n}{\sum }}{m}_i{\bar{r}}_i$ ,

Where $M=\underset{i=1}{\overset{n}{\sum }}{m}_i$ is the total mass of the system.

Velocity of the Center of Mass:

The position of the center of mass is differentiated with respect to time, giving its velocity:

Centre of mass velocity formula = ${\bar{v}}_{cm}=\frac{1}{M}\underset{i=1}{\overset{n}{\sum }}{m}_i{\bar{v}}_i$

The center of mass moves as if all the mass of the system were concentrated at that point.

Total Linear Momentum of a System:

The sum of the momenta of individual particles is the total linear momentum $\bar{P}$ of a system of particles. 

Total Momentum = $\bar{P}=\underset{i=1}{\overset{n}{\sum }}{m}_i{\bar{v}}_i$

Using the cente of mass velocity formula, we get, $\bar{P}=M{\bar{v}}_{cm}$

The law of conservation of momentum describes that the total momentum of a system remain constant if no external force act on it.
Mathematically,

If ${\bar{F}}_{ext}=0$

then $\frac{d\bar{P}}{dt}=0\Rightarrow \bar{P}=\text{constant}.$  This law is useful in solving the problems based on collisions and explosions.

 

Important Links: 

NCERT Class 11 Notes	Class 11 Physics NCERT notes
CBSE Class 11 Chemistry NCERT Notes	Maths Class 11 NCERT notes

Applications of Conservation of Momentum:

Key applications of rotational motion are 
1. Collision Analysis (Elastic and Inelastic Collisions):

• In a collision, the total linear momentum is conserved. This principle helps to calculate the final velocities of colliding objects.
• Example: Two cars colliding on a frictionless track.

2. Rocket Propulsion (Recoil Motion):

• The functionality of a rocket is based on the principle of conservation of linear momentum.
• The rocket experiences forward momentum when fuel is expelled backwards.
• The total momentum of the rocket fuel system remains conserved

3. Recoil of a Gun:

• The momentum of the bullet is balanced by the backwards momentum (recoil) of the gun when it is fired.
• This is an example of conservation of momentum in a two-body system.

4. Human Movement and Sports:

• When we walk, jump, or dive the movement of body's part control the overall momentum.

5. Traffic Accidents and Safety Devices:

• Car crashes can be analyze using the momentum principles.
• Airbags and seat belts reduce injuries by increasing the time over which momentum changes, reducing force. 

There are various application of rotational motion. Some of them are mentioned below:

• The size of stars decreases when they collapse into black hole or neutron stars. However, their rotation speed increases due to conservation of angular momentum. 
• Gyroscopes uses rotational motion to maintain orientation. It is used in aircraft, smartphones (motion sensing), ships and satellites.
• Rotational motion in bicycle and motorcycle provides gyroscopic stability to maintain the balance.
• Crankshafts, turbines, and flywheels store angular momentum and smooth out the delivery of power in machines.

 

Example 1: Collision of Two Particles

Two particles of masses 2 kg and 3 kg move with velocities ${\vec{v}}_1=(4\hat{i}+3\hat{j})\mathrm{m}/\mathrm{s}$ and $\overrightarrow{v_2}=(-2\hat{i}+5\hat{j})\mathrm{m}/\mathrm{s}$ , respectively. Find the total linear momentum of the system.

Solution: $\begin{array}{c}\vec{P}=m_1{\vec{v}}_1+m_2{\vec{v}}_2\\ =2(4\hat{i}+3\hat{j})+3(-2\hat{i}+5\hat{j})\\ =(8\hat{i}+6\hat{j})+(-6\hat{i}+15\hat{j})=(2\hat{i}+21\hat{j})\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}\end{array}$

The magnitude of the momentum is: $\left|\vec{P}\right|=\sqrt[]{2^2+{21}^2}=\sqrt[]{4+441}=\sqrt[]{445}\approx 21.1\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}$

 

Example 2: Conservation in a Collision

A particle of mass 1 kg moving with velocity $10\hat{i}\mathrm{m}/\mathrm{s}$ collides with another particle of mass 2 kg at rest. After the collision, they move together. Find their common velocity.

Solution:

Since no external forces act, linear momentum is conserved. Initial momentum: ${\vec{P}}_{\text{initial}}=(1\cdot 10\hat{i})+(2\cdot 0)=10\hat{i}\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}$

Final momentum (masses stick together, so common velocity $\vec{v}$ ): ${\vec{P}}_{\text{final}}=(1+2)\vec{v}=3\vec{v}$

By conservation: $3\vec{v}=10\hat{i}⟹\vec{v}=\frac{10}{3}\hat{i}\approx 3.33\hat{i}\mathrm{m}/\mathrm{s}$

 

Example 3: Center of Mass Velocity

Three particles of masses $1\mathrm{k}\mathrm{g},2\mathrm{k}\mathrm{g}$ , and 3 kg have velocities $2\hat{i}\mathrm{m}/\mathrm{s},3\hat{j}\mathrm{m}/\mathrm{s}$ , and $-4\hat{i}+$ $2\hat{j}\mathrm{m}/\mathrm{s}$ , respectively. Find the velocity of the center of mass.

Solution:

Total mass $M=1+2+3=6\mathrm{k}\mathrm{g}$ . Total momentum: $\begin{array}{c}\vec{P}=(1\cdot 2\hat{i})+(2\cdot 3\hat{j})+(3\cdot (-4\hat{i}+2\hat{j}\left)\right)\\ =2\hat{i}+6\hat{j}+(-12\hat{i}+6\hat{j})=(-10\hat{i}+12\hat{j})\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\mathrm{s}\end{array}$

Velocity of center of mass: ${\vec{v}}_{\mathrm{c}\mathrm{m}}=\frac{\vec{P}}{M}=\frac{-10\hat{i}+12\hat{j}}{6}=\left(-\frac{5}{3}\hat{i}+2\mathrm{m}/\mathrm{s}\right.$

1. Two particles of masses 4 kg and 6 kg move with velocities $3\hat{i}-2\hat{j}\mathrm{m}/\mathrm{s}$ and $2\hat{i}+$ $4\hat{j}\mathrm{m}/\mathrm{s}$ . Calculate the total linear momentum and its magnitude.

2. A system of two particles, each of mass 5 kg , moves such that their center of mass has a velocity of $4\hat{i}\mathrm{m}/\mathrm{s}$ . What is the total linear momentum?

3. A 2 kg particle moving at $5\hat{i}\mathrm{m}/\mathrm{s}$ collides with a 3 kg particle moving at $-3\hat{i}\mathrm{m}/\mathrm{s}$ . If they stick together, find the final velocity.

Here we have provided the NCERT notes link for the complete chapter of Class 11 Physics.

Units and Measurements Class 11 Notes	Mechanical Properties of Solids Class 11 Notes
Motion in a Straight Line Class 11 Notes	Mechanical Properties of Fluids Class 11 Notes
NCERT Class 11 Notes for Motion in a Plane	Thermal Properties of Matter Class 11 Notes
Laws of Motion Class 11 Notes	Thermodynamics Class 11 Notes
Work, Energy, and Power Class 11 Notes	Kinetic Theory of Gas Class 11 Notes
System of Particles and Rotational Motion Class 11 Notes	Oscillations Class 11 Notes
Gravitation Class 11 Notes	Waves Class 11 Notes