JEE Physics: System of Particles – Complete Guide

The concept of the system of particles is essential in mechanics, especially in understanding complex motion involving multiple interacting bodies. For JEE aspirants, mastering this topic can help solve a wide range of problems related to the motion of composite bodies, collisions, and conservation laws. This guide covers everything you need to know about the system of particles, including center of mass, motion, momentum, internal and external forces, and problem-solving techniques.

1. What is a System of Particles?

A system of particles is a collection of two or more particles that interact with each other through forces. Instead of analyzing each particle separately, we consider the system as a whole and study its overall motion and dynamics.

This approach simplifies many problems where the particles are connected or interacting, such as in rigid bodies, molecules, or colliding bodies.

2. Center of Mass (COM)

The center of mass is the weighted average position of all the particles in the system. It acts as the point where the total mass of the system can be considered to be concentrated for translational motion analysis.

2.1 Definition

For a system of particles with masses \( m_1, m_2, \ldots, m_n \) located at position vectors \( \vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n \), the position vector of the center of mass \( \vec{R} \) is:

\( \displaystyle \vec{R} = \frac{\sum_{i=1}^n m_i \vec{r}_i}{\sum_{i=1}^n m_i} \)

2.2 Properties of Center of Mass

3. Motion of the Center of Mass

The velocity of the center of mass \( \vec{V} \) is given by differentiating \( \vec{R} \) with respect to time:

\( \displaystyle \vec{V} = \frac{d \vec{R}}{dt} = \frac{\sum m_i \vec{v}_i}{\sum m_i} \)

where \( \vec{v}_i \) is the velocity of the \( i^{th} \) particle.

Similarly, acceleration of the center of mass \( \vec{A} \):

\( \displaystyle \vec{A} = \frac{d \vec{V}}{dt} = \frac{\sum m_i \vec{a}_i}{\sum m_i} \)

3.1 Interpretation

The motion of the system's center of mass behaves like a single particle of mass equal to total mass under the influence of the net external force acting on the system.

4. Momentum of a System of Particles

The total linear momentum \( \vec{P} \) of the system is the vector sum of the momenta of all individual particles:

\( \displaystyle \vec{P} = \sum_{i=1}^n m_i \vec{v}_i \)

Using center of mass velocity, it can be shown that:

\( \displaystyle \vec{P} = M \vec{V} \), where \( M = \sum_{i=1}^n m_i \)

This means total momentum equals total mass times velocity of center of mass.

5. Newton’s Laws for System of Particles

Applying Newton’s second law to each particle and summing over the system, we get:

\( \displaystyle \sum_{i=1}^n \vec{F}_i = \sum_{i=1}^n m_i \vec{a}_i \)

Forces \( \vec{F}_i \) can be split into external and internal forces:

\( \displaystyle \vec{F}_i = \vec{F}_{i}^{ext} + \sum_{j \neq i} \vec{F}_{ij}^{int} \)

where \( \vec{F}_{ij}^{int} \) is force on \( i^{th} \) particle due to \( j^{th} \) particle.

5.1 Internal Forces Cancel Out

By Newton’s third law, internal forces between pairs of particles are equal and opposite:

\( \displaystyle \vec{F}_{ij}^{int} = - \vec{F}_{ji}^{int} \)

Summing all internal forces over the system cancels out:

\( \displaystyle \sum_{i=1}^n \sum_{j \neq i} \vec{F}_{ij}^{int} = \vec{0} \)

Hence, total force on system equals sum of external forces only.

5.2 Equation of Motion for the System

Therefore,

\( \displaystyle \sum \vec{F}^{ext} = M \vec{A} \)

where \( M \) is total mass, \( \vec{A} \) acceleration of center of mass.

6. Work-Energy Theorem for System of Particles

Work done by external forces on the system changes the total kinetic energy of the system.

Total kinetic energy \( K \) is:

\( \displaystyle K = \frac{1}{2} M V^2 + K_{internal} \)

where \( \frac{1}{2} M V^2 \) is kinetic energy of center of mass motion and \( K_{internal} \) is kinetic energy due to motion relative to COM.

7. Collisions and Conservation Laws in Systems

Momentum conservation applies to a system of particles if no external net force acts on it.

Internal forces can change individual particle momenta but do not affect total system momentum.

7.1 Types of Collisions

7.2 Collision Problems in System of Particles

Often JEE problems involve multiple particle collisions where applying conservation laws to the system simplifies calculations.

8. Examples & Problem-Solving

Example 1: Center of Mass of Two Particles

Two particles of masses 2 kg and 3 kg are located at points \( (1, 2) \) m and \( (4, 6) \) m respectively. Find the coordinates of the center of mass.

Solution:

\( \displaystyle X_{cm} = \frac{2 \times 1 + 3 \times 4}{2+3} = \frac{2 + 12}{5} = \frac{14}{5} = 2.8 \, m \)
\( \displaystyle Y_{cm} = \frac{2 \times 2 + 3 \times 6}{2+3} = \frac{4 + 18}{5} = \frac{22}{5} = 4.4 \, m \)

Coordinates of COM = (2.8, 4.4) m

Example 2: Velocity of Center of Mass

Two particles move with velocities \( \vec{v}_1 = 3 \hat{i} + 4 \hat{j} \) m/s and \( \vec{v}_2 = -1 \hat{i} + 2 \hat{j} \) m/s and masses 2 kg and 3 kg. Find velocity of center of mass.

Solution:

\( \displaystyle \vec{V} = \frac{2 (3 \hat{i} + 4 \hat{j}) + 3 (-1 \hat{i} + 2 \hat{j})}{2 + 3} = \frac{(6 - 3) \hat{i} + (8 + 6) \hat{j}}{5} = \frac{3 \hat{i} + 14 \hat{j}}{5} \)

\( \displaystyle \vec{V} = 0.6 \hat{i} + 2.8 \hat{j} \, m/s \)

Example 3: Conservation of Momentum in Collision

Two particles of masses 5 kg and 3 kg move towards each other with speeds 6 m/s and 4 m/s respectively. They collide and stick together. Find velocity of combined mass after collision.

Solution:

Assuming 1D motion along x-axis, taking direction of 5 kg mass as positive:

\( \displaystyle v = \frac{5 \times 6 - 3 \times 4}{5 + 3} = \frac{30 - 12}{8} = \frac{18}{8} = 2.25 \, m/s \)

The combined mass moves with velocity 2.25 m/s in the direction of the 5 kg mass.

9. Internal and External Forces in Detail

Forces acting on particles of a system can be classified into:

Internal forces affect relative motion of particles but do not change total momentum of the system.

10. Mathematical Treatment of Internal Forces

For particles \( i \) and \( j \), internal force exerted by \( j \) on \( i \) is \( \vec{F}_{ij} \).

According to Newton’s third law:

\( \displaystyle \vec{F}_{ij} = - \vec{F}_{ji} \)

Summing over all internal forces in the system:

\( \displaystyle \sum_{i=1}^n \sum_{j \neq i} \vec{F}_{ij} = \vec{0} \)

This property simplifies the equation of motion for the system, as internal forces do not contribute to net external force.

11. Applications in JEE Problems