What is Centre of Mass in physics?

Centre of Mass (COM) is a unique point in a system where the entire mass can be considered concentrated. For a system of particles, x_cm = Σ(m_i·x_i)/Σm_i. It represents the average position of mass distribution and follows Newton's laws as a point particle.

What is the difference between centre of mass and centre of gravity?

Centre of Mass depends only on mass distribution while Centre of Gravity depends on gravitational field variation. In uniform gravitational field (near Earth's surface), COM and COG coincide. In non-uniform fields or in space (g=0), they differ. For all JEE problems, COM = COG.

What is the formula for centre of mass of two particles?

For two particles with masses m₁ and m₂ at positions x₁ and x₂, the centre of mass is: x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂). COM divides the line joining two masses in inverse ratio of masses: d₁/d₂ = m₂/m₁.

What is conservation of linear momentum?

If no external force acts on a system (F_ext = 0), the total linear momentum remains constant: P_initial = P_final. This means m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. This applies to collisions, explosions, and all closed systems.

What is rocket propulsion principle?

Rocket propulsion works on conservation of momentum. When gas is ejected backward, rocket moves forward. Thrust force = v_rel × |dm/dt| where v_rel is exhaust velocity relative to rocket. Final velocity is given by Tsiolkovsky equation: v = v_rel·ln(m₀/m).

How many questions come from Centre of Mass in JEE?

Centre of Mass typically has 8-12% weightage in JEE. In JEE Main, expect 2-3 questions (8-12 marks), and in JEE Advanced, expect 3-4 questions (12-16 marks). Most common topics are COM position, conservation of momentum, and rocket propulsion.

What is COM of semicircular disc?

The centre of mass of a semicircular disc of radius R is at distance 4R/3π from the diameter (on the axis of symmetry). This is approximately 0.424R from the flat edge. This must be memorized for JEE.

What is impulse in physics?

Impulse (J) is the product of force and time: J = F·Δt = Δp (change in momentum). For variable force, J = ∫F dt. Impulse equals the area under Force-Time graph. Unit is N·s or kg·m/s. Used in collision and impact problems.

Centre of Mass and Momentum Class 11 Physics Notes for JEE 2025

Centre of Mass and Collision JEE notes, Formulas, PYQs — Centre of Mass and Collision JEE Notes, Formulas, PYQs

Introduction to Centre of Mass

The Centre of Mass (COM) is a unique point in a system of particles or an extended body where the entire mass of the system can be considered to be concentrated. It is the point where the weighted position vectors of all the particles sum to zero.

1.1 What is Centre of Mass?

Definition

Centre of Mass is the point at which the entire mass of a body or system of particles can be assumed to be concentrated for analyzing translational motion.

Physical Significance

• Point where external force can be applied
• Represents average position of mass
• Follows Newton's Laws as a point particle

Key Properties

• May or may not lie inside the body
• Depends on mass distribution
• Unique for any given configuration

1.2 Centre of Mass vs Centre of Gravity

Aspect	Centre of Mass (COM)	Centre of Gravity (COG)
Definition	Point where total mass is concentrated	Point where total weight acts
Depends on	Mass distribution only	Gravitational field variation
Uniform g	COM = COG (Same point)
Non-uniform g	Remains same	Changes position
In space (g = 0)	Exists	Does not exist

💡 JEE Important Point

For all JEE problems near Earth's surface (uniform gravitational field), COM = COG. The distinction only matters for very large objects in non-uniform fields (like satellites extending over large heights).

1.3 Examples Where COM Lies Outside the Body

🍩

Ring/Torus

COM at geometric center

🌙

Hollow Hemisphere

COM inside hollow region

🔺

Boomerang

COM outside material

Position of Centre of Mass

2.1 Two Particle System

COM of Two Particles

x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}

Distance from m₁:

d_1 = \frac{m_2 \cdot d}{m_1 + m_2}

Distance from m₂:

d_2 = \frac{m_1 \cdot d}{m_1 + m_2}

💡 Quick Formula

COM divides the line joining two masses in inverse ratio of masses:

\frac{d_1}{d_2} = \frac{m_2}{m_1}

COM is always closer to the heavier mass!

2.2 System of n Particles (Discrete)

General Formula for n Particles

x-coordinate:

x_{cm} = \frac{\sum m_ix_i}{\sum m_i}

y-coordinate:

y_{cm} = \frac{\sum m_iy_i}{\sum m_i}

z-coordinate:

z_{cm} = \frac{\sum m_iz_i}{\sum m_i}

Vector Form:

\vec{r}_{cm} = \frac{\sum m_i\vec{r}_i}{\sum m_i} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + ... + m_n\vec{r}_n}{m_1 + m_2 + ... + m_n}

2.3 Continuous Mass Distribution

Integration Method

x-coordinate:

x_{cm} = \frac{\int x \, dm}{\int dm}

y-coordinate:

y_{cm} = \frac{\int y \, dm}{\int dm}

z-coordinate:

z_{cm} = \frac{\int z \, dm}{\int dm}

Mass Elements for Different Geometries:

Geometry	Linear Density (λ)	Mass Element (dm)
1D - Rod/Wire	λ = M/L (kg/m)	dm = λ dx
2D - Plate	σ = M/A (kg/m²)	dm = σ dA
3D - Solid	ρ = M/V (kg/m³)	dm = ρ dV

📝 Solved Example 1 (JEE Main Pattern)

Question: Three particles of masses 1 kg, 2 kg, and 3 kg are placed at (1, 2), (3, 4), and (5, 6) respectively. Find the position of COM.

Solution:

Total mass M = 1 + 2 + 3 = 6 kg

x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{M} = \frac{1(1) + 2(3) + 3(5)}{6} = \frac{1 + 6 + 15}{6} = \frac{22}{6} = \frac{11}{3}

y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{M} = \frac{1(2) + 2(4) + 3(6)}{6} = \frac{2 + 8 + 18}{6} = \frac{28}{6} = \frac{14}{3}

\text{COM} = \left(\frac{11}{3}, \frac{14}{3}\right) \approx (3.67, 4.67)

COM of Common Shapes

These standard results must be memorized for JEE. They save time and are directly applicable in many problems.

3.1 One-Dimensional Objects

Shape	Position of COM
Uniform Rod (length L)	At center: L/2 from either end
Semi-circular Wire (radius R)	y = 2R/π from center (on axis of symmetry)
Quarter circular arc	x = y = 2R/π from center

3.2 Two-Dimensional Objects (Laminas)

Shape	Position of COM
Triangular Lamina	At centroid: h/3 from base (intersection of medians)
Circular Disc	At geometric center
Semi-circular Disc (radius R)	y = 4R/3π from diameter
Quarter Disc	x = y = 4R/3π from center
Sector of Disc (angle 2α)	r = (2R sin α)/(3α) from center

3.3 Three-Dimensional Objects

Shape	Position of COM
Solid Sphere	At geometric center
Hollow Sphere	At geometric center
Solid Hemisphere (radius R)	y = 3R/8 from flat surface
Hollow Hemisphere (radius R)	y = R/2 from flat surface
Solid Cone (height h)	y = h/4 from base (3h/4 from apex)
Hollow Cone (height h)	y = h/3 from base (2h/3 from apex)

💡 Memory Pattern

Semi-shapes (2D vs Wire):

Semicircular wire: 2R/π

Semicircular disc: 4R/3π

Hemisphere (Solid vs Hollow):

Solid Hemisphere: 3R/8

Hollow Hemisphere: R/2

3.4 Cavity Problems (Negative Mass Method)

When a portion is removed from a body:

x_{cm} = \frac{M \cdot x_1 - m \cdot x_2}{M - m}

M = mass of complete body, m = mass of removed portion
x₁ = COM of complete body, x₂ = COM of removed portion

📝 Solved Example 2 (JEE Main 2022 Type)

Question: A circular disc of radius R has a circular hole of radius R/2 cut from it. If the center of hole is at R/2 from center of disc, find position of new COM from original center.

Solution:

Let σ = mass per unit area

Mass of complete disc: M = σπR²

Mass of removed part: m = σπ(R/2)² = σπR²/4 = M/4

Remaining mass: M - M/4 = 3M/4

COM of complete disc: x₁ = 0

COM of removed part: x₂ = R/2

x_{cm} = \frac{M \cdot 0 - (M/4) \cdot (R/2)}{M - M/4} = \frac{-MR/8}{3M/4} = \frac{-R}{6}

\text{COM shifts by } \frac{R}{6} \text{ away from the hole}

Motion of Centre of Mass

The COM of a system moves as if all the mass were concentrated at that point and all external forces were applied there. This is the most powerful concept in this chapter.

4.1 Velocity of COM

\vec{v}_{cm} = \frac{d\vec{r}_{cm}}{dt} = \frac{\sum m_i\vec{v}_i}{\sum m_i} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2 + ...}{M}

4.2 Acceleration of COM

\vec{a}_{cm} = \frac{d\vec{v}_{cm}}{dt} = \frac{\sum m_i\vec{a}_i}{\sum m_i}

4.3 Equation of Motion for COM

Newton's Second Law for System

\vec{F}_{ext} = M\vec{a}_{cm}

The COM accelerates only due to EXTERNAL forces. Internal forces cannot change COM motion!

⚠️ Key Insight (Most Important!)

If F_ext = 0:

a_cm = 0 → v_cm = constant
If system was at rest, COM remains stationary
Individual particles may move, but COM doesn't!

4.4 Applications

Man on Boat Problem

When a man walks on a boat (no external horizontal force):

m_1 \Delta x_1 + m_2 \Delta x_2 = 0

Man moves right → Boat moves left

Explosion in Mid-air

When a projectile explodes at highest point:

• COM continues parabolic path
• COM lands at same point as original
• Individual pieces may land elsewhere

📝 Solved Example 3 (JEE Advanced Pattern)

Question: A 60 kg man stands on one end of a 40 kg boat of length 10 m. He walks to the other end. How far does the boat move? (Neglect water resistance)

Solution:

No external horizontal force → COM remains stationary

Let boat move distance x to left when man moves to right end.

Man's displacement relative to ground = 10 - x (to right)

Using: m₁Δx₁ + m₂Δx₂ = 0

\[60(10 - x) + 40(-x) = 0\]

\[600 - 60x - 40x = 0\]

\[100x = 600\]

x = 6 \text{ m (boat moves 6 m opposite to man's motion)}

Linear Momentum

5.1 Definition

Linear Momentum

\vec{p} = m\vec{v}

SI Unit

kg·m/s or N·s

Dimensional Formula

[MLT⁻¹]

5.2 Momentum of System

\vec{P}_{system} = \sum m_i\vec{v}_i = M\vec{v}_{cm}

Total momentum of system = (Total mass) × (Velocity of COM)

5.3 Newton's Second Law in Momentum Form

\vec{F} = \frac{d\vec{p}}{dt}

Force = Rate of change of momentum

💡 Relation between p and KE

p = \sqrt{2mKE}

KE = \frac{p^2}{2m}

• Same KE → Heavier object has more momentum (p ∝ √m)

• Same momentum → Lighter object has more KE (KE ∝ 1/m)

Conservation of Linear Momentum

📜 Law of Conservation of Momentum

"If no external force acts on a system, the total linear momentum of the system remains constant."

Mathematical Form

\text{If } \vec{F}_{ext} = 0, \text{ then } \vec{P}_{initial} = \vec{P}_{final}

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

6.1 Conditions for Conservation

✓ Momentum IS Conserved When:

• No external force (F_ext = 0)
• External forces cancel out
• In a particular direction if F_ext ⊥ that direction

✗ Momentum NOT Conserved When:

• Net external force exists
• In direction of net external force

6.2 Component-wise Conservation

Momentum can be conserved in one direction but not another!

x-direction:

If F_x,ext = 0

Σp_x = constant

y-direction:

If F_y,ext = 0

Σp_y = constant

z-direction:

If F_z,ext = 0

Σp_z = constant

6.3 Applications

Scenario	Application
Collisions	Momentum conserved (both elastic and inelastic)
Explosions	Total momentum before = Total momentum after
Gun Recoil	m_bulletv_bullet = m_gunv_recoil
Rocket Propulsion	Momentum of gas = Momentum gained by rocket
Man jumping from boat	Horizontal momentum conserved

📝 Solved Example 4

Question: A 5 kg gun fires a 50 g bullet at 400 m/s. Find recoil velocity of gun.

Solution:

Initial momentum = 0 (system at rest)

By conservation of momentum:

m_{bullet}v_{bullet} + m_{gun}v_{gun} = 0

0.05 \times 400 + 5 \times v_{gun} = 0

v_{gun} = -\frac{20}{5} = -4 \text{ m/s}

\text{Recoil velocity} = 4 \text{ m/s (opposite to bullet)}

Variable Mass Systems & Rocket Propulsion

In systems where mass changes with time (like rockets), we cannot directly use F = ma. Instead, we must use the momentum form of Newton's Law. This is a JEE Advanced favorite topic!

7.1 General Equation for Variable Mass

Variable Mass Equation

\vec{F}_{ext} + \vec{v}_{rel}\frac{dm}{dt} = m\frac{d\vec{v}}{dt}

v_rel = velocity of ejected/added mass relative to main body

7.2 Rocket Propulsion

Rocket in Free Space (No Gravity)

Thrust Force:

F_{thrust} = v_{rel}\left|\frac{dm}{dt}\right| = v_{rel} \cdot \mu

μ = rate of fuel burn (kg/s)

Instantaneous Acceleration:

a = \frac{v_{rel} \cdot \mu}{m}

m = instantaneous mass

7.3 Rocket Velocity (Tsiolkovsky Equation)

Final Velocity of Rocket

v = v_0 + v_{rel} \ln\left(\frac{m_0}{m}\right)

m₀ = initial mass, m = final mass (after fuel burn)

If starting from rest (v₀ = 0):

v = v_{rel} \ln\left(\frac{m_0}{m}\right)

7.4 Rocket with Gravity

Vertical Launch (against gravity)

Net Force:

F_{net} = F_{thrust} - mg

Condition for liftoff:

F_{thrust} > m_0g

Velocity after time t:

v = v_{rel} \ln\left(\frac{m_0}{m_0 - \mu t}\right) - gt

⚠️ Important Sign Convention

dm/dt is negative for rocket (mass decreasing)
v_rel is velocity of exhaust relative to rocket (directed backward)
Thrust is always positive and opposite to exhaust direction

📝 Solved Example 5 (JEE Advanced Pattern)

Question: A rocket has initial mass 1000 kg. Gas is ejected at 2 km/s relative to rocket at rate 10 kg/s. Find (a) Initial thrust (b) Initial acceleration (c) Velocity after 50 seconds in free space.

Given: m₀ = 1000 kg, v_rel = 2000 m/s, μ = 10 kg/s

(a) Initial Thrust:

F_{thrust} = v_{rel} \times \mu = 2000 \times 10 = 20000 \text{ N} = 20 \text{ kN}

(b) Initial Acceleration:

a_0 = \frac{F_{thrust}}{m_0} = \frac{20000}{1000} = 20 \text{ m/s}^2

(c) Velocity after 50 s:

Mass after 50 s: m = 1000 - 10(50) = 500 kg

v = v_{rel} \ln\left(\frac{m_0}{m}\right) = 2000 \ln\left(\frac{1000}{500}\right)

v = 2000 \ln(2) = 2000 \times 0.693

v \approx 1386 \text{ m/s} \approx 1.4 \text{ km/s}

Impulse

8.1 Definition of Impulse

Impulse

\vec{J} = \vec{F} \cdot \Delta t = \Delta \vec{p} = m(\vec{v} - \vec{u})

SI Unit

N·s or kg·m/s

Dimensional Formula

[MLT⁻¹]

8.2 Impulse for Variable Force

\vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt = \text{Area under F-t curve}

8.3 Impulse-Momentum Theorem

Theorem

\vec{J} = \vec{p}_f - \vec{p}_i = \Delta \vec{p}

Impulse = Change in momentum

8.4 Applications of Impulse

Short Duration Collisions

When force acts for very short time:

J = F_{avg} \cdot \Delta t

Used in bat-ball, hammer-nail problems

Impulsive Forces

Very large forces acting for very short time:

• Collision forces
• Explosion forces
• Sudden string becoming taut

💡 Reducing Impact Force

Since J = F × Δt = constant (for same momentum change):

Increasing Δt decreases F

Airbags increase collision time → less force
Pulling hands back while catching → less impact
Cushioned shoes → reduce force on joints

📝 Solved Example 6

Question: A ball of mass 200 g moving at 10 m/s is hit by a bat. It returns at 15 m/s. If contact time is 0.02 s, find (a) Impulse (b) Average force.

Given: m = 0.2 kg, u = 10 m/s, v = -15 m/s (returns), Δt = 0.02 s

(a) Impulse:

J = m(v - u) = 0.2(-15 - 10) = 0.2 \times (-25) = -5 \text{ N·s}

Magnitude of Impulse = 5 N·s

(b) Average Force:

F_{avg} = \frac{J}{\Delta t} = \frac{5}{0.02} = 250 \text{ N}

J = 5 \text{ N·s}, \quad F_{avg} = 250 \text{ N}

📋 Complete Formula Sheet - Centre of Mass

Position of COM

x_cm = Σm_ix_i/Σm_i

For 2 particles: x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂)

Distance ratio: d₁/d₂ = m₂/m₁

Continuous: x_cm = ∫x dm / ∫dm

COM of Shapes

Semicircular wire: 2R/π

Semicircular disc: 4R/3π

Solid hemisphere: 3R/8

Hollow hemisphere: R/2

Solid cone: h/4 from base

Hollow cone: h/3 from base

Motion of COM

v_cm = Σm_iv_i/M

a_cm = Σm_ia_i/M

F_ext = Ma_cm

If F_ext = 0: v_cm = constant

Momentum

p = mv

P_system = Mv_cm

F = dp/dt

p = √(2mKE)

Rocket Propulsion

Thrust = v_rel × |dm/dt|

v = v₀ + v_rel ln(m₀/m)

a = v_relμ/m

With gravity: F_net = Thrust - mg

Impulse

J = FΔt = Δp

J = ∫F dt

J = m(v - u)

F_avg = J/Δt

Download Formula PDF

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Position of COM: 25%
✓ Conservation of Momentum: 30%
✓ Motion of COM: 20%
✓ Rocket/Variable Mass: 15%
✓ Impulse: 10%

JEE Advanced (Last 5 Years)

✓ COM of continuous bodies: 25%
✓ Variable mass systems: 25%
✓ Multi-body COM problems: 20%
✓ COM motion without external force: 20%
✓ Cavity problems: 10%

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)

JEE Advanced: 12-16 marks (3-4 questions)

Difficulty Level: Medium to Hard

Study Time Required: 12-15 hours

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Two particles of mass 3 kg and 5 kg are at (2, 3) and (4, 5). Find COM.
A uniform rod of length 4 m and mass 2 kg. Find position of COM from one end.
Find COM of semicircular wire of radius 10 cm.
Three particles of equal mass are at vertices of equilateral triangle. Find COM.
A 70 kg man and 30 kg cart. Man walks 5 m forward on cart. How far does cart move?
A bullet of mass 10 g with velocity 500 m/s hits a 5 kg block at rest. Find common velocity.
Calculate impulse when a 1 kg ball moving at 5 m/s is stopped.
A projectile explodes at highest point into two equal pieces. One falls vertically. Where does other land?

Level 2: Intermediate (JEE Main/Advanced)

Find COM of a non-uniform rod of length L with linear density λ = λ₀(1 + x/L).
A disc of radius R has a hole of radius R/3 at distance R/2 from center. Find new COM.
A hemisphere and cone of same base and height are joined. Find COM of combination.
Two blocks 2 kg and 3 kg connected by spring on smooth surface. 2 kg given 10 m/s velocity. Find velocity of COM and individual velocities when spring is maximum compressed.
A 1000 kg rocket ejects 5 kg/s at 1000 m/s. Find acceleration at t = 0 and t = 100 s.
A body explodes into 3 pieces of mass ratio 1:1:2. Two equal pieces fly perpendicular to each other with 30 m/s. Find velocity of third piece.
Sand falls on a conveyor belt at 2 kg/s. Belt moves at 5 m/s. Find force needed to maintain belt speed.
Find COM of a sector of disc with angle 60° and radius R.

Level 3: Advanced (JEE Advanced/Olympiad)

A triangular plate with vertices at (0,0), (a,0), (0,b) has density ρ = kxy. Find COM.
A sphere of radius R has spherical cavity of radius R/2 with center at R/2 from center. Find COM.
Two boats of mass M each carry persons of mass m. Person jumps from boat 1 to boat 2, then from 2 to 1. Find net displacement of boats.
A rocket in gravity-free space starts from rest. Half the mass is fuel. What fraction must be burnt to reach v = v_rel?
Chain of length L hanging from table starts falling when L/n hangs. Find velocity of chain when just leaving table.
Two identical blocks connected by spring are on smooth surface. One is given velocity v₀ towards other. Find maximum compression and time period of oscillation.
A rocket has initial mass M and burns fuel at rate μ with exhaust velocity u. Find velocity when mass becomes M/e.
Find COM of a hollow cone of height h and radius R.

Get Solutions PDF with Step-by-Step Working

📑 Quick Navigation - Centre of Mass

Centre of Mass JEE Main & Advanced 2025-26

Introduction to Centre of Mass

1.1 What is Centre of Mass?

Definition

1.2 Centre of Mass vs Centre of Gravity

💡 JEE Important Point

1.3 Examples Where COM Lies Outside the Body

Position of Centre of Mass

2.1 Two Particle System

COM of Two Particles

💡 Quick Formula

2.2 System of n Particles (Discrete)

General Formula for n Particles

2.3 Continuous Mass Distribution

Integration Method

Mass Elements for Different Geometries:

📝 Solved Example 1 (JEE Main Pattern)

COM of Common Shapes

3.1 One-Dimensional Objects

3.2 Two-Dimensional Objects (Laminas)

3.3 Three-Dimensional Objects

💡 Memory Pattern

3.4 Cavity Problems (Negative Mass Method)

When a portion is removed from a body:

📝 Solved Example 2 (JEE Main 2022 Type)

Motion of Centre of Mass

4.1 Velocity of COM

4.2 Acceleration of COM

4.3 Equation of Motion for COM

Newton's Second Law for System

⚠️ Key Insight (Most Important!)

4.4 Applications

Man on Boat Problem

Explosion in Mid-air

📝 Solved Example 3 (JEE Advanced Pattern)

Linear Momentum

5.1 Definition

Linear Momentum

5.2 Momentum of System

5.3 Newton's Second Law in Momentum Form

💡 Relation between p and KE

Conservation of Linear Momentum

📜 Law of Conservation of Momentum

Mathematical Form

6.1 Conditions for Conservation

6.2 Component-wise Conservation

Momentum can be conserved in one direction but not another!

6.3 Applications

📝 Solved Example 4

Variable Mass Systems & Rocket Propulsion

7.1 General Equation for Variable Mass

Variable Mass Equation

7.2 Rocket Propulsion

Rocket in Free Space (No Gravity)

7.3 Rocket Velocity (Tsiolkovsky Equation)

Final Velocity of Rocket

7.4 Rocket with Gravity

Vertical Launch (against gravity)

⚠️ Important Sign Convention

📝 Solved Example 5 (JEE Advanced Pattern)

Impulse

8.1 Definition of Impulse

Impulse

8.2 Impulse for Variable Force

8.3 Impulse-Momentum Theorem

Theorem

8.4 Applications of Impulse

Short Duration Collisions

Impulsive Forces

💡 Reducing Impact Force

📝 Solved Example 6

📋 Complete Formula Sheet - Centre of Mass

Position of COM

COM of Shapes

Motion of COM

Momentum

Rocket Propulsion

Impulse

📝 Previous Year Questions Analysis