What is the formula for Newton's Second Law?

Newton's Second Law is expressed as F = ma (Force equals mass times acceleration) or F = dp/dt (Force equals rate of change of momentum). This is the most important formula in Laws of Motion.

What is the difference between static and kinetic friction?

Static friction acts when an object is at rest and prevents motion (fs ≤ μsN), while kinetic friction acts when an object is already in motion (fk = μkN). The coefficient of static friction (μs) is always greater than the coefficient of kinetic friction (μk).

How many questions come from Laws of Motion in JEE?

Laws of Motion typically has 8-15% weightage in JEE. In JEE Main, expect 2-3 questions (8-12 marks), and in JEE Advanced, expect 3-4 questions (12-18 marks). Friction and connected bodies are the most frequently asked topics.

What is centripetal force formula?

Centripetal force formula is Fc = mv²/r or Fc = mrω². It is the net force directed toward the center required for circular motion. Centripetal force is not a new type of force but is provided by tension, friction, gravity, or normal force.

Laws of Motion JEE Physics - Notes, Important Questions, Formulas & PYQs

Q: What is Newton's First Law of Motion?

Newton's First Law states that every object continues in its state of rest or uniform motion in a straight line unless compelled by an external unbalanced force to change that state. This is also known as the Law of Inertia.

Laws of Motion JEE notes, Formulas, PYQs — Laws of Motion JEE Notes, Formulas, PYQs

Force and Inertia

Force and Inertia are the foundational concepts of classical mechanics. Understanding these concepts is essential for mastering Newton's Laws of Motion and solving complex JEE problems involving dynamics.

1.1 What is Force?

Force is a physical quantity that causes or tends to cause a change in the state of rest or uniform motion of an object. It is a vector quantity with both magnitude and direction.

SI Unit

Newton (N)

1 N = 1 kg·m/s²

Dimensional Formula

[MLT⁻²]

Mass × Length × Time⁻²

Characteristics of Force:

✓

Force is a vector quantity (has magnitude and direction)

✓

Multiple forces add using vector addition rules

✓

Can change speed, direction, or shape of objects

✓

Net force determines the acceleration of a body

1.2 What is Inertia?

Inertia is the natural tendency of an object to resist any change in its state of rest or uniform motion. It is a fundamental property of matter.

💡 Key Points about Inertia

Inertia is directly proportional to mass
Heavier objects have greater inertia
Inertia is NOT a force; it's a property of matter
Mass itself is the measure of inertia (no separate unit)

Types of Inertia:

1. Inertia of Rest

Tendency to remain at rest

Example: When a bus starts suddenly, passengers jerk backward

2. Inertia of Motion

Tendency to remain in motion

Example: When a bus stops suddenly, passengers jerk forward

3. Inertia of Direction

Tendency to maintain direction

Example: Mud flies off tangentially from a rotating wheel

Newton's First Law of Motion

📜 Statement (Law of Inertia)

"Every object continues in its state of rest or uniform motion in a straight line unless compelled by an external unbalanced force to change that state."

2.1 Mathematical Form

\text{If } \sum \vec{F} = 0, \text{ then } \vec{a} = 0

No net force → No acceleration → Constant velocity (including zero)

Important Implications:

Force is NOT required to maintain motion

Objects in motion stay in motion naturally

Force is required to CHANGE motion

Starting, stopping, or turning requires force

Zero net force ≠ Zero velocity

Velocity could be any constant value

Defines Inertial Reference Frames

First law is valid in inertial frames only

2.2 Inertial vs Non-Inertial Frames

Aspect	Inertial Frame	Non-Inertial Frame
Definition	Frame where First Law holds true	Frame that is accelerating
Acceleration	Zero or constant velocity	Non-zero acceleration
Newton's Laws	Apply directly	Need pseudo forces
Examples	Ground, train at constant velocity	Accelerating car, rotating platform

⚠️ Pseudo Force (JEE Advanced Important)

In a non-inertial frame accelerating with a₀, a pseudo force must be applied:

\vec{F}_{pseudo} = -m\vec{a}_0

Pseudo force acts opposite to the frame's acceleration and on every object in that frame.

Newton's Second Law of Motion

📜 Statement

"The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force."

3.1 Mathematical Formulation

Standard Form

\vec{F} = m\vec{a}

Valid when mass is constant

General Form

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

Valid for variable mass (rockets)

3.2 Derivation from Momentum Form

Step 1: Start with momentum: \(\vec{p} = m\vec{v}\)

Step 2: Apply Newton's statement: \(\vec{F} \propto \frac{d\vec{p}}{dt}\)

Step 3: \(\vec{F} = k\frac{d\vec{p}}{dt}\) where k = 1 in SI units

Step 4: \(\vec{F} = \frac{d(m\vec{v})}{dt}\)

Step 5: For constant mass: \(\vec{F} = m\frac{d\vec{v}}{dt}\)

\boxed{\vec{F} = m\vec{a}}

3.3 Impulse and Momentum

Impulse-Momentum Theorem

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

Impulse

F·Δt

Force × Time

Δp

Change in momentum

N·s or kg·m/s

SI Unit

Applications of Impulse:

🏏 Catching a Ball

Pulling hands back increases Δt, reducing F (since J = FΔt = constant)

🚗 Airbags

Airbags increase collision time, significantly reducing impact force

🏃 Sand Pit

Sand increases stopping time, reducing force on athlete's legs

📝 Solved Example 1 (JEE Main Pattern)

Question: A ball of mass 0.5 kg moving at 10 m/s is hit by a bat and returns at 15 m/s. If contact time is 0.01 s, find the average force.

Given: m = 0.5 kg, u = 10 m/s, v = -15 m/s (opposite direction), Δt = 0.01 s

Solution:

\Delta p = m(v - u) = 0.5(-15 - 10) = 0.5 \times (-25) = -12.5 \text{ kg·m/s}

F = \frac{\Delta p}{\Delta t} = \frac{-12.5}{0.01} = -1250 \text{ N}

\text{Magnitude of Force} = 1250 \text{ N}

Newton's Third Law of Motion

📜 Statement (Action-Reaction Law)

"To every action, there is always an equal and opposite reaction."

4.1 Mathematical Form

\vec{F}_{12} = -\vec{F}_{21}

F₁₂

Force on body 1 by body 2

F₂₁

Force on body 2 by body 1

4.2 Key Characteristics

✓ Act on DIFFERENT bodies

Never on the same object

✓ Equal in magnitude

|F₁₂| = |F₂₁| always

✓ Opposite in direction

180° apart

✓ Act simultaneously

No time delay

✓ Same type of force

Both gravitational, both contact, etc.

✓ Do NOT cancel out

They act on different bodies!

4.3 Action-Reaction Pairs

Situation	Action	Reaction
Book on table	Book pushes table down	Table pushes book up
Earth-Person	Earth pulls person down	Person pulls Earth up
Rocket	Rocket pushes gas downward	Gas pushes rocket upward
Swimming	Swimmer pushes water backward	Water pushes swimmer forward
Walking	Foot pushes ground backward	Ground pushes foot forward

❌ Common Misconception

"Action and reaction cancel each other" - This is WRONG!

They act on different bodies, so they cannot cancel. For forces to cancel, they must act on the same body.

Free Body Diagrams (FBD)

A Free Body Diagram is a simplified representation showing all external forces acting on a single isolated object. Mastering FBD is crucial for solving 70% of Laws of Motion problems in JEE.

⭐ Why FBD is Important for JEE

70% of Laws of Motion problems become straightforward once you draw the correct FBD. It's your roadmap to the solution - if your FBD is wrong, your entire solution will be wrong!

5.1 Steps to Draw FBD

Isolate the body

Draw only the object you're analyzing, remove everything else

Draw weight (mg) downward

Always acts vertically downward from center of mass

Draw Normal force perpendicular to surface

Always 90° to contact surface, pointing away from surface

Draw friction along the surface

Opposes motion or tendency of motion

Draw tension along strings/ropes

Always pulling, never pushing

Add any applied external forces

As given in the problem

5.2 Do's and Don'ts

✅ DO Include

✓ All external forces
✓ Weight (mg)
✓ Normal force (N)
✓ Friction (f)
✓ Tension (T)
✓ Applied forces

❌ DON'T Include

✗ Forces the body exerts on others
✗ Internal forces
✗ Velocity or acceleration vectors
✗ "ma" as a force
✗ Connected objects

5.3 Common FBD Cases

Block on Horizontal Surface

       ↑ N (Normal)
       |
   ←f  ■  → F (Applied)
       |
       ↓ mg (Weight)

N = mg (if no vertical force)

Block on Inclined Plane (θ)

     ↑ N (⊥ to surface)
    /
   ■ ← f (up incline)
  /|
 / |
   ↓ mg (vertical)

N = mg cos θ

Friction

Friction is the force that opposes relative motion or tendency of relative motion between two surfaces in contact. It's one of the most important topics for JEE with 3-4 questions expected every year.

6.1 Types of Friction

Static Friction (fs)

Prevents motion from starting

0 \leq f_s \leq \mu_s N

Self-adjusting force

Kinetic Friction (fk)

Acts during motion

f_k = \mu_k N

Constant magnitude

Rolling Friction

During rolling motion

f_r = \mu_r N

μr << μk << μs

⚠️ Important: μs > μk (Always!)

It takes more force to start motion than to maintain it. That's why static friction coefficient is always greater than kinetic friction coefficient.

6.2 Laws of Friction

1. Friction is proportional to Normal force: f ∝ N

2. Friction is independent of area of contact (for rigid bodies)

3. Kinetic friction is independent of velocity (for moderate speeds)

4. Static friction is self-adjusting up to a maximum value

6.3 Angle of Repose & Angle of Friction

Angle of Repose (θ)

Maximum angle of incline at which object stays at rest

\tan \theta = \mu_s

Angle of Friction (λ)

Angle between N and resultant of N and f

\tan \lambda = \mu

💡 Key Insight

Angle of Repose = Angle of Friction (θ = λ)
This is a frequently asked MCQ in JEE Main!

6.4 Motion on Inclined Plane

Key Formulas for Incline (angle θ)

Normal Force:

N = mg\cos\theta

Component along incline:

F_\parallel = mg\sin\theta

Acceleration (sliding down):

a = g(\sin\theta - \mu\cos\theta)

Condition for sliding:

\theta > \tan^{-1}(\mu_s)

📝 Solved Example 2 (JEE Main 2023 Type)

Question: A block of mass 10 kg is on a horizontal surface with μs = 0.4 and μk = 0.3. A horizontal force of 50 N is applied. Find acceleration. (g = 10 m/s²)

Step 1: Find Normal force

N = mg = 10 \times 10 = 100 \text{ N}

Step 2: Find maximum static friction

f_{s(max)} = \mu_s N = 0.4 \times 100 = 40 \text{ N}

Step 3: Check if block moves

Applied force (50 N) > fs(max) (40 N) ∴ Block moves!

Step 4: Find kinetic friction

f_k = \mu_k N = 0.3 \times 100 = 30 \text{ N}

Step 5: Apply F = ma

50 - 30 = 10 \times a

a = 2 \text{ m/s}^2

Circular Motion & Forces

When an object moves in a circle, it requires a net force directed toward the center (centripetal force). Understanding what provides this force is key to solving circular motion problems.

7.1 Centripetal Force

F_c = \frac{mv^2}{r} = mr\omega^2 = m \cdot a_c

Always directed toward center of circular path

⚠️ Important Concept

Centripetal force is NOT a new type of force! It's the NET force toward the center, provided by other forces like tension, friction, gravity, or normal force.

7.2 Sources of Centripetal Force

Situation	Centripetal Force Provided By
Stone tied to string (horizontal)	Tension in string
Car turning on flat road	Friction between tires and road
Car on banked road (no friction)	Horizontal component of Normal force
Satellite orbiting Earth	Gravitational force
Electron around nucleus	Electrostatic attraction

7.3 Banking of Roads

Optimal Banking Angle (No Friction)

\tan\theta = \frac{v^2}{rg}

θ = banking angle, v = speed, r = radius, g = gravity

With Friction (Maximum Speed)

v_{max} = \sqrt{rg\left(\frac{\mu + \tan\theta}{1 - \mu\tan\theta}\right)}

7.4 Centrifugal Force (Pseudo Force)

Inertial Frame

Only centripetal force exists (toward center)

Real force providing circular motion

Rotating Frame

Centrifugal force appears (outward)

Pseudo force = mv²/r (outward)

Connected Bodies & Pulley Systems

Problems involving multiple connected objects are JEE favorites. The key is to draw separate FBDs, write F = ma for each, and use constraint relations.

8.1 Strategy for Connected Bodies

Draw separate FBD for each body

Write F = ma for each body in appropriate direction

Identify constraint relations (same string = same |a|)

Solve simultaneous equations

8.2 Atwood Machine

    ╔═══╗
    ║   ║ (Pulley)
    ║   ║
   ─┴───┴─
   |     |
   |     |
  [m₁]  [m₂]
   ↓     ↓

Formulas (m₂ > m₁)

Acceleration:

a = \frac{(m_2 - m_1)g}{m_1 + m_2}

Tension:

T = \frac{2m_1m_2g}{m_1 + m_2}

8.3 Two Blocks on Horizontal Surface

F → [m₁] ─── [m₂]
    (Connected by string)

System Acceleration:

a = \frac{F}{m_1 + m_2}

Tension in String:

T = \frac{m_2 \cdot F}{m_1 + m_2}

📝 Solved Example 3 (JEE Advanced Pattern)

Question: Two masses 5 kg and 3 kg are connected by a string over a frictionless pulley. Find acceleration and tension. (g = 10 m/s²)

Given: m₁ = 5 kg, m₂ = 3 kg, g = 10 m/s²

Acceleration:

a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(5-3) \times 10}{5+3} = \frac{20}{8} = 2.5 \text{ m/s}^2

Tension:

T = \frac{2m_1m_2g}{m_1 + m_2} = \frac{2 \times 5 \times 3 \times 10}{8} = \frac{300}{8} = 37.5 \text{ N}

\boxed{a = 2.5 \text{ m/s}^2, \quad T = 37.5 \text{ N}}

📋 Complete Formula Sheet - Laws of Motion

Newton's Laws

First Law: ΣF = 0 ⟹ a = 0

Second Law: F = ma = dp/dt

Third Law: F₁₂ = -F₂₁

Momentum: p = mv

Impulse: J = FΔt = Δp

Friction

Static: fs ≤ μsN

Kinetic: fk = μkN

Angle of repose: tan θ = μs

Angle of friction: tan λ = μ

Inclined Plane

Normal: N = mg cos θ

Parallel: F∥ = mg sin θ

a (smooth): a = g sin θ

a (rough): a = g(sin θ - μ cos θ)

Circular Motion

Fc = mv²/r = mrω²

Banking: tan θ = v²/rg

Max speed: v = √(μrg)

Atwood Machine

a = (m₂-m₁)g/(m₁+m₂)

T = 2m₁m₂g/(m₁+m₂)

Pseudo Force

F_pseudo = -ma₀

(in non-inertial frame with acceleration a₀)

Download Formula PDF

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Friction Problems: 35%
✓ Connected Bodies/Pulleys: 25%
✓ Inclined Plane: 20%
✓ Circular Motion Forces: 15%
✓ Pseudo Force: 5%

JEE Advanced (Last 5 Years)

✓ Multi-body problems: 30%
✓ Constraint motion: 25%
✓ Variable mass systems: 15%
✓ Non-inertial frames: 20%
✓ Circular + friction: 10%

Weightage Analysis

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

JEE Advanced: 12-18 marks (3-4 questions)

Difficulty Level: Medium to Hard

Study Time Required: 15-20 hours

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

A force of 20 N acts on a 5 kg mass. Find the acceleration.
Find maximum static friction if μs = 0.5 and N = 100 N.
A 10 kg block is on a 30° incline. Find normal force. (g = 10 m/s²)
Calculate impulse when a 2 kg ball moving at 5 m/s is stopped.
Two forces 3 N and 4 N act perpendicular to each other. Find resultant.
A car of mass 1000 kg accelerates at 2 m/s². Find the net force.
What is the centripetal force on a 0.5 kg stone moving at 4 m/s in circle of radius 2 m?
An Atwood machine has masses 6 kg and 4 kg. Find acceleration. (g = 10 m/s²)

Level 2: Intermediate (JEE Main/Advanced)

A 5 kg block on 37° incline with μ = 0.3. Will it slide? Find acceleration if it does.
Two blocks 3 kg and 5 kg connected by string on smooth surface. 40 N force applied on 3 kg. Find tension.
Find minimum force to move a 20 kg block on rough surface (μ = 0.4). Force applied at 30° to horizontal.
A car takes a turn on flat road. If μ = 0.5 and r = 100 m, find maximum safe speed.
Block of 2 kg on 10 kg block. If μ between blocks = 0.3 and floor is smooth, find max force on lower block so upper doesn't slide.
A pendulum bob of 0.2 kg swings with speed 3 m/s at bottom. If length = 0.5 m, find tension at bottom.
Find acceleration of wedge-block system if wedge angle is 45° and all surfaces smooth.
A 1000 kg car on banked road of 30°. If radius = 50 m, find optimum speed.

Level 3: Advanced (JEE Advanced/Olympiad)

A block on accelerating wedge. If wedge acceleration a₀ and angle θ, find condition for block to remain stationary relative to wedge.
Two blocks A (5 kg) on B (10 kg). Force F applied on B. μ between A and B = 0.3, μ between B and ground = 0.1. Find range of F for no slipping.
A chain of length L and mass M hangs with L/4 over edge. If μ = 0.4, will it slide? Find time to fall if it does.
Double Atwood machine: One pulley connected to another. Find acceleration of all masses.
A conical pendulum makes angle θ with vertical. Derive expression for angular velocity.
Block on rotating disc at distance r from center. If μ and ω given, find condition for no slipping.
Rocket of initial mass M₀ ejects gas at rate dm/dt with velocity u relative to rocket. Find thrust and acceleration.
Three blocks on frictionless surface connected. Force F on first. Find acceleration and all tensions.

Get Solutions PDF with Step-by-Step Working

📑 Quick Navigation - Laws of Motion

Laws of Motion JEE Main & Advanced 2025-26

Force and Inertia

1.1 What is Force?

SI Unit

Dimensional Formula

Characteristics of Force:

1.2 What is Inertia?

💡 Key Points about Inertia

Types of Inertia:

1. Inertia of Rest

2. Inertia of Motion

3. Inertia of Direction

Newton's First Law of Motion

📜 Statement (Law of Inertia)

2.1 Mathematical Form

Important Implications:

2.2 Inertial vs Non-Inertial Frames

⚠️ Pseudo Force (JEE Advanced Important)

Newton's Second Law of Motion

📜 Statement

3.1 Mathematical Formulation

Standard Form

General Form

3.2 Derivation from Momentum Form

3.3 Impulse and Momentum

Impulse-Momentum Theorem

Applications of Impulse:

🏏 Catching a Ball

🚗 Airbags

🏃 Sand Pit

📝 Solved Example 1 (JEE Main Pattern)

Newton's Third Law of Motion

📜 Statement (Action-Reaction Law)

4.1 Mathematical Form

4.2 Key Characteristics

4.3 Action-Reaction Pairs

❌ Common Misconception

Free Body Diagrams (FBD)

⭐ Why FBD is Important for JEE

5.1 Steps to Draw FBD

5.2 Do's and Don'ts

✅ DO Include

❌ DON'T Include

5.3 Common FBD Cases

Block on Horizontal Surface

Block on Inclined Plane (θ)

Friction

6.1 Types of Friction

Static Friction (fs)

Kinetic Friction (fk)

Rolling Friction

⚠️ Important: μs > μk (Always!)

6.2 Laws of Friction

6.3 Angle of Repose & Angle of Friction

Angle of Repose (θ)

Angle of Friction (λ)

💡 Key Insight

6.4 Motion on Inclined Plane

Key Formulas for Incline (angle θ)

📝 Solved Example 2 (JEE Main 2023 Type)

Circular Motion & Forces

7.1 Centripetal Force

⚠️ Important Concept

7.2 Sources of Centripetal Force

7.3 Banking of Roads

Optimal Banking Angle (No Friction)

With Friction (Maximum Speed)

7.4 Centrifugal Force (Pseudo Force)

Inertial Frame

Rotating Frame

Connected Bodies & Pulley Systems

8.1 Strategy for Connected Bodies

8.2 Atwood Machine

Formulas (m₂ > m₁)

8.3 Two Blocks on Horizontal Surface

📝 Solved Example 3 (JEE Advanced Pattern)

📋 Complete Formula Sheet - Laws of Motion

Newton's Laws

Friction