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Laws of Motion

Master Laws of Motion for JEE Main, JEE Advanced, NEET & Board Exams. Covers Newton's three laws, friction, impulse, momentum conservation, FBD techniques, Atwood's machine, pseudo force & circular motion dynamics with solved examples.

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The Laws of Motion, formulated by Sir Isaac Newton in 1687, are the cornerstone of classical mechanics. They describe the relationship between the forces acting on a body and the motion that results. Every problem in mechanics — from a block on an inclined plane to a rocket launch — ultimately rests on these three laws. For JEE (Main & Advanced), NEET, and board exams, this is one of the highest-weightage topics, appearing directly in questions and indirectly in work-energy, rotational motion, and fluid mechanics problems.

1. Newton's Laws of Motion

Newton's First Law — Law of Inertia

A body continues to remain in its state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.

Inertia is the inherent property of a body to resist any change in its state of rest or motion. It is directly proportional to mass — heavier objects have more inertia.
The First Law defines what a force is: an agent that changes (or tends to change) the state of motion of a body.
It also defines inertial frames of reference — frames in which the First Law holds. A frame at rest or moving with constant velocity is inertial.

Newton's Second Law — Law of Force and Acceleration

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

$\vec{F} = \frac{d \vec{p}}{d t} = \frac{d (m \vec{v})}{d t}$

For constant mass: $\vec{F} = m \vec{a}$

Force is a vector. The net force ( ${\vec{F}}_{n e t}$ ) is the vector sum of all forces acting on the body.
The SI unit of force is the Newton (N): $1 N = 1 {kg·m/s}^{2}$ .
The Second Law is applicable instantaneously — it connects force and acceleration at the same instant.

Newton's Third Law — Law of Action and Reaction

For every action, there is an equal and opposite reaction. If body A exerts force ${\vec{F}}_{A B}$ on body B, then B exerts force ${\vec{F}}_{B A}$ on A such that:

${\vec{F}}_{A B} = - {\vec{F}}_{B A}$

Action and reaction always act on different bodies, so they never cancel each other.
They are simultaneous — neither is the "cause" and the other the "effect."
They are always equal in magnitude and opposite in direction, regardless of the masses or motion of the bodies.

2. Linear Momentum and Impulse

Linear Momentum

The linear momentum of a body of mass $m$ moving with velocity $\vec{v}$ is:

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

It is a vector quantity (SI unit: kg·m/s). Newton's Second Law in its most general form is $\vec{F} = \frac{d \vec{p}}{d t}$ .

Impulse

Impulse ( $\vec{J}$ ) is the product of force and the time interval over which it acts. For a constant force:

$\vec{J} = \vec{F} \cdot Δ t = Δ \vec{p} = m \vec{v} - m \vec{u}$

Impulse equals the change in momentum of the body. SI unit: N·s = kg·m/s.

Impulse is important when large forces act for very short durations — e.g., a bat hitting a ball, explosion of a bullet, collision forces.
Graphically, impulse = area under the $F$ – $t$ graph.

Conservation of Linear Momentum

If the net external force on a system is zero, the total linear momentum of the system remains constant:

${\vec{F}}_{e x t} = 0 ⟹ {\vec{p}}_{t o t a l} = constant$

This is one of the most powerful conservation laws in physics and is a direct consequence of Newton's Third Law. It holds even during collisions and explosions.

3. Types of Forces

Force	Symbol	Direction	Key Property
Weight	$W = m g$	Vertically downward	Acts at centre of gravity; always present
Normal Reaction	$N$	Perpendicular to contact surface	Contact force; adjusts to maintain equilibrium
Tension	$T$	Along string, away from body	Same throughout a massless string
Friction	$f$	Opposing relative motion/tendency	$f_{s} \leq μ_{s} N$ ; $f_{k} = μ_{k} N$
Spring Force	$F = k x$	Opposing deformation (restoring)	Proportional to extension/compression

4. Friction in Detail

Types of Friction

Static Friction ( $f_{s}$ ): Acts when there is no relative motion between surfaces. It is self-adjusting — it exactly balances the applied force up to a maximum value called limiting friction: $f_{s, m a x} = μ_{s} N$ .
Kinetic (Sliding) Friction ( $f_{k}$ ): Acts when surfaces slide over each other. $f_{k} = μ_{k} N$ . Always $μ_{k} < μ_{s}$ .
Rolling Friction ( $f_{r}$ ): Acts when a body rolls over a surface. $f_{r} ≪ f_{k}$ . This is why wheels are used.

Property	Static Friction	Kinetic Friction
Relative motion	Absent	Present
Magnitude	$0 \leq f_{s} \leq μ_{s} N$	$f_{k} = μ_{k} N$ (constant)
Coefficient	$μ_{s}$ (higher)	$μ_{k}$ (lower)
Depends on area?	No	No

Angle of Friction ( $λ$ ) and Angle of Repose ( $α$ )

Angle of friction $λ$ is the angle between the normal reaction $N$ and the resultant of $N$ and $f_{s, m a x}$ : $\tan λ = μ_{s}$ .
Angle of repose $α$ is the maximum angle of an inclined plane at which a body just begins to slide: $\tan α = μ_{s}$ .
Therefore $λ = α$ — a very useful and frequently tested result.

5. Free Body Diagram (FBD) — The Problem-Solving Tool

A Free Body Diagram is a diagram of a single body (isolated from its surroundings) showing all forces acting on it. It is the most important technique for solving Laws of Motion problems.

Steps to Draw an FBD and Solve:

Step 1: Identify the body (or system) whose motion you want to analyse.
Step 2: Draw the body as a point or a box. Represent all external forces acting on it as arrows — weight ( $m g$ ) downward, normal ( $N$ ) perpendicular to surface, tension ( $T$ ) along string, friction along surface, applied force, etc.
Step 3: Choose a convenient coordinate system (usually along and perpendicular to motion).
Step 4: Apply Newton's Second Law along each axis: $\sum F_{x} = m a_{x}$ and $\sum F_{y} = m a_{y}$ .
Step 5: Solve the simultaneous equations for unknowns.

Key Rule: Only external forces on the chosen body appear in its FBD. Internal forces (e.g., tension between two connected blocks treated as a system) cancel out.

6. Common System Problems

(i) Two blocks connected by a string on a horizontal surface

For a system of two blocks of masses $m_{1}$ and $m_{2}$ pulled by force $F$ (friction absent):

Common acceleration: $a = \frac{F}{m_{1} + m_{2}}$
Tension in the string: $T = \frac{m_{2} \cdot F}{m_{1} + m_{2}}$ (tension pulls the rear block forward)

(ii) Atwood's Machine

Two masses $m_{1}$ and $m_{2}$ ( $m_{1} > m_{2}$ ) connected by a string over a frictionless, massless pulley:

Acceleration: $a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}$
Tension: $T = \frac{2 m_{1} m_{2} g}{m_{1} + m_{2}}$
If $m_{1} = m_{2}$ : $a = 0$ and $T = m g$ (equilibrium).

(iii) Block on an Inclined Plane

For a block of mass $m$ on a smooth incline of angle $θ$ :

Along the plane: $m g \sin θ = m a$ ⟹ $a = g \sin θ$
Perpendicular to plane: $N = m g \cos θ$

With friction ( $μ_{k}$ ): $a = g (\sin θ - μ_{k} \cos θ)$ (sliding down)

Condition for the block to remain stationary: $\tan θ \leq μ_{s}$

(iv) Apparent Weight in a Lift

Condition	Apparent Weight ( $N$ )	Feels
Lift at rest or uniform velocity	$N = m g$	Normal
Lift accelerating upward ( $a$ )	$N = m (g + a)$	Heavier
Lift accelerating downward ( $a$ )	$N = m (g - a)$	Lighter
Lift in free fall ( $a = g$ )	$N = 0$	Weightless

7. Pseudo Force and Non-Inertial Frames

A non-inertial frame is one that accelerates with respect to an inertial frame (e.g., an accelerating car, a rotating frame). Newton's laws do not directly apply in such frames.

To apply Newton's laws in a non-inertial frame accelerating with ${\vec{a}}_{0}$ , a pseudo force (or fictitious force) ${\vec{F}}_{p s e u d o} = - m {\vec{a}}_{0}$ is introduced on every body of mass $m$ .

Pseudo force has no real physical origin — it arises purely due to the acceleration of the reference frame.
Direction: opposite to the acceleration of the frame.
Example: In a car braking suddenly, you feel pushed forward — this is the pseudo force in the car's (decelerating) frame.
Pendulum in an accelerating vehicle: The bob hangs at angle $θ = \tan^{- 1} (\frac{a}{g})$ from the vertical (towards the rear of the vehicle).

8. Dynamics of Circular Motion

When a body moves in a circle, it has a centripetal acceleration directed towards the centre. The net force towards the centre (centripetal force) is:

$F_{c} = \frac{m v^{2}}{r} = m r ω^{2}$

Centripetal force is not a new type of force — it is the name given to the net inward force provided by real forces (friction, tension, gravity, normal reaction, etc.).

Key Applications

Vehicle on a flat circular road: Friction provides centripetal force. Maximum speed: $v_{m a x} = \sqrt{μ_{s} r g}$ .
Vehicle on a banked road (no friction): Normal reaction provides centripetal force. $\tan θ = \frac{v^{2}}{r g}$ . Optimal speed: $v = \sqrt{r g \tan θ}$ .
Vehicle on a banked road (with friction):
- Maximum speed: $v_{m a x} = \sqrt{r g \frac{\tan θ + μ_{s}}{1 - μ_{s} \tan θ}}$
- Minimum speed: $v_{m i n} = \sqrt{r g \frac{\tan θ - μ_{s}}{1 + μ_{s} \tan θ}}$
Vertical circular motion (string): At the top of the circle, minimum speed for maintaining contact: $v_{m i n} = \sqrt{g R}$ . Tension at top $= 0$ at this minimum speed.
Normal force at the bottom of a vertical circle: $N = m g + \frac{m v^{2}}{R}$ (feels heavier).

Spring Scale and Massless String — Important Assumptions

A massless string transmits tension equally throughout, regardless of acceleration. If the string has mass $m_{s}$ , tension varies along its length.
A massless, frictionless pulley only changes the direction of tension, not its magnitude.
A spring scale reads the tension in the spring, which equals the apparent weight of the hanging body — not necessarily $m g$ .
When two springs of constants $k_{1}$ and $k_{2}$ are in series: $\frac{1}{k_{e f f}} = \frac{1}{k_{1}} + \frac{1}{k_{2}}$ . In parallel: $k_{e f f} = k_{1} + k_{2}$ .

JEE/NEET Power Tips — Avoid These Common Mistakes

Action-reaction pairs act on different bodies. Do not include both in the FBD of a single body — only forces acting ON the chosen body are drawn.
Normal force is NOT always equal to $m g$ . It changes when there is a vertical component of applied force or when the body accelerates vertically (e.g., in a lift).
Friction is not always $μ N$ . Static friction is self-adjusting and can be anywhere from $0$ to $μ_{s} N$ . Use $f = μ_{k} N$ only when the body is sliding.
Centripetal force is not an extra force to be added in the FBD. It is the resultant of real forces already acting — identify which real force(s) provide it.
In the $s_{n}$ formula and kinematics, use consistent signs. Choose one direction as positive and stick to it throughout the problem.
Momentum conservation requires zero net external force, not zero net force altogether. Internal forces (between parts of a system) always cancel in pairs.

Practice Questions (JEE / NEET / Board Level)

Q1: A body of mass $5$ kg is acted upon by two perpendicular forces of $8$ N and $6$ N. The acceleration of the body is:

A) $7$ m/s²
B) $2$ m/s²
C) $1.5$ m/s²
D) $0.5$ m/s²

Answer: B) 2 m/s². Net force $= \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$ N. Acceleration $= \frac{F}{m} = \frac{10}{5} = 2$ m/s².

Q2: In an Atwood's machine, masses $m_{1} = 4$ kg and $m_{2} = 2$ kg are connected over a massless, frictionless pulley. The tension in the string is ( $g = 10$ m/s²):

A) $\frac{80}{3}$ N
B) $20$ N
C) $40$ N
D) $\frac{40}{3}$ N

Answer: A) $\frac{80}{3}$ N.
$T = \frac{2 m_{1} m_{2} g}{m_{1} + m_{2}} = \frac{2 \times 4 \times 2 \times 10}{4 + 2}$
$= \frac{160}{6} = \frac{80}{3}$ N
$\approx 26.7$ N.

Q3: A block of mass 10 kg is placed on a rough horizontal surface with $μ_{s} = 0.4$ and $μ_{k} = 0.3$ . A horizontal force of 35 N is applied. What is the friction force on the block? ( $g = 10 {m/s}^{2}$ )

A) 35 N (static)
B) 40 N (static)
C) 30 N (kinetic)
D) 0 N

Answer: A) 35 N (static).

Explanation: First, determine the Normal force ( $N$ ). On a horizontal surface, $N = m g = 10 \times 10 = 100 N$ .

Next, calculate the limiting (maximum) static friction: $f_{s, max} = μ_{s} N = 0.4 \times 100 = 40 N$ .

The applied force is 35 N. Since the applied force is less than the limiting static friction ( $35 N < 40 N$ ), the block will not move. Because static friction is a self-adjusting force, it will exactly match the applied force to maintain equilibrium. Therefore, the actual friction force is 35 N (static).

Q4: A man of mass $60$ kg is in a lift accelerating downward at $2$ m/s². His apparent weight is ( $g = 10$ m/s²):

A) $600$ N
B) $720$ N
C) $480$ N
D) $0$ N

Answer: C) $480$ N.
$N = m (g - a)$
$= 60 (10 - 2) = 60 \times 8 = 480$ N.

Q5: A car of mass $1000$ kg moves on a circular road of radius $50$ m. If $μ_{s} = 0.5$ , the maximum speed at which it can negotiate the curve is ( $g = 10$ m/s²):

A) $25$ m/s
B) $5$ m/s
C) $15$ m/s
D) $\sqrt{250}$ m/s

Answer: D) $v_{m a x} = \sqrt{μ_{s} r g} = \sqrt{0.5 \times 50 \times 10}$
$= \sqrt{250}$ m/s.
Option D ( $\sqrt{250}$ ) is the answer.

Q6 (JEE Advanced type): Two blocks A ( $3$ kg) and B ( $2$ kg) are placed on a smooth surface, connected by a massless string. A horizontal force $F = 25$ N is applied on A. What is the tension in the string between A and B?

A) $25$ N
B) $15$ N
C) $10$ N
D) $5$ N

Answer: C) $10$ N. System acceleration $a = \frac{F}{m_{A} + m_{B}} = \frac{25}{3 + 2} = 5$ m/s². For block B alone: $T = m_{B} \times a = 2 \times 5 = 10$ N.

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Est. Read Time22 minutes

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Laws of Motion

1. Newton's Laws of Motion

Newton's First Law — Law of Inertia

Newton's Second Law — Law of Force and Acceleration

Newton's Third Law — Law of Action and Reaction

2. Linear Momentum and Impulse

Linear Momentum

Impulse

Conservation of Linear Momentum

3. Types of Forces

4. Friction in Detail

Types of Friction

Angle of Friction ( $λ$ ) and Angle of Repose ( $α$ )

5. Free Body Diagram (FBD) — The Problem-Solving Tool

Steps to Draw an FBD and Solve:

6. Common System Problems

(i) Two blocks connected by a string on a horizontal surface

(ii) Atwood's Machine

(iii) Block on an Inclined Plane

(iv) Apparent Weight in a Lift

7. Pseudo Force and Non-Inertial Frames

8. Dynamics of Circular Motion

Key Applications

Spring Scale and Massless String — Important Assumptions

Practice Questions (JEE / NEET / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

We Value Your Privacy

Edvaya Target

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Take a Free Mock Test

Target Subjects

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Subject Mastery

Laws of Motion

1. Newton's Laws of Motion

Newton's First Law — Law of Inertia

Newton's Second Law — Law of Force and Acceleration

Newton's Third Law — Law of Action and Reaction

2. Linear Momentum and Impulse

Linear Momentum

Impulse

Conservation of Linear Momentum

3. Types of Forces

4. Friction in Detail

Types of Friction

Angle of Friction (λ) and Angle of Repose (α)

5. Free Body Diagram (FBD) — The Problem-Solving Tool

Steps to Draw an FBD and Solve:

6. Common System Problems

(i) Two blocks connected by a string on a horizontal surface

(ii) Atwood's Machine

(iii) Block on an Inclined Plane

(iv) Apparent Weight in a Lift

7. Pseudo Force and Non-Inertial Frames

8. Dynamics of Circular Motion

Key Applications

Spring Scale and Massless String — Important Assumptions

Practice Questions (JEE / NEET / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

We Value Your Privacy

Angle of Friction ( $λ$ ) and Angle of Repose ( $α$ )