Laws of Motion Notes for NEET — Class 11 NCERT

Introduction

Newton's three laws are the foundation of all of mechanics. Get the free body diagram right, write the right form of $F = ma$ , and almost every NEET mechanics problem becomes routine.

For NEET 2027, expect 3 to 5 questions from this chapter. The favourites are friction on an inclined plane, the Atwood machine, conservation of momentum, banking of roads, and impulse. Master these patterns and you also unlock Work-Energy, Rotational Motion and Gravitation, which all build on Newton's laws.

Newton's first law (law of inertia)

An object at rest stays at rest, and an object in uniform motion stays in uniform motion, unless a net external force acts on it.

The first law tells you that motion does not need a continuous force to be maintained — only acceleration does. Inertia is the property of a body that resists any change to its state of motion. Mass is the quantitative measure of inertia.

Types of inertia

Inertia of rest: a stationary body resists being moved.
Inertia of motion: a moving body resists being stopped.
Inertia of direction: a body in motion resists changes in its direction of motion.

Linear momentum and the second law

Linear momentum

Momentum is the product of mass and velocity:

p = m v .

It is a vector quantity in the direction of velocity. SI unit: $kg m/s$ , dimensions $[M L T^{- 1}]$ .

Newton's second law

The rate of change of momentum is proportional to the applied force:

F = \frac{d p}{d t} .

For a body of constant mass, this reduces to the familiar form:

F = m a .

SI unit of force: newton (N), where $1 N = 1 kg m/s^{2}$ . Dimensions: $[M L T^{- 2}]$ .

Force F (N)

Mass m (kg)

Acceleration a (m/s²)

Answer

F = 10.000 N

F = m a = 5 \times 2 = 10.000 N

Impulse

Integrating the second law over time gives the impulse-momentum theorem:

J = \int F d t = Δ p .

For a constant force acting for time $Δ t$ :

J = F Δ t = m Δ v .

Impulse explains why catching a fast ball with a soft glove hurts less: extending the contact time reduces the average force needed to change the same momentum.

A ball of mass m hits a wall with velocity v and bounces back with the same speed. The contact time is Δt. See how the average force depends on the contact time.

Mass m: 0.50 kg

Speed v: 10 m/s

Contact time Δt: 50 ms (0.050 s)

Δp = J

10.00 N·s

Avg force F

200 N

Direction of Δp

Away from wall

J = Δ p = m (- v) - m v = - 2 m v = - 2 \times 0.5 \times 10 = - 10.00 N\cdots

F_{avg} = \frac{∣ J ∣}{Δ t} = \frac{10.00}{0.050} = 200 N

Newton's third law

For every action, there is an equal and opposite reaction. If body A exerts force $F_{A B}$ on body B, then B exerts $F_{B A} = - F_{A B}$ on A.

Action and reaction act on different bodies. They never cancel each other when you analyse a single body — that is the most common third-law trap in NEET.

Examples: walking (your foot pushes back on the floor, the floor pushes you forward); a rocket (exhaust gas is pushed down, gas pushes the rocket up); a swimmer pushing water back and being pushed forward.

Conservation of linear momentum

If the net external force on a system is zero, its total linear momentum is conserved:

\sum p_{i, before} = \sum p_{i, after} .

This is one of the most powerful tools in NEET physics. It is true even when internal forces between parts of the system are very complicated.

Recoil of a gun

If a gun of mass $M$ fires a bullet of mass $m$ with velocity $v$ , the gun recoils with velocity $V$ :

0 = m v + M V \Rightarrow V = - \frac{m v}{M} .

1D collisions

Two bodies collide along a line. Conservation of momentum always holds. Energy is conserved only in elastic collisions.

Perfectly elastic: momentum and KE both conserved. For equal masses, the velocities are exchanged.
Perfectly inelastic: bodies stick together. Final velocity $v_{f} = (m_{1} u_{1} + m_{2} u_{2}) / (m_{1} + m_{2})$ . KE is lost.

m₁: 2 kg

u₁: 6 m/s

m₂: 3 kg

u₂: 0 m/s

v₁ (after)

-1.20 m/s

v₂ (after)

4.80 m/s

Initial momentum

12.00 kg·m/s

Final momentum

12.00 kg·m/s

Initial KE

36.00 J

Final KE

36.00 J

no loss (elastic)

v_{1} = \frac{( m _{1} - m _{2} ) u _{1} + 2 m _{2} u _{2}}{m _{1} + m _{2}} = - 1.20

v_{2} = \frac{( m _{2} - m _{1} ) u _{2} + 2 m _{1} u _{1}}{m _{1} + m _{2}} = 4.80

Practice these on the timed test

Try a free 10-question NEET mock test on Laws of Motion — instant results, no sign-up needed.

Friction

Friction is a contact force that opposes the relative motion (or tendency of motion) between two surfaces. It is the most commonly tested non-trivial force in NEET.

Types of friction

Static friction $f_{s}$ — when surfaces are at rest with respect to each other. It adjusts itself up to a maximum value $(f_{s})_{max} = μ_{s} N$ .
Kinetic friction $f_{k} = μ_{k} N$ — when surfaces slide. Approximately constant during sliding.
Rolling friction — when one body rolls on another. Much smaller than kinetic friction.

Always $μ_{s} > μ_{k}$ : it takes more force to start motion than to maintain it.

Coefficient of friction

$μ$ depends only on the nature of the two surfaces in contact. It does not depend on the contact area or on the weight of the body. It is dimensionless.

Angle of repose

The angle of repose $θ_{r}$ is the maximum angle of an inclined plane at which a block just does not slide. At this angle, the gravitational pull along the incline equals the maximum static friction:

m g sin θ_{r} = μ_{s} m g cos θ_{r} \Rightarrow tan θ_{r} = μ_{s} .

This identity is asked nearly every NEET year.

Mass m: 5 kg

μₛ (static): 0.40

μₖ (kinetic): 0.30

Applied force F: 15 N

State

At rest (static friction balancing)

Friction force: 15.00 N

Normal N = m·g = 50 N
Max static friction = μₛ·N = 20.00 N
Kinetic friction = μₖ·N = 15.00 N
Angle of repose θᵣ = tan⁻¹(μₛ) ≈ 21.8°

(f_{s})_{max} = μ_{s} m g = 0.4 \times 5 \times 10 = 20.00 N

tan θ_{r} = μ_{s} \Rightarrow θ_{r} = 21.8°

Inclined plane with friction

For a block of mass $m$ on an incline of angle $θ$ with friction coefficient $μ$ :

Normal force: $N = m g cos θ$ .
Component of gravity along the incline: $m g sin θ$ .
Maximum static friction: $μ_{s} m g cos θ$ .

Will the block slide?

Compare $m g sin θ$ with $μ_{s} m g cos θ$ :

If $tan θ \leq μ_{s}$ , the block stays put. Static friction adjusts to balance gravity.
If $tan θ > μ_{s}$ , the block slides down and the kinetic friction acts.

Acceleration during sliding

a = g (sin θ - μ_{k} cos θ) .

For motion up the incline (e.g. block thrown upward), friction reverses direction and:

a_{up} = - g (sin θ + μ_{k} cos θ) .

Mass m: 2 kg

Incline angle θ: 30°

Friction μ: 0.30 (angle of repose ≈ 16.7°)

State

Sliding, a = 2.40 m/s²

m g sin θ = 10.00 N, μ m g cos θ = 5.20 N

a = g (sin θ - μ cos θ) = 2.40 m/s^{2}

The block slides whenever tan θ > μ. With your current μ that happens past θ ≈ 16.7°.

Connected systems and the Atwood machine

Two blocks on a horizontal surface, joined by a string

Pull on block 2 with force $F$ . Both blocks accelerate at the same rate $a = F / (m_{1} + m_{2})$ . The tension in the string is $T = m_{1} a$ .

Atwood machine

Two masses $m_{1}$ and $m_{2}$ hang on either side of a frictionless, massless pulley. Take the heavier mass $m_{1}$ . Both have the same acceleration magnitude:

a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}, T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}} .

These two formulas turn nearly every Atwood NEET problem into one-line plug-and-chug.

Mass m₁ (left): 5 kg

Mass m₂ (right): 3 kg

Acceleration |a|

2.50 m/s²

String tension T

37.50 N

Heavier mass on the left accelerates downward at 2.50 m/s²; the other goes up at the same rate.

a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( 5 - 3 ) \times 10}{8} = 2.50 m/s^{2}

T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}} = \frac{2 \times 5 \times 3 \times 10}{8} = 37.50 N

Circular motion: banking of roads

On a flat road, friction provides the centripetal force needed for a car to turn. On a banked road, the horizontal component of the normal force does the same job — friction is no longer essential.

Banked road, no friction

For a road banked at angle $θ$ , the maximum speed at which a car can take a turn of radius $r$ without friction is:

v_{ideal} = r g tan θ .

Banked road with friction

With friction (coefficient $μ$ ), the safe speed range widens:

v_{max} = \frac{r g ( tan θ + μ )}{1 - μ tan θ}, v_{min} = \frac{r g ( tan θ - μ )}{1 + μ tan θ} .

Turn radius r: 100 m

Banking angle θ: 15°

Friction μ: 0.20

Ideal speed (no friction)

16.37 m/s

Max safe speed (with friction)

22.24 m/s

Min safe speed (with friction)

8.03 m/s

(below this, car slips inward)

v_{ideal} = r g tan θ = 100 \times 10 \times 0.268 = 16.37 m/s

v_{max} = \frac{r g ( tan θ + μ )}{1 - μ tan θ} = 22.24 m/s

Pseudo forces (non-inertial frames)

In a frame accelerating at $a_{0}$ (e.g. inside an accelerating car), Newton's laws appear to be violated unless we add a fictitious pseudo force on every body:

F_{pseudo} = - m a_{0} .

With this added, $F_{net} = m a_{rel}$ works inside the accelerating frame just like in an inertial frame. Common NEET application: a pendulum in an accelerating train tilts toward the back at angle $tan θ = a_{0} / g$ .

Worked NEET problems

NEET-style problem · Inclined plane

Question

A block of mass

2 kg

slides down a frictionless incline of angle

30°

. Take

g = 10 m/s^{2}

. Its acceleration is:

Solution

On a frictionless incline, only the gravity component along the slope acts:

a = g sin θ = 10 \times sin 30° = 10 \times 0.5 = 5 m/s^{2} .

(Mass cancels, as always on an incline without friction.)

NEET-style problem · Atwood machine

Question

In an Atwood machine, masses

3 kg

and

2 kg

hang on either side of a frictionless pulley. Take

g = 10 m/s^{2}

. Find the acceleration and the tension in the string.

Solution

a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( 3 - 2 ) \times 10}{3 + 2} = 2 m/s^{2} .

T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}} = \frac{2 \times 3 \times 2 \times 10}{5} = 24 N .

NEET-style problem · Friction

Question

A block of mass

5 kg

rests on a horizontal floor. The coefficient of static friction is

μ_{s} = 0.4

. Take

g = 10 m/s^{2}

. The minimum horizontal force required to start moving the block is:

Solution

Maximum static friction: $(f_{s})_{max} = μ_{s} m g = 0.4 \times 5 \times 10 = 20 N$ .

The block starts to move when the applied force just exceeds this value, so the minimum is 20 N.

NEET-style problem · Conservation of momentum

Question

A bullet of mass

50 g

is fired from a gun of mass

5 kg

with velocity

200 m/s

. Find the recoil velocity of the gun.

Solution

Initial momentum is zero. After firing:

0 = m v + M V \Rightarrow V = - \frac{m v}{M} = - \frac{0.05 \times 200}{5} = - 2 m/s .

The gun recoils at 2 m/s in the direction opposite to the bullet.

NEET-style problem · Banking of roads

Question

A car takes a circular turn of radius

100 m

on a road banked at

tan θ = 0.2

. Take

g = 10 m/s^{2}

. The maximum speed without slipping (no friction) is:

Solution

v = r g tan θ = 100 \times 10 \times 0.2 = 200 \approx 14.14 m/s .

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Summary cheat sheet

First law: a body persists in rest or uniform motion unless a net external force acts.
Second law: $F = d p / d t = m a$ .
Third law: $F_{A B} = - F_{B A}$ . Action and reaction act on different bodies.
Impulse-momentum theorem: $J = \int F d t = Δ p$ .
Conservation of momentum: if net external force is zero, $\sum p$ is constant.
Friction: $(f_{s})_{max} = μ_{s} N$ , $f_{k} = μ_{k} N$ , $μ_{s} > μ_{k}$ .
Angle of repose: $tan θ_{r} = μ_{s}$ .
Block sliding down incline (with friction): $a = g (sin θ - μ_{k} cos θ)$ .
Atwood machine: $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ , $T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}$ .
Banking (no friction): $v_{ideal} = r g tan θ$ .
Pseudo force in accelerating frame: $F_{pseudo} = - m a_{0}$ .

Next: try the interactive widgets for Newton's laws, friction, Atwood machine and banking of roads, or work through the 30+ NEET PYQs with full solutions. To time yourself, take the free 10-question mock test.

Frequently asked questions

How many questions come from Laws of Motion in NEET 2027?

You can expect 3 to 5 questions from Laws of Motion in NEET 2027. The chapter has very high PYQ frequency. Friction on an inclined plane, the Atwood machine, conservation of momentum and banking of roads are the most heavily tested concepts.

Why is Laws of Motion considered the most important chapter in NEET Physics?

Almost every problem in mechanics, rotational motion, gravitation and even fluid mechanics applies Newton's laws somewhere. Once you can draw a clean free body diagram and write Newton's second law correctly, half the problem is already done.

What are the three laws of motion?

First law (inertia): an object stays at rest or in uniform motion unless a net force acts on it. Second law: F equals dp by dt, which simplifies to F equals m a for constant mass. Third law: every action has an equal and opposite reaction. The action and reaction act on different bodies.

What is the difference between static and kinetic friction?

Static friction acts when surfaces are not yet sliding and adjusts itself up to a maximum f_s = mu_s N. Kinetic friction acts during sliding and equals f_k = mu_k N (a constant). Always mu_s is greater than mu_k for the same surfaces, which is why it takes more force to start moving an object than to keep it moving.

How do I solve a block-on-an-incline-with-friction problem?

Resolve gravity along and perpendicular to the incline: m g sin theta along, m g cos theta perpendicular. Normal force N = m g cos theta. Maximum static friction = mu_s m g cos theta. Compare m g sin theta with maximum static friction: if larger, the block slides and net force = m g sin theta minus mu_k m g cos theta. Acceleration follows from F = m a.

What is the angle of repose?

The angle of repose is the maximum angle of an inclined surface at which a block does not slide. At this critical angle, m g sin theta exactly equals the maximum static friction mu_s m g cos theta, which gives tan theta = mu_s. This is one of the most-tested NEET facts in this chapter.

How do I find the safe speed for banking of roads?

For a road banked at angle theta with no friction, the maximum safe speed is v = square root of (r g tan theta). With friction (coefficient mu), the safe speed range is from v_min equals square root of (r g (tan theta minus mu) over (1 plus mu tan theta)) up to v_max equals square root of (r g (tan theta plus mu) over (1 minus mu tan theta)).

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Laws of MotionNEET Physics · Class 11 · NCERT Chapter 4

Introduction

Newton's first law (law of inertia)

Types of inertia

Linear momentum and the second law

Linear momentum

Newton's second law

Impulse

Newton's third law

Conservation of linear momentum

Recoil of a gun

1D collisions

Friction

Types of friction

Coefficient of friction

Angle of repose

Inclined plane with friction

Will the block slide?

Acceleration during sliding

Connected systems and the Atwood machine

Two blocks on a horizontal surface, joined by a string

Atwood machine

Circular motion: banking of roads

Banked road, no friction

Banked road with friction

Pseudo forces (non-inertial frames)

Worked NEET problems

Summary cheat sheet

Frequently asked questions

How many questions come from Laws of Motion in NEET 2027?

Why is Laws of Motion considered the most important chapter in NEET Physics?

What are the three laws of motion?

What is the difference between static and kinetic friction?

How do I solve a block-on-an-incline-with-friction problem?

What is the angle of repose?

How do I find the safe speed for banking of roads?

Continue with the next chapter notes

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