NEET Physics · NCERT Class 11

Laws of Motion notes

Force, Inertia & Newton's Laws, Momentum & Impulse, Equilibrium of Particles, Friction, Circular Motion Applications and Problem Solving — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 7 topics
  • 67 min read
  • 35 formulas
  • 47 PYQs
1

Chapter Overview

7 min

Laws of Motion explains why objects start moving, stop moving, change direction or remain at rest. Aristotle believed force is necessary to keep a body moving, but Newton showed that force is required to change the state of motion, not to maintain uniform motion. The chapter introduces inertia, mass, force, momentum, impulse, equilibrium, friction and circular motion applications. Newton’s three laws form the core of mechanics and are used in blocks, pulleys, elevators, inclined planes and vehicle motion. For NEET, this chapter is very important because it gives direct conceptual questions, free body diagram problems, friction numericals and multi-body Newton’s law applications.

Key points6

  • A body can move with constant velocity even when net force is zero.
  • Free body diagrams are the most important tool for solving Laws of Motion questions.
  • Internal forces cancel for a system, but external forces change total momentum.
  • Static friction is self-adjusting up to limiting friction.
  • Circular motion always requires inward centripetal force.
  • Most NEET mistakes happen due to wrong direction of friction, wrong normal reaction or missing pseudo-free body analysis.

Memory tricks2

Newton’s Laws in One Line: First: no net force, no change. Second: net force gives acceleration. Third: forces come in pairs.

FBD Rule: Draw only forces acting on the selected body, never forces exerted by that body on others.

Examples2

Daily Life Example

When a bus suddenly starts, passengers tend to fall backward due to inertia of rest.

NEET Concept Check

If a 2 kg body has net force 10 N, acceleration is F/m = 5 m s^-2.

Common mistakes3

Believing Force Is Needed for Constant Motion

A body moving with constant velocity has zero net force. Force is needed to change velocity.

Action-Reaction on Same Body

Newton’s third-law pair acts on two different bodies, so they do not cancel on a single free body diagram.

Ignoring Normal Reaction Changes

Normal reaction is not always equal to mg. It changes in elevators, inclined planes and circular motion.

Reference tables2

Quick revision8

Formulas5

Newton’s Second Law
Fnet=maF_{net} = ma

Net external force equals mass multiplied by acceleration for constant mass.

F_net
Resultant external force
m
Mass of body
a
Acceleration
Momentum
p=mvp = mv

Linear momentum is the product of mass and velocity.

p
Linear momentum
m
Mass
v
Velocity
Impulse
J=FΔt=ΔpJ = FΔt = Δp

Impulse is the change in momentum produced by force acting for a time interval.

J
Impulse
F
Average force
Δt
Time interval
Δp
Change in momentum
2 more formulas — sign up free

Diagrams3

2

Force, Inertia & Newton's Laws

10 min

Force is an interaction that can change the state of motion or shape of a body. Forces may be contact forces such as normal reaction, tension, friction and spring force, or non-contact forces such as gravitational, electrostatic and magnetic forces. Inertia is the natural tendency of a body to resist any change in rest or motion, and mass is its measure. Newton’s first law states that a body remains at rest or in uniform motion unless acted upon by a net external force. The second law relates net force to acceleration, F = ma. The third law states that every action has an equal and opposite reaction on another body.

Key points6

  • Balanced forces produce zero acceleration, not necessarily zero velocity.
  • Unbalanced force produces acceleration in the direction of net force.
  • Weight is gravitational force mg; mass is the measure of inertia.
  • Normal reaction is perpendicular to contact surface.
  • Tension acts along a string, pulling away from the body.
  • FBD should include weight, normal, tension, friction, applied force and spring force wherever applicable.

Memory tricks2

FBD Checklist: Remember WNTF: Weight, Normal, Tension, Friction or applied Force.

Third Law: Action and reaction are twins on different bodies: equal, opposite, separate.

Examples2

Real-Life Application

A seat belt prevents passengers from continuing forward due to inertia when a car stops suddenly.

Numerical Example

A 5 kg block is pulled by a 20 N horizontal force on a smooth surface. Acceleration = F/m = 20/5 = 4 m s^-2.

Common mistakes3

Drawing Action-Reaction on Same FBD

Do not draw the reaction exerted by the body on another object in its own FBD.

Assuming Normal Always Equals mg

Normal equals mg only in simple horizontal cases with no vertical acceleration and no extra vertical force.

Confusing Mass and Weight

Mass is inertia and remains same; weight is gravitational force and depends on g.

Reference tables2

Quick revision8

Formulas4

Newton’s Second Law
Fnet=maF_{net} = ma

The acceleration of a body is proportional to net external force and inversely proportional to mass.

F_net
Vector sum of all external forces
m
Mass of body
a
Acceleration
Weight
W=mgW = mg

Gravitational force exerted by Earth on a body near its surface.

W
Weight
m
Mass
g
Acceleration due to gravity
2 more formulas — sign up free

Diagrams3

3

Momentum & Impulse

9 min

Linear momentum measures the quantity of motion of a body and is defined as p = mv. It is a vector in the direction of velocity. For a system of particles, total momentum is the vector sum of individual momenta. If no external force acts on a system, its total momentum remains conserved. Impulse is the effect of a force acting for a short time and equals change in momentum. This idea explains why catching a ball softly reduces force by increasing stopping time. Collisions are interactions of short duration where momentum is conserved if external impulse is negligible. Elastic collisions conserve kinetic energy, while inelastic collisions do not.

Key points6

  • Conservation of momentum applies to isolated systems.
  • Internal forces cannot change total momentum of a system.
  • Impulse has the same dimensions as momentum.
  • Large force for small time and small force for large time can give same impulse.
  • In a perfectly inelastic collision, bodies stick together after collision.
  • Momentum conservation must be applied vectorially with signs.

Memory tricks2

Impulse: Impulse is force-time effect: push harder or push longer to change momentum.

Collision Rule: Momentum survives collisions; kinetic energy survives only elastic collisions.

Examples2

Impulse Numerical

A 0.2 kg ball changes velocity from 20 m/s to -10 m/s. Δp = 0.2(-10 - 20) = -6 kg m/s, so impulse magnitude is 6 N s.

Conservation Example

A 2 kg body at 4 m/s sticks to a 2 kg body at rest. Final speed = total momentum/total mass = 8/4 = 2 m/s.

Common mistakes3

Forgetting Momentum Is Vector

Use signs in one-dimensional collision problems. Opposite directions require opposite signs.

Assuming Kinetic Energy Always Conserved

Kinetic energy is conserved only in elastic collisions, not in general collisions.

Applying Momentum Conservation with External Force

Momentum conservation applies only when net external impulse is zero or negligible.

Reference tables2

Quick revision8

Formulas5

Linear Momentum
p=mvp = mv

Momentum is the product of mass and velocity.

p
Linear momentum
m
Mass
v
Velocity
Momentum of a System
P=p1+p2+...+pnP = p1 + p2 + ... + pn

Total momentum is vector sum of individual momenta.

P
Total momentum
p1, p2, pn
Momenta of individual particles
Conservation of Momentum
Σpinitial=ΣpfinalΣp_{\text{initial}} = Σp_{\text{final}}

Total momentum remains constant when net external force is zero.

Σp_initial
Total initial momentum
Σp_final
Total final momentum
2 more formulas — sign up free

Diagrams3

4

Equilibrium of Particles

8 min

A particle is in equilibrium when the vector sum of all forces acting on it is zero. This means the particle has no acceleration; it may be at rest or moving with constant velocity. For forces in a plane, equilibrium requires both horizontal and vertical force components to separately add to zero. Many NEET problems involve strings, weights, inclined forces and concurrent forces meeting at one point. Free body diagrams and force resolution are the safest methods. Lami’s theorem is a shortcut for three concurrent, coplanar and non-parallel forces in equilibrium. It relates each force to the sine of the angle between the other two forces.

Key points6

  • Equilibrium does not always mean rest; it means acceleration is zero.
  • If three forces keep a particle in equilibrium, their vector triangle closes.
  • For two forces to balance, they must be equal, opposite and collinear.
  • Resolve inclined forces into x and y components.
  • Lami’s theorem is useful for tensions in two strings supporting a weight.
  • Angles in Lami’s theorem are the angles between the other two forces, not arbitrary angles with axes.

Memory tricks2

Equilibrium Components: Balance east-west and north-south separately: ΣFx = 0 and ΣFy = 0.

Lami’s Theorem: Each force sits over sine of the angle opposite to it.

Examples2

Practice Question

A 10 N weight is held by two equal strings symmetrically. Vertical components of tensions add to 10 N; horizontal components cancel.

Lami Example

If three equal forces are in equilibrium, the angle between any two forces is 120 degrees.

Common mistakes2

Applying Lami’s Theorem Without Concurrent Forces

Lami’s theorem is valid only for three concurrent coplanar forces in equilibrium.

Using Forces Exerted by the Particle

In FBD, draw forces acting on the particle, not forces exerted by it.

Reference tables2

Quick revision8

Formulas4

Condition of Equilibrium
ΣF=0ΣF = 0

Vector sum of all external forces on a particle must be zero.

ΣF
Resultant force
Component Conditions
ΣFx=0ΣFx = 0
ΣFy=0ΣFy = 0

For equilibrium in a plane, net force in each perpendicular direction must be zero.

ΣFx
Sum of x-components of forces
ΣFy
Sum of y-components of forces
2 more formulas — sign up free

Diagrams3

5

Friction

11 min

Friction is a contact force that opposes relative motion or the tendency of relative motion between surfaces. Static friction acts when there is no slipping and adjusts itself according to need up to a maximum called limiting friction. Once sliding begins, kinetic friction acts and is usually slightly smaller than limiting friction. The coefficient of friction measures roughness and equals friction divided by normal reaction in limiting or kinetic cases. Angle of friction is related to the resultant contact force, while angle of repose is the minimum inclination at which a body just begins to slide. Friction is necessary for walking, writing, vehicle motion and braking, but also causes wear and energy loss.

Key points6

  • Friction does not always oppose motion; it opposes relative motion or tendency at contact.
  • For a body at rest, friction may be less than μ_sN.
  • Normal reaction depends on surface geometry and other forces.
  • On an incline, components of weight are mg sinθ along plane and mg cosθ normal to plane.
  • At impending motion down an incline, mg sinθ = μmg cosθ.
  • Rolling friction is much smaller than sliding friction.

Memory tricks2

Static Friction: Static friction is smart: it adjusts only as much as needed, up to its limit.

Angle of Repose: At repose angle, sliding is about to start, so tanθ = μ.

Examples2

Horizontal Surface Example

A 10 kg block has μ_s = 0.4. Maximum static friction = μ_smg = 0.4×10×10 = 40 N.

Angle of Repose Example

If μ = 1/√3, angle of repose θ satisfies tanθ = 1/√3, so θ = 30 degrees.

Common mistakes3

Always Writing f = μN for Static Friction

Static friction equals μ_sN only at limiting condition. Otherwise it may be smaller.

Wrong Friction Direction

Friction opposes relative motion or tendency at contact, not always the motion of the body relative to ground.

Taking N = mg on Incline

On an incline, normal reaction is mg cosθ if no other perpendicular forces exist.

Reference tables2

Quick revision8

Formulas6

Static Friction
fs≤μsNf_s ≤ μ_sN

Static friction adjusts from zero up to limiting value.

f_s
Static friction
μ_s
Coefficient of static friction
N
Normal reaction
Limiting Friction
flim=μsNf_{lim} = μ_sN

Maximum value of static friction just before slipping begins.

f_lim
Limiting friction
μ_s
Coefficient of static friction
N
Normal reaction
Kinetic Friction
fk=μkNf_k = μ_kN

Frictional force during sliding motion.

f_k
Kinetic friction
μ_k
Coefficient of kinetic friction
N
Normal reaction
3 more formulas — sign up free

Diagrams3

6

Circular Motion Applications

10 min

Circular motion applications use Newton’s second law toward the centre of the circle. A body moving in a circle requires centripetal force mv²/r, which is not a new force but the inward resultant of real forces such as tension, friction, gravity or normal reaction. In a vehicle moving on a curved road, friction or banking supplies the required centripetal force. Banking reduces dependence on friction by using a component of normal reaction. A conical pendulum moves in a horizontal circle, where horizontal tension gives centripetal force and vertical tension balances weight. In vertical circular motion, tension and weight change at different points, making top and bottom conditions important.

Key points6

  • Always choose radial inward direction for centripetal equation.
  • Tangential velocity is perpendicular to radius.
  • On a flat road, maximum safe speed is v = √(μrg).
  • Banked road design speed does not require friction when tanθ = v²/rg.
  • In vertical circle, weight may help or oppose centripetal requirement depending on position.
  • For a string just taut at top, tension at top becomes zero.

Memory tricks2

Centripetal Force: Centripetal means centre-seeking: always write radial equation toward the centre.

Banking: Banking tilts normal reaction so it can help turn the vehicle.

Examples2

Flat Road Example

If μ = 0.25 and r = 40 m, maximum speed = √(0.25×40×10) = 10 m/s.

Conical Pendulum Example

For a conical pendulum, dividing T sinθ = mv²/r by T cosθ = mg gives tanθ = v²/rg.

Common mistakes3

Calling Centripetal Force a Separate Force

Centripetal force is the net inward force, supplied by real forces like friction, tension or gravity.

Wrong Direction of Acceleration

Centripetal acceleration is always inward, not in the direction of velocity.

Using Same Vertical Circle Equation Everywhere

At top and bottom, radial directions differ, so tension equations are different.

Reference tables2

Quick revision8

Formulas5

Centripetal Force
Fc=mv2/rF_c = mv²/r

Net inward force required for circular motion.

F_c
Centripetal force
m
Mass
v
Speed
r
Radius
Flat Curved Road Maximum Speed
vmax=√(μrg)v_{max} = √(μrg)

Maximum speed on a flat road when friction provides centripetal force.

v_max
Maximum safe speed
μ
Coefficient of friction
r
Radius of curve
g
Acceleration due to gravity
Banking of Roads Without Friction
tanθ=v2/(rg)tanθ = v²/(rg)

Condition for safe turning on a banked road without relying on friction.

θ
Banking angle
v
Design speed
r
Radius of road curve
2 more formulas — sign up free

Diagrams3

7

Problem Solving

12 min

Problem solving in Laws of Motion depends on selecting the body or system, drawing a correct free body diagram and applying Newton’s second law along suitable axes. For a block on a horizontal surface, horizontal forces decide acceleration and vertical forces decide normal reaction. For an inclined plane, axes along and perpendicular to the plane simplify equations. Connected blocks and pulleys require separate FBDs and constraint relations because connected bodies often have related accelerations. Elevator problems modify apparent weight due to vertical acceleration. Friction-based problems require deciding whether friction is static, limiting or kinetic. Mixed NEET problems usually combine FBD, friction, pulley tension and acceleration equations.

Key points7

  • For a block on smooth horizontal surface, acceleration is applied force divided by mass.
  • For a block sliding down smooth incline, acceleration is g sinθ.
  • For connected blocks, tension is usually internal if the whole system is selected.
  • In a massless frictionless pulley, tension is same throughout the same string.
  • Elevator apparent weight is N = m(g + a) upward acceleration and N = m(g - a) downward acceleration.
  • Constraint relation means connected bodies may have equal or related accelerations.
  • If assumed friction direction gives negative value, actual friction direction is opposite.

Memory tricks3

Newton Problem Method: FBD first, formula later. Most mistakes disappear when the diagram is correct.

Incline Components: On incline: sin slides, cos presses.

Elevator: Accelerating up feels heavier; accelerating down feels lighter.

Examples4

Block on Incline

A block slides down a smooth 30 degree incline. Acceleration = g sin30 = 10 × 1/2 = 5 m s^-2.

Connected Blocks

Two blocks 2 kg and 3 kg on a smooth surface are pulled by 10 N. System acceleration = 10/(2+3) = 2 m s^-2.

Elevator Example

A 60 kg person in an elevator accelerating upward at 2 m s^-2 has apparent weight N = 60(10+2) = 720 N.

Friction Example

A 5 kg block has μ = 0.2 and is pulled by 20 N. Friction = 0.2×5×10 = 10 N, so acceleration = (20-10)/5 = 2 m s^-2.

Common mistakes4

Skipping FBD

Direct formula use fails in mixed problems. Always draw forces before equations.

Using Same Acceleration Without Constraint

Connected bodies may have equal acceleration only when string geometry demands it.

Ignoring Static Friction Check

Before using kinetic friction, verify whether the applied force exceeds maximum static friction.

Wrong Tension Assumption

Tension is same only in a massless string over a frictionless pulley.

Reference tables3

Quick revision8

Formulas6

Block on Horizontal Smooth Surface
a=F/ma = F/m

Acceleration when a net horizontal force F acts on mass m without friction.

a
Acceleration
F
Applied horizontal force
m
Mass
Block on Rough Horizontal Surface
a=(F−μmg)/ma = (F - μmg)/m

Acceleration of a sliding block pulled horizontally on a rough surface.

μ
Coefficient of kinetic friction
mg
Weight
Block on Smooth Inclined Plane
a=gsinθa = g sinθ

Acceleration down a smooth incline due to component of gravity.

θ
Angle of inclination
g
Acceleration due to gravity
3 more formulas — sign up free

Diagrams3

Videos for Laws of Motion

  • Class 11 | Newton’s Laws of MotionVedantu JEE English · English
  • Newton's Law of Motion - First, Second & Third - PhysicsThe Organic Chemistry Tutor · English
  • Newton's Laws Of Motion Class 11 | One Shot in English Unacademy JEE English · English
  • Forces and Newton's Laws of Motion [Overview]Mostly Physics · English
  • Types of forces and free body diagrams | PhysicsKhan Academy · English
  • Newton's first law of motion | Forces and Newton's laws of motion | PhysicsKhan Academy · English
Watch with a free account

NEET PYQs from Laws of Motion

The latest 3 of 47 previous-year questions from this chapter.

  1. NEET 2026Set 11EasyQ1

    The magnitude and direction of the acceleration produced in a body of mass 5 kg when two mutually perpendicular forces 8 N and 6 N act on it, are respectively:

    • A20 m s⁻²; tan⁻¹(4/3) with 8 N force
    • B2 m s⁻²; tan⁻¹(3/4) with 6 N force
    • C2 m s⁻²; tan⁻¹(4/3) with 8 N force
    • D2 m s⁻²; tan⁻¹(3/4) with 8 N force
  2. NEET 2026Set 50EasyQ2

    A car travels on a circular racetrack of radius $50\ \mathrm{m}$, which is banked at an angle $\theta$. If the car travels at a speed $10\ \mathrm{ms^{-1}}$, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be $10\ \mathrm{ms^{-2}}$, the value of $\theta$ is:

    • A$\tan^{-1}\left(\frac{1}{5}\right)$
    • B$\tan^{-1}\left(\frac{2}{5}\right)$
    • C$\tan^{-1}(\sqrt{3}/2)$
    • D$\tan^{-1}(2\sqrt{3})$
  3. NEET 2026Set 11EasyQ3

    A box of mass 15 kg is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is 0.12. Keeping the box in stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in m s⁻² is: (g = 10 m/s²)

    • A2.1
    • B1.8
    • C1.5
    • D1.2

Get the whole of Laws of Motion

A free account opens every formula, diagram, table and quick-revision list in this chapter, and tracks the topics you finish.

  • Free to start
  • Progress saved per topic
  • PYQ practice for Laws of Motion