Laws of Motion Class 11 Notes | CBSE Physics Chapter 4 (Free PDF)

Chapter summary

Laws of Motion builds on Newton’s three laws to explain why objects start, stop, speed up, or change direction, linking force to momentum through F = ma and F = dp/dt. It also covers friction, impulse, equilibrium, and conservation of momentum, the ideas behind everyday recoil, collisions, and motion on inclines. It is a high-yield mechanics chapter for NEET, supplying both direct concept questions and the force-balance setups you reuse across the whole physics paper.

Chapter notes

Table of Contents


Key Concepts

1. Force and Inertia

A force is a push or a pull that can change the state of rest or of uniform motion of a body, or change its shape. SI unit: newton (N). Force is a vector quantity.

Inertia is the natural tendency of a body to resist any change in its state of rest or of uniform motion. A heavier body has more inertia — that is why it is harder to push a loaded trolley than an empty one.

Three Types of Inertia

  • Inertia of rest: a body at rest stays at rest (dust falls off a carpet when you beat it).
  • Inertia of motion: a moving body keeps moving (a passenger lurches forward when a bus brakes suddenly).
  • Inertia of direction: a body resists change in its direction of motion (mud flies off tangentially from a spinning wheel).

2. Newton’s First Law of Motion (Law of Inertia)

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

The first law gives us the qualitative definition of force (force is what changes a body’s state of motion) and defines inertia. If the net force is zero, acceleration is zero — the body is in equilibrium.

Key idea: No net force is needed to keep a body moving at constant velocity — only to change its velocity.


3. Linear Momentum

Linear momentum (p) is the quantity of motion contained in a body. It is the product of mass and velocity.

p = mv

  • SI unit: kg·m/s (or N·s)
  • It is a vector quantity, in the direction of velocity.
  • A heavy slow truck and a light fast bullet can have the same momentum.

4. Newton’s Second Law of Motion

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.

F = dp/dt = d(mv)/dt = ma (for constant mass)

  • This is the quantitative definition of force.
  • 1 newton = the force that gives a 1 kg mass an acceleration of 1 m/s².
  • Acceleration is along the direction of the net force.

The second law contains the first law as a special case: if F = 0, then a = 0, so velocity is constant.


5. Impulse

Impulse is the product of a force and the time for which it acts. It equals the change in momentum of the body (the impulse-momentum theorem).

Impulse J = F × Δt = Δp = m(v − u)

  • SI unit: N·s or kg·m/s
  • It is a vector quantity.
  • This is why cricketers pull their hands back while catching a ball — increasing Δt reduces the force F felt for the same change in momentum.

6. Newton’s Third Law of Motion

To every action there is an equal and opposite reaction. Action and reaction act on two different bodies, are equal in magnitude, opposite in direction, and act simultaneously.

  • A gun recoils when a bullet is fired.
  • A swimmer pushes water backward; water pushes the swimmer forward.
  • A rocket expels gas downward; the gas pushes the rocket upward.

Important: Action and reaction never cancel out because they act on different bodies.


7. Law of Conservation of Linear Momentum

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

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

This law is a direct consequence of Newton’s second and third laws. It is the principle behind recoil of a gun, rocket propulsion, and collisions.

Recoil of a Gun

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

V = −mv/M (negative sign shows the gun moves opposite to the bullet)


8. Free-Body Diagrams (FBD)

A free-body diagram is a sketch showing a single body isolated from its surroundings with all the external forces acting on it drawn as arrows.

[DIAGRAM: A block on a table — weight mg downward, normal reaction N upward, applied force F horizontal, friction f opposite to motion.]

Common Forces in an FBD

  • Weight (W = mg): always acts vertically downward.
  • Normal reaction (N): perpendicular to the surface of contact.
  • Tension (T): along a string, pulling away from the body.
  • Friction (f): along the surface, opposing relative motion.

Method: Isolate the body → draw all forces → choose axes → apply ΣFₓ = maₓ and ΣF_y = ma_y.


9. Equilibrium of Concurrent Forces

A body is in equilibrium when the net force on it is zero, so it has zero acceleration (either at rest or moving with constant velocity).

Condition: ΣFₓ = 0 and ΣF_y = 0

For three concurrent forces in equilibrium, the Lami’s theorem applies:

F₁/sin α = F₂/sin β = F₃/sin γ

where α, β, γ are the angles opposite to forces F₁, F₂, F₃ respectively.


10. Friction

Friction is the force that opposes relative motion (or tendency of motion) between two surfaces in contact. It acts along the surfaces, opposite to the direction of motion.

Types of Friction

TypeDescriptionRelation
Static friction (fₛ)Acts when the body is at rest; self-adjusting up to a maximumfₛ ≤ μₛN
Limiting friction (fₘₐₓ)Maximum static friction, just before motion beginsfₘₐₓ = μₛN
Kinetic friction (f_k)Acts when the body is in motionf_k = μ_kN

Note: μₛ > μ_k, which is why it is harder to start sliding an object than to keep it sliding.

Laws of Friction

  • Friction is independent of the area of contact.
  • Friction is proportional to the normal reaction N.
  • Kinetic friction is nearly independent of speed.

Angle of Friction and Angle of Repose

  • Angle of friction (φ): tan φ = μ — the angle between the resultant of N and limiting friction, and the normal.
  • Angle of repose (θ): the minimum angle of an inclined plane at which a body just begins to slide. tan θ = μₛ. For a plane inclined at angle θ, the angle of repose equals the angle of friction.

11. Motion on an Inclined Plane

For a body of mass m on a frictionless incline of angle θ, gravity splits into two components.

  • Along the incline (down): mg sin θ → causes acceleration a = g sin θ
  • Perpendicular to incline: mg cos θ → balanced by normal reaction N = mg cos θ

With friction (body sliding down): a = g(sin θ − μ cos θ).


12. Dynamics of Circular Motion

When a body moves in a circle, its direction keeps changing, so it is accelerating even at constant speed. The acceleration points towards the centre — this is centripetal acceleration.

a_c = v²/r = ω²r

The net inward force causing this is the centripetal force:

F_c = mv²/r = mω²r

Important: Centripetal force is not a new kind of force — it is provided by tension, gravity, friction, or normal reaction depending on the situation.

Banking of Roads

On a curved road, banking (tilting the road inward) provides the centripetal force without relying entirely on friction.

  • Ignoring friction: tan θ = v²/(rg), so the safe speed is v = √(rg tan θ).
  • With friction (maximum speed): v_max = √[rg(tan θ + μ)/(1 − μ tan θ)]

Vehicle on a Level Curved Road

Here friction alone supplies the centripetal force, so the maximum safe speed is v_max = √(μrg).


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Weightage in Board & Entrance Exams

ExamTypical WeightageMost-Tested Areas
CBSE Board (Class 11)8–10 marksNewton’s laws, FBD, friction, conservation of momentum
JEE Main / Advanced2–4 questions (with Work-Energy)Pulley-block systems, friction on inclines, circular motion
NEET2–3 questionsMomentum, impulse, friction, banking of roads

[TABLE: Question-type split — VSA (1 mark): laws & definitions; SA (2–3 marks): FBD, friction numericals, impulse; LA (5 marks): conservation of momentum derivations, banking of roads.]


Important Definitions

TermDefinition
ForceAn external push or pull that changes a body’s state of rest or motion: F = ma
InertiaTendency of a body to resist change in its state of rest or uniform motion
Linear momentumQuantity of motion: p = mv (a vector)
Newton’s second lawF = dp/dt = ma — force equals rate of change of momentum
ImpulseProduct of force and time = change in momentum: J = FΔt = Δp
Conservation of momentumTotal momentum is constant if net external force is zero
Static frictionSelf-adjusting opposing force on a body at rest: fₛ ≤ μₛN
Kinetic frictionConstant opposing force on a moving body: f_k = μ_kN
Angle of reposeIncline angle at which a body just begins to slide: tan θ = μₛ
Centripetal forceInward force keeping a body in circular motion: F = mv²/r

Solved Examples

Example 1

A force of 20 N acts on a body of mass 4 kg initially at rest. Find the acceleration and the velocity after 5 s.

Answer: a = F/m = 20/4 = 5 m/s². v = u + at = 0 + 5 × 5 = 25 m/s.

Example 2

A cricket ball of mass 0.15 kg moving at 20 m/s is stopped by a fielder in 0.1 s. Find the average force exerted.

Answer: Impulse = Δp = m(v − u) = 0.15(0 − 20) = −3 N·s. F = Δp/Δt = −3/0.1 = −30 N (magnitude 30 N, opposing motion).

Example 3

A gun of mass 5 kg fires a bullet of mass 25 g with a velocity of 500 m/s. Find the recoil velocity of the gun.

Answer: By conservation of momentum: 0 = MV + mv. V = −mv/M = −(0.025 × 500)/5 = −12.5/5 = −2.5 m/s (gun recoils at 2.5 m/s).

Example 4

A block of mass 10 kg rests on a horizontal surface with μₛ = 0.4. Find the minimum horizontal force needed to just move it. (g = 10 m/s²)

Answer: fₘₐₓ = μₛN = μₛmg = 0.4 × 10 × 10 = 40 N.

Example 5

A car of mass 1000 kg takes a circular turn of radius 50 m at 10 m/s. Find the centripetal force required.

Answer: F = mv²/r = (1000 × 10²)/50 = 100000/50 = 2000 N.

Example 6

Two masses m₁ = 5 kg and m₂ = 3 kg are connected by a string over a frictionless pulley (Atwood machine). Find the acceleration. (g = 10 m/s²)

Answer: a = (m₁ − m₂)g/(m₁ + m₂) = (5 − 3)(10)/(5 + 3) = 20/8 = 2.5 m/s². Tension T = 2m₁m₂g/(m₁ + m₂) = 2 × 5 × 3 × 10/8 = 37.5 N.


Important Questions for Board Exams

1-Mark Questions (VSA)

  1. Define one newton of force.
  2. Why does a passenger jerk forward when a moving bus stops suddenly?
  3. What is the SI unit of linear momentum?
  4. Why is it easier to pull a lawn roller than to push it?
  5. Can a body be in equilibrium under the action of a single force? Justify.

2–3-Mark Questions (SA)

  1. State and explain the impulse-momentum theorem with one real-life example.
  2. Define angle of repose and show that it equals the angle of friction (tan θ = μ).
  3. A body slides down a rough inclined plane of angle θ. Derive an expression for its acceleration.
  4. Explain why action and reaction, though equal and opposite, do not cancel each other.

5-Mark Questions (LA)

  1. State Newton’s second law and derive F = ma from it. Hence define one newton.
  2. State the law of conservation of linear momentum and derive it from Newton’s third law. Apply it to explain the recoil of a gun.
  3. Derive the expression for the maximum safe speed of a vehicle on a banked road with friction.

Quick Revision Points

  • Inertia = resistance to change in motion; three types — rest, motion, direction
  • First law (inertia): no net force needed for constant velocity
  • Second law: F = dp/dt = ma; 1 N gives 1 kg an acceleration of 1 m/s²
  • Momentum p = mv (vector); impulse J = FΔt = Δp
  • Third law: action and reaction are equal, opposite, on different bodies
  • Conservation of momentum holds when net external force = 0
  • Friction: fₛ ≤ μₛN, f_k = μ_kN, with μₛ > μ_k
  • Angle of repose: tan θ = μₛ; equals the angle of friction
  • Incline (frictionless): a = g sin θ; with friction: a = g(sin θ − μ cos θ)
  • Centripetal force F = mv²/r; banking (no friction): tan θ = v²/rg
  • Level curve safe speed: v_max = √(μrg)

Next Chapter: Chapter 5 — Work, Energy and Power

🃏 Flash Cards: Laws of Motion

Class 11 Physics · Chapter 5 – swipe through all 9 cards to understand the whole chapter.

🧊Start here1/9

Inertia & Newton’s First Law

A body keeps doing what it is already doing until a net external force changes it.

ΣF = 0 → rest stays rest, motion stays uniform

Holds only in an inertial (non-accelerating) frame.

  • Inertia = resistance to a change in motion; more mass = more inertia.
  • No force is needed to keep moving at constant velocity, only to change it.
  • Three flavours: inertia of rest, of motion, of direction.
⚖️Core law2/9

Newton’s Second Law

The net force on a body equals its mass times its acceleration.

F_net = m · a (a ∝ 1/m for fixed F)

Unit: newton (N) = kg·m·s⁻2; keep mass in kg.

  • Always use the NET force — add/subtract opposing forces first.
  • 10 N on 2 kg → a = F/m = 5 m·s⁻2.
  • F and a are vectors and point the same way; velocity need not.
🎯Momentum form3/9

Momentum & the Deeper Law

Force is really the rate of change of momentum, the ‘quantity of motion’.

p = m · v , F_net = Δp / Δt

Collapses to F = ma when mass is constant.

  • Momentum unit: kg·m·s⁻1.
  • Use Δp/Δt for rockets, collisions and impulse problems.
  • Longer contact time → smaller force (why you ‘give way’ on a catch).
💥Impulse4/9

Impulse–Momentum Theorem

A force acting over a time interval produces a change in momentum.

J = F · Δt = Δp = m·v − m·u

Unit of impulse: N·s = kg·m·s⁻1 (same as momentum).

  • Same impulse, more time → less peak force (airbags, crumple zones).
  • Catching a ball: pull hands back to lengthen Δt and cut the force.
  • Impulse is a vector along the applied force.
🤝Core law5/9

Newton’s Third Law

Every action has an equal and opposite reaction, on a different body.

F_AB = − F_BA (equal size, opposite direction)

The pair acts on TWO different bodies, so it can never cancel.

  • Swimmer pushes water back; water pushes swimmer forward.
  • Trap: a book’s weight and the table’s normal force are NOT a pair (same body).
  • Equal forces, unequal masses → unequal accelerations (gun recoil).
🚀Key result6/9

Conservation of Momentum

With no external force, the total momentum of a system stays constant.

If ΣF_ext = 0: m1u1 + m2u2 = m1v1 + m2v2

Follows directly from F = Δp/Δt and the Third Law.

  • Explains recoil: gun and bullet gain equal and opposite momentum.
  • Rocket pushes gas down → rocket gains upward momentum.
  • Holds in collisions and explosions, elastic or not.
🧱Equilibrium7/9

Equilibrium of Forces

A body is in equilibrium when all the forces on it add to zero.

ΣF = 0 ⇒ a = 0 (rest OR constant velocity)

Zero net force ≠ at rest; it can move uniformly.

  • Resolve forces into perpendicular components; each direction sums to zero.
  • Balanced forces and ‘no force’ give identical motion.
  • A body at constant velocity has zero acceleration, so zero net force.
🛑Contact force8/9

Friction

A contact force that opposes the relative sliding of two surfaces.

f_s ≤ μ_s·N , f_k = μ_k·N (with μ_s > μ_k)

μ is dimensionless; on flat ground N = mg.

  • Static friction self-adjusts up to a ceiling f_max = μ_s·N.
  • Kinetic friction is fixed at μ_k·N once sliding starts.
  • Rolling friction ≪ sliding — why we use wheels and bearings.
⛰️Apply it9/9

Friction on an Incline

On a slope the weight splits into components and the normal force shrinks.

N = mg·cosθ , driving force = mg·sinθ

Body slips when mg·sinθ > μ_s·mg·cosθ, i.e. tanθ > μ_s.

  • N is NOT mg on an incline; it is mg·cosθ, so friction drops too.
  • Angle of repose: tanθ = μ_s, the steepest slope before sliding.
  • Net force down the slope = mg·sinθ − f, then a = F_net/m.
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📝 Practice Laws of Motion — 10 NEET PYQs
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Q1NEET 2020
A truck is stationary and has a bob suspended by a light string, in a frame attached to the truck. The truck suddenly moves to the right with an acceleration of a. The pendulum will tilt
Correct answer: B. Due to inertia the bob lags left; in the truck frame a pseudo force ma acts horizontally so tanθ = ma/mg = a/g, giving θ = tan⁻¹(a/g).
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Q2NEET 2020
Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless. The acceleration of the system in terms of acceleration due to gravity g is
Correct answer: B. For an Atwood machine a = (m₂ − m₁)g/(m₁ + m₂) = (6−4)g/(4+6) = g/5.
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Q3NEET 2020
Calculate the acceleration of the block and trolly system shown in the figure. The coefficient of kinetic friction between the trolly (10 kg) and the surface is 0.05. The hanging block is 2 kg. (g = 10 m/s², mass of the string is negligible and no other friction exists)
Correct answer: A. For trolly: T − μ(10)g = 10a; for block: 2g − T = 2a. Adding: 20 − 0.05×10×10 = 12a → 15 = 12a → a = 1.25 m/s².
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Q4NEET 2019
A particle moving with velocity v is acted upon by three forces shown by the vector triangle PQR. The velocity of the particle will
Correct answer: B. The three forces form the sides of a triangle taken in order, so their resultant is zero. Net force = 0 means acceleration = 0, so velocity stays constant.
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Q5NEET 2019
A body of mass m is kept on a rough horizontal surface (coefficient of friction = μ). Horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by F, where F is
Correct answer: C. The resultant of N = mg and friction f (≤ μmg) is √(N² + f²). Since the body is not moving, f ≤ μmg, so |F| ≤ mg√(1 + μ²).
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Q6NEET 2018
A block of mass m is placed on a smooth inclined wedge ABC of inclination θ. The wedge is given an acceleration a towards the right. The relation between a and θ for the block to remain stationary on the wedge is
Correct answer: D. In the wedge frame a pseudo force ma keeps the block stationary: N sinθ = ma and N cosθ = mg. Dividing gives tanθ = a/g, so a = g tanθ.
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Q7NEET 2017
Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut are respectively
Correct answer: B. Spring force kx = 4mg stays the same instant the string is cut. For 3m: (4mg − 3mg) = 3m·a_A → a_A = g/3 (up). For m: only gravity acts → a_B = g (down).
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Q8NEET 2013
Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m? (g is the acceleration due to gravity)
Correct answer: A. The blocks move at constant velocity, so acceleration is zero; by Newton’s first law the net force on every block, including the 2m block, is zero.
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Q9NEET 2010
A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor, the distance of the man above the floor will be
Correct answer: B. By momentum conservation the man recoils up: m₁r₁ = m₂r₂ → man moves r = (0.5×10)/50 = 0.1 m up while stone falls 10 m, so the man is 10 + 0.1 = 10.1 m above floor.
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Q10NEET 1995
If the force on a rocket moving with a velocity of 300 m/s is 345 N, then the rate of combustion of the fuel is
Correct answer: C. Thrust on rocket = vᵣ(−dm/dt) and equals the reaction force, so −dm/dt = F/u = 345/300 = 1.15 kg/s.
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Frequently Asked Questions

What is Newton’s first law of motion?

Newton’s first law states that a body stays at rest, or keeps moving in a straight line at constant speed, unless a net external force acts on it. This tendency to resist any change in its state of motion is called inertia, and a body with more mass has more inertia.

What is the formula for Newton’s second law and what does it mean?

Newton’s second law is F_net = m times a, meaning the net force on a body equals its mass times its acceleration. More generally force is the rate of change of momentum, F = dp/dt, and the unit of force is the newton (N), equal to kg times m per second squared.

Is Laws of Motion important for NEET?

Yes, Laws of Motion is part of the NEET physics syllabus and is a high-yield mechanics chapter. It is tested directly and its force-balance and free-body-diagram skills carry over into work-energy, circular motion, and many other numerical problems, so it gives strong returns for the effort.

What is the difference between static and kinetic friction?

Static friction acts before sliding begins and self-adjusts up to a maximum of mu_s times N, balancing the applied force until that limit. Kinetic friction acts while the surfaces are sliding and stays fixed at mu_k times N opposite to the motion, and since mu_s is greater than mu_k it takes more force to start sliding than to keep it going.

Why are the weight of a book and the normal force from the table not an action-reaction pair?

A Newton’s third law action-reaction pair must act on two different bodies, but the book’s weight and the table’s normal force both act on the book itself, so they are not a pair. The true reaction to Earth pulling the book down is the book pulling Earth up.

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