Laws of Motion — NEET Notes

Inertia & Newton's first law

Before Galileo, the dominant view — inherited from Aristotle — was that a body needs a continuous external push to keep moving. It is a natural conclusion to draw from everyday life: a toy car on a floor coasts to rest the moment you stop dragging the string. Galileo realised that the rest is not the natural state at all; it is the frictionless state that is the true reference. Rolling balls down a double-inclined plane, he extrapolated to the idealised case where friction vanishes — and concluded that a ball released from one slope must rise to the same height on the other, no matter how shallow the second slope. In the limit of a horizontal second plane, the ball travels forever. Uniform motion needs no force; only changes to motion do.

Newton crystallised Galileo's insight into the first law of motion: every body continues in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise. Re-read carefully — the law does not distinguish between rest and uniform motion. Both are states of zero acceleration. A spaceship drifting through interstellar space with rockets off is in exactly the same dynamical state as a book lying on your desk. In both, the net external force is zero.

The property described here — a body's tendency to resist any change in its state of motion — is called inertia. Inertia is measured by mass. The reason you lurch forward when a bus stops suddenly is inertia: your feet are coupled to the floor by friction and decelerate with it, but the rest of you carries on at the bus's earlier velocity until restoring muscular forces (or the seat in front) intervene. The first law, then, is more than a statement about isolated objects in deep space — it is the everyday physics behind whiplash, child-seat regulations, and why you stir cocoa for thirty seconds after taking the spoon out.

There are three useful flavours of inertia. Inertia of rest — a stationary bus jerks forward and the passengers fall backward. Inertia of motion — a moving bus brakes hard and the passengers pitch forward. Inertia of direction — a car takes a sharp left turn and the passengers slide right. NEET 2023's football-player question (Q.4) was a textbook test of inertia of direction: changing direction at constant speed still requires a force, because momentum is a vector and its direction changed.

The first law also quietly defines what a frame of reference must be like for Newton's laws to apply. A frame in which a force-free body moves at constant velocity is called an inertial frame. A frame fastened to the ground is approximately inertial for everyday mechanics; a frame fastened to an accelerating car or rotating turntable is non-inertial, and Newton's laws appear to fail there unless one adds pseudo-forces like centrifugal or Coriolis. For NEET, the safe default is to do all mechanics in the ground frame.

Newton's second law — F = ma and impulse

The first law handles the case when the net external force is zero. The second law tells you what happens when it is not. The momentum of a body is defined as p = mv — the product of mass and velocity, a vector quantity. Newton's second law in its most general form says that the rate of change of momentum is directly proportional to the applied force and takes place in the direction in which the force acts:

F = dp/dt

Newton's second law in modern form — valid even when mass varies with time

For a body of fixed mass m, this collapses to the familiar F = ma, because dp/dt = m(dv/dt) = ma. The SI unit of force is the newton: 1 N is the force that gives 1 kg an acceleration of 1 m/s². Three observations every NEET student must internalise: the second law is a vector equation, applying component-wise along x, y and z; F refers to the net external force on a single particle (or the centre of mass of a system); and the relation is local — force here and now produces acceleration here and now, with no memory of past motion.

Sometimes a large force acts for a very short time — a bat striking a ball, a hammer driving a nail, a football boot meeting a stationary ball. The force itself is enormous and hard to measure, and so is the contact duration. But the product of the two is the change in momentum, which is straightforward to compute from initial and final velocities. That product is called impulse:

The impulse principle explains much of safety engineering. Crumple zones, air bags, foam packaging, gym mats, and a cricketer's wrist-curl on a catch all stretch out Δt to reduce the force needed for a given Δp. Δp is fixed by speed and mass; only the time is negotiable.

Three points to file away. Vector nature: F = ma is three scalar equations, one per axis. A force off-axis from velocity changes only the parallel component — which is exactly why a projectile's horizontal velocity is constant under gravity. Single-particle origin: for an extended body, F = ma applies to the centre of mass, with internal forces cancelling in third-law pairs. Locality: acceleration here and now is set by force here and now. A stone dropped out of an accelerating train carries no memory of the train's acceleration.

A clean numerical from NCERT example 4.2: a 0.04 kg bullet at 90 m/s stops in 60 cm. Retardation a = u²/(2s) = 6750 m/s²; average resistive force F = ma = 270 N — several thousand times the bullet's weight of 0.4 N. The "average" qualifier matters because the actual force peaks even higher during the brief penetration.

Newton's third law & action-reaction pairs

Where does the external force on a body come from? Newton's answer is that every force is an interaction between two bodies — and these interactions come in equal-and-opposite pairs. To every action, there is always an equal and opposite reaction. Stated more carefully: the force on body A by body B is equal in magnitude and opposite in direction to the force on body B by body A. The two forces act at the same instant on different bodies. There is no cause-and-effect ordering — neither force is the "first" one.

This is the single most misquoted law in physics. The most common mistake is to identify two forces acting on the same body and call them an action-reaction pair. They are not. Look at a book resting on a table. The forces on the book are its weight W (gravity, pulling it down) and the normal reaction R (table, pushing it up). W and R are equal and opposite, but they are not a Newton's-third-law pair, because they act on the same body and they are different kinds of force (gravitational and contact). The actual third-law partners are: the gravitational pull of the Earth on the book is paired with the gravitational pull of the book on the Earth (acting at the Earth's centre); and the table's push on the book is paired with the book's push on the table. NEET tests this constantly through statement-based questions and "which of the following is an action-reaction pair" framings.

Newton himself was careful to point out that the words "action" and "reaction" are just labels for two halves of a single mutual interaction. Either force can be called the action; the other is then the reaction. This symmetric phrasing matters in problems involving a rocket and its exhaust, a swimmer pushing against water, or a person walking on the ground — each of which is fundamentally a Newton's-third-law setup.

Canonical third-law pairs to recite cold. Walking: foot pushes ground back, ground pushes person forward. Rowing: oar pushes water back, water pushes boat forward. Rocket propulsion: rocket ejects gases backward, gases push rocket forward — and because the pair is between rocket and its own exhaust, rockets work in vacuum. Bird flight: wings push air down, air pushes wings up. Recoil of a gun: gases push the bullet forward, gases push the gun back — the analysis we apply next.

Two diagnostic questions help separate real third-law pairs from impostors. Are the two forces on different bodies? Would removing one body remove the other force? Both checks must say yes. Apply them to "weight of book ↔ normal reaction from table" and both fail: same body, and the weight would persist even without the table. So weight and normal reaction are not a third-law pair, even though they happen to be equal and opposite in this equilibrium.

Conservation of linear momentum

The second and third laws together imply one of the deepest results in physics: conservation of linear momentum. Consider a gun firing a bullet. By the second law, F · Δt is the impulse on the bullet and hence its change in momentum. By the third law, the force on the gun by the bullet is −F over the same Δt, giving an equal and opposite change in momentum on the gun. Since both were initially at rest, the bullet's forward momentum p_b exactly equals the gun's backward (recoil) momentum p_g in magnitude, so p_b + p_g = 0. The total momentum of the (bullet + gun) system is unchanged by the firing.

Generalising: in an isolated system — one with no net external force — the total linear momentum is conserved. Internal forces between particles come in third-law pairs and cancel when summed; only external forces can change the system's total momentum. This single principle handles a vast range of NEET problems — collisions (elastic and inelastic), recoil of guns, explosions, decays of nuclei at rest, and rocket propulsion. Note that momentum conservation applies whether or not kinetic energy is conserved; elastic and inelastic collisions both conserve momentum, but only elastic collisions conserve kinetic energy.

A concrete NCERT example: a 100 kg gun fires a 0.020 kg shell at 80 m/s. By momentum conservation, 0 = 100 v_gun + 0.020 × 80, giving v_gun = −0.016 m/s. Small but non-zero — the recoil kick the shooter feels. Another canonical setup: a stationary nucleus decays into two fragments. The fragments fly off in opposite directions with m₁v₁ = m₂v₂ in magnitude — the lighter fragment moves faster, which is why alpha particles emerge much faster than the recoiling daughter nucleus.

Equilibrium of a particle

A particle is in mechanical equilibrium when the net external force on it is zero. By the first law, this means it is either at rest or moving with uniform velocity. For two forces, equilibrium demands F₁ = −F₂ — equal magnitudes, opposite directions, same line of action. For three concurrent forces, F₁ + F₂ + F₃ = 0, meaning the three force vectors can be arranged tip-to-tail to form a closed triangle. More generally, any number of concurrent forces in equilibrium can be represented by the sides of a closed polygon traversed in order.

In practice, equilibrium problems are solved by resolving the forces into components along chosen axes and demanding that each component sum to zero independently. A bob hung from the ceiling, displaced sideways by a horizontal force; a mass on an inclined plane; a sign suspended by two cables — all reduce to two scalar equations, ΣF_x = 0 and ΣF_y = 0. This is the bread-and-butter problem-solving recipe of statics.

A subtlety NCERT flags: equilibrium of a particle needs only that the net force be zero. Equilibrium of an extended body needs an extra condition — the net torque about any point must also vanish, so the body neither translates nor rotates. NEET problems in Class 11 mostly treat the bodies as point particles, but as soon as a rigid rod, a hinged board, or a wheel appears, you must remember to balance torques as well. Torque is treated formally in the chapter on rotational motion, but the lead-up begins here.

Common forces in mechanics

Mechanics deals with a small zoo of common forces. Gravity is the pervasive non-contact force: every mass attracts every other mass; near the Earth's surface the gravitational pull on a body of mass m is its weight W = mg, directed vertically downward. Normal reaction is the component of the contact force perpendicular to the surfaces in contact — a self-adjusting force that prevents bodies from interpenetrating. Friction is the component of contact force parallel to the surfaces; we treat it in detail in the next section. Tension in a string is the restoring force that a stretched string transmits along its length; for a light (massless) inextensible string, tension has the same magnitude throughout. Spring force is the restoring force generated when a spring is compressed or extended, F = −kx for small displacements (Hooke's law).

NCERT notes that all contact forces — friction, tension, normal reaction, spring force — fundamentally arise from electrostatic interactions between the charged constituents of matter. Modern physics recognises only four fundamental forces (gravitational, electromagnetic, weak and strong nuclear); the contact forces of everyday mechanics are all electromagnetic in origin. Mass is invariant but weight W = mg is not (g varies with latitude and altitude). Normal reaction is self-adjusting — including in lifts, where it equals m(g ± a). Tension is one of the unknowns you solve for. Spring force F = −kx holds only within the elastic limit.

Friction — static, kinetic, rolling

Friction is the contact force component parallel to the surfaces in contact, and it opposes relative motion (or impending relative motion) between them. Three distinct regimes exist. Static friction is the friction between two surfaces that are not sliding. It is self-adjusting: it grows just large enough to prevent relative motion, up to a maximum value f_s,max = μ_sN. Kinetic friction (also called sliding friction) acts when the surfaces are sliding, with magnitude f_k = μ_kN. Rolling friction acts when a body rolls without slipping, and is far smaller still. Coefficients of friction are dimensionless — this dimensional fact was tested literally in NEET 2018.

A useful piece of geometry: place a block on a plank and tilt the plank slowly upward. At a critical angle θ_max, the block just begins to slide. Balancing forces along the incline gives mg sin θ_max = μ_smg cos θ_max, so μ_s = tan θ_max. This is the angle of friction — a single tilt experiment measures the coefficient. NEET 2018's wedge problem (block on a smooth wedge that accelerates sideways) used the same geometry without friction: balancing forces in the wedge's frame gave a = g tan θ.

The "opposing relative motion" clause is the source of more confusion than any other line in the chapter. Friction opposes the tendency to slip, not motion in any absolute sense. Static friction on a tyre of an accelerating car points forward (the wheel would slip backward without it); on the feet of a walker, forward; on a box in an accelerating train, forward. In all three cases friction produces forward acceleration — the very reverse of "opposing motion".

Free-body diagrams — the universal method

Almost every NEET problem in this chapter reduces to a single workflow: isolate the body, draw every force acting on it as an arrow originating from the body, choose convenient axes, write Newton's second law as ΣF = ma along each axis, and solve. The drawing — a free-body diagram, or FBD — is the most important step. Mistakes in FBDs are the single largest source of lost marks on mechanics problems in NEET. Here is the procedure NCERT teaches.

A worked illustration. A 6 kg mass hangs from the ceiling on a 2 m rope, and a horizontal 50 N force is applied at the midpoint P. Free-body diagram of the weight: tension T₂ pulling up, weight 60 N pulling down — so T₂ = 60 N. Free-body diagram of point P: tension T₁ up the upper rope, tension T₂ down, applied 50 N horizontal. Resolving T₁ into components: T₁ cos θ = 60 N (vertical balance), T₁ sin θ = 50 N (horizontal balance). Dividing, tan θ = 50/60, giving θ ≈ 40°. Notice the answer does not depend on the rope's length or where the force is applied — a feature of equilibrium for a massless rope.

Pulleys and connected-body problems

When two or more bodies are connected by a light inextensible string passing over a smooth pulley, three things follow automatically. First, the tension is the same on both sides of the string (because the string is massless and the pulley is frictionless). Second, the magnitude of acceleration is the same for both bodies (because the string is inextensible — one end can only move as much as the other). Third, you treat each body separately with its own FBD, then combine the equations.

The textbook Atwood machine: masses m₁ and m₂ on either side of a frictionless pulley with m₂ > m₁. For m₂ (descending) with acceleration a downward: m₂g − T = m₂a. For m₁ (ascending) with the same magnitude a upward: T − m₁g = m₁a. Adding: (m₂ − m₁)g = (m₁ + m₂)a, so a = (m₂ − m₁)g / (m₁ + m₂). NEET 2020 set this verbatim — 4 kg and 6 kg masses on a frictionless pulley — giving a = (6 − 4)g/10 = g/5. Substituting back, T = 2m₁m₂g / (m₁ + m₂).

A subtler problem: two blocks A (mass 3m) and B (mass m) hang one below the other from a spring, joined by a string. The spring stretches by x with kx = 4mg (supporting both blocks). When the string between A and B is cut, what are their instantaneous accelerations? For B, only gravity acts now — a_B = g downward. For A, the spring is still stretched (the spring's extension cannot change instantaneously), so it still pulls up with kx = 4mg, while gravity pulls down with 3mg. Net upward force on A is 4mg − 3mg = mg, so a_A = g/3 upward. NEET 2017 asked precisely this question — the trap is to assume the spring relaxes the moment the string is cut. Springs deform over time; massless ropes adjust instantly.

The general "constraint" insight: an inextensible string forces equal end-speeds; a massless string carries uniform tension; a smooth pulley redirects the string without changing it. For mixed systems — table block + hanging block over a pulley — the recipe is unchanged: draw two FBDs, write F = ma along each block's natural direction, combine.

Circular motion & centripetal force

A body moving in a circle of radius R at uniform speed v has an acceleration of magnitude v²/R directed toward the centre — even though the speed is constant. By the second law, this acceleration must be produced by a net inward force F_c = mv²/R, called the centripetal force. The centripetal force is not a new kind of force — it is a role played by whatever ordinary force happens to be supplying the inward pull: tension for a whirled stone, gravity for an orbiting planet, friction for a car on a flat curved road, the horizontal component of the normal reaction for a car on a banked curve, electrostatic attraction for an electron around a nucleus.

On a level road, friction alone supplies the centripetal force: mv²/R ≤ μ_sN = μ_smg, giving a maximum cornering speed of v_max = √(μ_sRg). The maximum is set by the road–tyre interaction and is independent of the car's mass. NCERT's worked example: a cyclist at 18 km/h (5 m/s) on a turn of radius 3 m with μ_s = 0.1 needs v² = 25 m²/s², but μ_sRg = 2.94 m²/s² — far short. The cyclist slips.

The centripetal acceleration v²/R is perpendicular to the velocity, so it changes the direction of velocity but not its magnitude — that is uniform circular motion. If the speed is also changing, there is an additional tangential component along the velocity. For a vertical circle, gravity must be included in the force equation. At the top, T + mg = mv²_top/R, so the minimum speed at the top (T = 0) is v_min = √(gR). At the bottom, T − mg = mv²_bot/R and the tension peaks. The maximum-minus-minimum tension across the loop is exactly 6mg — a standard NEET-trap result.

Banking of roads

To handle higher cornering speeds without relying on friction, roads on curves are tilted inward — banked. The normal reaction now has a horizontal component that points toward the centre of the turn, and at one particular speed this component alone supplies the centripetal force, with no friction required. From the geometry: vertical balance gives N cos θ = mg, horizontal balance gives N sin θ = mv²/(Rg). Dividing one by the other yields the foundational banking-angle equation.

If friction is also available — with coefficient μ_s — the maximum safe speed on the banked road is increased to v_max = √[Rg(μ_s + tan θ)/(1 − μ_s tan θ)]. This was the precise formula asked in NEET 2016 (Q.149), word for word. The derivation: at the maximum speed, friction has reached its limit μ_sN and acts down the slope. Setting up the two equilibrium equations and eliminating N gives the formula. Setting μ_s = 0 recovers the optimum-speed result; setting θ = 0 recovers the flat-road maximum.

NCERT's worked example 4.11 puts numbers on this. A 300 m racetrack banked at 15° with μ_s = 0.2 has optimum speed v₀ ≈ 28.1 m/s (~100 km/h) and maximum permissible speed v_max ≈ 38.1 m/s (~137 km/h). Banking buys both safety and tyre life — at v₀ the tyres do not grip sideways at all. Below v₀, the car tends to slide down the bank so friction acts up the slope; above v₀, the car tends to slide up so friction acts down. A car can be parked on a banked road only if tan θ ≤ μ_s — the standard angle-of-repose condition.

Apparent weight in lifts

Stand on a weighing scale in a stationary lift; the scale reads your true weight W = mg. Now let the lift accelerate. The reading on the scale is the normal reaction N the floor exerts on you — and that, applying Newton's second law to your body in the ground frame, is what changes. With upward chosen as positive: N − mg = ma, so N = m(g + a). Four cases tell the whole story.

NCERT's worked example 4.13 frames this with concrete numbers: a 70 kg man in a lift moving up with a = 5 m/s² records N = 70(10 + 5) = 1050 N on the scale, compared with his true weight of 700 N. The astronaut who steps outside an accelerating spaceship in deep space (NCERT example 4.1) is in the opposite regime — no nearby massive body means no gravitational force, so by the first law his acceleration is zero the instant he is outside.

True weight is the gravitational force on the body; the scale reads the normal reaction. The two coincide only in ground-frame equilibrium. In free fall — astronaut in orbit, lift cut loose — the scale reads zero, the famous "apparent weightlessness", even though gravity on the body is essentially unchanged (the ISS orbits in ~90% of ground gravity).