1🗓️ How to Prepare
NEET weightage: 1–2 questions directly, but the chapter is the engine behind Work-Energy, Rotation, Gravitation and Fluids. Most questions are a pulley/incline/lift setup or a friction-or-banking calculation. The single biggest scorer here is a clean free-body diagram — draw it and the algebra takes 30 seconds.
Day 1 — Laws
Aristotle's fallacy → Galileo's inertia → the three laws in NCERT's exact words. Momentum, impulse, and why F = ma is a vector, local law.
Day 2 — FBDs
Equilibrium of a particle, normal reaction, tension, spring force. Draw 15 free-body diagrams: lift, two blocks, pulley, string on a wall.
Day 3 — Friction
Static (self-adjusting) vs kinetic, μₛ > μₖ, angle of repose, blocks on inclines. 12 problems with the "is it moving yet?" check.
Day 4 — Circular
Centripetal force is a name, not a new force. Level road, banked road (with and without friction), v_max formulas.
Day 5 — Mock 🧪
Timed 20-question chapter test mixing pulleys, inclines, lifts and banking. Wrong answers → error notes.
2🧠 Concepts
1 · Aristotle's fallacy statement question
Aristotle held that an external force is required to keep a body in uniform motion. That is wrong. The truth NCERT states: a force is needed in practice only to counter the opposing force of friction. Remove friction and no force is needed to keep moving.
2 · Galileo & the law of inertia
Galileo's inclined-plane extrapolation: a ball rolling down one plane and up another reaches nearly the same height; make the second plane horizontal and the ball would roll forever. Hence: a body in motion needs no force to keep moving — the law of inertia. Newton's First Law is the same law rephrased.
3 · Newton's First Law — quote it exactly
"Everybody continues to be in its state of rest or of uniform motion in a straight line, unless compelled by some external force to act otherwise."
In simple terms: if the external force on a body is zero, its acceleration is zero. Note it defines inertia (mass is the measure of inertia) and tells you nothing about how much a force accelerates a body — that's the Second Law's job.
4 · Momentum and the Second Law most tested
Momentum p = mv — a vector, SI unit kg m s⁻¹. Second law: "The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts."
F = dp/dt = ma (k = 1 in SI) 1 N = 1 kg m s⁻²
Four properties NCERT lists — all four are examinable:
- (a) Consistent with the First Law: F = 0 ⟹ a = 0.
- (b) It is a vector equation — it holds component-wise (Fₓ = maₓ, and so on).
- (c) Applies to a particle and to a system, provided F is the total external force and a is the acceleration of the system as a whole.
- (d) It is a local law: F at a point at an instant fixes a at that same point and instant. Acceleration does not depend on the history of motion.
5 · Impulse
Impulse = F × Δt = change in momentum (J = Δp). Useful when a large force acts for a short time — a bat hitting a ball, a catch, a collision. Because the time is so short you may assume the body does not appreciably move during the impulse. SI unit: N s (= kg m s⁻¹).
This is why a cricketer draws his hands back while catching: the same Δp spread over a longer Δt means a smaller force.
6 · Newton's Third Law — the four riders
"To every action, there is always an equal and opposite reaction." Better phrasing: Force on A by B = − Force on B by A. The examinable riders:
- Action and reaction are simultaneous — there is no cause-effect relation, neither precedes the other.
- Either one may be called "action"; the labels are arbitrary.
- They act on different bodies, so they never cancel.
- Internal action-reaction pairs within one body do sum to zero.
7 · Conservation of momentum
The total momentum of an isolated system of particles is conserved. It follows from the second and third laws together (internal forces cancel in pairs; zero external force ⟹ dp/dt = 0).
Standard cases: recoil of a gun, explosion of a shell, collisions, a person walking on a boat, rocket propulsion.
8 · Equilibrium of a particle
Net external force zero ⟹ the particle is at rest or in uniform motion (First Law). Two forces: F₁ = −F₂. Three concurrent forces: F₁ + F₂ + F₃ = 0 — so they form a closed triangle, and each component sum vanishes separately (ΣFₓ = 0, ΣF_y = 0).
Note: full equilibrium of a body also needs zero net torque (rotational equilibrium) — that comes in Chapter 6.
9 · The forces you will actually meet
- Gravity — acts at a distance, needs no medium; weight W = mg.
- Normal reaction (N) — component of the contact force perpendicular to the surfaces.
- Friction (f) — component of the same contact force parallel to the surfaces.
- Tension (T) — restoring force in a string; constant throughout a massless string; an inextensible string has a very high force constant.
- Spring force — F = −kx, the minus sign meaning it opposes the displacement from the unstretched state.
- Buoyancy, viscous drag, air resistance — contact forces from fluids.
All contact forces ultimately arise from electrical forces between charged constituents — of the four fundamental forces, only gravitational and electrical matter in mechanics.
3🧩 The Free-Body Diagram Method
Almost every mechanics problem in NEET is solved by the same five steps. Do them in order and never in your head.
Step 1 · Isolate one body
Draw a box around exactly one object (or one clearly-defined system). Everything outside that box can only act on it.
Step 2 · Draw only external forces
Weight (always), normal reaction (wherever it touches a surface), tension (along every string), friction (along the surface, opposing relative/impending motion), applied force. Never draw ma — that is the effect, not a force.
Step 3 · Choose axes cleverly
On an incline, take x along the slope and y perpendicular to it — then only mg needs resolving (mg sin θ down-slope, mg cos θ into the surface).
Step 4 · Write ΣF = ma per axis
In the direction of no acceleration write ΣF = 0. In the direction of motion write ΣF = ma. For connected bodies, the magnitude of a is shared.
Step 5 · Repeat for each body, then solve
N bodies give N equation-sets; the string tension and the common a are the unknowns tying them together. Third-law pairs appear with opposite signs in the two diagrams — that's your check.
4🧮 Formula Bank
| Situation | Formula | Note |
|---|---|---|
| Momentum | p = mv | vector, kg m s⁻¹ |
| Second law | F = dp/dt = ma | 1 N = 1 kg m s⁻² |
| Impulse | J = F·Δt = Δp = m(v − u) | N s; large force, short time |
| Variable-mass form | F = m(dv/dt) + v(dm/dt) | rocket-type problems |
| Conservation of momentum | m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ | isolated system, F_ext = 0 |
| Recoil of gun | V = −(m/M)v | from p_initial = 0 |
| Equilibrium | ΣFₓ = 0, ΣF_y = 0 | three forces ⟹ closed triangle |
| Spring force | F = −kx | restoring; k = force constant |
| Static friction | fₛ ≤ (fₛ)ₘₐₓ = μₛN | self-adjusting — inequality! |
| Kinetic friction | f_k = μ_k N | equality; μ_k < μₛ |
| Angle of repose / friction | tan θ = μₛ | block just begins to slide |
| Block on rough incline (sliding down) | a = g(sin θ − μ_k cos θ) | up the incline: sin θ + μ_k cos θ |
| Lift (apparent weight) | N = m(g ± a) | + going up/accelerating up, − down; free fall N = 0 |
| Two blocks, force F on m₁ | a = F/(m₁+m₂), contact = m₂F/(m₁+m₂) | contact force acts on the far block |
| Atwood machine | a = (m₁−m₂)g/(m₁+m₂), T = 2m₁m₂g/(m₁+m₂) | massless, frictionless pulley |
| Centripetal force | f_c = mv²/R = mω²R | a name, not a new force |
| Level road, max speed | v_max = √(μₛRg) | independent of mass |
| Banked road, optimum speed | v₀ = √(Rg tan θ) | friction not needed at this speed |
| Banked road with friction | v_max = √[Rg(μₛ + tan θ)/(1 − μₛ tan θ)] | > flat-road value; μₛ = 0 gives v₀ |
| Parking on a bank | tan θ ≤ μₛ | else the car slides down |
5📋 One-Glance Formula Sheet
| Setup | Acceleration | Tension / Normal |
|---|---|---|
| Lift accelerating up (a) | a (up) | N = m(g + a) — feels heavier |
| Lift accelerating down (a) | a (down) | N = m(g − a) — feels lighter |
| Lift in free fall | g (down) | N = 0 — weightlessness |
| Lift at constant velocity | 0 | N = mg — same as at rest |
| Two blocks on a smooth floor, push F | F/(m₁+m₂) | contact force m₂F/(m₁+m₂) |
| Atwood (m₁ > m₂) | (m₁−m₂)g/(m₁+m₂) | T = 2m₁m₂g/(m₁+m₂) |
| Block on table + hanging block | m₂g/(m₁+m₂) (smooth) | T = m₁m₂g/(m₁+m₂) |
| Smooth incline, angle θ | g sin θ | N = mg cos θ |
| Rough incline, sliding down | g(sin θ − μ_k cos θ) | N = mg cos θ |
| Body just about to slide | 0 | tan θ = μₛ (angle of repose) |
6🛑 Friction — the whole story
Static friction is self-adjusting most missed
With no applied force there is no static friction at all. It appears the moment you push, and it grows to exactly match your applied force — keeping the body at rest — until it hits its ceiling (fₛ)ₘₐₓ = μₛN. That is why the law is an inequality: fₛ ≤ μₛN.
Static friction opposes impending motion — the motion that would occur if friction were absent.
Kinetic friction
Once the body moves, friction drops to a fixed value f_k = μ_k N and opposes the actual relative motion. Experimentally μ_k < μₛ — which is why a stuck object suddenly jerks forward when it finally breaks free.
Both laws are approximate and share two features
- Independent of the area of contact — a brick on its face or its edge has the same limiting friction.
- Proportional to the normal force N — not to weight. On an incline N = mg cos θ, not mg; in a lift N changes with the lift's acceleration.
- μₛ and μ_k depend only on the nature of the pair of surfaces — they are dimensionless.
Reducing friction
NCERT's methods: ball bearings between moving parts (rolling replaces sliding), a compressed cushion of air between surfaces, and lubrication. Rolling friction is much smaller than sliding friction.
Friction is not always the villain
Without friction you could not walk (your foot pushes back, friction pushes you forward), a car could not accelerate or turn, and knots and nails would not hold. NCERT's point: an animate body needs an external force to accelerate exactly like an inanimate one — and on the ground, that force is friction.
7🔄 Circular Motion — level and banked roads
Centripetal force is a name, not a new force
A body moving in a circle of radius R with uniform speed v has acceleration v²/R directed towards the centre. By the second law the force causing it is f_c = mv²/R — called the centripetal force. Always identify the real force playing that role: tension for a stone on a string, gravitation for a planet, friction for a car on a flat road, the normal force's component on a banked road.
Car on a level road
Three forces: weight mg, normal N, friction f. Vertically N = mg. The centripetal force comes entirely from static friction, so mv²/R ≤ μₛN = μₛmg, giving
v_max = √(μₛRg)
Independent of the mass of the car — the classic one-mark trap. Also note it is static friction (the tyre doesn't slide sideways), opposing the impending motion away from the circle.
Car on a banked road
Banking lets the horizontal component of N supply centripetal force, so less is asked of friction. With friction included:
N cos θ = mg + f sin θ · N sin θ + f cos θ = mv²/R
Setting f = μₛN and solving:
v_max = √[ Rg (μₛ + tan θ)/(1 − μₛ tan θ) ]
Two readings of this formula that NEET loves:
- Put μₛ = 0 and you get the optimum speed v₀ = √(Rg tan θ) — the speed at which no friction is needed at all, so tyre wear is minimal.
- The banked v_max is always greater than the flat-road √(μₛRg).
- For v < v₀ friction acts up the slope; a car can be parked on the bank only if tan θ ≤ μₛ.
NCERT's worked example: R = 300 m, θ = 15°, μₛ = 0.2 ⟹ v₀ = 28.1 m s⁻¹ and v_max ≈ 38.1 m s⁻¹.
8🎯 Problem Types — how NEET asks this chapter
T1 · Lift / apparent weight
Man in a lift, spring balance reading. N = m(g ± a); free fall ⟹ N = 0. Read the acceleration's direction, not the velocity's.
T2 · Two or three blocks pushed together
Whole system for a, then one block for the contact force. Contact force is always smaller than F.
T3 · Pulley (Atwood) systems
Same |a| for both, T equal throughout a massless string. a = (m₁−m₂)g/(m₁+m₂).
T4 · Block on an incline, smooth or rough
Resolve mg only. a = g(sin θ − μ_k cos θ) sliding down. Angle of repose when a = 0.
T5 · "Will it move?" friction check
Compare the applied force with (fₛ)ₘₐₓ = μₛN before assuming motion. If F < μₛN, then f = F and a = 0.
T6 · Impulse / catching / collision
Force from Δp/Δt. Longer contact time ⟹ smaller force. Watch the sign reversal when a ball rebounds: Δp = m(v + u).
T7 · Recoil, explosion, rocket
Momentum conservation from zero total: MV = −mv. Fragments' momenta must add to the original.
T8 · Three concurrent forces / string on a wall
Resolve into components, ΣFₓ = ΣF_y = 0. Often ends in tan θ = F_horizontal / W.
T9 · Level or banked road
v_max = √(μₛRg) or the banked formula. Ask whether the question wants optimum (no friction) or maximum (friction at limit).
T10 · Statement / assertion on the three laws
Aristotle wrong, action-reaction on different bodies, F = ma is local and vector, ma is not a force. Pure NCERT-line recall.
9🔢 Standard Values & Units
| Quantity | Symbol / value | Dimensions |
|---|---|---|
| Force | 1 N = 1 kg m s⁻² | [M L T⁻²] |
| Momentum | kg m s⁻¹ = N s | [M L T⁻¹] |
| Impulse | N s (same as momentum) | [M L T⁻¹] |
| Spring constant k | N m⁻¹ | [M T⁻²] |
| Coefficients μₛ, μ_k | dimensionless, μ_k < μₛ | [M⁰L⁰T⁰] |
| g in NCERT exercises | 10 m s⁻² (stated), else 9.8 | [L T⁻²] |
| Angle of repose | θ = tan⁻¹ μₛ | — |
| NCERT banking example | R = 300 m, θ = 15°, μₛ = 0.2 ⟹ v₀ = 28.1 m s⁻¹ | — |
10⚡ Shortcuts
For connected blocks, get a from the whole system (internal forces cancel), then isolate one block for tension or contact force. Two lines instead of six.
Acceleration up (going up speeding, or coming down slowing) ⟹ heavier, N = m(g+a). Acceleration down ⟹ lighter. Velocity direction is irrelevant.
Smooth incline a = g sin θ, level-road v_max = √(μₛRg), angle of repose tan θ = μₛ, free-fall weightlessness — all mass-free. If your answer still has m in it, re-check.
"Which way would it slide if the surface were suddenly ice?" Friction points the opposite way. Works for static and kinetic alike.
Numerator μₛ + tan θ, denominator 1 − μₛ tan θ — it is the tan addition formula in disguise: v_max = √(Rg·tan(θ + φ)) where tan φ = μₛ.
When the question mentions a time interval and a velocity change, use Δp/Δt directly and never find a separately.
11⚠️ Traps
In F = ma, F is the net force of all external material agencies; ma is its effect. Never add ma to a free-body diagram.
Static friction is self-adjusting: fₛ ≤ μₛN. Use the equality only when the body is on the verge of sliding.
They act on different bodies. Only internal pairs within one body cancel. A book on a table: mg and N are not a third-law pair — they act on the same body.
True only in equilibrium. In an accelerating lift, on an incline (N = mg cos θ), or with an extra push, N ≠ mg. And this equality has no connection with the third law.
A ball at the top of its flight has v = 0 but weight mg still acts and a = g. Momentarily at rest ≠ in equilibrium.
Don't add it alongside tension/friction/gravity — it is one of them. And there is no outward "centrifugal force" in an inertial frame.
Doubling the contact area changes nothing. On an incline use N = mg cos θ.
F may be along v, opposite to v, perpendicular (circular motion) or at any angle. It is always parallel to acceleration.
Release a stone from an accelerating train and the stone's acceleration is g downward — it does not "remember" the train's acceleration.
A truck and a scooter can round the same curve at the same maximum speed. Options offering a mass-dependent answer are wrong.
Optimum v₀ = √(Rg tan θ) uses no friction; maximum uses friction at its limit. Read which one is asked.
A ball hitting a wall at u and rebounding at v has |Δp| = m(u + v), not m(u − v). The sign reversal doubles many answers.
12🧵 Mnemonics
1st = inertia (no force, no change) · 2nd = impact (F = ma, how much) · 3rd = interaction (pairs on different bodies).
Weight, Normal, Tension, Friction, Applied. If a force isn't one of these five, ask which body applies it.
μₛtatic > μkinetic, always. Breaking free is harder than staying moving.
a up ⟹ N = m(g+a) · a down ⟹ m(g−a) · free fall ⟹ N = 0.
v₀ = √(Rg tan θ) is the friction-free speed; anything faster needs friction and gives the big formula.
Reduce every force to "force on A by B" and third-law confusion disappears.
13💡 Points to Ponder (NCERT's own list)
Quoted almost verbatim in statement questions
- Force is not always along the motion — it may be along v, opposite, normal, or at any angle; but always parallel to acceleration.
- v = 0 at an instant does not mean force or acceleration are zero (ball at maximum height).
- Force is determined by the situation at that place and time — it is not carried over from earlier motion.
- In F = ma, F is the net force of all external material agencies; ma is not another force.
- Centripetal force is not a new kind of force — always look for the real force providing it.
- Static friction is self-adjusting up to μₛN — don't set fₛ = μₛN without checking.
- mg = R holds only in equilibrium, and has no connection with the third law.
- Action and reaction are simultaneous mutual forces — action does not precede or cause reaction, and they act on different bodies.
- Friction, normal reaction, tension, air resistance, viscous drag, thrust, buoyancy, weight and centripetal force are all just forces — reduce each to "force on A by B".
- There is no conceptual distinction between animate and inanimate objects in F = ma — without friction you could not walk.
- The physics concept of force is not the subjective "feeling of force" — on a merry-go-round the real force is inward though you feel pushed outward.
14🚨 Exceptions & edge cases
N = 0 though gravity is fully acting. "Weightless" means no contact force, not no weight — same reason astronauts float in orbit.
Tension is constant throughout only because the string is massless; give the pulley mass and the two tensions differ.
Ball at maximum height, mass at the extreme of an oscillation — v = 0 but a ≠ 0.
Walking, a car accelerating, a conveyor belt — here friction acts forward, in the direction of motion.
It does not "exist by itself" — it appears only in response to an applied force.
Possible only if tan θ ≤ μₛ; steeper than that and the stationary car slides down.
A particle needs only ΣF = 0; a rigid body also needs Στ = 0 — a bar can have zero net force yet still rotate.