Laws of Motion – JEE Main Physics Formula Sheet & Class 11 Notes | Newton's Laws, Friction, Pulley, Constraint Relations & All Motion Formulas

Newton's Three Laws, Momentum, Impulse, and Equilibrium

Why Newton's Laws and Lami's Theorem Are the Starting Point of All Laws of Motion JEE Main Questions

Newton's First Law of Motion (from Aakash PDF — Laws of Motion):

A body continues in its state of rest or of uniform motion in a straight line unless acted upon by a net external force. This law defines inertia — the tendency of a body to resist change in its state of motion. A body at rest has inertia of rest; a body in motion has inertia of motion. First law also defines the condition of equilibrium: F_net = 0 ↔ a = 0 (body either at rest or moving with constant velocity).

Newton's Second Law of Motion (from Aakash PDF — Laws of Motion JEE Main):

The rate of change of momentum of a body is proportional to the applied net external force and takes place in the direction of the net force.

F_net = ma = dp/dt

For constant mass: F = ma (magnitude: F = ma, direction: same as a).

Linear Momentum: p = mv (vector, direction same as velocity).

Impulse: J = F·Δt = Δp = m(v–u)

Impulse = change in momentum. For variable force: J = ∫F dt = area under F-t graph. SI unit of impulse = N·s = kg·m/s = same as momentum.

Newton's Third Law of Motion (from Aakash PDF — Laws of Motion):

For every action, there is an equal and opposite reaction. If body A exerts force F_AB on body B, then body B exerts F_BA = –F_AB on body A. Action and reaction are: equal in magnitude, opposite in direction, act on DIFFERENT bodies (so they do not cancel each other), of the same type (both contact or both gravitational, etc.).

Equilibrium of Concurrent Forces (from Aakash PDF — Laws of Motion JEE Main):

If three forces P, Q, R act on a body such that it is in equilibrium (F_net = 0), and the forces are concurrent, then by Lami's Theorem:

P/sinα = Q/sinβ = R/sinγ

where α, β, γ are the angles between the forces: α = angle between Q and R (i.e., opposite to P), β = angle between P and R (opposite to Q), γ = angle between P and Q (opposite to R). The three angles satisfy α+β+γ = 360°.

Alternative: For three concurrent forces in equilibrium, they can be represented as three sides of a triangle (triangle law of equilibrium). Download the Free PDF for Laws of Motion for all Newton's law examples and Lami's theorem applications for JEE Main.

Constraint Relations for Pulleys — All Three Cases

Why Constraint Relations Are the Most-Tested Setup in Laws of Motion JEE Main Questions

Constraint Relations (from Aakash PDF — Laws of Motion JEE Main): In a pulley system, the string is inextensible (fixed length). This imposes a fixed mathematical relation between the displacements, velocities, and accelerations of the ends of the string. These are called constraint relations.

Case I — Fixed Middle End (from Aakash PDF — Laws of Motion):

When the middle end of the string over a pulley is fixed and the two outer masses hang on either side:

x₁ = x₂; v₁ = v₂; a₁ = a₂

(Both masses move identically — their displacements, velocities and accelerations are equal in magnitude.)

Case II — Fixed Side End (from Aakash PDF — Laws of Motion):

When one side end of the string is fixed (attached to ceiling or wall), and a mass hangs from the movable pulley:

x₂ = 2x₁; v₂ = 2v₁; a₂ = 2a₁

(The hanging mass moves at half the speed/acceleration of the string's free end — or equivalently, the mass attached to the movable pulley moves half as fast as the free end.)

Case III — All Three Ends Free to Move (from Aakash PDF — Laws of Motion JEE Main):

When all three ends of the string are free (two masses hang from the ends, and the pulley itself moves):

(x₁+x₂)/2 = x₃; (v₁+v₂)/2 = v₃; a₁+a₂ = 2a₃

Equivalently: a₁+a₂ = 2a₃ (the average acceleration of the two end masses equals the acceleration of the pulley).

Sign convention: downward is taken positive. If any direction is upward, incorporate a negative sign in the corresponding term.

General Constraint Method: Let the string have total length L (constant). Differentiate the geometric sum of all string segments twice with respect to time. The resulting equation is the constraint relation. Download the Free PDF for Laws of Motion for all constraint relation examples for JEE Main.

Pulley Systems — Atwood's Machine, Moving Pulley, Inclined Plane

Why Atwood's Machine and Pulley Acceleration Formulas Are Direct Laws of Motion JEE Main Questions

Atwood's Machine — Stationary Pulley (from Aakash PDF — Laws of Motion JEE Main):

Two masses M₁ and M₂ (M₂ > M₁) hang over a fixed, massless, frictionless pulley. The string is massless and inextensible.

Acceleration: a = (M₂–M₁)g / (M₁+M₂)

Tension: T = 2M₁M₂g / (M₁+M₂)

Note: T lies between M₁g and M₂g (always). If M₁=M₂: a=0, T=M₁g=M₂g (no acceleration). The heavier mass M₂ accelerates down; M₁ accelerates up.

Atwood's Machine — Pulley Moving Upward with Acceleration a₀ (from Aakash PDF — Laws of Motion JEE Main):

When the pulley itself accelerates upward with a₀ (using pseudo force in pulley's frame — replace g by g+a₀):

T = 2M₁M₂(g+a₀) / (M₁+M₂)

Relative acceleration between masses (w.r.t. pulley): aᵣ = (M₂–M₁)(g+a₀)/(M₁+M₂)

Absolute accelerations: a₁ = –(a₀+aᵣ); a₂ = aᵣ–a₀ (taking upward positive).

Inclined Plane Pulley System (from Aakash PDF — Laws of Motion JEE Main):

Mass M₁ on incline of angle α, mass M₂ on incline of angle β, connected by string over pulley at the top:

a = (M₂ sinβ – M₁ sinα)g / (M₁+M₂)

(assuming M₂ sinβ > M₁ sinα for the motion shown, and smooth inclines.)

T = M₁M₂(sinα+sinβ)g / (M₁+M₂)

Mass on a String from Ceiling (from Aakash PDF — Laws of Motion JEE Main):

Mass M hangs from ceiling by string. (a) Stationary or uniform velocity: T = Mg. (b) System accelerates UP with a: T = M(g+a). (c) System accelerates DOWN with a: T = M(g–a). Note: if accelerating down with a=g (free fall), T=0 (apparent weightlessness).

Four standard configurations (from Aakash PDF — Laws of Motion JEE Main): All of these share the same formula structure:

Linear acceleration: a = F_net / (effective mass + I_cm/R²)

For all four standard pulley setups: a = F/(m + I_cm/R²) and T or friction = I_cm·a/R²

Download the Free PDF for Laws of Motion for all pulley system examples for JEE Main.

Contact Forces, Machine Gun, Liquid Jet, Rope Tension

Why Force on Blocks, Bullets on Walls, and Rope Tension Are Direct Laws of Motion JEE Main Questions

Machine Gun / Bullet Force (from Aakash PDF — Laws of Motion JEE Main):

A machine gun fires n bullets per second, each of mass m with speed u.

Force required to keep gun stationary: F = nmv (reaction force = rate of change of momentum of bullets)

This is also the recoil force on the gun.

Bullets hitting a wall at speed v (n per second):

(i) Bullets come to rest: F_wall = nmv

(ii) Bullets rebound elastically: F_wall = 2nmv (change in momentum = 2mv per bullet)

Liquid Jet Force (from Aakash PDF — Laws of Motion JEE Main):

Liquid jet of cross-sectional area A moving with speed v hits a wall. Density of liquid = ρ. Mass hitting per second = ρAv.

(i) Force required by pump to move liquid: F = (dm/dt)v = ρAv² (rate of momentum supply)

(ii) Jet hits wall, does NOT rebound: F_wall = ρAv²

(iii) Jet rebounds elastically: F_wall = 2ρAv²

(iv) Oblique impact at angle θ to wall: F_wall = 2ρAv²cosθ (only normal component changes)

Multi-Block Contact Forces (from Aakash PDF — Laws of Motion JEE Main):

Three blocks M₁, M₂, M₃ pushed by force F on a smooth surface (M₁ in front, F on M₁): all blocks accelerate together with:

a = F/(M₁+M₂+M₃)

Contact force between M₁ and M₂ (force F₁ on M₂ from M₁ by Newton's 3rd law):

F₁ = (M₂+M₃)a = (M₂+M₃)F/(M₁+M₂+M₃)

Contact force between M₂ and M₃: F₂ = M₃a = M₃F/(M₁+M₂+M₃)

String Tension at Distance x from Left End (from Aakash PDF — Laws of Motion JEE Main):

For a uniform rod/string of mass M and length L pulled by force F from left end:

Tₓ = Fx/L (tension at distance x from end where F is applied, mass to the right pulls back)

Tₓ = F(L–x)/L (tension at distance x from end where F is NOT applied)

For F₁ applied at left, F₂ applied at right: Tₓ = [F₁x/L] + [F₂(L–x)/L]

Uniform Rope Tension (from Aakash PDF — Laws of Motion JEE Main):

Uniform rope of total mass Mₛ, length L, hanging from ceiling or pulled vertically. Tension at distance x from the free end:

(a) Stationary: Tₓ = Mₛgx/L (weight of rope below the point)

(b) Accelerating upward with a: Tₓ = Mₛx(g+a)/L

(c) Accelerating downward with a: Tₓ = Mₛx(g–a)/L (if a=g: T=0, free fall)

Download the Free PDF for Laws of Motion for all contact force and rope tension examples for JEE Main.

Friction — Static, Kinetic, Angle of Repose, Minimum Force, Two-Block Systems

Why Friction Formulas Are the Most Calculation-Heavy Laws of Motion Topics in JEE Main

Types of Friction (from Aakash PDF — Laws of Motion JEE Main):

Static friction (fₛ): Friction between two surfaces when there is no relative motion. It is self-adjusting — it adjusts to oppose the tendency of relative motion. Maximum static friction = limiting friction = μₛN.

0 ≤ fₛ ≤ μₛN (static friction can be anywhere from 0 to maximum)

Kinetic friction (fₖ): Friction when there IS relative sliding motion between surfaces.

fₖ = μₖN (fixed value, independent of area and speed)

μₖ < μₛ (kinetic friction coefficient is always less than static)

Angle of Friction and Angle of Repose (from Aakash PDF — Laws of Motion JEE Main):

Consider a block on an inclined plane with friction μₛ. The maximum angle of incline for which the block stays at rest:

Angle of repose: λ = tan⁻¹(μₛ)

At angle of repose: N = mg cosλ; fₛ = mg sinλ; fₛ/N = tanλ = μₛ.

Minimum Force to Move Block on Rough Surface (from Aakash PDF — Laws of Motion JEE Main):

For a block of mass m on a horizontal rough surface (μ), a force F applied at angle θ above horizontal:

For motion: F cosθ > μ(mg – F sinθ) [normal force reduces due to upward component of F]

Normal force: N = mg – F sinθ; Friction: f = μN = μ(mg–F sinθ)

For motion: F cosθ ≥ μ(mg–F sinθ) → F(cosθ + μ sinθ) ≥ μmg

F ≥ μmg/(cosθ + μ sinθ)

This is minimum when (cosθ + μ sinθ) is maximum: d/dθ(cosθ+μ sinθ)=0 → –sinθ+μ cosθ=0 → tanθ=μ → θ = tan⁻¹(μ)

Fmin = mg·μ/√(1+μ²) = mg sinλ at angle θ = tan⁻¹(μ) = λ (angle of friction).

Wedge — Block on Accelerating Wedge (from Aakash PDF — Laws of Motion JEE Main):

Case I: Block m on smooth wedge M, no sliding — both move together with acceleration a. Force F required:

F = (M+m)g tanθ; acceleration a = g tanθ (from pseudo force analysis of block on wedge)

Contact force between block and wedge: R = mg/cosθ

Case II: Minimum F for block to fall freely (contact force between m and M = 0): F = Mg cotθ. At this F: wedge accelerates at g cotθ; block falls vertically with a = g.

Two-Block Systems — Friction (from Aakash PDF — Laws of Motion JEE Main):

Mass M₁ on bottom (on smooth ground), mass M₂ on top. Friction between M₁ and M₂ = μ. Force F on LOWER block M₁:

(a) If F ≤ (M₁+M₂)μg: both blocks move together, a = F/(M₁+M₂), maximum: amax = μg.

(b) If F > (M₁+M₂)μg: M₂ slips backward on M₁. M₂ has acceleration a₂ = μg. M₁ has acceleration a₁ = (F–M₂μg)/M₁.

Force F on UPPER block M₂:

(a) If F ≤ (M₁+M₂)M₂μg/M₁: both move together with amax = M₂μg/M₁.

(b) If F > above: M₂ slips forward on M₁. M₂: a₂ = (F–M₂μg)/M₂; M₁: a₁ = μg (friction from M₂ drives M₁). Download the Free PDF for Laws of Motion for all friction and two-block examples for JEE Main.

Dynamics of Circular Motion — Conical Pendulum, Banked Road

Why Conical Pendulum Period and Banked Road Maximum Speed Are Key Laws of Motion JEE Main Formulas

Circular Motion Dynamics — General (from Aakash PDF — Laws of Motion JEE Main):

For a body moving in a horizontal circle of radius r at speed v, the net inward (centripetal) force must equal mv²/r. This centripetal force is provided by the real forces on the body (tension, normal force, friction, gravity component).

Neglecting gravity: T = centripetal force = mv²/r = mω²r

Conical Pendulum (from Aakash PDF — Laws of Motion JEE Main):

A bob of mass m attached to a string of length l, swinging in a horizontal circle. The string makes angle θ with vertical. Vertical height h = l cosθ.

Resolving string tension T: T cosθ = mg (vertical equilibrium) and T sinθ = mω²r = mω²l sinθ (centripetal).

From these: T = mg/cosθ and ω² = g/h = g/(l cosθ)

Time period of conical pendulum: τ = 2π/ω = 2π√(h/g) = 2π√(l cosθ/g)

Key result from PDF: For θ to be 90° (string horizontal), T would need to be infinite → string CANNOT be horizontal. Maximum θ < 90°.

T sinθ = mω²l sinθ → T = mω²l (from centripetal equation).

Banked Road — Maximum Speed (from Aakash PDF — Laws of Motion JEE Main):

A vehicle negotiates a curve of radius r on a road banked at angle θ. Coefficient of friction = μ.

Maximum safe speed: vmax = √[rg(μ+tanθ)/(1–μtanθ)]

Special Cases from PDF:

(i) Smooth banked road (μ=0): v = √(rg tanθ) — the optimum speed for zero friction.

(ii) Horizontal rough road (θ=0): vmax = √(μrg) — friction provides all centripetal force.

Minimum speed for banked road: vmin = √[rg(tanθ–μ)/(1+μtanθ)]

Between vmin and vmax, the vehicle can negotiate the curve without slipping. Download the Free PDF for Laws of Motion for all circular motion dynamics examples for JEE Main.

Motion in a Vertical Circle — All Conditions and Formulas

Why Vertical Circle Minimum Speed and TL–TH=6mg Are the Most Important Laws of Motion Formulas

Vertical Circle Setup (from Aakash PDF — Laws of Motion JEE Main):

A particle of mass m is tied to a string of length l, and revolves in a vertical circle. At the lowest point L, it is given speed vL (horizontal). At an arbitrary point P (angle θ from bottom), its speed is vP. At the highest point H, its speed is vH.

Energy Conservation in Vertical Circle (from Aakash PDF — Laws of Motion):

Using conservation of energy from L to P: ½mvL² = ½mvP² + mg(l–l cosθ) = ½mvP² + mgl(1–cosθ)

So: vP² = vL² – 2gl(1–cosθ)

From L to H (height = 2l): vH² = vL² – 4gl

Tension in the String at Any Point P (from Aakash PDF — Laws of Motion JEE Main):

At point P (angle θ from bottom), net force toward center = Tₚ – mg cosθ = mvP²/l:

Tₚ = mvP²/l + mg cosθ [Wait — at angle θ from bottom, component of mg toward center depends on position]

More precisely: at point P, if θ is measured from the BOTTOM (lowest point):

The component of mg along the string toward center = mg cos(π–θ) = –mg cosθ for upper half. Let's use the angle from VERTICAL (from top):

At angle φ from the top (highest point): T – mg cosφ = mv²/l → T = mv²/l + mg cosφ

At highest point H (φ=0, cosφ=1): TH = mvH²/l – mg [mg acts downward = toward center at top]

At lowest point L (φ=π, cosφ=–1): TL = mvL²/l + mg [mg acts downward = away from center at bottom]

Key Result — TL – TH = 6mg (from Aakash PDF — Laws of Motion JEE Main):

TL – TH = (mvL²/l + mg) – (mvH²/l – mg) = m(vL²–vH²)/l + 2mg = m(4gl)/l + 2mg = 4mg + 2mg = 6mg

This result TL–TH = 6mg is always true regardless of the values of vL and vH.

Three conditions for vertical circle (from Aakash PDF — Laws of Motion JEE Main):

Condition 1: If vL ≤ √(2gl): particle oscillates between M and M' (does not reach the top). The string goes slack before reaching H.

Condition 2: If √(2gl) < vL < √(5gl): particle leaves the circular path somewhere between M (horizontal point) and H (top). String goes slack at some point in upper half.

Condition 3: If vL ≥ √(5gl): particle COMPLETES the vertical circle. Minimum speed at top: vH_min = √(gl) (at vH=√gl, TH=0).

Minimum speed at bottom to complete: vL_min = √(5gl)

At vL = √(5gl): vH = √(vL²–4gl) = √(5gl–4gl) = √(gl) ✓

At points M (horizontal, height l): vM² = vL²–2gl. Tension at M: TM = mvM²/l = m(vL²–2gl)/l.

Download the Free PDF for Laws of Motion for all vertical circle examples for JEE Main.

Download Free PDF — Laws of Motion JEE Main Physics Formula Sheet

All Laws of Motion formulas from the Aakash Rapid Revision PDF: Newton's 1st Law (inertia, F_net=0↔a=0), Newton's 2nd Law (F=ma=dp/dt, p=mv, impulse J=FΔt=Δp), Newton's 3rd Law (F_AB=–F_BA, action-reaction on different bodies), Lami's theorem (P/sinα=Q/sinβ=R/sinγ for 3-force equilibrium), constraint relations (Case I fixed middle: a₁=a₂; Case II fixed side: a₂=2a₁; Case III all free: a₁+a₂=2a₃), Atwood's machine (a=(M₂–M₁)g/(M₁+M₂); T=2M₁M₂g/(M₁+M₂)), moving pulley (T=2M₁M₂(g+a₀)/(M₁+M₂); aᵣ=(M₂–M₁)(g+a₀)/(M₁+M₂)), inclined plane pulley (a=(M₂sinβ–M₁sinα)g/(M₁+M₂); T=M₁M₂g(sinα+sinβ)/(M₁+M₂)), mass on string (stationary T=Mg; up T=M(g+a); down T=M(g–a); free fall T=0), machine gun force (F=nmv; wall: stop nmv, rebound 2nmv), liquid jet force (ρAv²; rebound 2ρAv²; oblique 2ρAv²cosθ), multi-block contact forces (a=F/ΣM; F₂=M₃a; F₁=(M₂+M₃)a), string tension at x (Tₓ=Fx/L), uniform rope tension (stationary Mₛgx/L; up Mₛx(g+a)/L; down Mₛx(g–a)/L), static friction (fₛ≤μₛN, self-adjusting), kinetic friction (fₖ=μₖN, μₖ<μₛ), angle of repose (λ=tan⁻¹μₛ), minimum force Fmin=mg/√(1+μ²) at θ=tan⁻¹(μ), wedge-block (F=(M+m)g tanθ, a=g tanθ, R=mg/cosθ), min F for free fall (Mg cotθ), two-block systems (F on lower: amax=μg; F on upper: amax=M₂μg/M₁), conical pendulum (T=mg/cosθ; τ=2π√(l cosθ/g)=2π√(h/g); string cannot be horizontal), banked road (vmax=√[rg(μ+tanθ)/(1–μtanθ)]; smooth: √(rg tanθ); horizontal rough: √(μrg)), vertical circle (TH=mvH²/l–mg; TL=mvL²/l+mg; TL–TH=6mg; conditions: vL≤√(2gl) oscillate; √(2gl)

Why Laws of Motion Is the Highest-Impact Chapter for JEE Main Physics Rank

Newton's Second Law F=ma applied through FBD is the single most-used technique in JEE Main Physics. It appears in Laws of Motion questions, but also in Work-Energy (net force × displacement = ΔKE), Rotational Motion (τ=Iα), Electrostatics (qE=ma), and everywhere else. The FBD technique — isolate each body, draw all real forces, apply F=ma — is the universal solving algorithm for JEE Main mechanics.

Atwood's machine formulas a=(M₂–M₁)g/(M₁+M₂) and T=2M₁M₂g/(M₁+M₂) are direct 4-mark JEE Main answers. Every pulley question reduces to identifying the effective driving force and the total inertia: a = (net force down)/(total mass). The tension formula comes from applying F=ma to either block alone: T–M₁g=M₁a → T=M₁(g+a)=M₁g+M₁a=M₁(g+(M₂–M₁)g/(M₁+M₂)) = 2M₁M₂g/(M₁+M₂).

The vertical circle result TL–TH=6mg is the most elegant formula in Laws of Motion for JEE Main. It is always true, regardless of the actual values of vL and vH. Derived purely from Newton's law and energy conservation, it can be used to immediately find TH if TL is given, or vice versa. Combined with the three conditions (vL<√(2gl): oscillates; between √(2gl) and √(5gl): leaves path; ≥√(5gl): completes circle), every vertical circle question in JEE Main is solved.

Banking road formula vmax=√[rg(μ+tanθ)/(1–μtanθ)] covers all road-friction-angle scenarios. The two special cases — smooth bank (μ=0) giving v=√(rg tanθ) and flat road (θ=0) giving v=√(μrg) — are both directly tested. The minimum speed formula vmin=√[rg(tanθ–μ)/(1+μtanθ)] is also testable. Download the Free PDF for Laws of Motion to have all formulas ready.

Who Should Use This Laws of Motion Formula Sheet?

Learning Outcomes After Completing Laws of Motion

After working through Laws of Motion using this formula sheet, a student should confidently accomplish the following for JEE Main physics. On Newton's Laws: state and apply all three Newton's laws; draw correct free-body diagrams for any system; apply Lami's theorem for three-force equilibrium; compute impulse and change in momentum.

On constraint relations: identify the type of pulley setup (fixed middle, fixed side, all free); write the correct constraint equation; relate accelerations of connected masses; apply constraint along with Newton's 2nd law to find acceleration and tension in any Atwood-type system including inclined planes and moving pulleys.

On contact forces: compute force on gun/wall from bullets; compute force from liquid jet (normal and oblique impact); find contact forces between blocks in multi-block systems; find tension at any point in a string or uniform rope under gravity and acceleration.

On friction: distinguish static (self-adjusting) from kinetic (fixed at μₖN); find angle of repose; compute minimum force to move block at optimal angle; solve wedge-block problems for force and acceleration; solve two-block problems for maximum F before slipping in both F-on-lower and F-on-upper configurations.

On circular dynamics: apply centripetal force equation for any horizontal circular motion; solve conical pendulum for T, ω, and time period; apply banked road formula for vmax and vmin; solve horizontal rough road for maximum speed. On vertical circle: apply energy conservation to find speed at any point; compute string tension at any point; state and use TL–TH=6mg; identify which of the three conditions applies given vL. Download the Free PDF for Laws of Motion to test all outcomes before your JEE Main exam.

Get the Free PDF for Laws of Motion — JEE Main Quick Revision

The Aakash Rapid Revision & Formula Bank PDF for Laws of Motion contains all Newton's law results, all pulley system formulas, all friction results, all circular motion dynamics formulas, and vertical circle conditions in one structured JEE Main physics reference.

Conclusion — Laws of Motion: The Force Framework of JEE Main Physics

Laws of Motion is the chapter that transforms kinematics (description of motion) into dynamics (explanation of motion). Newton's three laws — inertia, F=ma, action-reaction — are the complete framework for analysing any mechanical system. The FBD method, constraint relations, and Newton's 2nd law together can solve any pulley, inclined plane, or multi-block problem. Friction converts the ideally smooth world of kinematics into the realistic world of surfaces and resistance.

The four most-tested results in Laws of Motion for JEE Main: (1) Atwood's machine a=(M₂–M₁)g/(M₁+M₂) and T=2M₁M₂g/(M₁+M₂) — memorise both; (2) TL–TH=6mg and vertical circle three conditions — highest difficulty, highest marks; (3) Banked road vmax=√[rg(μ+tanθ)/(1–μtanθ)] — pure formula substitution; (4) Two-block amax=μg (lower) and amax=M₂μg/M₁ (upper) — direct substitution. Use this page and the Free PDF Download for Laws of Motion as your complete JEE Main revision foundation.

Frequently Asked Questions — Laws of Motion Formulas

Related Formula Sheets — JEE Main Physics

This is a free educational resource for JEE Main 2025–2026, JEE Advanced, NEET, Class 11 CBSE, and BITSAT Physics preparation. The PDF formula sheet for Laws of Motion is sourced from Aakash's Rapid Revision & Formula Bank. All Laws of Motion formulas and content are for academic reference only.