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Physics · NDA

Mechanics

📖 Chapter PN01 · NDA Class 11–12 Level 🎯 NDA Level : High Priority

Mechanics is the backbone of NDA Physics — it is the study of motion, forces, and energy. This chapter covers everything from how we measure physical quantities (units & dimensions) to how objects move (kinematics), what causes motion (Newton's laws), how energy is exchanged, and the beautiful mathematics of rotating bodies and gravity. NDA consistently tests this chapter across all sub-topics, making it the single most rewarding chapter to master.

📌 What to expect in NDA (based on 2022–2025 pattern):
(1) Dimensional analysis — finding dimensions of derived quantities, checking equations;
(2) Equations of motion — finding velocity, distance, time under uniform acceleration;
(3) Projectile motion — range, maximum height, time of flight;
(4) Newton's laws — especially impulse-momentum, friction problems;
(5) Work-energy theorem, conservation of energy — block on incline, spring problems;
(6) Conservation of linear momentum — collisions (elastic/inelastic);
(7) Rotational motion — moment of inertia, torque, angular velocity;
(8) Gravitation — orbital velocity, escape velocity, satellite period.

Topics at a Glance

① Physical World & Measurement

SI units, dimensions, dimensional analysis

② Kinematics

Equations of motion, projectile, circular motion

③ Laws of Motion

Newton's 3 laws, friction, impulse, momentum

④ Work, Energy & Power

Work-energy theorem, KE, PE, conservation

⑤ Rotational Motion

Torque, MI, angular momentum, CM

⑥ Gravitation

Newton's law, g, escape velocity, satellites

1. Physical World & Measurement

1.1

Fundamental & Derived Units (SI System)

Seven base quantities form the foundation of all measurement

The International System of Units (SI) defines 7 fundamental (base) quantities. Every other physical quantity is derived from these using mathematical relationships.

Fundamental Quantity	SI Unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric Current	ampere	A
Temperature	kelvin	K
Luminous Intensity	candela	cd
Amount of Substance	mole	mol

💡 Memory Trick: “Let Me Time Every King's Coat Monocle” → Length, Mass, Time, Electric current, (temperature) Kelvin, Candela, Mole.

1.2

Dimensional Analysis

Express quantities in terms of M, L, T — the core NDA tool

Dimensions show how a physical quantity relates to the fundamental quantities M (mass), L (length), T (time), I (current), θ (temperature).

⚡ Key Dimensional Formulae

Force (F) = M L T⁻² Energy / Work (E) = M L² T⁻² Power (P) = M L² T⁻³ Pressure (P) = M L⁻¹ T⁻² Momentum (p) = M L T⁻¹ Impulse = M L T⁻¹ Velocity = L T⁻¹ Acceleration = L T⁻² Gravitational Constant G = M⁻¹ L³ T⁻² Planck's Constant h = M L² T⁻¹

Square brackets [ ] denote "dimension of". Dimensionless quantities: angle (radian), strain, refractive index, all pure numbers.

⚡ Uses of Dimensional Analysis

1. Checking correctness of an equation — both sides must have same dimensions. 2. Deriving relationships — express quantity in terms of relevant physical quantities. 3. Converting units — use dimensional formula to find conversion factors. LIMITATION: Cannot find dimensionless constants (like ½, 2π, etc.)

Worked Example — Dimensional Check

Check if v = u + at is dimensionally correct.

LHS: [v] = LT⁻¹. RHS: [u] = LT⁻¹, [at] = LT⁻² × T = LT⁻¹. All terms match. ✓ Equation is dimensionally correct.

📝 TOPIC-WISE PYQ

Units & Dimensions — NDA Pattern Questions

Q1. The dimensional formula of angular momentum is:

(a) ML²T⁻¹ (b) ML²T⁻² (c) MLT⁻¹ (d) ML²T⁻³

Answer: (a) ML²T⁻¹
Angular momentum L = Iω = (ML²)(T⁻¹) = ML²T⁻¹. (Same as Planck's constant — a frequently tested fact.)

Q2. Which of the following pairs have the same dimensions?

(a) Force and Momentum (b) Work and Power (c) Work and Torque (d) Pressure and Force

Answer: (c) Work and Torque
Both Work = F·d and Torque = F·r have dimension ML²T⁻². (Note: they have the same dimension but different physical meaning — a NDA favourite trap!)

Q3. If velocity (V), acceleration (A) and force (F) are taken as fundamental units, what is the dimension of mass?

(a) FA⁻¹V⁰ (b) FV⁻¹A (c) FA⁻¹ (d) FV⁻¹

Answer: (c) FA⁻¹
From F = ma → m = F/a = FA⁻¹. Mass has dimension FA⁻¹ in this new system.

2. Kinematics

2.1

Scalars, Vectors & Equations of Motion

Describing motion — with and without direction

Scalars have magnitude only (speed, distance, mass, time, energy). Vectors have magnitude and direction (velocity, displacement, force, acceleration, momentum).

⚡ Equations of Motion (Uniform Acceleration)

v = u + at … (1) velocity–time s = ut + ½at² … (2) displacement–time v² = u² + 2as … (3) velocity–displacement sₙ = u + a(2n−1)/2 … (4) nth second formula Where: u = initial velocity (m/s) v = final velocity (m/s) a = acceleration (m/s²) s = displacement (m) t = time (s) n = nth second

For free fall: replace a with g (≈ 9.8 m/s² ≈ 10 m/s²). For upward throw: a = −g. Body thrown down: a = +g.

📊 Graphical Interpretation

Slope of s–t graph = velocity
Slope of v–t graph = acceleration
Area under v–t graph = displacement
Area under a–t graph = change in velocity

🔸 Common Pitfalls

Distance ≠ displacement (path vs straight line)
Speed = |velocity| (always positive)
Use v² = u² + 2as when time is not given
At max height of vertical throw: v = 0

2.2

Projectile Motion

Motion under gravity in two dimensions — a regular NDA topic

A projectile is launched with initial velocity u at angle θ to the horizontal. Horizontal motion is uniform; vertical motion has constant downward acceleration g.

⚡ Projectile Formulae

Horizontal component: uₓ = u cos θ Vertical component: uᵧ = u sin θ Time of flight: T = 2u sin θ / g Maximum height: H = u² sin²θ / 2g Range: R = u² sin 2θ / g Maximum range occurs at θ = 45° → Rₘₐₓ = u²/g At θ = 45°: H = R/4 (height is one quarter of range) At any time t: x = u cos θ · t y = u sin θ · t − ½gt²

Complementary angles (θ and 90°−θ) give the same range. Range is same for 30° and 60°, for 15° and 75°.

Fig. 1 — Projectile trajectory showing Range (R), Maximum Height (H), and launch angle θ

2.3

Uniform Circular Motion

Constant speed, changing direction — acceleration always points to centre

⚡ Circular Motion Formulae

Centripetal acceleration: a = v²/r = ω²r (directed towards centre) Centripetal force: F = mv²/r = mω²r Angular velocity: ω = 2π/T = 2πf (rad/s) Linear velocity: v = ωr Time period: T = 2πr/v Where: r = radius, m = mass, T = time period, f = frequency

Key point: In UCM, speed is constant but velocity changes (direction changes). Hence acceleration ≠ 0. No tangential acceleration — only centripetal (radial) acceleration.

⚠ NDA Exam Trap: Centripetal force is not a new force — it is the net inward force provided by existing forces (gravity for satellites, tension for circular string, normal force for banked roads, friction for turning cars). Always identify what provides centripetal force in a given problem.

📝 TOPIC-WISE PYQ

Kinematics — NDA Pattern Questions

Q1. A body is thrown vertically upward with velocity 20 m/s. What is the maximum height reached? (g = 10 m/s²)

(a) 10 m (b) 20 m (c) 40 m (d) 5 m

Answer: (b) 20 m
At max height, v = 0. Use v² = u² − 2gH → 0 = 400 − 2(10)H → H = 20 m.

Q2. A projectile is fired at 45° with initial velocity 20 m/s. What is its range? (g = 10 m/s²)

(a) 20 m (b) 40 m (c) 10 m (d) 80 m

Answer: (b) 40 m
R = u²sin2θ/g = (400 × sin90°)/10 = 400/10 = 40 m. (sin 90° = 1 for θ = 45°)

Q3. The horizontal range is same for angles of projection:

(a) 30° and 45° (b) 20° and 80° (c) 30° and 60° (d) 40° and 55°

Answer: (c) 30° and 60°
Range R = u²sin2θ/g. Complementary angles (sum = 90°) give same range. 30° + 60° = 90°. ✓

🤔 TRICKY QUESTIONS

Kinematics — Watch Out!

T1. A ball is dropped from the top of a building. Another ball is thrown horizontally simultaneously from the same point. Which ball reaches the ground first?

Answer: Both reach at the same time.
The horizontal velocity has no effect on vertical motion. Both balls have the same vertical acceleration g and start with zero vertical velocity. Hence both reach the ground in the same time t = √(2H/g). This is a classic NDA concept trap.

T2. The displacement of a body is given by s = 4t² + 3t. What is the acceleration?

Answer: 8 m/s²
v = ds/dt = 8t + 3. a = dv/dt = 8 m/s². (Acceleration = coefficient of t² term × 2). Constant — not dependent on t.

3. Laws of Motion

3.1

Newton's Three Laws

The foundation of all classical mechanics

① First Law (Inertia)

A body at rest stays at rest; body in motion stays in motion
Until net external force acts on it
Defines inertia — resistance to change in motion
Greater mass = greater inertia

② Second Law (F = ma)

F = ma — net force = mass × acceleration
Force direction = acceleration direction
Impulse = F·t = change in momentum
Rate of change of momentum = net force

③ Third Law (Action-Reaction)

Every action has equal & opposite reaction
Forces act on different bodies
Explain: gun recoil, rocket propulsion, swimming
Cannot cancel each other (different bodies)

⚡ Impulse & Momentum

Momentum: p = mv (kg m/s = MLT⁻¹) Impulse: J = F × t = Δp = m(v−u) Conservation of Linear Momentum: If net external force = 0 → total momentum is constant m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (always, for any collision) Law of Conservation of Momentum applies to ALL collisions.

Impulse = area under F-t graph. Large force for small time (collision, explosion) gives same impulse as small force for long time.

3.2

Friction

Static, kinetic, rolling — the force that opposes relative motion

⚡ Friction Formulae

Limiting static friction: fₛ = μₛ × N (maximum static friction) Kinetic friction: fₖ = μₖ × N (while sliding) Rolling friction: fᵣ ≈ μᵣ × N (μᵣ << μₖ << μₛ) Where N = Normal reaction force, μ = coefficient of friction Angle of friction (λ): tan λ = μ Angle of repose (θ): tan θ = μₛ (body just starts sliding on incline) On inclined plane (angle θ): N = mg cosθ Component along plane = mg sinθ Friction force = μ mg cosθ (opposing motion)

🔸 Types of Friction

Static friction: prevents motion; adjustable up to limiting value
Kinetic friction: acts during sliding; less than max static
Rolling friction: least of all; enables wheels to work
μₛ > μₖ > μᵣ (always)

⚠ Friction Exam Traps

Friction ≠ always kinetic. It's static when no sliding
Ball bearings: reduce friction using rolling friction
Normal force on slope = mg cosθ, not mg
Friction is self-adjusting up to limiting value

📝 TOPIC-WISE PYQ

Laws of Motion & Friction — NDA Pattern Questions

Q1. A bullet of mass 10 g moving at 400 m/s enters a sandbag and comes to rest in 0.05 s. The force exerted by the sandbag on the bullet is:

(a) 40 N (b) 80 N (c) 100 N (d) 8000 N

Answer: (b) 80 N
F = m×(v−u)/t = 0.01×(0−400)/0.05 = 0.01×(−8000) = −80 N. Magnitude = 80 N.

Q2. A person of 60 kg stands on a weighing machine in a lift. The lift accelerates upward at 5 m/s². The reading of the machine is: (g = 10 m/s²)

(a) 600 N (b) 300 N (c) 900 N (d) 60 N

Answer: (c) 900 N
Apparent weight = m(g + a) = 60(10 + 5) = 60 × 15 = 900 N. In accelerating lift (upward), apparent weight increases. (Apparent weight decreases if lift accelerates downward.)

Q3. Coefficient of static friction is 0.4. What is the angle of repose?

(a) tan⁻¹(0.2) (b) tan⁻¹(0.4) (c) 40° (d) sin⁻¹(0.4)

Answer: (b) tan⁻¹(0.4)
Angle of repose θ satisfies tanθ = μₛ = 0.4. Hence θ = tan⁻¹(0.4) ≈ 21.8°.

🤔 TRICKY QUESTIONS

Laws of Motion — Classic Traps

T1. A gun fires a bullet. Does the gun recoil with the same speed as the bullet?

No — same momentum, different speeds.
By conservation of momentum: m_bullet × v_bullet = m_gun × v_gun. Since m_gun >> m_bullet, v_gun << v_bullet. Same impulse (force × time) acts on both, but heavy gun gains less velocity.

T2. A horse pulls a cart forward. By Newton's 3rd law, the cart pulls the horse backward with equal force. Then how does the system move forward?

Action-reaction pairs act on different bodies.
The horse's hooves push the ground backward; the ground (external surface) pushes the horse-cart system forward. This external friction from ground is not cancelled by any reaction within the system. Net external force on the horse-cart system = ground friction forward − air drag. System moves forward.

4. Work, Energy & Power

4.1

Work, Kinetic & Potential Energy

Energy is the capacity to do work; work is the transfer of energy

⚡ Work, Energy & Power Formulae

Work done: W = F·d·cosθ (θ = angle between F and displacement) W = F·d (when F ∥ d, i.e. θ = 0°) Kinetic Energy: KE = ½mv² Potential Energy: PE = mgh (gravitational) PE = ½kx² (spring: k = spring constant, x = extension) Work-Energy Theorem: Net Work = ΔKE = ½mv² − ½mu² Power: P = W/t = F·v (watts = J/s = ML²T⁻³) 1 horsepower (hp) = 746 W Conservation of Energy: KE + PE = constant (no friction) ½mv² + mgh = constant

🔸 Signs of Work

Work is positive: F and d in same direction (0° < 90°)
Work is zero: F ⊥ d (e.g., carrying load horizontally, circular motion)
Work is negative: F and d in opposite direction (friction)

🔸 Elastic vs Inelastic Collision

Elastic: momentum + KE both conserved
Inelastic: only momentum conserved; KE lost
Perfectly inelastic: bodies stick together after collision
Coefficient of restitution: e = 1 (elastic), 0 (perfectly inelastic)

📝 TOPIC-WISE PYQ

Work, Energy & Power — NDA Pattern Questions

Q1. A body of mass 2 kg moving at 4 m/s collides with a stationary body of mass 2 kg and sticks to it. What is the velocity after collision?

(a) 4 m/s (b) 2 m/s (c) 8 m/s (d) 1 m/s

Answer: (b) 2 m/s
Perfectly inelastic collision. By conservation of momentum: 2×4 + 2×0 = (2+2)×v → v = 8/4 = 2 m/s.

Q2. A force of 10 N acts on a body at 60° to its displacement of 5 m. Work done is:

(a) 50 J (b) 25 J (c) 43.3 J (d) 0 J

Answer: (b) 25 J
W = F·d·cosθ = 10 × 5 × cos60° = 50 × 0.5 = 25 J.

Q3. A machine gun fires 10 bullets per second each of mass 10 g at 500 m/s. The average force on the gun is:

(a) 50 N (b) 500 N (c) 5 N (d) 100 N

Answer: (a) 50 N
F = n×m×v = 10 × 0.01 × 500 = 50 N. (Impulse per second = rate of change of momentum = force.)

5. Rotational Motion

5.1

Torque, Moment of Inertia & Angular Momentum

Rotation is to angular quantities what linear motion is to linear quantities

Every linear quantity has a rotational analogue. Understanding this parallel is the fastest way to learn rotational mechanics.

Linear	Symbol	Rotational	Symbol
Mass	m	Moment of Inertia	I = ∑mr²
Force (F)	F = ma	Torque	τ = Iα
Velocity	v	Angular velocity	ω
Momentum	p = mv	Angular momentum	L = Iω
KE = ½mv²		Rotational KE = ½Iω²

⚡ Key Rotational Formulae

Torque: τ = r × F = rF sinθ (N·m = ML²T⁻²) Moment of Inertia: I = Σmr² (varies with axis and shape) Angular Momentum: L = Iω (conserved if τ_ext = 0) Rotational KE: KE = ½Iω² Theorem of Parallel Axes: I = Icm + Md² (Icm = MI about CM axis; d = distance between axes) Theorem of Perpendicular Axes (flat lamina only): Iz = Ix + Iy Centre of Mass: x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂)

Conservation of Angular Momentum: A spinning skater brings arms in → I decreases → ω increases (constant L). Same principle as a diver tucking in.

🔸 MI of Common Bodies

Solid sphere (about diameter): I = 2/5 MR²
Hollow sphere (about diameter): I = 2/3 MR²
Solid disc (about axis): I = ½MR²
Ring (about axis): I = MR²
Thin rod (about centre): I = ML²/12
Thin rod (about end): I = ML²/3

🔸 Rolling on Incline

Rolling body has both KE_trans + KE_rot
a = g sinθ / (1 + I/MR²)
Solid sphere rolls fastest (smallest I/MR² = 2/5)
Ring rolls slowest (I/MR² = 1)
Hollow sphere faster than disc, slower than solid sphere

📝 TOPIC-WISE PYQ

Rotational Motion — NDA Pattern Questions

Q1. A disc and a ring of same mass and radius roll down an incline. Which reaches the bottom first?

(a) Ring (b) Disc (c) Both simultaneously (d) Depends on incline angle

Answer: (b) Disc
a = g sinθ/(1 + I/MR²). Disc: I = MR²/2 → a = g sinθ/(1.5). Ring: I = MR² → a = g sinθ/2. Disc has larger acceleration, reaches bottom first.

Q2. A diver tucks her body while diving. Her angular velocity:

(a) Increases (b) Decreases (c) Remains same (d) Becomes zero

Answer: (a) Increases
No external torque acts on the diver. Angular momentum L = Iω is conserved. Tucking reduces r → I decreases → ω increases to keep L constant.

6. Gravitation

6.1

Newton's Law of Gravitation & g

Every mass attracts every other mass in the universe

⚡ Gravitation Formulae

Newton's Law: F = G M m / r² (G = 6.674 × 10⁻¹¹ N m² kg⁻²) Acceleration due to gravity: g = GM/R² (at surface; R = radius of Earth) g' = g(1 − 2h/R) (at height h above surface; h << R) g' = g(1 − d/R) (at depth d below surface) g at poles > g at equator (Earth is flattened at poles) Orbital velocity: v₀ = √(GM/r) = √(gR²/r) At surface: v₀ = √(gR) ≈ 7.9 km/s Escape velocity: vₑ = √(2GM/R) = √(2gR) ≈ 11.2 km/s Note: vₑ = √2 × v₀ (escape = √2 × orbital, at surface) Time period of satellite: T = 2π√(r³/GM) Geostationary orbit: T = 24 hrs, height ≈ 36,000 km Gravitational PE: U = −GMm/r (negative — bound system)

G ≠ g. G is universal gravitational constant (fixed everywhere). g is acceleration due to gravity (varies with location, height, depth).

Fig. 2 — Satellite in circular orbit. Centripetal force = gravitational force. Escape velocity = √2 × orbital velocity.

🌎 Variation of g

g decreases as we go above surface (altitude)
g decreases as we go below surface (depth)
g = 0 at centre of Earth
g_poles > g_equator (Earth is oblate spheroid)
g on Moon ≈ g_Earth / 6

🚀 Satellites & Orbits

Geostationary: T = 24 h, same direction as Earth's rotation
Polar satellite: passes over poles, T ≈ 100 min
No fuel needed for satellite orbit (gravity = centripetal)
Weightlessness: gravity provides centripetal, no normal force

📝 TOPIC-WISE PYQ

Gravitation — NDA Pattern Questions

Q1. The escape velocity from the Earth's surface is 11.2 km/s. What is the escape velocity from a planet of same mass but double the radius?

(a) 22.4 km/s (b) 5.6 km/s (c) 11.2 km/s (d) 7.9 km/s

Answer: (b) 5.6 km/s
vₑ = √(2GM/R). If R doubles (M same), vₑ = √(2GM/2R) = vₑ/√2 = 11.2/√2 ≈ 7.92 km/s. Closest answer: 5.6 km/s if mass is also halved. With only R doubled: vₑ ∝ 1/√R → vₑ = 11.2/√2 ≈ 7.9 km/s.

Q2. A satellite is orbiting close to Earth's surface. What is its approximate orbital velocity? (R = 6400 km, g = 10 m/s²)

(a) 7.9 km/s (b) 11.2 km/s (c) 3.5 km/s (d) 5.6 km/s

Answer: (a) 7.9 km/s
v₀ = √(gR) = √(10 × 6.4×10⁶) = √(6.4×10⁷) ≈ 8000 m/s ≈ 7.9 km/s (standard value to remember).

Q3. At what depth below the Earth's surface is g equal to half its surface value?

(a) R/4 (b) R/2 (c) R/3 (d) 3R/4

Answer: (b) R/2
g' = g(1 − d/R). For g' = g/2: 1 − d/R = ½ → d = R/2.

🤔 TRICKY QUESTIONS

Gravitation & Rotational — Exam Surprises

T1. If Earth suddenly stops rotating, what happens to g at the equator?

g increases at equator.
Currently at equator, part of g provides centripetal force for Earth's rotation. Effective g_equator = g_actual − ω²R. If Earth stops rotating (ω = 0), this reduction disappears and g increases. At poles, rotation has no effect (centripetal force = 0 at poles), so g at poles remains unchanged.

T2. A body weighs 72 N on the surface of Earth. What is its weight at height h = R (one Earth radius above surface)?

Answer: 18 N
At height R above surface, distance from centre = 2R. g' = GM/(2R)² = GM/4R² = g/4. Weight = mg/4 = 72/4 = 18 N. (Weight ∝ 1/r² from centre.)

⚡ High-Yield Formula Sheet — PN01 Mechanics

📊 Equations of Motion

v = u + at
s = ut + ½at²
v² = u² + 2as
sₙ = u + a(2n−1)/2
Free fall: u=0, a=g, s=½gt²

🌠 Projectile Motion

T = 2u sinθ/g
H = u²sin²θ/2g
R = u²sin2θ/g
R_max at θ = 45° = u²/g
Same R: complementary angles (θ + 90°−θ)

👪 Newton & Friction

F = ma; p = mv; J = Δp
Momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Friction: f = μN
Angle of repose: tanθ = μ
Lift: N = m(g±a)

⚡ Work, Energy, Power

W = Fd cosθ
KE = ½mv²; PE = mgh
Work-Energy: W_net = ΔKE
P = W/t = Fv
1 hp = 746 W

🔄 Rotational Motion

τ = rF sinθ; L = Iω
I_sphere = 2/5 MR²; Ring = MR²; Disc = MR²/2
Parallel axis: I = Icm + Md²
v₀ = ωr; a_c = v²/r = ω²r
KE_rot = ½Iω²

🌎 Gravitation

F = GMm/r²; g = GM/R²
v₀ = √(GM/r) ≈ 7.9 km/s (surface)
vₑ = √(2gR) ≈ 11.2 km/s
T = 2π√(r³/GM)
g at depth d: g'= g(1−d/R)

📌 Dimensions to Memorise: Force = MLT⁻², Energy = ML²T⁻², Power = ML²T⁻³, Momentum = MLT⁻¹, Angular Momentum = ML²T⁻¹, Pressure = ML⁻¹T⁻², G = M⁻¹L³T⁻².

⚡ Quick Revision Booster — PN01 Mechanics

📚 Units Tricks

7 fundamental SI units: m, kg, s, A, K, cd, mol
Joule = N·m = kg m² s⁻²
Same dims: Work & Torque (ML²T⁻²)
Same dims: Angular momentum & Planck's h (ML²T⁻¹)
Dimensionless: angle, strain, relative density

🌠 Projectile Facts

Horizontal velocity = constant (no air resistance)
At max height: vertical v = 0, horizontal v = u cosθ
Same range for 30° and 60°, also 15° and 75°
T ∝ sinθ; H ∝ sin²θ; R ∝ sin2θ
Max range at 45°

👪 Newton's Laws

Inertia ∝ mass (heavier = harder to start/stop)
Impulse = area under F-t graph
Friction μ: s > k > r (static > kinetic > rolling)
Lift up (a↑): W_app = m(g+a); Lift down: m(g−a)
Free fall: weight = 0 (weightlessness)

⚡ Energy Rules

W=0 if F⊥s (carrying load, UCM)
Elastic collision: both KE & momentum conserved
Inelastic: only momentum conserved
Spring PE = ½kx² (x = extension from natural length)
Power = Fv (most useful form)

🔄 Rotation Quick Facts

Solid sphere (2/5) fastest on incline; ring slowest
Skater tucks in: I↓, ω↑ (L constant)
Torque = 0 if F passes through axis of rotation
Perpendicular axis theorem: only for flat lamina
ω = 2πf = 2π/T

🚨 Gravity Must-Know

vₑ = √2 × v₀ (escape = √2 × orbital)
g = 0 at centre of Earth; max at poles
Geostationary orbit: T = 24 h, height ≈ 36,000 km
Moon: g_moon = g_earth/6
G is universal constant; g varies with location

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