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v = v₀ + at x = x₀ + v₀t + ½at²v² = v₀² + 2a(x − x₀) x = x₀ + ½(v₀ + v)tUse when acceleration is constant. Pick the equation with the variables you have + need; only one variable is missing.
| Have | Need | Use |
|---|---|---|
| v₀, a, t | v | v = v₀ + at |
| v₀, a, t | x | x = v₀t + ½at² |
| v₀, v, a (no t) | x | v² = v₀² + 2a·Δx |
| v₀, v, t (no a) | x | x = ½(v₀+v)t |
Free fall: a = −g = −9.8 m/s² (or −10 for quick estimates). Air resistance ignored unless stated. Sign convention: up positive.
The horizontal motion is independent of vertical. Don't put g in the x-equation. Don't put v_x in the y-equation. Each axis has its own kinematic equation set, linked only by the shared time variable.
ac = v²/r = ω²r Fc = m·ac = mv²/r T = 2π/ωCentripetal acceleration always points TOWARD the center. The 'centripetal force' is whatever provides this — tension, friction, gravity, normal — not a separate force.
F = G m₁ m₂ / r² G = 6.67 × 10⁻¹¹ N·m²/kg²| Concept | Formula |
|---|---|
| g at planet surface | g = GM/R² |
| Orbital velocity | v = √(GM/r) |
| Period (Kepler) | T² = (4π²/GM) r³ |
| Escape velocity | vesc = √(2GM/R) |
Kepler's 3 laws: (1) ellipses, sun at focus. (2) equal areas in equal times. (3) T² ∝ r³ (semi-major axis cubed).
There's no outward 'centrifugal force' in inertial frames. The thing pushing the ball outward is its inertia; the only real force is the centripetal one pulling it in. Saying 'centrifugal force' on AP loses points.
| Law | Statement |
|---|---|
| 1st | ΣF = 0 ⇒ v = const (inertia) |
| 2nd | F = ma (vector equation) |
| 3rd | FA on B = −FB on A (action-reaction) |
1. Isolate the body. Draw a dot.
2. Draw EVERY force as an arrow from the dot.
3. Pick coordinate axes (align one with motion / incline).
4. Sum forces per axis: ΣFx = max, ΣFy = may.
5. Solve.
Tension in massless rope: same throughout. Pulleys (ideal) just redirect tension.Atwood machine: two masses on a pulley. Take system as a whole: a = (m₁ − m₂)g / (m₁ + m₂) when both hang.
On flat ground, N = mg. On an incline, N = mg cos θ. In an elevator accelerating up, N = m(g+a). Friction = μN, so getting N wrong cascades into wrong friction.
| Linear | Rotational | Relation |
|---|---|---|
| x | θ | x = rθ |
| v | ω | v = rω |
| a | α | a = rα |
| m | I (moment of inertia) | see table |
| F | τ = rF sin θ | torque |
| p = mv | L = Iω | angular momentum |
| F = ma | τ = Iα | rotational 2nd law |
| KE = ½mv² | KE_rot = ½Iω² | rotational KE |
Rolling without slipping: v = Rω, a = Rα, friction is static (does no work). KE_total = ½Mv² + ½Iω². Hollow rolls slower down a ramp than solid.
Conservation: L_initial = L_final if Στ_external = 0A solid object rolling without slipping has KE split between translation (½Mv²) and rotation (½Iω²). Sliding without rolling: only ½Mv². Rolling reaches the bottom slower because some KE is in rotation.
x(t) = A cos(ωt + φ) v(t) = −Aω sin(ωt + φ)ω = √(k/m) (spring) ω = √(g/L) (pendulum, small angle)| Quantity | Formula | Note |
|---|---|---|
| Period T | 2π/ω | indep of amplitude (SHM) |
| Frequency f | 1/T | Hz = cycles/sec |
| Total energy | ½kA² | fixed amplitude |
| v_max | Aω | at equilibrium |
| a_max | Aω² | at extremes |
v = fλ. v fixed by medium (string tension, sound air density). f and λ trade off inversely.Sound: intensity I = P/area, level β = 10 log(I/I₀) in dB. Doubling distance → I drops by ¼ (3 dB drop).
Doppler: moving source / observer changes observed frequency. f_obs = f_source · (v ± v_obs)/(v ∓ v_source).
The period of a spring or pendulum SHM does not depend on amplitude (small-angle approx). Bigger swing = same time. Many students assume bigger amplitude = longer period; wrong for SHM.
W = F · d · cos θ W_net = ΔKE = ½mv² − ½mv₀²Net work equals change in kinetic energy. Sign of W comes from cos θ: same direction = +, opposite = −, perpendicular = 0.
| Energy | Formula | Where |
|---|---|---|
| KE | ½ mv² | moving |
| PE_grav | mgh | height (near Earth) |
| PE_spring | ½ kx² | compressed/stretched |
| Mechanical | KE + PE | conserved if no friction |
Conservation: KE₁ + PE₁ = KE₂ + PE₂ (no nonconservative forces)Pendulum: at top all PE (mgh), at bottom all KE (½mv²). Set equal → v = √(2gh).
Normal force is perpendicular to motion (cos 90° = 0), so W_N = 0. Same for tension on a swinging pendulum (always perp to velocity). Only forces with a component along motion do work.
p = mv J = ΔP = F · Δt = ∫F dtImpulse equals change in momentum. Spread the time → smaller force (airbags, knee bend on landing).
| Type | Momentum | KE | Note |
|---|---|---|---|
| Elastic | conserved | conserved | billiards, ideal |
| Inelastic | conserved | not conserved | cars crumple |
| Perfectly inelastic | conserved | max KE lost | they stick together |
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (always for closed system)2D collisions: conserve momentum per axis (x and y separately). Vector decomposition is the whole game.
Momentum is always conserved in collisions (no external impulse). KE is conserved only in elastic collisions. In inelastic, some KE → heat / sound / deformation. Don't write 'energy is conserved' for a car crash.
| Question says… | Use § from | Approach |
|---|---|---|
| 'how far / how fast / how long' (constant a) | § ① | kinematic equations — pick the one with right vars |
| projectile, launched, landed | § ① | decompose v₀ into x and y; t links them |
| FBD, ΣF = ma, given forces | § ② | Newton's 2nd, draw FBD |
| 'incline, friction, find a' | § ② | tilt axes; mg sin θ along, mg cos θ perp |
| two masses on pulley | § ② | Atwood: a = (m₁−m₂)g/(m₁+m₂) |
| 'find v at height / spring stretch' | § ③ | energy conservation: KE+PE |
| friction present, height change | § ③ | energy with W_friction = −μ_k mg d |
| 'power required' | § ③ | P = Fv or W/t |
| collision, two objects | § ④ | momentum conservation; KE if elastic |
| 'they stick' / 'embed' | § ④ | perfectly inelastic: v = (m₁u₁+m₂u₂)/(m₁+m₂) |
| 'force during collision' | § ④ | impulse: F·Δt = Δp |
| 'rolling object', 'angular momentum' | § ⑤ | I, ω; rolling: v = Rω; KE_total = ½Mv² + ½Iω² |
| 'skater pulls in arms' | § ⑤ | L = Iω conserved → ω increases |
| 'moment of inertia about' | § ⑤ | standard table + parallel-axis if shifted |
| circular motion, banked turn, satellite | § ⑥ | F_c = mv²/r; gravity: F = GMm/r² |
| orbit: 'find period / speed / r' | § ⑥ | Kepler T² ∝ r³ or v = √(GM/r) |
| 'spring period, pendulum period' | § ⑦ | T = 2π√(m/k) or 2π√(L/g) |
| 'wave on string, frequency' | § ⑦ | v = fλ; standing: f_n = nv/(2L) |
| 'Doppler shift' | § ⑦ | f' = f(v ± v_o)/(v ∓ v_s) |
Energy in inelastic collisions = wrong tool. Momentum in dropping-from-height = wrong (gravity provides external impulse). Pick the conserved quantity that matches the situation — momentum for collisions, energy for height/spring changes (no friction).
Force, velocity, acceleration, momentum — all vectors. Energy, work, power, mass — scalars. Adding speeds for opposite directions (signs!) is the most common arithmetic loss. Always check directions before computing.