• Every non-zero digit counts.
• Zero between two non-zero digits counts.
• Leading zeros never count.
• Trailing zeros count only if there’s a decimal point.

1. Multiply/divide: keep as many sig figs as the least-precise input.
2. Add/subtract: keep as many decimal places as the least-precise input.

Velocity, average velocity, speed: all dimensionally identical, $[M^0 L T^{-1}]$ — same kind of quantity regardless of value.

4700 m written unambiguously: $4.700 \times 10^3\text{ m}$ — the exponent form fixes the sig-fig count that trailing zeros alone can’t.

1. Dropped digit $< 5$: round down.
2. Dropped digit $> 5$ (or exactly 5 with non-zero digits after): round up.
3. Dropped digit exactly 5, nothing after: round so the preceding digit becomes even.

E.g. $2.735 \to 2.74$ (5 becomes even), $2.745 \to 2.74$ (4 was already even).

Multiplying/dividing measured quantities: percentage uncertainties add.

Sides $16.2 \pm 0.1$ cm ($0.6\%$) and $10.1 \pm 0.1$ cm ($1\%$) → area uncertainty ≈ $1.6\%$.

• Length — metre (m)
• Mass — kilogram (kg)
• Time — second (s)
• Electric current — ampere (A)
• Thermodynamic temperature — kelvin (K)
• Amount of substance — mole (mol)
• Luminous intensity — candela (cd)

Plane angle (radian) and solid angle (steradian) are dimensionless — both are ratios of lengths/areas.

Every term in a valid equation must carry identical dimensions, on both sides.

Passing this check only proves an equation isn’t obviously wrong — it can never catch a missing dimensionless constant (e.g. the $\tfrac{1}{2}$ in $s = ut + \tfrac12 at^2$).

Found by matching powers of $L$ and $T$ in $a = k\,v^x r^y$; the value of $k$ needs the full derivation.

1. Can never recover a dimensionless constant.
2. Can’t handle a sum of several same-dimensioned terms.
3. Says nothing about relations built from sines, logarithms, or exponentials.

$u$ = velocity at $t=0$, $v$ = velocity at time $t$. Hold only while $a$ is truly constant over the interval — re-derive (or split into constant-$a$ pieces) otherwise.

Displacement: $\Delta x = x_2 - x_1$, signed, net change only. Distance (path length): always positive, only ever adds up.

1. No reversal: distance = |displacement|.
2. Velocity reverses: distance = sum of |Δx| each leg; displacement = net Δx. Distance ≥ |displacement|, always.

Average speed = path length / $\Delta t$ (not displacement / $\Delta t$). Always $\text{avg speed} \ge |\bar v|$.

1. Round trip (ends where it started): displacement = 0, so $\bar v = 0$ — even though average speed is clearly nonzero.

Slope of the tangent to the $x\text{-}t$ graph. Positive while $x$ increases, negative while decreasing, zero at a momentary flat (e.g. top of a vertical throw).

1. Instantaneous speed = $|v|$ — exactly, unlike the averaged versions, since over a vanishing interval the path can’t double back on itself.

SI unit: $\text{m/s}^2$. Geometrically, slope of the $v\text{-}t$ graph.

1. Same sign as $v$ (both + or both −): speed is increasing, regardless of whether $a$ itself reads positive or negative.
2. Opposite sign to $v$: speed is decreasing — this is true deceleration.
3. Never judge from the sign of $a$ alone — always compare it against the sign of $v$.

Apply the same sign convention consistently to both $v_A$ and $v_B$.

1. Same direction (chasing): relative speed is reduced — A appears slower to B than its ground speed.
2. Opposite directions (closing in): relative speed adds — A appears to rush past much faster than its ground speed.

• Slope of $x\text{-}t$ graph = $v$ (instantaneous velocity).
• Slope of $v\text{-}t$ graph = $a$ (acceleration).
• Area under $v\text{-}t$ graph, between two instants = displacement $s$ over that interval.

• Scalar: magnitude only (distance, mass, time). Combines by ordinary algebra.
• Vector: magnitude and direction (displacement, velocity, force). Combines by the triangle/parallelogram law.
• Two vectors are equal only if magnitude and direction both match.

Path length $\geq$ $|\text{displacement}|$, equal only when the object never changes direction.

$\mathbf{A}+(-\mathbf{A})=\mathbf{0}$ (no defined direction). Subtraction: $\mathbf{A}-\mathbf{B}=\mathbf{A}+(-\mathbf{B})$.

1. Resultant of two velocities on one object: $\mathbf{v}=\mathbf{v}_1+\mathbf{v}_2$ — addition.
2. Relative velocity of 1 w.r.t. 2: $\mathbf{v}_{1/2}=\mathbf{v}_1-\mathbf{v}_2$ — subtraction. Easy to conflate with the above.

Sign of each component depends only on which quadrant $\theta$ falls in:

1. Quadrant I $(0^\circ\text{-}90^\circ)$: $A_x>0,\; A_y>0$.
2. Quadrant II $(90^\circ\text{-}180^\circ)$: $A_x<0,\; A_y>0$.
3. Quadrant III $(180^\circ\text{-}270^\circ)$: $A_x<0,\; A_y<0$.
4. Quadrant IV $(270^\circ\text{-}360^\circ)$: $A_x>0,\; A_y<0$.

Velocity is always tangent to the path. Unlike straight-line motion, the angle between $\mathbf{v}$ and $\mathbf{a}$ can be anywhere from $0^\circ$ to $180^\circ$.

The two axes are independent: two ordinary 1D problems running at once.

1. Maximum range: at $\theta_0=45^\circ$ (fixed launch speed).
2. Equal range: $\theta_0$ and $90^\circ-\theta_0$ give the same $R$ (same $\sin 2\theta_0$).
3. Caveat: both assume equal launch and landing height — see the concept page for the asymmetric case, where 45° is no longer optimal.

$\omega = \Delta\theta/\Delta t$, and since arc length $=R\,\Delta\theta$, $v=\omega R$.

$\mathbf{a}_c$ always points toward the centre; its magnitude is constant but its direction keeps changing, so it is not a constant vector — the constant-acceleration equations of §5 do not apply to circular motion.

• First: no net force ⟹ no change in velocity (rest or uniform straight-line motion).
• Second: $\mathbf{p}=m\mathbf{v}$, and

• Vector equation, applies per-component; local and instantaneous (no memory of past motion); for a system, $\mathbf{F}$ is the total external force only.
• Third: $\mathbf{F}_{AB} = -\mathbf{F}_{BA}$, simultaneous, on two different bodies — never a pair on the same body.

A large force acting for a short time — bat on ball, foot on ground — is just an ordinary force that happens to be brief:

Third law + second law, for two bodies A and B interacting over the same contact time:

Isolated system, no external force ⟹ total momentum constant — internal forces only shuffle it between particles. (Bullet + gun: zero total momentum before and after.)

• Two forces: $\mathbf{F}_1 = -\mathbf{F}_2$.
• Three or more concurrent forces:

Geometrically: forces drawn head-to-tail close into a polygon. In components, 3 independent scalar conditions, all must hold at once.

1. No sliding tendency: static friction $f_s = 0$.
2. Applied force below the limit: $f_s$ self-adjusts to exactly cancel it (body stays at rest).
3. At the ceiling: $f_s \le \mu_s N$, about to slip.
4. Sliding: friction drops to kinetic, $f_k = \mu_k N$, with $\mu_k$ reliably less than $\mu_s$.

Both coefficients depend only on the two surfaces in contact, never on contact area.

Only gravity and inter-molecular electrical forces are truly fundamental here. Normal reaction, tension, buoyancy, air resistance, friction, and the spring force $F=-kx$ are all electrical in origin, just expressed macroscopically.

Centripetal force is not a new force — just the name for whichever real force supplies the inward pull (tension, gravity, or friction):

Independent of the car’s mass.

At $v_0$, zero friction is needed — least tyre wear. Below it friction acts up the slope, above it (up to $v_{max}$) down the slope.

1. θ < 90°: force has a component along the motion, work is positive (gravity on a falling stone).
2. θ = 180°: force opposes the motion, work is negative (friction on a sliding block, gravity on a rising ball).
3. θ = 90°: force is perpendicular to the motion, work is zero (tension in a conical pendulum, normal force on level motion).

Area under the F–x graph:

Hooke’s law: $F=-kx$. Work to stretch a spring by $x$ from natural length:

Kinetic energy: $K=\tfrac12 mv^2$.

Net work by all forces equals the change in KE — holds for constant or variable force.

1. Conservative force (e.g. gravity): work between two points is path-independent; work around a closed loop is zero.
2. Non-conservative / dissipative (e.g. friction): work depends on path length, lost as heat, cannot be recovered.

1. No friction (only conservative forces act): mechanical energy K + U stays constant.
2. With friction (rough incline, angle θ, coefficient μ): slides only if $\tan\theta > \mu$; mechanical energy visibly shrinks as it converts to heat, but mechanical + heat together stays constant.

Momentum conserved (every collision):

1. Relative velocity of separation = relative velocity of approach: $u_1+v_1=u_2+v_2$.
2. Equal masses (m₁ = m₂): velocities simply exchange, $v_1=u_2,\ v_2=u_1$.

Bodies stick together, move off with one common velocity v — maximum KE loss momentum conservation allows:

Heat produced (KE lost):

The CM moves exactly as if the whole mass were concentrated there and the whole external force acted there — internal forces always cancel in pairs (Newton’s third law).

Both $\boldsymbol{\tau}$ and $\mathbf{L}$ are defined only relative to a chosen origin. $\tau = rF\sin\theta$, direction by the right-hand rule.

• Ring about central (symmetry) axis: $I = MR^2$
• Disc about central axis: $I = \tfrac{1}{2}MR^2$
• Solid sphere about a diameter: $I = \tfrac{2}{5}MR^2$
• Thin rod about centre, ⊥ to length: $I = \tfrac{1}{12}ML^2$

$r_i$ is the perpendicular distance from the axis; $K$ (radius of gyration) is where all the mass would sit, as a point, to give the same $I$.

1. Perpendicular axes: planar lamina only; $x,y$ in the plane, $z$ through their intersection, normal to the plane.
2. Parallel axes: any rigid body; $I_{cm}$ is the minimum over all parallel axes — moving away only adds $Md^2 \ge 0$.

1. Skater pulls arms in: $I$ drops, $\omega$ shoots up to keep $L$ fixed, and $KE$ rises — that extra energy comes from muscular work, not for free.

1. Smaller $K^2/R^2$ (mass closer to the axis) needs less energy budget to spin up, so more is left for translation — it arrives faster, regardless of mass or radius.
2. Solid sphere $(K^2/R^2=2/5)$ beats a hollow cylinder $(K^2/R^2=1)$ down the same incline, every time.

1. First law (orbits): every planet moves in an ellipse, Sun at one focus.
2. Second law (areas): the Sun–planet line sweeps equal areas in equal times — angular momentum about the Sun stays constant.
3. Third law (periods): $T^2 \propto a^3$, same constant for every planet around the Sun.

$G = 6.674\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$ — always attractive, always inverse-square.

1. Height h (h≪R): $g(h) \approx g\left(1-\dfrac{2h}{R}\right)$ — falls off twice as fast per metre as depth.
2. Depth d: $g(d) = g\left(1-\dfrac{d}{R}\right)$ — a straight line, zero at the centre.

Independent of launched mass $m$ and of launch direction — only launch speed matters.

Every bound circular orbit: total energy negative, equals $-K$, equals $U/2$ — for any $r$, $m$, or $M$.

1. Launch speed = $v_{\text{orbital}}$: circular orbit.
2. Between $v_{\text{orbital}}$ and $v_e$: elliptical (bound) orbit.
3. At or above $v_e$ (= $\sqrt{2}\,v_{\text{orbital}}$): orbit opens up — a one-way escape, not a closed path.

1. Geostationary: equatorial plane + west-to-east + $T=24\text{ h}$ → one unique $r$ (altitude ≈ 35,900 km).
2. Polar: low altitude, period ≈ 100 min, passes near both poles — images the whole planet over many orbits.
3. Weightlessness: satellite and contents in free fall together — normal force vanishes, scale reads zero.

Units Pa (N/m²), dimensions $[ML^{-1}T^{-2}]$. Strain is a ratio of lengths/volumes — no units, no dimensions.

1. Longitudinal (tensile/compressive): force along the length; strain $= \Delta L/L$.
2. Shearing: tangential force on one face, opposite face fixed; strain $= \theta \approx \Delta x/L$, no volume change.
3. Volume (hydraulic): uniform inward pressure; stress $= \Delta P$, strain $= \Delta V/V$.

Holds only within the elastic limit; $k$ is whichever modulus (Y, G, or B) matches the kind of stress/strain in play.

1. O to A (elastic limit/yield point): Hooke’s law holds, fully elastic — loading and unloading retrace the same line, nothing left behind.
2. Past A (plastic flow): strain grows faster than stress; release the load here and unloading follows a NEW line parallel to OA (same slope $Y$) — leaves a permanent set.
3. At B (ultimate tensile strength): maximum stress the material can bear; beyond B the wire keeps stretching (necking) while the stress it sustains drops.
4. At E (fracture point): the wire snaps.
5. Ductile vs brittle: large plastic region (A to E) → ductile (most metals); little/none → brittle (glass, cast iron, ceramics).

Depends on the highest strain ever reached, not the current one — the wire “remembers” its worst stretch.

Compressibility $= 1/B$. Steel $Y \approx 2\times10^{11}\text{ Pa}$ (stiff — cables, rails, beams); rubber $Y \approx 10^{5}\text{–}10^{6}\text{ Pa}$ (large reversible deformation). $G$ is always smaller than $Y$ for the same material, typically by a factor of 2–3.

Typically 0.2–0.4 (~0.3 for metals); theory restricts $-1 < \sigma < 0.5$, with $\sigma = 0.5$ meaning constant volume while stretching (rubber, nearly).

Know any two of $Y$, $G$, $\sigma$ and the third is fixed.

The second form is energy stored per unit volume — applies alike to a stretched wire, a twisted shaft, or a compressed block.

1. Cables/beams are operated well inside the elastic region (a factor of safety), never near the ultimate tensile strength.
2. Loading past the elastic limit even once leaves a permanent set and weakens the material; repeating it causes fatigue failure.
3. Metal beats wood for beams: larger $Y$ (less sag) + higher elastic limit (more load before permanent bending).
4. Hollow box / I-section beams concentrate material at the top and bottom edges, where bending stress is largest — more resistance for less weight.

Pressure is a scalar; depth alone sets it, not shape or volume.

Pascal’s law: pressure applied anywhere in an enclosed fluid is transmitted undiminished everywhere. Work done is equal on both pistons.

Volume flow rate $Av$ stays fixed along a pipe; narrower section ⟹ faster flow.

Valid for steady, incompressible, non-viscous flow. Horizontal pipe: faster flow ⟹ lower pressure.

Throat (narrow section) always has the lowest manometer column — lowest pressure, highest speed.

Gas viscosity rises with temperature; liquid viscosity falls.

$\rho$ = sphere density, $\sigma$ = fluid density. $v_t \propto r^2$ — bigger particles settle faster.

1. Laminar: $Re \lesssim 1000\text{–}2000$
2. Turbulent: $Re$ above that range

Large $Re$: inertial forces dominate over viscous damping.

1. Liquid drop (1 surface): $\Delta P = 2S/r$
2. Soap bubble (2 surfaces): $\Delta P = 4S/r$

Water wets glass ($\theta$ small, $\cos\theta>0$): rises. Mercury ($\theta>90^\circ$, $\cos\theta<0$): depressed. Rise is inversely proportional to tube radius $r$.

Molar specific heat $C = Mc$. Water: $c_{\text{water}} = 4186\text{ J/(kg·K)}$.

Insulated mixture, no heat escapes to surroundings.

For water at 1 atm: $L_f = 333\text{ kJ/kg}$ (fusion, at 0°C), $L_v = 2260\text{ kJ/kg}$ (vaporisation, at 100°C).

1. Melting plateau (0°C): temperature flat — all heat breaks bonds, none raises KE.
2. Boiling plateau (100°C): flat again, but far WIDER than the melting plateau — $Q_{\text{boil}}/Q_{\text{melt}} = 2260/333 \approx 6.8$ for the same mass.

$k$ = thermal conductivity, W/(m·K). Thermal resistance $R = x/(kA)$, so $dQ/dt = \Delta T/R$ (Ohm’s-law analogy).

1. Natural convection: heated fluid expands, rises; cooler fluid sinks — sets up its own circulating current (land/sea breeze, boiling pot, weather).
2. Forced convection: fluid pushed mechanically (fan, pump) — car radiator, blood circulation.

$\sigma = 5.67\times10^{-8}\text{ W/(m}^2\text{K}^4)$. Doubling absolute temperature multiplies radiated power by 16.

Linear approximation of radiative loss, valid only for small excess over surroundings $T_0$. For large excess, true $T^4$ behaviour makes cooling faster than this predicts.

Two systems each in thermal equilibrium with a third are in thermal equilibrium with each other. This is what makes a thermometer valid — touch each object to the same thermometer in turn, never to each other.

Q positive = heat absorbed by the system; W positive = work done by the system on its surroundings. Get a sign backwards and every later result flips.

= area under the P-V curve, path-dependent. $\Delta U$ is a state variable (endpoints only); Q absorbs whatever path-dependence is left over.

1. Isothermal (ΔT = 0, reservoir contact): Boyle’s law $PV = nRT = \text{constant}$; since $\Delta U = 0$, first law gives $Q = W$.
2. Adiabatic (Q = 0, insulated or fast): $PV^{\gamma} = \text{constant}$ (also $TV^{\gamma-1} = \text{constant}$); $\Delta U = -W$ — compression does work ON the gas with zero heat input, so T rises with no heater anywhere (diesel-engine effect).

$\gamma = C_p/C_v$ is always $> 1$, so the adiabatic curve through any point is always steeper than the isothermal curve through that same point.

Cp always exceeds Cv: the same ΔT costs more heat at constant pressure because some of it leaks out as expansion work instead of staying behind as ΔU.

Kelvin-Planck statement (second law): no process is possible whose sole result is absorbing heat from a reservoir and converting it completely into work — $\eta = 1$ is forbidden; every real engine rejects some heat to a cold sink.

T always in Kelvin. Four reversible steps between hot reservoir T₁ and cold reservoir T₂:

1. Isothermal expansion at T₁ — absorbs Q₁ from the hot reservoir.
2. Adiabatic expansion — cools T₁ → T₂, no heat exchange.
3. Isothermal compression at T₂ — rejects Q₂ to the cold reservoir.
4. Adiabatic compression — returns T₂ → T₁, closing the cycle.

These give $Q_2/Q_1 = T_2/T_1$. No engine beats $\eta_{Carnot}$ between the same two temperatures; it reaches 1 only if $T_2 = 0\text{ K}$ — unreachable (third law).

• 1 cal = 4.186 J
• 1 kPa × 1 L = 1 J — convenient shortcut for P-V work
• $\gamma = C_p/C_v$, always greater than 1 for an ideal gas
• T must be in Kelvin in every engine/Carnot formula

$\mu$ = moles, $N$ = number of molecules, $R = 8.314\text{ J/mol·K}$.

$\rho = Nm/V$ is mass density, $\langle v^2\rangle$ the mean square speed. Derived from wall collisions alone — never required molecules to collide with each other.

Depends on $T$ and molar mass $M$ ONLY — never on $P$ or $V$ separately. Squeezing a gas into a smaller box at fixed $T$ raises pressure via collision rate, not molecular speed.

Each independent quadratic degree of freedom contributes an average $\tfrac{1}{2}k_BT$, regardless of its nature or the temperature.

1. Monatomic (3 translational dof): $U = \mu\left(\tfrac{3}{2}N_Ak_BT\right) = \tfrac{3}{2}\mu RT$
2. Diatomic, rigid, moderate $T$ (3 translational + 2 rotational = 5 dof): $U = \tfrac{5}{2}\mu RT$

Vibration adds 2 more dof (7 total) at high $T$, but is quantised and stays frozen out at ordinary temperatures.

1. Monatomic: $C_v=\tfrac{3}{2}R \approx 12.5\text{ J/mol·K},\ C_p=\tfrac{5}{2}R,\ \gamma=\dfrac{5}{3}$
2. Diatomic (rigid): $C_v=\tfrac{5}{2}R \approx 20.8\text{ J/mol·K},\ C_p=\tfrac{7}{2}R,\ \gamma=\dfrac{7}{5}$

Holds for any ideal gas — the extra $R$ is the work done pushing back surroundings at constant pressure.

$d$ = molecular diameter, $n = N/V$ number density. Shrinks as pressure rises, grows with temperature at fixed pressure.

Periodic: repeats identically after a fixed period $T$ (frequency $\nu = 1/T$, in Hz). Oscillatory: periodic and to-and-fro about a fixed mean position. Every oscillation is periodic; not every periodic motion (e.g. uniform circular motion) is oscillatory.

$\omega = 2\pi/T = 2\pi\nu$. Speed max ($A\omega$) at the mean position; acceleration max ($A\omega^2$) at the extremes.

SHM = projection of uniform circular motion onto a diameter.

Any linear restoring force $F=-kx$ guarantees SHM with $\omega=\sqrt{k/m}$, whatever the system.

Constant for all time. $K$ and $U$ each oscillate at twice the frequency of $x(t)$, out of step: $U$ peaks at $x=\pm A$, $K$ peaks at $x=0$. They cross at $x=\pm A/\sqrt2$.

Pendulum derivation needs $\sin\theta\approx\theta$, valid only for small $\theta$ (roughly under 10°).

1. Spring-block: period is independent of amplitude — always, exactly, since $F=-kx$ is exactly linear at any amplitude within the linear range.
2. Simple pendulum, large angle: isochronism breaks down — period creeps up with amplitude, because $\sin\theta$ falls measurably below $\theta$ and the restoring force is only approximately linear, and only at small amplitude.

Damping force $F_d=-bv$; lowered frequency $\omega'=\sqrt{k/m-b^2/4m^2}$. Energy decays as $E(t)=\tfrac12kA^2e^{-bt/m}$ — faster for larger $b$.

Driven at frequency $\omega_d$, a damped oscillator settles into steady oscillation at $\omega_d$ itself, not its natural frequency $\omega_0=\sqrt{k/m}$. As $\omega_d \to \omega_0$, amplitude grows sharply — resonance, limited only by damping.

1. Transverse: particle motion ⊥ wave direction (e.g. string wave). Needs shear rigidity — travels in solids and along surfaces, not through the bulk of a fluid.
2. Longitudinal: particle motion ∥ wave direction (e.g. sound, as compressions and rarefactions). Needs only bulk elasticity — travels through solids, liquids, and gases.

Wave speed $v$ is the speed of the pattern, not the particle. Transverse particle velocity is $\partial y/\partial t = -A\omega\cos(kx-\omega t+\phi)$ — a different quantity, depends on amplitude and frequency.

T = tension, μ = m/L = linear mass density. This is the wave’s propagation speed, not the transverse particle velocity.

1. Fixed end: reflected pulse inverted (phase change of $\pi$).
2. Free end: reflected pulse upright (no phase change).

Nodes: where $\sin(kx)=0$ (no motion). Antinodes: where $\sin(kx)=\pm1$ (swings through $2A$). Consecutive nodes (or antinodes) are $\lambda/2$ apart; a node sits $\lambda/4$ from the nearest antinode. A standing wave transports no energy — only stores it.

Carrier term oscillates at the average frequency $(f_1+f_2)/2$; the envelope $2A\cos(\pi(f_1-f_2)t)$ swells and fades. Loudness depends on the envelope’s size, which peaks twice per cycle — doubling $(f_1-f_2)/2$ back to $|f_1-f_2|$.

1. Source approaching stationary observer (speed $v_s$): $f_{\text{approach}} = f_0\dfrac{v}{v-v_s}$ — wavefronts crowd closer, frequency rises.
2. Source receding: $f_{\text{recede}} = f_0\dfrac{v}{v+v_s}$ — frequency falls.
3. Observer moving (speed $v_o$) toward/away from a stationary source: $f' = f_0\dfrac{v\pm v_o}{v}$ — wavefront spacing is unchanged; only the interception rate changes.