IB Physics HL — All Flashcards

The **velocity** (rate of change of displacement).

The **acceleration** (rate of change of velocity).

The **displacement**.

The **change in velocity**, Δv.

The **straight-line distance and direction** from start to finish — a vector (m).

Distance = total path length (scalar); displacement = straight-line start→end with direction (vector).

Speed = rate of distance (scalar); velocity = rate of displacement, with direction (vector).

Velocity: **m s⁻¹**. Acceleration: **m s⁻²**.

Constant velocity (zero acceleration → zero gradient).

Motion in the **negative** direction → negative (subtracted) displacement.

$s = \tfrac{1}{2}(u+v)t$ — the area of the trapezium under the line.

The **rate of change of velocity** — how much the velocity changes each second. Unit: m s⁻².

**m s⁻²** (metres per second, every second).

The **acceleration**.

The **change in velocity**. From rest, that area is the velocity reached.

Constant velocity → **zero** acceleration.

When its **acceleration is opposite to its velocity**. (If positive is the direction of motion, that means a negative acceleration.)

The velocity is **decreasing**. If the velocity is positive the object is slowing down; if it is already negative it is speeding up in the negative direction.

The **acceleration** (not the velocity). The object may still be moving.

**Yes** — velocity includes direction, so changing direction changes the velocity.

$a = \dfrac{v - u}{t}$ — change in velocity ÷ time.

The **displacement** — how far the object travels.

The **acceleration**. (Area = displacement, slope = acceleration — don't swap them.)

$s = \dfrac{u + v}{2}\,t$ — average velocity × time (the trapezium area).

The **average velocity** — halfway between the start velocity u and the final velocity v.

**½ × base × height** = ½ × time × final velocity.

**speed × time** — a constant velocity gives a rectangular area.

**Split it** into a rectangle + a triangle, work out each, then **add** them.

**Negative** displacement — the object is moving backwards. Subtract it from the forward area for the net displacement.

Rectangle area = 10 × 3.0 = **30 m**.

Triangle area = ½ × 4.0 × 12 = **24 m**.

Height (m s⁻¹) × width (s) = m s⁻¹ × s = **m** — exactly a displacement.

The five constant-acceleration quantities: **s** displacement, **u** initial velocity, **v** final velocity, **a** acceleration, **t** time.

Only when the **acceleration is constant** (a straight velocity–time line).

v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u + v)t — all four are **given** in the data booklet.

Write your **three knowns** + the unknown, then pick the equation that contains those four letters and **leaves out the fifth**.

**v² = u² + 2as** — use it when the time is unknown (e.g. stopping distance).

**s = ut + ½at²** — use it to find displacement from time.

**s = ½(u + v)t** — displacement from the average of the two speeds.

The **final velocity v = 0**.

**a = −5 m s⁻²** (negative, because the object is slowing down).

The **initial velocity u = 0** (it kills the ut term in s = ut + ½at²).

Use v² = u² + 2as: 0 = 400 − 10s → s = 40 m.

The equations come from a **straight** v–t line; a changing acceleration curves the line, so they no longer hold.

Motion where **gravity is the only force** acting — air resistance is ignored.

**g = 9.81 m s⁻²**, directed **downward** (given on the data booklet).

**No** — with no air resistance every object accelerates at the same g = 9.81 m s⁻².

It is constant-acceleration motion with **a = g**. Take up as positive, so a = −9.81 m s⁻².

**Velocity = 0** for an instant; **acceleration = 9.81 m s⁻² downward** (still g).

Time up to the top = time back down. Total flight time = **2 × time to the top**.

**Zero** — it ends where it started; it lands at the **same speed**, moving downward.

Same speed **u**, but downward: velocity = **−u** (up positive).

$v = gt$ — e.g. after 2.0 s, v = 9.81 × 2.0 ≈ 20 m s⁻¹.

Going up the velocity is positive; at the top it is zero; coming down it is negative — same slope (g) throughout.

An object moving through the air with **only gravity** acting on it (e.g. a thrown ball). Air resistance is ignored at SL.

Split it into **two independent parts**: horizontal (constant velocity) and vertical (free fall, a = g). They share the same time.

It stays **constant** — there is no sideways force.

It **increases** downward at g = 9.8 m s⁻² (free fall).

The **time** — it is the **same** for both columns.

From the **vertical** drop only: use s = u_y t + ½gt² (with u_y = 0 for a horizontal launch).

**Range = horizontal velocity × time of flight** (R = u_x·t), using the time from the vertical part.

**Together** — same height and same vertical start, so identical fall time. The throw only adds sideways distance.

**No** — horizontal speed adds range but does not change the vertical fall time.

A **parabola** — constant horizontal steps combined with growing vertical drops.

Both gain the same **vertical** speed, but the horizontal launch also keeps its **horizontal** velocity, so the combined speed is bigger.

A friction-like force from the air or liquid an object moves through. It always acts **against the motion** and **grows with speed**.

The **constant** velocity a falling object reaches when the **drag equals its weight**, so the resultant force (and acceleration) is zero.

It **increases** — drag grows with speed.

**Drag = weight** → resultant force = 0 → acceleration = 0.

**Zero** — weight and drag are equal and opposite, so they cancel.

**No** — weight and drag both act; they are **balanced**, so they cancel.

It **starts steep**, then **bends over and goes flat** — the flat value is the terminal velocity.

Just after release it is **near g** (drag tiny); at terminal velocity it has fallen to **zero**.

$F_g = mg$ — mass × gravitational field strength.

**Lower** — going up, drag adds to gravity, so the ball decelerates faster and rises less far.

**No** — the weight stays mg the whole way down; it is the **drag** that grows to match it.

The **velocity** — the rate of change of displacement.

The **acceleration** — the rate of change of velocity.

The **displacement** travelled in that time.

The **change in velocity** over that time.

**No** — it has no physical meaning.

The object is **at rest** — the displacement is not changing, so the velocity is zero.

**Constant velocity** — the velocity is not changing, so the acceleration is zero.

**Constant velocity** — equal displacement in equal times. The steeper the line, the faster.

The velocity is **changing** (accelerating). Getting steeper = speeding up; flattening off = slowing down.

Draw a **tangent** at that point and find its gradient.

**Gradient goes down** (s → v → a); **area comes back up** (a → v → s).

A sketch of **one object as a dot**, with an **arrow for every force acting ON it** (and nothing it pushes on other things).

The **net (resultant) force is zero**, so the object stays at rest or moves at **constant velocity**.

A **vector** — it has a size (in newtons) **and** a direction.

Horizontal $A_{H} = A\cos\theta$, vertical $A_{V} = A\sin\theta$. **Given** in the data booklet.

Split it into a **horizontal** and a **vertical** part that together do the same job.

**cos** = the side **next to** the angle (horizontal); **sin** = the side **opposite** it (vertical).

A **pull along a rope or string**, acting on the object **away** from it along the rope.

Resolve every force, then set the total to **zero in each direction separately** (left = right, up = down).

Only its **small vertical part** ($A\sin\theta$) holds the weight, so the **full tension** must be huge.

$F_g = mg$ — mass × gravitational field strength (g = 9.8 N kg⁻¹). **Given** in the data booklet.

**No.** At rest is one case; moving at **constant velocity** is also equilibrium (net force still zero).

The **horizontal** ($A\cos 50°$) is slightly larger, since cos 50° > sin 50° — but check the angle's reference each time.

With **zero net force**, an object stays at rest or keeps moving at **constant velocity**. (Motion needs no force — only a change in motion does.)

The **net force** equals mass × acceleration: **F = ma**, with the acceleration in the same direction as the net force.

If A exerts a force on B, then **B exerts an equal and opposite force on A**. The pair acts on **different objects**.

The **newton (N)**. 1 N = 1 kg m s⁻² (the force that gives a 1 kg mass an acceleration of 1 m s⁻²).

The **net (resultant)** force — every force on the object added together, with direction.

Because they act on **different objects**. Two forces only cancel when they act on the **same** object.

The **same acceleration** — connected bodies move together.

Apply **F = ma** to **one** of the masses on its own: tension = that mass × the shared acceleration.

**Bigger** — the cable must support the weight **and** provide the extra net force to accelerate it up (T − mg = ma).

$F = ma = \dfrac{\Delta p}{\Delta t}$ — net force = mass × acceleration = rate of change of momentum.

a = F ÷ m = 20 ÷ 5.0 = **4.0 m s⁻²**.

A **single** force is just one push/pull; the **net** force is all of them combined. Only the net force goes into F = ma.

The force that **resists sliding** between two surfaces in contact; it always **opposes the motion** (or attempted motion).

**Static** acts while the object is **still** (grows to match the push, up to μ_s R). **Dynamic** acts while it is **sliding** (a fixed μ_d R).

$F_f \le \mu_s R$ — friction can be anything up to a maximum of μ_s R.

$F_f = \mu_d R$ — a fixed value while the object moves.

The **support force** from the surface, perpendicular to it. On flat ground **R = mg**. Also written F_N.

**μ_s R** — you must beat the **maximum static** friction (use μ_s, not μ_d).

The **maximum static** friction — that's why it's harder to start something moving than to keep it moving.

μ = F_f ÷ R is a **force ÷ a force**, so the newtons cancel — it has **no unit** (a pure number).

R = mg = 98 N, so F_f = μ_d R = 0.20 × 98 = 19.6 ≈ 20 N.

**No** (in this model) — it depends on μ and the normal force R, not on how big the contact patch is.

Usually between **0 and 1** (e.g. ~0.3 for many everyday surfaces); it can exceed 1 for very grippy surfaces.

The **buoyancy (upthrust) force** on an object equals the **weight of the fluid it pushes aside** (displaces).

The **upward force** a fluid exerts on an object, because the fluid presses harder underneath than on top.

$F_b = \rho V g$ — fluid density × displaced (submerged) volume × g. **Given** in the data booklet.

The **fluid's** density — not the object's.

The **submerged** volume — the volume of fluid pushed aside.

When it is **less dense** than the fluid, so the buoyancy can balance its weight.

Buoyancy = weight: $\rho_{fluid} V_{sub}\, g = \rho_{obj} V_{total}\, g$.

The **density ratio**: ρ_object ÷ ρ_fluid.

$\rho = \dfrac{m}{V}$ — mass ÷ volume. **Given** in the data booklet.

Buoyancy ∝ submerged volume (F_b = ρVg), so the ratio of upthrusts = ratio of submerged volumes.

Ice (≈9.2 × 10²) is only slightly less dense than seawater (≈1.03 × 10³), so the submerged fraction ≈ 0.89.

Using the **object's** density for ρ, or the **whole** volume when only part is submerged.

A **resistive force** a fluid (air or liquid) exerts on an object moving through it. It points **against the motion** and **grows with speed**.

The **steady (constant) speed** a falling object reaches when the **drag balances the weight**, so the net force — and the acceleration — is zero.

How **thick or sticky** a fluid is (symbol η, unit Pa s). Honey has high viscosity; water has low viscosity.

$F_d = 6\pi\eta r v$ — drag grows with viscosity η, radius r and speed v. **Given** in the data booklet.

**Weight = drag**: $mg = 6\pi\eta r v$ (net force zero, so steady speed).

About **g** — there's no drag yet because the speed is zero.

It **starts near g and decreases to zero** as drag builds up — it is **not** constant.

The **terminal velocity** — speed constant, acceleration zero, drag = weight.

**4× faster.** The weight goes up 8 times (volume), but the drag only 2 times — so it must fall 4 times faster before drag balances it.

Assuming the acceleration is **constant** while falling. It isn't — it falls from ≈ g to zero as drag grows.

Constant speed ⇒ no acceleration ⇒ net force = 0, so the upward drag exactly balances the downward weight.

**Downwards.** Drag always opposes the motion, and the bubble is moving up — so the upthrust has to balance the weight **plus** the drag.

The **net (resultant) force** that points **toward the centre** of a circle and keeps an object moving in that circle.

**Toward the centre**, along the radius — never along the direction of motion.

**Yes** — its direction keeps changing, so its velocity changes (it accelerates toward the centre).

$F_c = \dfrac{mv^2}{r}$ — from $F = ma$ with $a = \dfrac{v^2}{r}$.

$a = \dfrac{v^2}{r} = \omega^2 r = \dfrac{4\pi^2 r}{T^2}$ (in the data booklet).

$v = \dfrac{2\pi r}{T} = \omega r$ (in the data booklet).

It becomes **4× bigger** — because $F_c \propto v^2$.

$T - mg = \dfrac{mv^2}{r}$, so $T = mg + \dfrac{mv^2}{r}$ — the tension is **greater** than the weight.

**Friction** between the tyres and the road (pointing toward the centre).

**No** — F_c is the **net** of the real forces (friction, tension, gravity, normal). Never draw it as a separate arrow.

$F_c = \dfrac{1.5 \times 4.0^2}{2.0} = 12$ N.

The **mass × velocity** of an object — how much motion it has. p = mv, unit kg m s⁻¹. It is a **vector** (has direction).

The **average force × the time** it acts for, J = FΔt. It equals the **change in momentum** (Δp). Unit: N s.

**kg m s⁻¹**. Impulse uses **N s**, which is the same unit.

$p = mv$ — mass × velocity (given in the data booklet).

$J = F\Delta t = \Delta p$ — force × time = change in momentum (given).

$F = \dfrac{\Delta p}{\Delta t}$ — the change in momentum ÷ the contact time (given form of Newton's 2nd law).

The **impulse** — which equals the **change in momentum**.

**No** — the direction flips, so Δp = m(v + u) = 2mu. Bouncing changes momentum more than stopping.

They **increase the contact time** Δt. Since F = Δp/Δt, a longer time means a **smaller force** for the same change in momentum.

Δp = J = 6.0 kg m s⁻¹, so v = p/m = 6.0 ÷ 0.50 = 12 m s⁻¹.

Impulse gives momentum p = J; then $E_k = \dfrac{p^2}{2m} = \dfrac{1}{2}mv^2$ once you have the speed.

**Momentum p = mv** — mass × velocity. It is a **vector** (has direction). Unit: kg m s⁻¹.

If no external force acts, the **total momentum before = total momentum after** a collision or explosion.

**Yes** — momentum is conserved in **every** collision (with no outside force), elastic or inelastic.

One where the **total kinetic energy is also conserved** (KE before = KE after). Objects bounce cleanly.

One where the objects **stick together** and move as one. Momentum is conserved, but the **most kinetic energy is lost** (to heat/sound).

Compare **total KE before** and **total KE after** (E_k = ½mv²). If they're equal, it's elastic.

Velocity has direction — objects moving opposite ways get opposite signs, or the momentum total is wrong.

As **one combined mass** at one common velocity: (m₁ + m₂)v.

$p = mv$ (given in the data booklet).

$E_k = \tfrac{1}{2}mv^2$ (given) — used to test elasticity.

**No** — only in an **elastic** collision. In an inelastic one some KE becomes heat/sound.

(KE before − KE after) ÷ KE before. It's never zero for a sticking (perfectly inelastic) collision.

The **energy transferred** when a **force moves something through a distance**. Unit: the **joule (J)**.

$W = Fs\cos\theta$ — force × distance × the cosine of the angle between them. (Given in the data booklet.)

The **angle between the force and the direction of motion**. If the force is along the motion, θ = 0 and cos 0 = 1, so W = Fs.

**Zero** — cos 90° = 0, so W = 0. (e.g. the normal force on a block sliding along a floor.)

The **work done** by the force.

The **joule (J)** — the same unit as all forms of energy.

**Zero** — no movement means no distance, so no work, however hard you push.

The work becomes kinetic energy: set **W = ½mv²** and solve for **v**.

The energy of a moving object: $E_k = \tfrac{1}{2}mv^{2}$ (m = mass, v = speed). Given in the data booklet.

W = Fs = 9.0 × 4.0 = **36 J** (which equals the kinetic energy gained).

The energy an object has because it is **moving**. It depends on the mass and the **speed squared**. Unit: the joule (J).

$E_k = \tfrac{1}{2}mv^{2} = \dfrac{p^{2}}{2m}$ — use ½mv² with the speed, or p²/2m with the momentum.

It becomes **four times** as big, because the speed is squared (2² = 4).

The **net work done** on an object equals its **change in kinetic energy**: W_{net} = ΔE_k.

Friction does **negative work**, removing kinetic energy. The object stops when all its E_k is used up.

Set **friction force × distance = E_k**, then **distance = E_k ÷ friction force**.

The **joule (J)** — the same unit as work and all other forms of energy.

When you're **given the momentum** p (= mv) instead of the speed — both forms give the same energy.

E_k = ½ × 3.0 × 4.0² = ½ × 3.0 × 16 = 24 J.

distance = E_k ÷ friction = 90 ÷ 18 = 5.0 m.

Kinetic energy ½mv² is a **scalar** (no direction) measured in joules; momentum mv is a **vector** measured in kg m s⁻¹.

The energy an object has **because of its height** in a gravitational field. It increases when the object is raised.

The energy an object has **because of its motion**. The faster it moves, the more KE it has.

$\Delta E_p = mg\Delta h$ — mass × gravitational field strength × change in height. (Given in the data booklet.)

$E_k = \tfrac{1}{2}mv^{2}$ — half × mass × speed². (Given in the data booklet.)

With no air resistance, **PE + KE stays constant**: the PE lost equals the KE gained.

$mg\Delta h = \tfrac{1}{2}mv^{2}$ — set the PE lost equal to the KE gained.

**All PE, no KE** — it is at maximum height and not yet moving.

**All KE, no PE** (taking the bottom as the reference height) — all the PE has converted to KE.

**No** — in mgΔh = ½mv² the mass cancels, so heavy and light objects reach the same speed (no air resistance).

**A quarter** — KE gained = PE lost, so the fraction of height fallen = the fraction now KE.

**Half-way down** — there it has lost half its PE, which has become KE, so PE = KE.

The **joule (J)**. PE and KE are both measured in joules.

The energy **stored in a spring** (or springy material) when it is **stretched or squashed**. Unit: the joule (J).

$E_H = \tfrac{1}{2}k\,\Delta x^{2}$ — half × spring constant × extension squared. (Also written E_p = ½kx².)

How **stiff** a spring is — the force needed per metre of stretch. Unit: N m⁻¹ (newtons per metre).

The **extension or compression** — how far the spring is stretched or squashed from its natural length, in metres.

It becomes **four times** as big, because the extension is squared (2² = 4).

By conservation of energy: **E_H = kinetic energy before − kinetic energy of the combined motion**.

From **conservation of momentum**: total momentum before = (combined mass) × common speed.

The **joule (J)** — the same unit as all other forms of energy.

E_H = ½ × 300 × 0.020² = ½ × 300 × 0.0004 = 0.060 J.

Power = energy ÷ time = 0.60 ÷ 0.015 = 40 W.

Elastic PE (½kΔx²) is stored by **stretching/squashing** a spring; gravitational PE (mgΔh) is stored by **lifting** a mass to a height. Both are in joules.

The **rate of energy transfer** — the energy transferred (or work done) each **second**. Unit: the watt (W).

**1 watt = 1 joule per second** (1 W = 1 J s⁻¹).

$P = \dfrac{\Delta W}{\Delta t} = Fv$ — energy ÷ time, or force × speed.

**P = Fv**, where F is the **resistive (drag) force** — it equals the driving force at constant speed.

**Average** = total energy ÷ total time (ΔW/Δt). **Instantaneous** = the power at one instant, using Fv with the speed right now.

The fraction of the energy put in that comes out as **useful** energy: η = useful out ÷ total in (× 100 for a %). It has no unit.

**No** — you can never get more useful energy out than you put in; some is always wasted (mostly as heat).

Mostly to **thermal energy (heat)**, plus some sound — energy spread out and no longer useful.

P = ΔW/Δt = 600 ÷ 5.0 = 120 W.

P = Fv = 400 × 30 = 12 000 W = 12 kW.

At constant speed P = Fv = cv², so c = P ÷ v². Its SI unit is kg s⁻¹.

Energy cannot be created or destroyed — it is only **transferred** from one store to another. The total amount stays the same.

It has been transferred to a **less useful** store — almost always **thermal energy (heat)** — that spreads out and can't easily be reused. It is NOT destroyed.

The **useful fraction** of the energy (or power) supplied: η = useful output ÷ total input. It has no unit and is often given as a %.

$\eta = \dfrac{\text{useful out}}{\text{total in}}$ — useful energy (or power) out ÷ total energy (or power) in.

An arrow diagram showing how the input energy splits into useful and wasted branches; the **width** of each arrow shows the amount of energy.

The useful and wasted branches **add up to the input arrow** — energy is conserved, so nothing is missing.

**Thermal energy (heat)** — and sometimes **sound** in moving parts. It spreads into the surroundings.

No — the useful output can never be larger than the total input, so efficiency is always between **0 and 1** (0–100%).

wasted = total energy in − useful energy out.

η = useful ÷ total = 350 ÷ 500 = 0.70 = 70%.

Wasted = 60 − 9 = 51 J, transferred as thermal energy (heat).

Because energy is **conserved** — it isn't lost, only **transferred** to a less useful store (heat). The total is unchanged.

The **angle turned per second** — the rate of change of θ. Unit: **rad s⁻¹**.

The **rate of change of angular velocity** ω. Unit: **rad s⁻²**.

The angle whose **arc length equals the radius**. A full turn = **2π rad = 360°**.

Slope = **angular acceleration** α; area = **angle turned** θ.

$\omega = \omega_0 + \alpha t$.

Multiply by **2π** (one revolution = 2π rad).

The **turning effect** of a force: $\tau = Fr\sin\theta$. Unit: **N m**.

When it acts **through the pivot** (θ = 0, sin θ = 0).

The **total torque about any point is zero** (clockwise = anticlockwise).

A point on an **unknown force's line**, so that force has zero torque.

Bigger **r** → bigger torque for the same force.

Both are N m, but torque is **N m** (a turning effect); energy is the **joule**.

Rotation's version of **mass** — resistance to angular acceleration: $I = \sum m r^{2}$. Unit: **kg m²**.

Because r is **squared** in I = Σmr² — doubling the distance quadruples that part's contribution.

$\tau = I\alpha$ (torque = moment of inertia × angular acceleration).

The **hoop** (I = MR²); the disc is ½MR².

$I = \tfrac{1}{2}MR^{2}$ (given in the question).

$I = MR^{2}$ — all the mass is at radius R.

No — the **exam gives them**; recognise and substitute.

$\alpha = \tau / I$ — rearranged from τ = Iα.

**Torque** τ.

Yes — the same object has different I about different axes.

**kg m²**.

Rotation's version of momentum: $L = I\omega$. Unit: **kg m² s⁻¹**.

When there is **no external torque** on the system.

$I_1\omega_1 = I_2\omega_2$.

I decreases, so ω increases to keep **L = Iω** constant.

$E_k = \tfrac{1}{2}I\omega^{2}$ (the rotational ½mv²).

$\tfrac{1}{2}mv^{2} + \tfrac{1}{2}I\omega^{2}$ — translational **plus** rotational.

It **quadruples** (E_k ∝ ω²).

**No** — angular momentum is conserved, but some kinetic energy is lost.

$L = I\omega$.

ω **decreases** (I up, L constant).

**kg m² s⁻¹** (or equivalently N m s).

A coordinate grid and clock you measure motion **against**. All motion is **relative** to a chosen frame.

A frame moving at **constant velocity** (no acceleration). Newton's first law holds in it.

Inertial: a train cruising in a straight line at steady speed. Non-inertial: a car going round a bend.

$u' = u - v$ — the object's velocity in the moving frame equals its ground velocity minus the frame's velocity.

$x' = x - vt$ — position in the moving frame, where v is the frame's speed.

Same direction ⇒ **subtract** the speeds; opposite directions ⇒ the speeds **add**.

Same direction ⇒ add: $12 + 1.5 = 13.5$ m s⁻¹.

$u' = u - v = 20 - 30 = -10$ m s⁻¹, i.e. 10 m s⁻¹ westward.

The **laws of mechanics are the same in every inertial frame** — no experiment can detect uniform motion.

**No.** All inertial frames are equivalent; 'at rest' only ever means 'relative to something'.

Near the **speed of light** — light travels at the same speed in every frame, so simple addition fails (→ special relativity).

Opposite directions ⇒ add: $25 + 30 = 55$ m s⁻¹.

The **laws of physics are the same in all inertial (non-accelerating) reference frames**.

The **speed of light in a vacuum is the same for all inertial observers**, regardless of the motion of the source or observer.

$c = 3.00 \times 10^{8}$ m s⁻¹ — the same for every inertial observer.

A frame moving at **constant velocity** — no acceleration (no speeding up, slowing down, or turning).

Exactly **c**, not 1.5c — by postulate 2 light's speed never adds on the source's speed.

Classically speeds add; **relativistically light always measures c** for everyone, so they do not add.

**c** — the postulates make it impossible for anything with mass to reach or exceed the speed of light.

Whether two events happen **'at the same time' depends on the observer's motion** — observers in relative motion can disagree.

Because **c is the speed limit**, so any relative speed must stay **below c**; velocities do not add the everyday way near c.

**No** — lengths and time intervals depend on the observer's motion; only the speed of light c is the same for all.

Because **both measure light at the same speed c**, they are forced to disagree about **when** events happen.

'The speed of light in a vacuum is the same for **all inertial observers, regardless of the motion of the source or observer**.'

$\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}$ — and it is **always ≥ 1**.

The time between two events measured by a **single clock present at both** — the **shortest** possible time.

The length of an object measured **in its own rest frame** — the **longest** possible length.

$\Delta t = \gamma\,\Delta t_0$. Since γ ≥ 1, **moving clocks run slow**.

$L = \dfrac{L_0}{\gamma}$. Since γ ≥ 1, **moving objects contract** along the motion.

**Multiply** the proper time by γ ($\Delta t = \gamma\,\Delta t_0$) — the time gets bigger.

**Divide** the proper length by γ ($L = L_0/\gamma$) — the length gets smaller.

$\gamma = \dfrac{1}{\sqrt{1 - 0.80^2}} = \dfrac{1}{\sqrt{0.36}} = 1.67$.

Only the dimension **along the direction of motion**; width and height are unchanged.

$u' = \dfrac{u - v}{1 - uv/c^2}$ — it always keeps the result **below c**.

$\dfrac{1.00c}{1 + 0.25} = 0.80c$, **not** 1.0c.

Time **stretches** (Δt = γΔt₀, multiply); length **shrinks** (L = L₀/γ, divide). Both use the same γ.

**ct** (speed of light × time) up the **vertical** axis, position **x** along the **horizontal** axis.

The **path an object traces** on a space-time diagram — its position at every instant.

A single **point** — a definite **place at a definite time**.

A **vertical** line — x stays fixed while ct keeps climbing.

At **45°**, because light travels $x = ct$, so equal steps in x and ct.

**More tilted toward the x-axis** — the faster it goes, the further it leans (but never past 45°).

$v = c\,\dfrac{\Delta x}{\Delta(ct)}$ — the more horizontal the line, the faster the object.

$(\Delta s)^2 = (c\Delta t)^2 - (\Delta x)^2$ — the same in every inertial frame.

It is **invariant**: all inertial observers measure the **same Δs**, even though Δt and Δx differ.

$(c\Delta t)^2 = 2.25\times10^6$, $(\Delta x)^2 = 8.1\times10^5$, so $(\Delta s)^2 = 1.44\times10^6$ and **Δs = 1200 m**.

**No** — events simultaneous in one frame need not be in another; the line of 'now' **tilts** for a moving observer.

The **space-time interval** Δs, the cause-and-effect order of events, and that **light travels at 45°** (speed c).

The **total random kinetic energy** of all the particles **plus** the **total intermolecular potential energy** of all the particles.

**Random KE** (the particles' motion) and **intermolecular PE** (energy in the forces between particles).

Both the **random KE** of the particles **and** the **intermolecular PE** (a real gas has weak forces, so the PE part is not zero).

The **average random kinetic energy** of the particles (not the potential energy).

During a **change of state** (melting, boiling) — the spacing of the particles changes there.

The particles are packed **closer together** in the solid, so there is **more mass per volume**.

$\rho = \dfrac{m}{V}$ — mass ÷ volume. **Given** in the data booklet.

**kg m⁻³** (kilograms per cubic metre).

About **4 °C** — water's density anomaly.

Ice (and water below 4 °C) is **less dense** than water at 4 °C, so it rises and floats.

Ponds freeze **top-down**; the ice insulates the ≈4 °C water below, so fish survive the winter.

A **real gas** has KE **and** intermolecular PE; an **ideal gas** is modelled with no forces, so its internal energy is the **KE only**.

The **energy needed to raise the temperature of 1 kg of a substance by 1 degree** (1 K). Unit: J kg⁻¹ K⁻¹.

**J kg⁻¹ K⁻¹** (joules per kilogram per kelvin: the energy to raise 1 kg by 1 K, i.e. 1 °C).

$Q = mc\Delta T$ — mass × specific heat capacity × temperature change. **Given** in the data booklet.

The temperature **change** = final temperature − start temperature (Δ means 'change in').

**No** — a change of 1 K is the same size as a change of 1 degree C, so either works. Never convert ΔT to kelvin.

$c = \dfrac{Q}{m\,\Delta T}$ — energy ÷ (mass × temperature change).

$m = \dfrac{Q}{c\,\Delta T}$.

Is **hard to heat** — it needs lots of energy per degree, so it warms and cools **slowly** (like water).

It has a **very large** specific heat capacity (about 4200 J kg⁻¹ K⁻¹), so it absorbs a lot of energy with only a small temperature rise.

During a **change of state** (melting/boiling), where the temperature stays constant — use $Q = mL$ instead.

Putting the **actual temperature** into ΔT instead of the **change** (final − start).

The added energy goes into **breaking the bonds** between particles (latent heat), not into their kinetic energy — so the temperature does not change.

The **energy needed to change the state of 1 kg** of a substance with **no temperature change**. Unit: J kg⁻¹.

$Q = mL$ — energy = mass × specific latent heat. **Given** in the data booklet. Used for the flat parts (state change).

$Q = mc\Delta T$ — mass × specific heat capacity × temperature change. **Given** in the data booklet. Used for the sloping parts.

**Fusion (Lf)** = melting/freezing. **Vaporisation (Lv)** = boiling/condensing. For one substance, **Lv ≫ Lf**.

A **state change** (melting or boiling) at **constant temperature** — use $Q = mL$.

The **temperature is changing** (warming or cooling) — use $Q = mc\Delta T$.

Vaporising fully separates the particles, needing far more energy than melting (Lv ≫ Lf), so it takes longer at a steady heating rate.

**Energy lost by the hot object = energy gained by the cold object.** Add one Q-term per step (warm, melt, warm…).

Use a **separate Q-term for each step**: $Q = mc\Delta T$ for each temperature change and $Q = mL$ for each state change, then add them.

Some thermal energy is **lost to the surroundings** or absorbed by the **container**, which the ideal 'no losses' calculation ignores.

$Q = mL = 0.50 \times 3.3\times10^{5} = 1.65\times10^{5}$ J (≈ 1.7 × 10⁵ J).

**Conduction**, **convection** and **radiation**. Heat always flows from hotter to colder.

Faster-vibrating hot particles jostle their cooler neighbours, passing **energy** along while the particles stay put. In metals, free **electrons** also carry it (so metals conduct best).

The **hot fluid itself** moves: warmed fluid expands, becomes less dense and **rises**, carrying its energy with it (only in liquids and gases).

As **infrared electromagnetic waves**, needing **no material** — so it is the only method that works through a **vacuum** (e.g. the Sun → Earth).

**Radiation** only — conduction and convection both need particles/material.

$\dfrac{\Delta Q}{\Delta t} = kA\dfrac{\Delta T}{\Delta x}$ — rate = conductivity × area × (temperature difference ÷ thickness). **Given** in the data booklet.

The **watt** (W), i.e. joules per second (J s⁻¹) — it is a rate of energy transfer.

A bigger thickness **Δx** (on the bottom) **slows** conduction: rate ∝ 1 ÷ Δx, so doubling the thickness halves the rate.

A larger conductivity **k**, larger area **A**, or a larger temperature difference **ΔT**.

The object cools toward room temperature, so the **temperature difference** driving the heat loss shrinks — a smaller difference means a slower rate, i.e. a flatter graph.

They contain **free electrons** that move quickly through the metal and carry thermal energy, on top of the usual particle-to-particle vibration.

From a **hotter** region to a **colder** one, until they reach the same temperature (thermal equilibrium).

The radiation **power received per unit area** (perpendicular to the rays). Unit: **W m⁻²**.

$I = \dfrac{P}{A}$ — power ÷ area. **Given** in the data booklet.

**W m⁻²** (watts per square metre).

$I = \dfrac{P}{4\pi d^{2}}$ — the power spread over a sphere of radius d (so I ∝ 1/d²).

The **intensity of the Sun's radiation arriving at Earth's distance** (just above the atmosphere): **S = 1.36 × 10³ W m⁻²**.

**1.36 × 10³ W m⁻²** — given in the data booklet.

A fixed power spreads over an ever-larger **sphere** (A = 4πd²); same power ÷ bigger area = smaller intensity.

It drops to a **quarter** (× 1/4), because I ∝ 1/d² (inverse-square law).

Multiply by the whole sphere area: **P = I × 4πd²**.

Incident **intensity × panel area × efficiency** (efficiency as a decimal).

**Per m²** — it is an intensity (W m⁻²) at Earth's distance, not the Sun's total power (W).

A perfect **absorber and emitter** of radiation — it absorbs every wavelength that hits it and, when hot, radiates over all wavelengths. Stars are a good model.

The total power (luminosity) radiated by a black body is $L = \sigma A T^4$ — surface area × temperature to the fourth power × the Stefan-Boltzmann constant. **Given** in the data booklet.

The **Stefan-Boltzmann constant**, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (given).

As **T⁴** — doubling the kelvin temperature multiplies the power by 2⁴ = **16**.

The peak wavelength and absolute temperature multiply to a constant: $\lambda_{max} T = 2.9 \times 10^{-3}$ m K. **Given** in the data booklet.

They are **inversely** related — a **hotter** body has a **shorter** peak wavelength (bluer light).

**Kelvin (K)** — never °C. Convert with K = °C + 273.

It gets **taller** (more total power, Stefan-Boltzmann) and its **peak shifts to a shorter wavelength** (Wien).

$\lambda_{max} = \dfrac{2.9 \times 10^{-3}}{5800} = 5.0 \times 10^{-7}$ m (500 nm).

Write $L = \sigma A T^4$ for each and **divide** one by the other — σ cancels, leaving a ratio of areas and T⁴.

Rising T shifts the spectrum's peak to shorter wavelengths (Wien) and adds power across all wavelengths (Stefan-Boltzmann), so the visible colour shifts red → orange → white.

The sphere's surface area, $A = 4\pi r^2$, so $L \propto r^2 T^4$.

The **fraction of incident sunlight that a surface reflects** (scatters back). A number between 0 and 1, with no unit.

$\text{albedo} = \dfrac{\text{total scattered power}}{\text{total incident power}}$ — the reflected fraction. **Given** in the data booklet.

**0.70** — the absorbed fraction is 1 − albedo.

About **0.30** — roughly 30% of sunlight is reflected back to space.

**High:** fresh snow/ice (~0.8), thick cloud (~0.7). **Low:** dark ocean (~0.06), forest/asphalt (~0.1–0.2).

Sunlight lands on the **disc** Earth shows the Sun (πr²) but is shared over the whole **sphere** (4πr²): πr² ÷ 4πr² = 1/4.

About **240 W m⁻²**: (1 − 0.30) × (S ÷ 4) = 0.70 × 340 ≈ 240 W m⁻².

At a steady temperature the power **absorbed** from the Sun equals the power **radiated** away. Energy in = energy out.

The **surface** (ice/cloud high, ocean/forest low), **cloud cover**, and the Sun's angle — so **latitude** and **time of day**.

Treating albedo as the **absorbed** fraction. Albedo is the **reflected** fraction; absorbed = 1 − albedo.

A real surface radiates **emissivity ×** the black-body value. Use it when the surface is not a perfect black body (emissivity < 1).

Greenhouse gases let **sunlight in** but absorb the **infrared** the warm surface radiates out, sending some **back down** — so the surface stays **warmer**.

**Carbon dioxide (CO₂)**, **methane (CH₄)**, **water vapour (H₂O)** and **nitrous oxide (N₂O)**.

**Outgoing infrared** from the warm surface. Incoming sunlight (mostly visible) passes straight through.

Greenhouse gases **absorb** the **infrared** the surface emits, then **re-emit** it in all directions, so some returns **back down** to the surface, keeping it warmer.

CO₂/CH₄ bonds **resonate** (vibrate) at infrared frequencies, so they absorb infrared; the simple N₂/O₂ bonds do not.

The infrared radiation's frequency **matches** the natural vibration frequency of the gas molecule's bonds, so the bond absorbs the energy.

**Natural** = warming from gases always present (Earth ~33 °C warmer, needed for life). **Enhanced** = **extra** warming from human-added gases.

**Burning fossil fuels** (coal, oil, gas), which releases extra **CO₂**.

About **33 °C** warmer than it would be with no atmosphere — without it, Earth would be far too cold for life.

Saying the gases block **incoming sunlight**. They don't — sunlight passes in; the gases trap the **outgoing infrared**.

**Farming** (cattle), **rice fields**, **landfill** and **gas leaks** — a strong infrared absorber per molecule.

At **constant temperature**, the pressure and volume of a fixed mass of gas obey **P V = constant** (inversely proportional).

At **constant pressure**, the volume of a fixed mass of gas obeys **V ÷ T = constant** — volume is proportional to the absolute (kelvin) temperature.

At **constant volume**, the pressure of a fixed mass of gas obeys **P ÷ T = constant** — pressure is proportional to the absolute (kelvin) temperature.

$\dfrac{PV}{T} = \text{constant}$ — so $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$. **Given** in the data booklet.

**T (K) = θ (°C) + 273.** Always do this before using a gas law.

The laws count temperature from **absolute zero** (−273 °C = 0 K); only the kelvin scale makes V and P truly proportional to T.

A **curve** (hyperbola) that sweeps down to the right, because P V is constant.

A **straight line through the origin**, because P = K(1/V); its **slope is the constant K**.

Pressure × volume = **Pa × m³ = J** (the joule).

Volume is fixed, so **P ÷ T = constant**: the pressure rises in proportion to the kelvin temperature.

Leaving the temperature in **°C** — every gas-law T must be in **kelvin** (°C + 273).

The **combined gas law** P V ÷ T = constant: fix T → Boyle, fix P → Charles, fix V → Gay-Lussac.

$PV = nRT = N k_B T$ — given in the data booklet. T must be in **kelvin**.

A fixed-size 'pack' of particles: one mole = **6.02 × 10²³** particles (the **Avogadro constant** N_A).

$N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$ — the number of particles in one mole. **Given**.

$n = \dfrac{N}{N_A}$, so $N = n\,N_A$. **Given** in the data booklet.

Use **R** (8.31) with the amount in **moles n**; use **k_B** (1.38 × 10⁻²³) with the **number of molecules N**. Never mix them.

**Kelvin** (K). Convert from Celsius by adding 273.

**Equal N** — same P, V, T means the same number of molecules, whatever the gas.

Write $PV = NkT$ for each and **divide** one by the other — any equal quantity (P, V or T) cancels, leaving a simple ratio.

$R = 8.31\ \text{J K}^{-1}\,\text{mol}^{-1}$ — used with the amount in moles. **Given**.

$k_B = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ — used with the number of molecules N. **Given**.

$n = \dfrac{PV}{RT}$ — with P in Pa, V in m³, T in K.

Leaving **T in Celsius** (must be kelvin), or mixing **n with k_B** / **N with R**.

Gas particles **colliding with the walls** of the container — each collision pushes on the wall.

The **average kinetic energy** of its particles — hotter gas means faster particles.

It is **proportional to the absolute temperature**: average KE ∝ T (T in kelvin).

$\overline{E_k} = \tfrac{3}{2}k_B T$ — **given** in the data booklet (T in kelvin).

The **Boltzmann constant**, 1.38 × 10⁻²³ J K⁻¹ — it links energy to temperature for one particle.

The relation average KE ∝ T only works from **absolute zero** (0 K); convert Celsius with **+ 273**.

**Equal** — average kinetic energy depends only on the temperature, not the gas or particle mass.

The piston **does work** on the gas, raising the particles' average kinetic energy, so they move faster.

It is **zero** — the particles have the least possible motion.

Particles are tiny points with negligible volume; there are **no forces between them** except during (elastic) collisions.

**Only kinetic** energy — no intermolecular potential energy (no forces between particles).

All gases have the **same average KE** at a given temperature, so heavier particles need a **lower speed** to have that energy.

The **total energy of all the particles**: their random **kinetic energy** + the **potential energy** of the forces between them.

**Temperature only** — an ideal gas has no inter-particle PE, so U is fixed by the random KE of the particles.

$Q = \Delta U + W$ — the heat **added** equals the rise in **internal energy** plus the **work done by** the gas.

$\Delta U = Q - W$ (heat in **minus** work done by the gas).

**Negative** — Q is the heat **added** to the gas, so heat leaving makes Q < 0.

**Negative** — W is the work done **by** the gas; on compression the surroundings do work on it, so W < 0.

$W = P\,\Delta V$ — pressure times the change in volume.

P in **pascals (Pa)**, ΔV in **cubic metres (m³)**, giving W in **joules (J)**.

**Internal energy** is energy a gas **already has** inside; **heat** is energy **flowing** in or out due to a temperature difference.

**ΔU = 0** — U depends on temperature alone, so no temperature change means no change in internal energy.

$\Delta U = Q - W = 500 - 200 = 300$ J (the gas warms).

Write each sign in **words** first ('heat removed → Q negative', 'gas compressed → W negative'), then plug into $\Delta U = Q - W$.

The **disorder** of a system — the **number of microstates** (microscopic arrangements) available. Unit: **J K⁻¹**.

One specific microscopic arrangement of the particles that gives the same overall (macroscopic) state. **More microstates ⇒ higher entropy**.

$\Delta S = \dfrac{\Delta Q}{T}$, with **T in kelvin**.

$\Delta S$ in **J K⁻¹**, $\Delta Q$ in **J**, $T$ in **K**.

Heat **in** ⇒ ΔQ is **positive** (entropy rises); heat **out** ⇒ ΔQ is **negative** (entropy falls).

The **entropy of an isolated system never decreases** — it increases for any irreversible (real) process.

Yes — but only if another part gains **more**, so the **total** entropy of the isolated system still does not decrease.

Because it **increases the total entropy** of the universe ($\Delta S_{total} > 0$); the reverse would decrease it, so it never happens unaided.

The **direction** of time set by the second law: real processes always run the way that **increases total entropy**.

Calculate $\Delta S_{total}$ for the isolated system. If it is **positive**, the process can occur (and is irreversible).

$\Delta S = \Delta Q/T$, and the **cold** body has the **smaller T**, so for the same ΔQ it gains **more** entropy than the hot body loses.

Entropy is the **joule per kelvin (J K⁻¹)**; energy is the **joule (J)** — do not confuse them.

$\Delta U = Q - W$, where **W** is the work done **by** the gas. Internal energy U depends only on temperature.

**T** is constant, so $\Delta U = 0$ and therefore $Q = W$.

**P** is constant; the work done by the gas is $W = P\,\Delta V$.

**V** is constant, so $W = 0$ and therefore $Q = \Delta U$.

**Q = 0** (no heat flows), so $\Delta U = -W$.

The **area under the curve** between the start and end volumes.

Takes in **Q_in** from the hot reservoir, does useful **work W**, and rejects **Q_out** to the cold reservoir. $W = Q_{in} - Q_{out}$.

$\eta = \dfrac{\text{useful work}}{\text{energy input}} = 1 - \dfrac{Q_{out}}{Q_{in}}$.

$\eta_{Carnot} = 1 - \dfrac{T_{cold}}{T_{hot}}$, with both temperatures in **kelvin**.

Friction, turbulence and unwanted heat loss waste energy, so the real efficiency is always **lower** than the Carnot ceiling.

$\eta = 1 - \dfrac{600}{800} = 0.25$, i.e. **25%**.

$\eta_{Carnot} = 1 - \dfrac{300}{500} = 0.40$, i.e. **40%**.

The **rate of flow of charge** — the charge passing a point each second. Unit: ampere (A).

The **energy given to each coulomb of charge** as it passes through a component. Unit: volt (V).

The **coulomb (C)**.

The **ampere (A)** — one ampere is one coulomb of charge per second.

$I = \dfrac{\Delta q}{\Delta t}$ — charge ÷ time. **Given** in the data booklet.

$V = \dfrac{W}{q}$ — energy ÷ charge. **Given** in the data booklet.

**1 joule of energy given to every 1 coulomb of charge** (1 V = 1 J C⁻¹).

$\Delta q = I \times \Delta t$ — current × time.

$W = V \times q$ — voltage × charge.

**Through** it — an ammeter goes in series (in the line).

**Across** it — a voltmeter goes in parallel.

I = Δq/Δt = 0.80 ÷ 5.0 = 0.16 A.

How hard it is to push current through a component: $R = \dfrac{V}{I}$ (voltage across it ÷ current through it). Unit: the **ohm (Ω)**.

The voltage across a component equals the current through it times its resistance: $V = IR$. Given in the data booklet as R = V ÷ I.

The **ohm (Ω)**.

**R = V ÷ I** at a point on the graph. For a straight line through the origin, R is the same at every point.

A **straight line through the origin** — current is proportional to voltage, so R is constant.

A **curve** — R = V ÷ I changes from point to point, so the resistance is not constant.

As the current increases the filament gets **hotter**, and a hotter metal wire has a **higher resistance**, so the I–V graph curves over.

$R = \dfrac{\rho L}{A}$ — resistivity × length ÷ cross-sectional area. Given in the data booklet (as ρ = RA ÷ L).

The **resistivity** of the material (unit Ω m) — a property of the material itself, independent of the wire's shape.

R **doubles** — resistance is proportional to length (R ∝ L).

R **halves** — resistance is inversely proportional to area (R ∝ 1/A).

R = V ÷ I = 12 ÷ 4.0 = 3.0 Ω.

Components joined in **one single loop**, end to end — only **one path** for the charge.

Components joined **side by side** on separate branches — the charge has a **choice of paths**.

The **current** — one loop means one current everywhere.

The **potential difference (voltage)** — every branch sits across the same two points.

They **add**: $R_s = R_1 + R_2 + \ldots$ — **given** in the data booklet. Total is bigger than any one.

Add the reciprocals then flip: $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots$ — **given**. Total is smaller than any one.

**R ÷ 2** (half of one). N equal resistors in parallel give R ÷ N.

It **splits** between the resistors **in proportion to their resistance**; the separate p.d.s add up to the supply.

It **splits** between the branches; the **smaller** resistance carries the **larger** current. The branch currents add up to the total.

**Combine** the resistors into one equivalent R, then use **I = V/R**.

Forgetting to **flip** 1/R_p back to R_p — or just adding the values as if in series.

**Lowers** it — an extra path makes it easier for charge to flow.

The **rate at which electrical energy is transferred** (turned into heat, light, motion). Unit: the **watt (W)** = 1 joule per second.

The **watt (W)**. 1 W = **1 joule of energy every second** (1 J s⁻¹).

$P = IV = I^{2}R = \dfrac{V^{2}}{R}$ — all **given** in the data booklet.

**P = IV** — current × voltage, the simplest form.

**P = I²R** — avoids having to find V first.

**P = V²/R** — avoids having to find I first.

**P = V²/R**, so P ∝ 1/R — **more** resistance means **less** power (e.g. double R → half the power).

**P = I²R**, so P ∝ R — **more** resistance means **more** power.

$E = Pt$ — energy = power × time. **Given** in the data booklet.

The energy a **1 kW** appliance uses in **1 hour** (= 3.6 × 10⁶ J). Energy bills are charged per kWh.

Energy in **kWh** (power in kW × time in hours), then **× the price per kWh**.

It **doubles** — for a uniform wire R ∝ length (L).

The **energy given to each coulomb** of charge by the cell — its 'pushing voltage'. Unit: **volt (V)**.

The **resistance inside the cell itself** (symbol r). The current flows through it, so some energy is lost inside the cell.

The voltage **actually delivered** across the cell's terminals to the circuit: $V = \varepsilon - Ir$.

$\varepsilon = I(R + r)$ — emf = current × (load + internal resistance). **Given** in the data booklet.

$V = \varepsilon - Ir$ — emf minus the lost volts (Ir).

**Ir** — the volts used up inside the cell by its internal resistance. They grow as the current grows.

Because some of the emf is used to push current through the **internal resistance r**, losing Ir volts inside the cell.

Lost volts = ε − V = Ir, so $r = \dfrac{\varepsilon - V}{I}$.

It **drops** — bigger I means bigger lost volts Ir, so less voltage reaches the circuit.

When the internal resistance **r is negligible** (or the current is very small), so Ir ≈ 0.

The simple **ε = IR** — the emf is just current × external resistance.

The **terminal p.d.** V = ε − Ir (the same as IR, the voltage across the load).

Oscillation in which the **acceleration is proportional to the displacement** from equilibrium and is always directed **back toward equilibrium**.

$a = -\omega^{2}x$ — **given** in the data booklet. a = acceleration, ω = angular frequency, x = displacement.

The acceleration points **opposite to the displacement** — always back toward equilibrium (the restoring direction).

A force that always acts to push or pull the object **back toward its equilibrium (resting) position**.

The central resting position where the object would sit still — where the displacement x = 0.

1. Acceleration **proportional to** displacement. 2. Acceleration directed **back toward equilibrium** (opposite to x).

A **straight line through the origin** with a **negative slope** equal to −ω².

There is a **restoring force** (and acceleration) directed **back to equilibrium** that is **proportional to the displacement** — exactly the condition a = −ω²x.

The steady loss of energy from an oscillation (to friction or drag), so each successive swing has a **smaller amplitude**.

The **amplitude slowly decreases** over many cycles while the **period stays almost the same**.

Yes — same form as a = −ω²x, so ω² = 25 → **ω = 5.0 rad s⁻¹**.

The **time for one complete oscillation** (one full cycle), measured in seconds.

The **number of oscillations per second**, measured in hertz (Hz). f = 1 ÷ T.

$f = \dfrac{1}{T}$ and $T = \dfrac{1}{f}$ — they are reciprocals. **Given** in the data booklet (T = 1/f).

$T = 2\pi\sqrt{\dfrac{m}{k}}$ — depends on the mass m and spring constant k. **Given**.

$T = 2\pi\sqrt{\dfrac{l}{g}}$ — depends on the length l and gravity g. **Given**.

**No** — mass does not appear in T = 2π√(l/g), so the period is unchanged.

**No** — g does not appear in T = 2π√(m/k); only the mass and stiffness matter.

**×√4 = ×2** — the period doubles, because T ∝ √l.

**×1/√2 ≈ 0.71** — a stiffer spring oscillates faster, so a shorter period (T ∝ 1/√k).

How fast the cycle turns (2π radians per cycle): **ω = 2πf = 2π ÷ T**. Unit: rad s⁻¹.

Both period formulas have a √, so a quantity ×4 inside the root comes out as ×√4 = ×2.

The spring's **stiffness** — a bigger k means a stiffer spring that pulls back harder and oscillates faster.

All three are **sinusoids** (smooth waves), but **shifted** in phase relative to one another.

Velocity **leads** displacement by a **quarter-cycle (90°)** — v is biggest at the centre, zero at the ends.

They are **antiphase (180° apart)** — a is the mirror image of x. This is the rule **a = -ω²x**.

At the **centre** (equilibrium, x = 0). It is **zero** at the turning points (maximum displacement).

At the **turning points** (maximum displacement). It is **zero** at the centre, because a = -ω²x.

The acceleration always points **back toward the equilibrium position** (a restoring acceleration), opposite to the displacement.

**T/4** — one quarter of the period (each quarter-cycle takes the same time).

**T/2** — half a period (two quarter-cycles, passing through the centre).

$a = -\omega^{2}x$ — acceleration proportional to displacement and directed back toward equilibrium.

$T = \dfrac{1}{f} = \dfrac{2\pi}{\omega}$ — both given in the data booklet.

Thinking it is greatest at the ends — it is greatest at the **centre** and **zero** at the ends.

**Four** quarter-periods, each lasting **T/4**.

**Kinetic energy** (of motion) and **potential energy** (stored, e.g. in a stretched spring). They swap back and forth as it oscillates.

It **stays constant** — KE and PE just trade places, but their sum never changes.

At the **centre** (equilibrium position), where the object moves **fastest**.

At the **ends** (the amplitude), where the object is **momentarily at rest**.

The **greatest displacement** from the centre — the turning point where the object briefly stops.

$E_{total} = \tfrac{1}{2}kA^{2}$ — set by the amplitude A. **Not** in the data booklet, so remember it.

At the amplitude the object is at rest (KE = 0), so all the energy is the elastic PE stored at the biggest stretch, ½kx² with x = A.

Set the **maximum KE equal to the total energy**: ½mv_{max}² = ½kA², then solve for v_{max}.

It becomes **four times larger**, because E_{total} = ½kA² depends on A² (the amplitude squared).

It **equals the total energy** — at the centre the PE is zero, so all the energy is kinetic.

**PE** is an upward parabola (min at the centre); **KE** is a downward parabola (max at the centre); their sum is a flat line.

Thinking the speed is greatest at the ends — it is greatest at the **centre**; at the ends the object is momentarily still.

$E_T = \tfrac{1}{2}m\omega^2 x_0^2$ — a **constant**, set by the amplitude. Unit: **J**.

$E_K = \tfrac{1}{2}m\omega^2(x_0^2 - x^2)$ — biggest at the centre. Unit: **J**.

$E_P = E_T - E_K = \tfrac{1}{2}m\omega^2 x^2$ — biggest at the extremes. Unit: **J**.

$E_T \propto x_0^2$ — **double the amplitude → four times** the energy.

$E_T \propto \omega^2$ — **double ω → four times** the energy.

At the **centre** (x = 0), where it equals the total energy E_T.

At the **extremes** (x = ±x₀), where it equals the total energy E_T.

The **total energy** E_T. Kinetic and potential just **interchange**: E_K + E_P = E_T always.

Subtract: $E_P = E_T - E_K$ — don't recompute from scratch.

Both formulas use **x²**, so x = +0.06 m and x = −0.06 m give the **same** energies.

$E_T = \tfrac{1}{2}(0.20)(25)(0.010) = 0.025$ J.

Use $E_K = \tfrac{1}{2}mv^2$ and solve for $v = \sqrt{2E_K/m}$.

A disturbance that carries **energy** from place to place **without** the medium itself travelling along with it.

The length of **one full wave** (e.g. crest to crest). Read it off a displacement-**distance** graph. Unit: metre (m).

The **maximum displacement** from the middle (rest) position — NOT crest to trough. Unit: metre (m).

The **time for one full wave** to pass a point. Read it off a displacement-**time** graph. Unit: second (s).

The **number of waves per second**. Unit: hertz (Hz), where 1 Hz = 1 per second.

$v = f\lambda$ — speed = frequency × wavelength. **Given** in the data booklet.

They are reciprocals: **f = 1/T** (and T = 1/f). Both are **given** in the data booklet.

$v = \dfrac{\lambda}{T}$ — since f = 1/T, speed = wavelength ÷ period.

Wavelength from a displacement-**distance** graph; period from a displacement-**time** graph. Always check the axis label.

v = f λ = 200 × 1.7 = 340 m s⁻¹.

v = λ ÷ T = 0.50 ÷ 0.0020 = 250 m s⁻¹.

Reading the wavelength off a **time** graph (or the period off a **distance** graph) — and forgetting to convert ms or kHz before substituting.

A wave in which the particles vibrate **perpendicular** (at right angles) to the direction the wave travels. Example: light.

A wave in which the particles vibrate **parallel** (back and forth along the same line) to the direction the wave travels. Example: sound.

Transverse: **light** (and a wave on a rope). Longitudinal: **sound** (and a push-pull on a spring).

**Crests** (highest points) and **troughs** (lowest points).

**Compressions** (particles bunched together, high pressure) and **rarefactions** (particles spread apart, low pressure).

**No** — the particles vibrate on the spot about their rest position; only the **energy** moves along.

The **amplitude** (peak displacement) and the **period T** (the repeat time). Then f = 1/T.

The **amplitude** and the **wavelength λ** (one full repeat along the distance axis).

Check the **x-axis**: distance → snapshot → read the **wavelength**; time → one particle → read the **period**.

The point copies the displacement of the point just **behind** it (the side the wave came from). Wave moving right → look just to the **left**.

$v = f\lambda = \dfrac{\lambda}{T}$ — **given** in the data booklet.

About **four amplitudes** (rest → top → rest → bottom → rest), so average particle speed ≈ 4 × amplitude ÷ T.

A **transverse** wave of vibrating electric and magnetic fields (light is one example). It needs **no medium** and travels through a vacuum.

They **all** travel at the same speed, **c = 3.00 × 10⁸ m s⁻¹** (the speed of light), whatever their region.

**Transverse** — the fields oscillate at right angles to the direction the wave travels.

**Radio → microwave → infrared → visible → ultraviolet → X-ray → gamma.** Wavelength falls, frequency and energy rise.

**Radio** — longest wavelength, lowest frequency, lowest energy. **Gamma** is the opposite end.

$c = f\lambda$ — speed of light = frequency × wavelength. Rearrange to $f = c/\lambda$ or $\lambda = c/f$. **Given** in the data booklet.

Sound **needs a medium** (it is mechanical); an EM wave **crosses a vacuum**. Also: sound is longitudinal, EM is transverse; EM is far faster.

EM wave: **electric and magnetic fields**. Sound wave: the **particles of the medium** (e.g. air molecules).

An **X-ray** (very short wavelength ⇒ very high frequency, about 10¹⁸ Hz).

Thinking different colours or regions travel at **different speeds** — in a vacuum they all travel at c.

The **speed of light c** — because $c = f\lambda$ rearranges to $f = c(1/\lambda)$, a straight line through the origin.

A line (or surface) joining all the points of a wave that are **in phase** — for example, all the crests.

A line showing the **direction in which the wave travels**, drawn at **right angles** to the wavefronts.

Two points are **in phase** if they are at the **same point in their cycle** at the same time (e.g. both at a crest).

The ray is always **perpendicular (90°)** to the wavefronts — the wave advances along the ray.

Exactly **one wavelength, λ** (crest to next crest).

Measure the gap between **two neighbouring wavefronts** — that distance is λ.

**Circles** (or spheres in 3D) spreading out from the source.

Straight, parallel lines — called **plane wavefronts**.

$v = f\lambda$ — wavelength λ (the spacing) × frequency f = wave speed v. **Given** in the data booklet.

λ = 0.50 m, f = 3.0 Hz, so v = fλ = 3.0 × 0.50 = 1.5 m s⁻¹.

Drawing it **along** a wavefront instead of **across** it — the ray must cross the wavefronts at 90°.

The **bending** of a wave as it crosses from one medium into another, caused by a **change in its speed** at the boundary.

How much the material **slows light down**: **n = c ÷ v**. A bigger n means slower light and more bending (an optically 'denser' medium).

**n₁ sinθ₁ = n₂ sinθ₂** — the indices and angles (measured from the normal) on each side of a boundary are linked this way. **Given** in the data booklet.

From the **normal** — the line drawn at 90° to the surface — not from the surface itself.

**Toward** the normal — the angle gets smaller.

**Away** from the normal — the angle gets bigger.

When light hitting a boundary is **completely reflected back** into the medium it started in, instead of refracting through. None of it escapes.

The angle of incidence at which the refraction angle is exactly **90°**. Above θc you get total internal reflection.

$\sin\theta_c = \dfrac{n_2}{n_1}$ — derived from Snell's law by setting θ₂ = 90°. n₁ is the denser medium.

1) Light going from a **denser to a less-dense** medium, and 2) an angle of incidence **above the critical angle**.

Rearrange **n = c ÷ v** to **v = c ÷ n**, using c = 3.0 × 10⁸ m s⁻¹.

Measuring the angle from the **surface** instead of from the **normal** — always use the dashed normal line.

Where two (or more) waves overlap, you **add their displacements** at every point to get the resultant.

Waves arrive **in step** (in phase) so their **amplitudes add**, giving a bigger wave (bright / loud).

Waves arrive **half a cycle out of step** (antiphase); equal **amplitudes cancel**, giving zero (dark / quiet).

The **extra distance** one wave travels compared with the other to reach a point, in metres.

**nλ** — a whole number of wavelengths (0, λ, 2λ, …). **Given** in the data booklet.

**(n + ½)λ** — a whole number of wavelengths plus a half. **Given** in the data booklet.

The sources keep a **constant phase difference** (same wavelength, fixed step). Needed for a steady, observable pattern.

So the constructive and destructive points **stay in fixed places**; a drifting phase difference would smear the pattern away.

**Zero.** 1.5λ = (1 + ½)λ is destructive, and equal amplitudes cancel completely.

**2A** — in step, so the amplitudes add.

Only when the two **amplitudes are equal**; otherwise just part of one wave cancels.

Divide by λ: a **whole number** → constructive (nλ); a whole number **+ ½** → destructive ((n+½)λ).

Light of one wavelength through **two close, coherent slits** overlaps on a screen to make a row of **equally spaced bright and dark fringes**.

One of the **bright or dark bands** on the screen in a double-slit (or similar interference) pattern.

The gap from one **bright** fringe to the **next** bright fringe — the same all the way across the screen.

$s = \dfrac{\lambda D}{d}$ — **given** in the data booklet (s spacing, λ wavelength, D slit-to-screen distance, d slit separation).

**s** fringe spacing, **λ** wavelength, **D** slits-to-screen distance, **d** slit separation — all in metres.

s gets **bigger** — d is on the bottom of s = λD/d, so closer slits give wider fringes.

s gets **bigger** — λ is on the top, so a longer wavelength widens the fringes.

They must give light of the **same wavelength** with a **fixed phase relationship**, so the pattern is stable instead of flickering.

About **θ ≈ λ/d** radians, from d sin θ = nλ with sin θ ≈ θ for small angles.

The energy 'missing' at the dark fringes is **redistributed into the bright fringes**; the total energy over the whole screen is unchanged.

**Mixing units** — convert every length to metres (mm = 10⁻³ m, nm = 10⁻⁹ m) before substituting into s = λD/d.

s = λD/d = (6.0×10⁻⁷ × 2.0) / (0.5×10⁻³) = 2.4×10⁻³ m = 2.4 mm.

The **spreading out** of a wave as it passes **through a gap** or **around an edge**.

When the **gap is about the same size as the wavelength** (gap ≈ λ).

Very **little** spreading — the wave carries almost straight on; only the edges curl in.

A **longer** wavelength — it is closer to the gap size, so it spreads more.

A **lower** frequency — lower frequency means a longer wavelength, which spreads more.

**All** of them — water, sound and light (every wave diffracts).

Sound's wavelength (~1 m) is about the size of a doorway (gap ≈ λ → strong diffraction); light's wavelength is far smaller, so it barely spreads.

The length of **one full wave** — for example from one crest to the next.

$v = f\lambda$ (given in the data booklet). Rearranged: $\lambda = \dfrac{v}{f}$.

Thinking a **higher** frequency spreads more — it's the opposite. Higher frequency → shorter λ → **less** diffraction.

At $\theta = \lambda/b$, where b is the slit width (small-angle approximation).

**Inversely** — narrower slit (smaller b) ⇒ **wider** central maximum, since $\theta = \lambda/b$.

**Directly** — longer wavelength (red) ⇒ wider spread, since $\theta = \lambda/b$.

$d\sin\theta = n\lambda$, with n = 0, 1, 2, … the order of the maximum.

Take the **reciprocal of the lines per metre**: $d = 1/(\text{lines per metre})$.

500 lines/mm = 500×10³ lines/m, so $d = 1/(500\times10^3) = 2.0\times10^{-6}$ m.

Many slits add in phase only at very precise angles, so each maximum is **narrow and bright**.

The whole number of wavelengths of path difference between **adjacent** slits ($d\sin\theta = n\lambda$).

The **grating** equation has sinθ ($d\sin\theta = n\lambda$); the single-slit small-angle result is just $\theta = \lambda/b$.

sinθ ≤ 1, so $n\lambda \le d$ — orders with $n > d/\lambda$ cannot exist.

**Twice** the first-minimum angle: a full width of about $2\lambda/b$.

Precisely **measuring wavelengths** (spectroscopy), because its sharp maxima pin down θ accurately.

The fixed pattern made when **two identical waves travel in opposite directions** and superpose — it does not move along.

When two waves overlap, you **add their displacements** at every point to get the total wave.

A point on a standing wave that **never moves** (zero displacement) — the two waves always cancel there.

A point on a standing wave that swings with the **largest amplitude**, halfway between two nodes.

**Half a wavelength (λ/2).** So λ = 2 × the node-to-node spacing. (Not in the data booklet — remember it.)

**No** — there is no net energy transfer along a standing wave; the energy stays stored in place.

They move **in phase** (all together). Points on opposite sides of a node move in **antiphase** (exactly opposite).

Standing: points are only ever **in phase or antiphase**. Travelling: the phase shifts **smoothly** from point to point.

A wave **reflects off a fixed end** and meets itself coming back — two identical opposite waves that superpose.

Microwaves reflect off the walls and form a **standing wave**; the field is strongest at the **antinodes**, so it melts there and stays solid at the nodes.

Spots are one antinode apart = λ/2, so λ = 2 × 0.060 = 0.12 m.

Thinking it **carries energy along** the string, or halving (instead of doubling) the node spacing to get the wavelength.

A point on a standing wave that **never moves** (zero amplitude).

A point on a standing wave that swings with the **largest** amplitude.

The **lowest** frequency at which a string or air column resonates — the standing-wave pattern with the fewest loops.

When a system is driven at one of its **natural frequencies** and vibrates with a large amplitude — that is what makes a harmonic loud.

**λ = 2L/n** for n = 1, 2, 3, … — n half-wavelengths fit into the length L.

**λ = 4L/n** with **n = 1, 3, 5, …** (odd harmonics only — node at the closed end, antinode at the open end).

**No** — you must memorise them. Only the wave equation v = fλ is given.

Its ends are different (node at the closed end, antinode at the open end), so only odd numbers of quarter-wavelengths fit: λ = 4L/n, n = 1, 3, 5, …

**Half a wavelength.** So λ = 2 × the node-to-node spacing.

Use the given wave equation **v = fλ**, rearranged to **f = v ÷ λ** (v is the wave speed — the speed of sound for a pipe).

λ = 2L = 1.3 m; f = v/λ = 260/1.3 = 200 Hz.

The spots (antinodes) are half a wavelength apart; double the spacing for λ, then f = c/λ.

The change in the **frequency (pitch) a listener hears** when the source (or listener) is **moving** — higher on approach, lower on recession.

**Higher** pitch — the wavefronts bunch up, shortening the wavelength and raising the frequency you hear.

**Lower** pitch — the wavefronts stretch out, lengthening the wavelength and lowering the frequency you hear.

**No** — the source always emits the same f. Only the **observed** frequency f' changes.

$f' = f\left(\dfrac{v}{v \pm v_{s}}\right)$ — **minus** approaching, **plus** receding. **Given** in the data booklet.

The **minus** sign — it makes the denominator smaller, so f' is **larger** (higher pitch).

The source moves forward between emitting each crest, so the **wavefronts ahead bunch together** → shorter wavelength → higher frequency.

**High and flat** (approaching) → a **sharp step down** (passing) → **low and flat** (receding). Not a smooth slope.

You hear it **above** its true pitch while it approaches, then a **sudden drop** to **below** its true pitch as it passes and recedes.

Drawing the heard pitch **sliding down smoothly**. It actually steps **down sharply** at the instant the source passes.

The change in the **wavelength (and frequency)** of light you receive when its **source moves toward or away** from you.

The observed wavelength is **stretched longer** (shifted toward red) because the source is **receding** (moving away).

The observed wavelength is **squashed shorter** (shifted toward blue) because the source is **approaching** (moving toward you).

$\dfrac{\Delta f}{f} = \dfrac{\Delta\lambda}{\lambda} \approx \dfrac{v}{c}$ — **given** in the data booklet (valid for v ≪ c).

Rearrange to **v = (Δλ ÷ λ) × c**, where Δλ = observed − laboratory wavelength and c = 3.0 × 10⁸ m s⁻¹.

The **change** in wavelength: observed wavelength − laboratory (true) wavelength — NOT the whole wavelength.

**Red = Receding** (away, longer λ); **Blue = approaching** (toward, shorter λ).

The **Universe is expanding**, so distant galaxies are **receding** from us — their light is shifted to longer (redder) wavelengths.

The **approaching edge is blueshifted** (shorter λ) and the **receding edge is redshifted** (longer λ) at the same time.

A **bigger** shift means a **faster** source — Δλ is proportional to v (Δλ/λ = v/c).

Only when the source speed **v is much smaller than c** (the speed of light).

Using the **whole observed wavelength** instead of the **change Δλ** (observed − lab) on the top of the fraction.

The **calculation** of the shift: the observed frequency f' (sound) or the fractional shift Δf/f (light), not just the qualitative 'higher/lower'.

$f' = f\,\dfrac{v}{v \mp u_s}$ — use **−** approaching, **+** receding.

The **minus** sign (smaller denominator) ⇒ bigger fraction ⇒ **higher** f'.

$\dfrac{\Delta f}{f} = \dfrac{\Delta\lambda}{\lambda} \approx \dfrac{v}{c}$.

The wavelength is **longer**, so the source is **receding** (moving away).

The wavelength is **shorter**, so the source is **approaching**.

Light needs **no medium**, so there is no medium speed v of sound; use Δλ/λ ≈ v/c instead.

Only when the source speed is **much less than c** ($v \ll c$); it is an approximation.

$f' = 512 \times \dfrac{340}{340-30} = 561$ Hz (higher).

$v = c \times 0.02 = 6 \times 10^{6}$ m s⁻¹ (about 0.02c).

Decide **physically**: approaching ⇒ higher f' (red/blue: blueshift = approaching). Then match the algebra to it.

Sound gives the **actual** f'; light gives the **fractional** shift Δλ/λ (multiply by λ or f for the change).

Every two masses attract each other with a force $F = G\dfrac{m_{1}m_{2}}{r^{2}}$ — proportional to each mass and to the inverse square of the distance r between their centres.

The gravitational **force per unit mass** on a small mass placed in the field: $g = \dfrac{F}{m}$. Unit: **N kg⁻¹**.

$g = G\dfrac{M}{r^{2}}$ — given in the data booklet. M is the source mass, r the distance from its centre.

**N kg⁻¹** — numerically the same as the free-fall acceleration in m s⁻².

Because F = mg and F = ma, so a = g. The falling mass cancels, so g is the acceleration — independent of the mass that falls.

**Inward**, towards the mass — gravity is always **attractive**.

g is divided by **3² = 9** (g is proportional to 1/r², the inverse-square law).

**No** (ignoring air resistance) — the acceleration g = GM/r² doesn't depend on the falling mass, so all masses fall equally fast.

The **gravitational constant**, G = 6.67 × 10⁻¹¹ N m² kg⁻² — the same everywhere in the universe.

About **9.8 N kg⁻¹** (or 9.8 m s⁻²) — found from g = GM/r² using Earth's mass and radius.

g is **inversely proportional to r²** (inverse-square): double r → quarter g; triple r → one-ninth g.

The **square** of a planet's orbital period is **proportional** to the **cube** of its orbital radius: $T^{2} \propto r^{3}$.

Each planet moves in an **ellipse** with the Sun at one **focus** of the ellipse.

A planet moves **faster when nearer the Sun** and **slower when farther away** (it sweeps out equal areas in equal times).

$T^{2} = \dfrac{4\pi^{2}r^{3}}{GM}$ — derived from $g = GM/r^{2}$ and $a = 4\pi^{2}r/T^{2}$. M is the mass being orbited.

Use $\dfrac{T_{A}^{2}}{r_{A}^{3}} = \dfrac{T_{B}^{2}}{r_{B}^{3}}$ — the constant $4\pi^{2}/(GM)$ cancels.