IB Physics HL — All Flashcards

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 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%**.

How much magnetic field threads through a loop: $\Phi = BA\cos\theta$. Unit: **weber (Wb)**.

The angle between **B** and the **normal** to the loop (not the surface). Square-on ⇒ θ = 0.

When the loop is **edge-on** to the field (θ = 90°, cos 90° = 0).

The induced emf equals the **rate of change** of flux linkage: $\varepsilon = -N\,\Delta\Phi/\Delta t$.

A **changing** flux. A steady flux — however strong — induces **no** emf.

An induced current flows so as to **oppose the change** in flux that produced it.

It is **Lenz's law** — the induced effect opposes the change. This is **conservation of energy**.

**Faraday** gives the **size** of the emf; **Lenz** gives its **direction**.

$\varepsilon = BvL$ — for a rod of length L moving at speed v perpendicular to field B.

$\varepsilon = BvL = 0.50\times3.0\times0.40 = 0.60$ V.

As it moves it **sweeps out new area**, so the flux through the circuit changes — that change induces the emf.

Apply **Lenz's law**: the current opposes the change in flux (it tries to keep the flux the same).

A **coil is spun** in a magnetic field. The changing flux induces a **sinusoidal emf** — alternating current (AC).

When the coil is **edge-on** to the field — the flux is changing **fastest** there.

$\varepsilon_0 = BAN\omega$ — increase any of **B**, **A**, **N** or **ω** to raise it.

**B** flux density, **A** coil area, **N** turns, **ω** angular frequency of rotation.

The **steady DC value** that delivers the **same average power** (same heating) as the AC.

$V_{rms} = \dfrac{V_0}{\sqrt{2}}$ and $I_{rms} = \dfrac{I_0}{\sqrt{2}}$ — divide the peak by √2 (≈ 1.41).

**Smaller** — rms = peak ÷ √2. The mains "230 V" is an **rms** value.

Changes an **AC voltage** up or down using two coils on a shared iron core.

$\dfrac{\varepsilon_p}{\varepsilon_s} = \dfrac{N_p}{N_s}$ — the **voltage ratio equals the turns ratio**.

**Step-up**: more secondary turns ⇒ higher V, lower I. **Step-down**: fewer secondary turns ⇒ lower V, higher I.

**Power**: $\varepsilon_p I_p = \varepsilon_s I_s$. That is why the **current ratio is inverted**.

$V_s = V_p \times \dfrac{N_s}{N_p}$ — multiply the primary voltage by the turns ratio.

A **quantum (packet) of light energy**, E = hf.

$E = hf$ (or $E = hc/\lambda$); h = 6.63×10⁻³⁴ J s.

Light ejecting **electrons** from a metal surface.

$E_{\max} = hf - \Phi$ (max electron KE = photon energy − work function).

The **minimum energy** needed to free an electron from the metal.

The lowest frequency that ejects electrons: $f_0 = \Phi/h$ (where E_max = 0).

**More** electrons ejected per second; their max KE is **unchanged**.

Max KE **increases** (E_max = hf − Φ).

One electron absorbs **one photon**; a sharp threshold can't be explained by a smooth wave.

**Diffraction** and **interference**.

The **photoelectric effect**.

Light (and matter) behaves as **both** a wave and a particle depending on the experiment.

The **wave** behaviour of a moving particle, with wavelength λ = h/p.

$\lambda = h/p$ (p = mv for a slow particle).

**Inversely** — bigger momentum gives a **shorter** wavelength.

**Electron diffraction** — electrons make diffraction patterns off crystals.

Its momentum is huge, so λ = h/p is far too small (~10⁻³⁴ m) to notice.

Find **p = mv** first, then **λ = h/p**.

$\Delta x\,\Delta p \ge h/4\pi$ — position and momentum can't both be exact.

**No** — it is a fundamental limit of nature, not a measurement fault.

Their λ (~10⁻¹⁰ m) matches the **atomic spacing**.

λ is **halved** (p doubles, λ = h/p).

**metres (m)**.

Particles like electrons show **both** particle and wave behaviour.