国際バカロレア · IB Diploma · Physics · SL

A.1 運動学の完全な網羅（SLとHL共有；A.1 に追加のHL内容はない）。取得した最初の評価-2025 ガイド（SHA 4884de32bd20、ガイド pp. 32–33）から拡張され、最初の評価 2025 が行われた。データブックレットおよびサンプル評価の要項は記録された保留状態を維持し、以下の方程式はすべてガイド本文内にある。

前提条件：指定された数量と単位を読み取り、算術を使用し、以下のモデル条件に従う。各授業は独立した転移前に独自の方法を発展させる。

これらは元々または明示的に架空の教学例であり、実際の測定値や完了した評価対象学習者調査ではない。

この観察を説明するのは何か？

• スピードメーターは特定の瞬間の速度を示すのみである。旅の平均速度は、停車や急激な速度変化を隠してしまうことがある。
• まず予測を立てる。比較したい量や特徴を明記し、2つの説明を区別できる証拠は何かを決定せよ。

モデルを構築する

• 変位には方向が含まれ、距離は全経路の長さを数える。速度は時間あたりの変位の増加量である。加速度は時間あたりの速度の増加量である。
• 速度: 変位の時間変化率; 加速度: 速度の時間変化率。

モデルを検証できる証拠を選ぶ

• 変位-時間グラフの傾きは速度である。速度-時間グラフ下の面積は変位を表す。一定加速度の公式は、その仮定が妥当な場合にのみ有効である。
• 正の方向を選び、それを明記する。反復可能な運動測定には、光ゲートまたは既知スケールとフレーム間隔を持つ映像を使用する。非常に短い間隔での手動タイミングは正確ではないと仮定しないよう注意する。

既知の値から始める

• 既知の値と単位を明記する。このケースに適応する仮定を持つ関係式を選び、代入前に変形する。
• 既知：速度は 2 から 10 m/s に 4 s で均一に上昇する。 a = (v-u)/t。 a = (10-2)/4 = 2 m/s²。変位は台形の面積： s = (u+v)t/2 = (2+10)×4/2 = 24 m。

例：

速度が秒3から秒15まで、秒6で変化する。加速度を求めよ。同じ手順を用いる：既知量 → モデル → 関係式 → 代入 → 単位と解釈。

結論とその限界を確認する

• 負の速度は選定された符号則に基づく方向を示すものであり、必ずしも減速を意味するわけではない。
• 元の観察に戻り、結果が何を裏付けているか、どのような条件を前提としているかを説明し、対立する説明を検証する方法を一つ示す。

警告：

負の速度は常に物体が減速していることを意味する。この主張は誤りである: 負の速度は選定された正負の慣例に基づく方向を示すのみであり、必ずしも減速を意味するわけではない。

一様等加速度運動の場合、ガイドには4つの方程式が示されている。これらすべてを速度–時間グラフから導出または検証し、次に自分が知っている情報に基づいて選択せよ：

計算例。 ボールを 15 m/s で真上に投げる（g = 9.8 m/s²、上向きを正とする）：最高点までの時間 $t = (0-15)/(-9.8) = 1.53$ s；$v^2 = u^2 + 2as$ からの最大高：$s = (0-225)/(2 \times (-9.8)) = 11.5$ m。

運動は 非一様 であることもあり、加速度が変化する場合、一定加速度の方程式を強制的に用いず、グラフを描くか積分を行う。

流体抵抗がない場合、水平運動と垂直運動は独立しており、水平方向は一定速度、垂直方向は一定加速度となる。射出速度を分解し、それぞれの軸に対して運動方程式を適用した後に結果を統合する。

** worked check.** 球を 12 m/s、水平面から 35° の角度で蹴り、水平地面から発射した（重力加速度 g = 9.8 m/s²）：$u_x = 12\cos 35° = 9.83$ m/s；$u_y = 12\sin 35° = 6.88$ m/s。 滞空時間 $t = 2u_y/g = 2(6.88)/9.8 = 1.40$ s；飛翔距離 $= u_x t = 9.83 \times 1.40 = 13.8$ m；最大高度 $= u_y^2/(2g) = 6.88^2/19.6 = 2.4$ m。

流体抵抗がある場合、軌道は放物線就不再であり、滞空時間、飛翔距離、終端速度の挙動がすべて変化する。物体が減速するにつれ経路は急になり、落下物体は抗力が重力と釣り合うと終端速度に近づく。ガイドでは定性的な効果のみを求めている。

重要：

運動グラフと加速度: 変位-時間グラフの傾きは速度を表す。速度-時間グラフの下の面積は変位を与える。等加速度の公式は、その仮定が妥当な場合にのみ有効である。

Full A.2 Forces and momentum coverage (SL and HL shared; A.2 has no additional HL content), extended from the acquired first-assessment-2025 guide (SHA 4884de32bd20, guide pp. 33–35). First assessment 2025. Quantitative collisions: one dimension at SL, two at HL. Data-booklet and sample-assessment demands keep their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A passenger continues moving when a vehicle brakes. The seat belt provides the force needed to change the passenger momentum.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Resultant force 合力 causes acceleration, not motion itself. Momentum is mass multiplied by velocity. For an isolated system, total momentum is conserved even when kinetic energy is not.
• momentum: Mass multiplied by velocity; resultant force: The vector sum of forces on an object.

Choose evidence that can test it

• Impulse equals momentum change. Increasing stopping time for the same momentum change reduces average force. Identify external forces before applying momentum conservation.
• Draw a free-body diagram containing only forces on the selected object. For spring measurements, add loads in steps within the elastic range and measure extension from the unloaded position.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a 60 kg passenger slows from 10 to 0 metres per second in 0.50 s. Δp = m(v-u) = 60(0-10) = -600 kg metres per second. Average F = Δp/Δt = -600/0.50 = -1,200 N. The sign shows the force opposes the initial motion.

Example:

A 2 kg trolley travels at 3 metres per second. Find its momentum. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Balanced forces do not require the object to be at rest. Mass and weight have different units and meanings.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

An object moving at constant velocity must have a nonzero resultant force. This claim is false: Balanced forces do not require the object to be at rest. Mass and weight have different units and meanings.

The guide names each force with a symbol and, where it applies, an equation. Contact forces 接触力 act where bodies touch: the normal force $F_N$ acts perpendicular to the surface; friction 摩擦力 $F_f$ acts parallel to the contact plane — static $F_f \le \mu_s F_N$ (up to a limit) and dynamic $F_f = \mu_d F_N$; tension 张力 pulls along a rope or string; an elastic restoring force follows Hooke's law $F_H = -kx$; viscous drag on a small sphere is $F_d = 6\pi\eta rv$; and buoyancy 浮力 is $F_b = \rho V g$ on the displaced fluid's volume. Field forces 场力 act without contact: weight $F_g = mg$, electric force $F_e$, magnetic force $F_m$.

Worked check (friction). A 2.0 kg block slides on a table ($\mu_d = 0.30$, g = 9.8 m/s²): $F_f = 0.30 \times 2.0 \times 9.8 = 5.9$ N. A 5 N horizontal pull cannot even start it if $\mu_s = 0.40$: the static limit is $0.40 \times 19.6 = 7.8$ N.

Free-body diagrams isolate one body and show every force on it; resolve to find the resultant. Newton's first law: zero resultant means constant velocity (translational equilibrium 平移平衡). Second law: $F = ma$ for constant mass, or more generally:

The impulse 冲量 $J$ delivered by a resultant force equals the momentum change it causes. Total momentum stays constant unless a resultant external force acts — so collisions and explosions conserve momentum in isolated systems.

Worked check. A 0.15 kg ball arrives at 12 m/s and leaves at 12 m/s the opposite way: $\Delta p = 0.15 \times 24 = 3.6$ kg·m/s; if contact lasts 0.020 s the average force is $3.6/0.020 = 180$ N. The guide treats quantitative collisions in one dimension at SL and two dimensions at HL; simultaneous energy-and-momentum equation pairs are not required.

Key:

Forces, momentum and safe stopping: Impulse equals momentum change. Increasing stopping time for the same momentum change reduces average force. Identify external forces before applying momentum conservation.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A motor can transfer some input energy to lifting and the rest to heating. Useful output is part of the total energy transfer.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Work done by a constant force parallel to displacement is force multiplied by distance. Kinetic energy depends on speed squared. Energy is conserved when all transfers and stores are included.
• power 功率: Energy transferred per unit time; efficiency: Useful output divided by total input.

Choose evidence that can test it

• Define the system and useful output before calculating efficiency. Doubling speed quadruples kinetic energy at constant mass. Power describes transfer per time, not total energy.
• Measure a lifting height and load, time the lift, and record electrical input with suitable instruments. Repeat trials and account for heating or friction as transfers, not missing energy.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a device receives 600 J and transfers 420 J usefully. Efficiency = useful output / total input. Efficiency = 420/600 = 0.70 = 70%. Over 3 s, useful power = useful energy/time = 420/3 = 140 W.

Example:

A device receives 250 J and gives 150 J useful output. Find efficiency as a percentage. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Efficiency cannot exceed 100% for a properly defined energy balance. Energy dissipated by heating is still conserved.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Dissipated energy is destroyed. This claim is false: Efficiency cannot exceed 100% for a properly defined energy balance. Energy dissipated by heating is still conserved.

Work done by a constant force depends on the component of the force along the displacement:

The net work done on a particle equals its change in kinetic energy. Mechanical energy is kinetic + gravitational potential + elastic potential; without frictional or resistive forces it is conserved, and work is the amount transformed between mechanical forms:

Worked check. A 2.0 kg crate is pulled 5.0 m by a 20 N rope at 30° above the horizontal: $W = 20 \times 5.0 \times \cos 30° = 86.6$ J. If it starts at rest on a frictionless floor, $E_k$ gained is 86.6 J, so $v = \sqrt{2 \times 86.6/2.0} = 9.3$ m/s.

Power is the rate of energy transfer, and efficiency compares output with input:

Worked check (power two ways). A car drives at a steady 25 m/s against 600 N of resistive force: $P = Fv = 600 \times 25 = 1.5 \times 10^4$ W. If the engine inputs 20 kW, $\eta = 15/20 = 0.75 = 75\%$.

The guide adds energy density of fuel sources as a comparing quantity (energy per unit mass or volume of fuel). The guidance point to carry into evaluation: interpret any change in total mechanical energy as the work done on the system by non-conservative forces — friction, drag, driving forces that are not stored-and-returnable.

Key:

Energy stores, work and efficiency: Define the system and useful output before calculating efficiency. Doubling speed quadruples kinetic energy at constant mass. Power describes transfer per time, not total energy.

Full B.1 Thermal energy transfers coverage (SL and HL shared; B.1 has no additional HL content), extended from the acquired first-assessment-2025 guide (SHA 4884de32bd20, guide pp. 39–41). Kelvin–Celsius conversion required. Data-booklet constants keep their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• Two equal masses receive the same energy but show different temperature rises. Material properties determine how energy transfer changes temperature.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Specific heat capacity 比热容 is energy needed to raise the temperature of unit mass by one degree. Specific latent heat 潜热 relates energy to change of state without temperature change for the idealized process.
• specific heat capacity: Energy per mass per temperature rise; latent heat: Energy associated with a change of state.

Choose evidence that can test it

• Temperature relates to particle motion in a model; internal energy includes kinetic and potential contributions. During a change of state, energy can change particle arrangements rather than temperature.
• Measure mass, electrical input and temperature change for an insulated block. Ensure the temperature sensor has good contact, allow time for equilibration, and consider energy transferred to the surroundings.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a 0.50 kg block gains 2,000 J and rises 10 °C. Use E = mcΔT. Rearranging gives c = E/(mΔT). c = 2,000/(0.50×10) = 400 J per kilogram per degree. Heat loss would make the value inferred from electrical input too large.

Example:

A 2 kg sample with c=500 J per kilogram per degree warms by 3 °C. Find energy gain. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• A flat section of a heating curve can show a phase change, not absence of energy transfer. Do not substitute temperature for a temperature difference in E = mcΔT.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A temperature plateau always means zero energy transfer. This claim is false: A flat section of a heating curve can show a phase change, not absence of energy transfer. Do not substitute temperature for a temperature difference in E = mcΔT.

For a monatomic ideal gas, absolute temperature is proportional to mean translational kinetic energy: $E_k = \tfrac32 k_B T$. Internal energy is the total intermolecular potential energy plus the total random kinetic energy of the molecules; a phase change changes particle behaviour at constant temperature. Heating quantities:

Worked check. Melting then warming 0.40 kg of ice at 0 °C to 30 °C ($c = 4186$, $L_f = 3.34\times10^5$ J/kg): $Q_1 = 0.40 \times 3.34\times10^5 = 1.34\times10^5$ J; $Q_2 = 0.40 \times 4186 \times 30 = 5.0\times10^4$ J; total $1.84\times10^5$ J.

Conduction transfers energy down a temperature gradient at a rate set by the material, area and gradient:

Worked check. A window ($k = 0.84$, area 1.5 m², thickness 4.0 mm) with a 20 K inside–outside difference: $\Delta Q/\Delta t = 0.84 \times 1.5 \times 20/0.0040 = 6.3\times10^3$ W.

Radiation from a black body follows the Stefan–Boltzmann law $L = \sigma A T^4$; the apparent brightness falls with distance as $b = L/(4\pi d^2)$, and Wien's displacement law $\lambda_\text{max}T = 2.9\times10^{-3}\ \mathrm{m\,K}$ locates the peak wavelength.

Worked check (Wien). The Sun's spectrum peaks at 580 nm: $T = 2.9\times10^{-3}/5.8\times10^{-7} = 5.0\times10^3$ K.

Worked check (Stefan–Boltzmann). A person (area 1.8 m², T = 303 K, $\sigma = 5.7\times10^{-8}$): $L = 5.7\times10^{-8} \times 1.8 \times 303^4 \approx 8.7\times10^2$ W radiated.

Emissivity compares a real surface with a black body at the same temperature: emissivity = (power radiated per unit area)$/\sigma T^4$. The guide's guidance notes: Kelvin–Celsius conversion is required, and $\Delta T$ is identical on both scales.

Key:

Thermal measurements and particle models: Temperature relates to particle motion in a model; internal energy includes kinetic and potential contributions. During a change of state, energy can change particle arrangements rather than temperature.

Full B.2 Greenhouse effect coverage (SL and HL shared; B.2 has no additional HL content), extended from the acquired first-assessment-2025 guide (SHA 4884de32bd20, guide pp. 41–42). Equilibrium-temperature estimation by energy balance (albedo, emissivity, solar constant) is a stated problem type.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• One cold day does not disprove a warming climate. Weather describes short-term conditions; climate describes distributions over longer times and regions.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• The Earth energy balance includes incoming solar radiation, reflection, absorption and outgoing infrared radiation. Greenhouse gases absorb and emit infrared radiation. Feedback can alter the response to an initial forcing.
• mitigation 减缓: Actions addressing causes of environmental change; adaptation 适应: Actions reducing harm from environmental impacts.

Choose evidence that can test it

• Distinguish mitigation, which addresses drivers, from adaptation, which reduces harm from impacts. A policy assessment needs evidence about effectiveness, cost, equity and uncertainty; one criterion is not the entire decision.
• Compare multi-year data using consistent baselines. State the region, timescale and uncertainty. At HL, connect a management decision to law, economics and ethics rather than treating these lenses as extra definitions only.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a surface receives 200 power units and reflects 50. Absorbed input = incoming-reflected = 200-50 = 150. Reflected fraction = 50/200 = 0.25 = 25%. A change in reflectivity alters the absorbed budget under this model.

Example:

Incoming energy is 240 units and reflected energy 72. Find reflected percentage. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• The greenhouse effect is not the same process as ozone depletion. A carbon footprint estimate depends on its system boundary.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

One local weather observation establishes a global climate trend. This claim is false: The greenhouse effect is not the same process as ozone depletion. A carbon footprint estimate depends on its system boundary.

Emissivity compares a surface's radiated power per unit area with a black body's $\sigma T^4$; albedo is the fraction of incident power scattered (reflected):

Earth's albedo varies daily with cloud and latitude. The solar constant S is the incoming intensity; averaged over the sphere, the mean incoming intensity is $S/4$ (a planet presents a disc $\pi R^2$ but spreads the power over $4\pi R^2$). Equilibrium comes from energy balance between incoming and outgoing radiation — the guide's stated problem type:

Worked check (equilibrium temperature). With $S = 1361\ \mathrm{W/m^2}$ and albedo 0.30, absorbed mean intensity $= 0.70 \times 1361/4 = 238\ \mathrm{W/m^2}$. Setting emission equal: $\sigma T^4 = 238$, so $T = (238/5.7\times10^{-8})^{1/4} = 255$ K — the effective radiating temperature; the surface is warmer because of the greenhouse effect.

The greenhouse gases — methane CH₄, water vapour, carbon dioxide CO₂ and nitrous oxide N₂O — have natural and human origins. Each absorbs infrared through transitions between molecular energy levels, then re-emits in all directions, returning part of the outgoing radiation to the surface. The effect can be described by a resonance model or by molecular energy levels; human enhancement of it is the enhanced greenhouse effect.

Key:

Climate evidence, energy budgets and policy: Distinguish mitigation, which addresses drivers, from adaptation, which reduces harm from impacts. A policy assessment needs evidence about effectiveness, cost, equity and uncertainty; one criterion is not the entire decision.

Full B.3 Gas laws coverage (SL and HL shared; B.3 has no additional HL content), extended from the acquired first-assessment-2025 guide (SHA 4884de32bd20, guide pp. 42–43). Gas laws limited to constant-V, constant-T, constant-P and the ideal gas law; pressure–volume diagrams included.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A sealed gas container changes pressure when heated. Celsius ratios cannot predict the pressure change because the gas model uses absolute temperature.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• In a gas model, particles move randomly and pressure results from momentum transfer at walls. The ideal gas 理想气体 equation connects pressure, volume, amount and absolute temperature.
• absolute temperature: Temperature on the kelvin scale; ideal gas: A gas model with specified simplifying assumptions.

Choose evidence that can test it

• At fixed amount and volume, pressure is proportional to kelvin temperature. At fixed temperature and amount, pressure is inversely proportional to volume. State which quantities are fixed before choosing a relationship.
• Use approved apparatus with a temperature range and pressure limit set by the teacher. Allow thermal equilibrium and record pressure against kelvin temperature. Never heat an improvised sealed vessel.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: pressure is 100 kPa at 300 K, with fixed volume and amount. At 330 K, p2/p1=T2/T1. p2=p1 T2/T1=100×330/300=110 kPa. A 30 °C rise is a 30 K change, but the temperature ratio must use kelvin.

Example:

At fixed volume, pressure is 120 kPa at 300 K. Find pressure at 350 K. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• An ideal gas is a model with conditions of validity. Celsius zero is not zero molecular motion, and internal energy is not determined by pressure alone.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A gas pressure ratio can always use Celsius temperatures. This claim is false: An ideal gas is a model with conditions of validity. Celsius zero is not zero molecular motion, and internal energy is not determined by pressure alone.

Pressure is force per area; the amount of substance counts molecules: $n = N/N_A$. The ideal gas is a modelled system that approximates a real gas; its law follows from the empirical gas laws:

Molecular collisions give pressure, relating it to the mean molecular speed: $P = \tfrac13 \rho \langle v^2\rangle$. The internal energy of an ideal monatomic gas is purely kinetic:

Worked check. A 2.0 mol sample at 300 K, 1.0 atm ($1.01\times10^5$ Pa): $V = nRT/P = 2.0 \times 8.31 \times 300/1.01\times10^5 = 4.9\times10^{-2}\ \mathrm{m^3}$; $U = \tfrac32 \times 2.0 \times 8.31 \times 300 = 7.5\times10^3$ J.

Changes of state can be drawn on pressure–volume diagrams (isobaric, isochoric, isothermal lines). An ideal gas approximates a real gas best at high temperature and low pressure (low density), far from condensation; near liquefaction the molecules' own volume and attractions matter.

Key:

Gas models and absolute temperature: At fixed amount and volume, pressure is proportional to kelvin temperature. At fixed temperature and amount, pressure is inversely proportional to volume. State which quantities are fixed before choosing a relationship.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A lamp becomes dimmer when another is added in series. Current and energy transfer depend on the whole circuit, not only one lamp.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Current is charge passing per time. Potential difference is energy transferred per charge. Resistance is potential difference divided by current for a stated operating point.
• current: Rate of flow of charge; potential difference: Energy transferred per unit charge.

Choose evidence that can test it

• Current is the same through components in series. Potential differences add around the series path. In parallel, branches share the same potential difference, while branch currents sum at a junction.
• Place an ammeter in series and a voltmeter in parallel. For an I-V investigation, change voltage in steps, reverse polarity when appropriate and limit current to reduce heating.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a resistor has 6.0 V across it and carries 0.30 A. Use R = V/I. R = 6.0/0.30 = 20 ohms. Power = VI. Power = 6.0×0.30 = 1.8 W. These quantities describe the same operating point.

Example:

A component has 12 V across it and carries 0.40 A. Find resistance. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Current is not used up by a lamp. A filament heats up, so its resistance need not remain constant as voltage changes.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A lamp uses up electric current as charge passes through it. This claim is false: Current is not used up by a lamp. A filament heats up, so its resistance need not remain constant as voltage changes.

A cell is a source of emf $\varepsilon$ — the energy it gives each coulomb of charge. Chemical cells convert stored chemical energy; solar cells convert radiant energy. Comparing sources is part of the guide: solar cells are renewable and need no fuel but work only in daylight and their manufacture costs energy; chemical cells are portable and reliable anywhere but store a finite amount of energy and become waste.

The full B.5 equation set:

where $r$ is the cell's internal resistance: the terminal pd is $V = \varepsilon - Ir = IR$, lower than the emf whenever current flows.

Resistivity $\rho$ is the material property behind a wire's resistance: $R = \rho L / A$, so long thin wires resist more, short thick ones less. Conduction is decided by the mobility of charge carriers, not their mere presence: a metal's free electrons move easily; an insulator's charges are locked in place.

Ohm's law: at constant temperature a metal conductor's I–V graph is a straight line through the origin — ohmic behaviour. A filament lamp heats as it conducts, so its curve bends upward in V — non-ohmic; a diode conducts one way only. Circuit symbols come from the Physics data booklet; an ideal ammeter (in series) has zero resistance, an ideal voltmeter (in parallel) infinite resistance.

Variable resistors: a thermistor has resistance that falls as temperature rises; an LDR has resistance that falls as light intensity rises; a potentiometer divides a pd through a sliding contact.

Worked check (emf and terminal pd). A cell of emf $1.5\ \mathrm{V}$ and internal resistance $0.50\ \Omega$ drives a $7.5\ \Omega$ resistor: $I = 1.5/(7.5 + 0.50) = 0.19\ \mathrm{A}$; terminal pd $= 1.5 - 0.19 \times 0.50 = 1.4\ \mathrm{V}$, which equals $IR = 0.19 \times 7.5 = 1.4\ \mathrm{V}$. The cell delivers $P = \varepsilon I = 0.28\ \mathrm{W}$, of which $I^2 r = 1.8\times10^{-2}\ \mathrm{W}$ heats the cell itself.

Worked check (resistivity). Copper has $\rho = 1.7\times10^{-8}\ \Omega\,\mathrm{m}$. A wire of length 1.5 m and cross-section $0.20\ \mathrm{mm^2} = 2.0\times10^{-7}\ \mathrm{m^2}$ has $R = \rho L/A = (1.7\times10^{-8} \times 1.5)/(2.0\times10^{-7}) = 0.13\ \Omega$.

Worked check (series–parallel). $6.0\ \Omega$ and $3.0\ \Omega$ in parallel: $1/R_p = 1/6.0 + 1/3.0 = 0.50$, so $R_p = 2.0\ \Omega$; in series they give $9.0\ \Omega$. Across a $6.0\ \mathrm{V}$ ideal battery the parallel pair draws $I = 6.0/2.0 = 3.0\ \mathrm{A}$, splitting into $1.0\ \mathrm{A}$ and $2.0\ \mathrm{A}$ — the smaller resistor carries the larger share.

Key:

Cells, resistivity and internal resistance: series shares one current and adds potential differences; parallel shares one potential difference and adds currents. The emf drives the whole loop resistance including the cell's own internal resistance, and the terminal pd falls below the emf by exactly $Ir$.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A swing gains a large amplitude when pushes arrive with the right timing. The forcing frequency and damping 阻尼 help determine the response.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Simple harmonic motion has acceleration proportional to displacement and directed toward equilibrium. Period is time per cycle; frequency is its reciprocal. Damping transfers energy away from the oscillating system.
• resonance: A large response to periodic forcing near a natural frequency; damping: Energy transfer out of an oscillating system.

Choose evidence that can test it

• Velocity is greatest near equilibrium for ideal SHM, while acceleration magnitude is greatest at extreme displacement. Resonance can occur near the natural frequency under periodic driving, with amplitude limited by damping.
• Measure time for several complete oscillations and divide. Define a cycle consistently and use a small displacement when the model requires it. Keep pendulum paths clear and record damping effects rather than assuming perfect motion.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: 20 cycles take 32 s. T=total time/cycles=32/20=1.6 s. Frequency=1/T=1/1.6=0.625 Hz. A shorter total timing interval would make reaction-time error a larger fraction of the measurement.

Example:

30 cycles take 45 s. Find period. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Not every repeated motion is SHM. A force proportional to displacement but directed away from equilibrium does not produce SHM.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Every periodic motion satisfies the condition for simple harmonic motion. This claim is false: Not every repeated motion is SHM. A force proportional to displacement but directed away from equilibrium does not produce SHM.

Simple harmonic motion (SHM) needs a restoring force proportional to displacement, directed back toward the equilibrium position. The defining equation folds that into acceleration:

The minus sign is the physics: acceleration always points against the displacement, so the particle is always pulled home. One oscillation is described by its amplitude $x_0$, period $T$, frequency $f$ and angular frequency $\omega$, linked by:

An ideal mass–spring system gives SHM; a simple pendulum approximates SHM for small angles. Their periods are:

Energy swaps between kinetic and potential through the cycle (qualitatively): total energy constant, kinetic energy greatest as the particle passes equilibrium at maximum speed, potential energy greatest at the extremes where the particle momentarily stops.

Worked check (mass–spring). A 0.50 kg mass on a spring of stiffness $20\ \mathrm{N/m}$: $T = 2\pi\sqrt{0.50/20} = 2\pi(0.158) = 0.99\ \mathrm{s}$, so $f \approx 1.0\ \mathrm{Hz}$ and $\omega = 2\pi/0.99 = 6.3\ \mathrm{rad/s}$.

Worked check (pendulum). A pendulum of length 1.0 m ($g = 9.8\ \mathrm{m/s^2}$): $T = 2\pi\sqrt{1.0/9.8} = 2.0\ \mathrm{s}$, $f = 0.50\ \mathrm{Hz}$.

Worked check (maximum acceleration). Amplitude $0.050\ \mathrm{m}$, period $0.50\ \mathrm{s}$: $\omega = 2\pi/0.50 = 12.6\ \mathrm{rad/s}$, and $a_{max} = \omega^2 x_0 = 158 \times 0.050 = 7.9\ \mathrm{m/s^2}$ — at the extremes, where displacement equals the amplitude.

Key:

Simple harmonic motion: the restoring condition $a = -\omega^2 x$ with $T = 1/f = 2\pi/\omega$; mass–spring $T = 2\pi\sqrt{m/k}$ and pendulum $T = 2\pi\sqrt{l/g}$; energy trades between kinetic (max at equilibrium) and potential (max at the extremes) while the total stays constant.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A wave can carry energy across water while a floating marker mainly oscillates. Energy transfer and bulk transfer of matter are different.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Frequency 频率 is oscillations per time; wavelength 波长 is distance between successive points in phase. Wave speed equals frequency multiplied by wavelength. Transverse oscillations are perpendicular to propagation.
• wavelength: Distance between successive points in phase; frequency: Number of oscillations per unit time.

Choose evidence that can test it

• At a boundary, frequency stays fixed by the source. A change of speed changes wavelength. Refraction follows from speed differences; angles are measured from the normal.
• Measure several wavelengths and divide to reduce fractional reading uncertainty. Use a ray box with a normal drawn at the boundary. Keep the beam away from eyes and record incident and refracted angles clearly.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: frequency 5 Hz and wavelength 0.40 m. Use v = fλ. v = 5×0.40 = 2.0 metres per second. If speed falls to 1.5 metres per second at the same frequency, wavelength = v/f = 1.5/5 = 0.30 m.

Example:

Frequency is 8 Hz and wavelength is 0.25 m. Find wave speed. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• The wave frequency does not change simply because the medium changes. A longitudinal wave has oscillations parallel to propagation.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Wave frequency always changes when a wave enters a new medium. This claim is false: the frequency is set by the source and stays fixed; it is the speed and the wavelength that change together.

A travelling wave carries energy through a medium by local disturbance, with no net displacement of the medium itself. Two geometries:

• Transverse wave — particle oscillations perpendicular to propagation (waves on a rope, all electromagnetic waves).
• Longitudinal wave — oscillations parallel to propagation, as compressions and rarefactions (sound).

Particle motion is described two ways, and the guide requires both: displacement against position along the wave at one instant (a snapshot of the wave profile) and displacement against time at one fixed point (how a single particle oscillates as the wave passes).

Sound is mechanical and longitudinal: it needs a medium, and its speed depends on that medium (about $340\ \mathrm{m/s}$ in air, faster in water, faster still in steel). Electromagnetic waves are transverse oscillations of electric and magnetic fields, perpendicular to each other and to the direction of travel; they need no medium and move through vacuum at $c = 3.0\times10^8\ \mathrm{m/s}$. That is the mechanical-versus-electromagnetic divide: matter versus field, medium versus vacuum.

The electromagnetic spectrum runs from radio through microwave, infrared, visible (about $400$–$700\ \mathrm{nm}$), ultraviolet and X-rays to gamma rays, each band an order of magnitude or more in wavelength — the data booklet lists the approximate values.

Worked check (sound). A $170\ \mathrm{Hz}$ note in air ($v = 340\ \mathrm{m/s}$): $\lambda = v/f = 340/170 = 2.0\ \mathrm{m}$, and the period $T = 1/f = 5.9\times10^{-3}\ \mathrm{s}$.

Worked check (radio wave). An FM station at $100\ \mathrm{MHz}$: $\lambda = c/f = (3.0\times10^8)/(1.0\times10^8) = 3.0\ \mathrm{m}$ — a radio wavelength.

Worked check (change of medium). A water wave of frequency $2.0\ \mathrm{Hz}$ passes from deep water ($v = 0.60\ \mathrm{m/s}$, $\lambda = 0.30\ \mathrm{m}$) to shallow ($v = 0.40\ \mathrm{m/s}$). Frequency is set by the source and unchanged, so the wavelength falls to $\lambda = v/f = 0.40/2.0 = 0.20\ \mathrm{m}$.

Key:

Travelling waves: transverse oscillate across the propagation direction, longitudinal along it; $v = f\lambda = \lambda/T$ with frequency fixed by the source when the medium changes; sound is mechanical and longitudinal, electromagnetic waves are transverse fields that cross a vacuum at $c$.

Full guide C.3 scope (Wave phenomena), first assessment 2025. Read directly from the acquired Physics guide, pages 47–48. Prerequisites: C.2 (v = fλ, transverse and longitudinal waves) and basic trigonometry.

These are original teaching examples, not actual measurements or completed assessed learner investigations.

In two and three dimensions a wave is drawn two ways: wavefronts — lines joining points in step (in phase), one wavelength apart — and rays — arrows perpendicular to the wavefronts showing the direction of travel. Sketches in this topic are limited to incident, reflected and transmitted waves.

At a boundary between media a wave does some or all of three things:

• Reflection — the wavefront returns from the boundary (angle of incidence = angle of reflection, both measured from the normal).
• Transmission — part of the wave passes into the second medium.
• Refraction — the transmitted wave changes direction because its speed changes: toward the normal entering a slower medium, away from it entering a faster one. Frequency is fixed by the source; wavelength changes with speed.

The refraction rule in ratio form:

where $n$ is the refractive index of each medium and $\theta$ the angle between the ray and the normal. Equivalently $n_1\sin\theta_1 = n_2\sin\theta_2$; the larger refractive index means the slower speed in that medium.

Travelling from a slower to a faster medium (glass to air, $n_1 > n_2$), there is an angle beyond which no ray can escape: at the critical angle the refracted ray runs along the boundary,

and beyond it all the light reflects totally back into the medium — total internal reflection (TIR), the principle of optical fibres.

Worked check (refraction into glass). Air to glass, $n_{air} = 1.0$, $n_{glass} = 1.5$, incidence $40°$: $\sin\theta_2 = (n_1/n_2)\sin\theta_1 = (1.0/1.5)(0.643) = 0.429$, so $\theta_2 = 25°$ — bent toward the normal, as expected entering a slower medium. The speeds follow the same ratio: $v_{glass} = c/1.5 = 2.0\times10^8\ \mathrm{m/s}$.

Worked check (critical angle). The same glass against air: $\sin\theta_c = 1.0/1.5 = 0.667$, so $\theta_c = 42°$. Any ray inside the glass hitting the wall at more than $42°$ to the normal is totally internally reflected.

A wave meeting an obstacle or an opening does not travel in a perfectly straight line: it bends and spreads — diffraction. The spreading is strongest when the aperture is comparable to or smaller than the wavelength, which is why sound bends around doorways (metre wavelengths) while light (hundreds of nanometres) appears to travel straight. A wavefront-ray diagram shows plane wavefronts entering a slit and leaving as curved, spreading fronts.

When two waves meet at a point, displacements add — superposition. For a steady pattern the sources must be coherent: same frequency and a constant phase difference. Then the outcome depends on the path difference from the two sources:

with $n$ an integer. Constructive interference builds a maximum (two crests arrive together); destructive cancels to a minimum (a crest meets a trough).

Worked check. Two coherent sources in phase, wavelength 0.50 m. A point 3.00 m from one source and 3.75 m from the other has path difference 0.75 m $= 1.5\lambda = (1 + \tfrac{1}{2})\lambda$ — destructive: a minimum. Moving the point until the path difference is 1.00 m $= 2\lambda$ restores a maximum.

Light through one narrow slit then two narrow slits (coherent by construction) produces bright and dark fringes on a screen — interference, the evidence that light is a wave. The fringe separation is

where $d$ is the slit separation and $D$ the slit-to-screen distance.

Worked check (measuring a wavelength). Slits 0.20 mm apart, screen 2.4 m away, and 12 bright fringes counted across 55 mm (11 gaps, so $s = 55/11 = 5.0\ \mathrm{mm}$): $\lambda = sd/D = (5.0\times10^{-3})(0.20\times10^{-3})/2.4 = 4.2\times10^{-7}\ \mathrm{m}$ — violet-blue light.

1. State what is meant by a wavefront and how a ray is related to it.
2. A ray crosses from air into water ($n = 1.3$) at $50°$ to the normal. Calculate the refraction angle.
3. State the condition for total internal reflection, and why a fish tank can mirror-image objects at glancing angles.
4. Compare the diffraction of sound and of light through a doorway, with a reason.
5. Two coherent sources 0.040 m apart emit 0.060 m waves in phase. A point lies 0.30 m from one source along a line perpendicular to the source line; state its path difference class (constructive or destructive) if it is 0.33 m from the other source.
6. In a double-slit set-up the slit separation is halved and the screen moved twice as far away. What happens to the fringe separation?
7. A double-slit with green light gives fringes 4.0 mm apart. Estimate the new separation with blue light of half the wavelength.

Answers: 1 a line of points in phase, one wavelength apart; the ray is everywhere perpendicular to the wavefronts and shows travel direction; 2 $\sin\theta_2 = (1.0/1.3)\sin 50° = 0.588$, $\theta_2 = 36°$; 3 travel from a slower (higher-$n$) medium toward a faster one at more than the critical angle; beyond it the surface mirrors totally; 4 sound (metre wavelengths, comparable with the doorway) diffracts strongly around the opening; light (hundreds of nanometres) spreads negligibly; 5 path difference 0.03 m $= \tfrac{1}{2}(0.060\ \mathrm{m})$ — destructive; 6 $s \propto D/d$: four times wider; 7 $s \propto \lambda$, so 2.0 mm.

Full guide C.4 scope (Standing waves and resonance), first assessment 2025. Read directly from the acquired Physics guide, pages 48–49. Prerequisites: C.1 (oscillation vocabulary) and C.3 (superposition).

These are original teaching examples, not actual measurements or completed assessed learner investigations.

Two identical waves travelling in opposite directions — usually a wave and its reflection — superpose into a standing wave: the pattern stays put while the medium oscillates inside it. Where the two always cancel there is a node (zero displacement); where they always reinforce there is an antinode (maximum amplitude). Two points between neighbouring nodes oscillate in phase; points on opposite sides of a node are in antiphase — unlike a travelling wave, the phase does not march along.

The ends decide which wavelengths fit. A fixed string end (or the closed end of a pipe) must be a displacement node; a free string end (or the open end of a pipe) must be an antinode. Pipes are described in displacement terms only (pressure nodes are not required, and no end corrections).

The lowest-frequency mode is the first harmonic (the terms "fundamental" and "overtone" are not used in this course). Determining the wavelength and frequency of the $n$th harmonic from the length and wave speed is a required skill.

Worked check (string). A string of length 0.60 m carries waves at 240 m/s. First harmonic: $\lambda_1 = 2L = 1.2\ \mathrm{m}$, $f_1 = v/2L = 240/1.2 = 200\ \mathrm{Hz}$; the second harmonic doubles to 400 Hz, the third to 600 Hz.

Worked check (closed–open pipe). A pipe closed at one end, open at the other, length 0.40 m, air at 340 m/s: $f_1 = v/4L = 340/1.6 = 213\ \mathrm{Hz}$; the next harmonic is the third, $f_3 = 3f_1 = 638\ \mathrm{Hz}$ — even harmonics cannot form.

Worked check (open–open pipe). The same pipe opened at both ends: $f_1 = v/2L = 340/0.80 = 425\ \mathrm{Hz}$, an octave above the closed–open first harmonic, and every harmonic is allowed.

Every elastic system has a natural frequency — the rate at which it oscillates when left alone. Drive it with a periodic force and the response depends on the driving frequency: when the driving frequency approaches the natural frequency the amplitude grows sharply — resonance. Energy arrives at exactly the rate the system can absorb it, so the amplitude at resonance is limited only by how fast energy dissipates.

Damping — the removal of oscillation energy — reshapes the response: it lowers the maximum amplitude, shifts the resonant peak slightly below the natural frequency, and broadens it.

• Light damping — a tall, sharp peak; large resonant amplitude (a swing pushed gently at its natural frequency).
• Critical damping — the oscillation dies away in the shortest time without passing the equilibrium point; the resonance peak has effectively vanished (car suspension).
• Heavy damping — slow, sluggish return; the system barely responds at any driving frequency (a heavy door closer).

Resonance is wanted and engineered (musical instruments, radio tuning, MRI) and unwanted or destructive (the Tacoma Narrows bridge, machinery shaking itself apart, soldiers breaking step on a bridge) — the guide asks for both sides.

1. State how a standing wave forms, and why nodes stay fixed while the medium between them moves.
2. Two points on a standing wave lie either side of a node, each 0.15 m from it. State their phase difference.
3. A string fixed at both ends is 0.75 m long and carries waves at 150 m/s. Calculate the first and second harmonic frequencies.
4. Explain why a closed–open pipe produces only odd harmonics.
5. A pipe of length 0.50 m is open at both ends; the air speed is 340 m/s. Calculate the first harmonic frequency and the wavelength of the second harmonic.
6. Sketch or describe how the resonance curve of a lightly damped system changes as damping increases.
7. Give one useful and one destructive effect of resonance, with the physical reason each happens.

Answers: 1 two identical waves travelling in opposite directions superpose; at a node their displacements always cancel, so it never moves, while between nodes the superposition oscillates; 2 antiphase — 180° (π rad) apart; 3 $f_1 = v/2L = 150/1.5 = 100\ \mathrm{Hz}$, $f_2 = 200\ \mathrm{Hz}$; 4 the closed end must be a node and the open end an antinode, so only an odd number of quarter-wavelengths fits — even harmonics cannot meet both conditions; 5 $f_1 = v/2L = 340/1.0 = 340\ \mathrm{Hz}$; $\lambda_2 = L = 0.50\ \mathrm{m}$; 6 the peak falls, broadens, and shifts slightly below the natural frequency; 7 useful — a radio circuit tuned to resonance picks one station out of many because it responds only near its natural frequency; destructive — driving at the natural frequency feeds energy in faster than it dissipates, so the amplitude grows until the structure fails.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A recognizable spectral line appears at a longer wavelength in a distant source. Its shift can provide evidence of recession along the line of sight.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Relative source-observer motion changes observed frequency or wavelength. For light at a relative speed much smaller than c, the magnitude of the fractional wavelength shift is approximately v/c. Recession corresponds to longer wavelength, and approach to shorter wavelength.
• redshift 红移: Shift of an identified line toward longer wavelength; rest wavelength 静止波长: Wavelength of the line measured with no relative source motion.

Choose evidence that can test it

• Compare a identified line with its laboratory rest wavelength. Use z = (observed−rest)/rest, then v approximately cz in the stated low-speed model. Do not apply this approximation without checking the regime or treating a cosmological redshift as a simple exact velocity.
• Use attributed spectra with calibration and line identification. Compare several lines for a consistent shift. State uncertainty and distinguish relative line-of-sight motion from an unmeasured transverse component.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a 500 nm rest line is observed at 501 nm. z = (501−500)/500 = 0.002. With c=3.0×10⁸ m/s, v approximately 6.0×10⁵ m/s away. The sound-source denominator formula is not the required SL light approximation.

Example:

A 600 nm rest line is observed at 601.2 nm. Find z=(observed−rest)/rest. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• The approximate shift relation does not show that light travels faster from an approaching source. A single unidentified line cannot safely establish a redshift.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

An observed redshift means light speed increased. This claim is false: The approximate shift relation does not show that light travels faster from an approaching source. A single unidentified line cannot safely establish a redshift.

Doppler effect: relative motion between a wave source and an observer changes the observed frequency (and wavelength) while the source keeps emitting steadily. In a wavefront diagram the moving source chases the circles it emits: wavefronts bunch ahead of it and stretch behind it, so an observer ahead receives more fronts per second (higher frequency) and one behind receives fewer.

For sound, the effect is qualitative at this level: pitch rises as a siren approaches and falls as it passes and recedes — the classic slide as an ambulance goes by. A moving observer meets the same wavefronts at a different rate and hears the same kind of shift.

For light, when the relative speed $v$ is much smaller than $c$, the fractional shift is

Approaching compresses the light (shorter wavelength — blueshift); receding stretches it (redshift). Because each element emits and absorbs at known rest wavelengths, a spectrum taken from a star or galaxy can be matched line by line: the whole pattern shifted long means the source recedes, and the shift size gives the radial speed.

Worked check (a galaxy line). A hydrogen line with rest wavelength 656 nm is observed at 659 nm: $\Delta\lambda/\lambda = 3/656 = 4.6\times10^{-3}$, so $v = 0.0046c = 1.4\times10^6\ \mathrm{m/s}$ away from us — a redshift, hence recession.

The effect is put to work everywhere: radar speed guns time the frequency shift of reflected radio waves, medical Doppler ultrasound measures blood-flow speed from the shift of echoes, and astronomy reads galaxy redshifts — the larger the shift, the faster the recession.

Key:

Doppler effect: wavefronts bunch ahead of a moving source and stretch behind it, so approach raises the observed frequency and recession lowers it; for light at v much less than c the fractional shift is v/c, read from spectral lines to find the radial speeds of stars and galaxies, with radar and medical ultrasound as everyday applications.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• An orbiting satellite is continuously falling while moving sideways. Being in orbit does not require gravity to vanish.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Gravitational field 引力场 strength is force per unit mass. A gravitational force can provide the centripetal force 向心力 for a circular orbit. Field and potential describe different quantities.
• gravitational field: A description of gravitational force per unit mass; centripetal force: Net force toward the centre of a curved path.

Choose evidence that can test it

• For a point mass or outside a spherical mass, field strength follows an inverse-square distance dependence. Use distance from the centre, not height above the surface alone.
• State the circular-orbit approximation and ignore atmospheric drag only when justified. Draw the force toward the central body and velocity tangential to the orbit. Do not add an outward force merely because the path is circular.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: field strength is 8 units at distance r from a spherical source. At 2r, g2/g1=(r/2r)²=1/4. g2=8/4=2 units. The gravitational force on a fixed test mass decreases in the same ratio.

Example:

Field strength is 18 units at r. Find it at 3r for an inverse-square field. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Weightlessness in an orbiting spacecraft means apparent weight is small in free fall, not that there is no gravitational field.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Gravity is zero everywhere a spacecraft experiences apparent weightlessness. This claim is false: Weightlessness in an orbiting spacecraft means apparent weight is small in free fall, not that there is no gravitational field.

Newton's universal law of gravitation treats masses as points:

An extended body acts as a point mass when it is spherically symmetric (uniform density spheres behave as if all mass sat at the centre) or when the separation hugely exceeds its size. The gravitational field strength at a point is the force per unit mass on a small test mass there:

Gravitational field lines point inward toward the mass, never cross, and crowd together where the field is stronger: a radial pattern around a sphere, near-parallel lines close to a large surface. Where two bodies act, consider resultant-field calculations at points on the line joining them — the individual fields add as vectors, and somewhere between the bodies they cancel exactly.

Kepler's three laws of orbital motion:

1. Orbits are ellipses with the central body at one focus (a circle is the special case used in calculations).
2. A line from the central body to the orbiting body sweeps equal areas in equal times — faster when closer, slower when farther.
3. $T^2 \propto r^3$: the square of the period scales with the cube of the orbital radius.

Worked check (surface gravity). Earth: $M = 6.0\times10^{24}\ \mathrm{kg}$, $r = 6.4\times10^6\ \mathrm{m}$: $g = GM/r^2 = (6.67\times10^{-11} \times 6.0\times10^{24})/(6.4\times10^6)^2 = 9.8\ \mathrm{m/s^2}$. At twice that distance from the centre the inverse square gives $g = 9.8/4 = 2.5\ \mathrm{m/s^2}$.

Worked check (Kepler's third law). Mars orbits at 1.52 AU: $T = 1.52^{3/2} = 1.87$ Earth years — cube the radius ratio, take the square root.

Key:

Kepler and Newton: every pair of masses pulls with $F = Gm_1m_2/r^2$, giving $g = GM/r^2$ around a point or spherical mass; field lines run inward and crowd where the field strengthens; Kepler's laws govern the orbits — ellipses, equal areas in equal times, and $T^2 \propto r^3$.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A negatively charged particle moves opposite to the direction of the electric field. Field direction is defined using a positive test charge.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Electric field strength 电场强度 is force per unit positive test charge. Like charges repel and unlike charges attract. Field lines point away from positive charges and toward negative charges. Closely spaced lines represent a stronger field within a consistently drawn diagram.
• electric field strength: Electric force per unit positive test charge; test charge: A small charge used to probe an electric field.

Choose evidence that can test it

• In a uniform field between ideal parallel plates, E = V/d away from edge effects. The force is F = qE, so its direction reverses for a negative charge. Coulomb force between ideal point charges decreases with separation squared.
• Sketch labelled plate polarities, field arrows and the charge before calculating. Convert separation to metres. In school, use simulations or approved low-voltage electrostatic models rather than exposed high-voltage equipment.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: plate potential difference is 120 V and separation is 0.020 m. E = V/d = 6,000 V/m. A +2.0 microcoulomb charge experiences F = qE = 0.012 N along the field; a −2.0 microcoulomb charge experiences the same magnitude opposite to it.

Example:

Use E = V/d. Find E for 100 V across 0.025 m. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Field strength is a vector. Potential difference is energy per charge and is not the same quantity. A field-line drawing is a model, not a set of physical threads.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Every charged particle experiences a force along the electric field direction. This claim is false: only a positive charge is pushed along the field — a negative charge feels the force opposite to the field direction. A field-line drawing is a model, not a set of physical threads.

Coulomb's law mirrors gravitation with charge in place of mass — but with two signs:

Like charges repel, unlike attract. Charge is conserved — it moves, never appears or vanishes — and quantized: Millikan's oil-drop experiment (weight balanced by an electric force, $qE = mg$) found every drop's charge a whole-number multiple of $e = 1.6\times10^{-19}\ \mathrm{C}$, the electron charge.

Charge moves between bodies by friction (rubbing transfers electrons), by contact (shared on touching), and by induction — a charged body brought near a conductor separates its charges, and grounding (earthing) lets one sign drain away so the conductor is left with the opposite sign, without the charged body ever touching it.

The electric field strength is force per unit positive test charge, $E = F/q$, with field lines running from positive to negative, never crossing, denser where the field is stronger: radial around a point charge and outside a sphere, straight and equally spaced between opposite parallel plates where

Magnetic field lines close on themselves (no monopoles): a bar magnet runs N-out S-in; a straight wire carries concentric circles whose direction follows the right-hand grip rule (thumb along the current); a circular coil loops through its centre; a solenoid's interior is near-uniform, its outside like a bar magnet's.

Worked check (Coulomb). Two $+1.0\ \mu\mathrm{C}$ charges $0.050\ \mathrm{m}$ apart: $F = (9.0\times10^9 \times 1.0\times10^{-12})/(0.050)^2 = 3.6\ \mathrm{N}$ — repulsion, falling to one quarter in magnitude per doubling of separation.

Worked check (parallel plates). $12\ \mathrm{V}$ across $4.0\ \mathrm{mm}$: $E = V/d = 3000\ \mathrm{V/m}$, and the force on an electron between them is $F = eE = 4.8\times10^{-16}\ \mathrm{N}$.

Worked check (Millikan). A drop of mass $3.2\times10^{-15}\ \mathrm{kg}$ carrying three excess electrons floats when $qE = mg$: $E = mg/q = (3.2\times10^{-15} \times 9.8)/(4.8\times10^{-19}) = 6.5\times10^4\ \mathrm{V/m}$ — the whole-number charge pattern across many drops is the quantization evidence.

Key:

Electric and magnetic fields: Coulomb's $F = kq_1q_2/r^2$ with conserved, quantized charge (Millikan); $E = F/q$ read from field-line density, uniform $E = V/d$ between plates; charge moves by friction, contact or induction with grounding; magnetic field lines close on themselves — wire circles by the right-hand rule, solenoid like a bar magnet.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A charged particle curves in a uniform magnetic field. Its direction changes although the magnetic force 磁力 alone does not increase its kinetic energy.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• The magnetic force magnitude is F = |q|vB sinθ. It is perpendicular to velocity and magnetic field, with direction determined by charge sign. For perpendicular entry into a uniform field, it can provide the centripetal force for circular motion.
• magnetic force: Force due to a magnetic field on a moving charge or current; momentum 动量: Mass multiplied by velocity in the classical model.

Choose evidence that can test it

• Equate magnetic force with mv²/r to obtain r = mv/(|q|B) for perpendicular motion. A stronger field makes a smaller radius at fixed momentum. Parallel entry gives zero magnetic force in this model.
• Draw velocity, field and force as separate arrows using stated into/out-of-page conventions. Use supplied beam data or a simulation. Do not treat a current-carrying wire direction as identical to electron-motion direction.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a particle has m = 2.0×10⁻²⁷ kg, v = 3.0×10⁶ m/s, |q| = 1.0×10⁻¹⁹ C and B = 0.20 T. r = mv/(|q|B) = 0.30 m. The calculation assumes non-relativistic motion and perpendicular entry.

Example:

Use r = mv/(|q|B). For m=2×10⁻²⁷ kg, v=2×10⁶ m/s, |q|=1×10⁻¹⁹ C and B=0.20 T, find r. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• A magnetic force can change momentum direction without doing work. Adding an electric field can change the energy, so conclusions about a magnetic field alone do not cover every combined-field apparatus.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A magnetic force alone continuously increases the kinetic energy of a charged particle. This claim is false: A magnetic force can change momentum direction without doing work. Adding an electric field can change the energy, so conclusions about a magnetic field alone do not cover every combined-field apparatus.

Three motions cover this topic. In a uniform electric field the force $qE$ is constant, so a charge shot across the plates follows a parabola — accelerated along the field lines, unchanged across them. In a uniform magnetic field the force

acts perpendicular to velocity, so it bends the path into a circle without changing speed — kinetic energy stays constant. The radius comes from circular motion:

so a measured path radius yields the charge-to-mass ratio $q/m = v/(rB)$ — how the nature of a particle is read from its track. With perpendicularly crossed fields ($E$ and $B$), the electric and magnetic forces can cancel for one particular speed, $qE = qvB$:

a velocity selector: only that speed passes straight.

Currents feel the field too:

Parallel wires carrying currents the same way attract; opposite ways repel.

Worked check (proton circle). A proton at $v = 2.0\times10^6\ \mathrm{m/s}$ in $B = 0.50\ \mathrm{T}$: $r = mv/qB = (1.67\times10^{-27} \times 2.0\times10^6)/(1.6\times10^{-19} \times 0.50) = 4.2\times10^{-2}\ \mathrm{m}$. Reading the same radius backward gives $q/m = v/(rB) = 9.5\times10^7\ \mathrm{C/kg}$ — the proton's signature.

Worked check (velocity selector). $E = 1.5\times10^4\ \mathrm{V/m}$ crossed with $B = 0.050\ \mathrm{T}$: only $v = E/B = 3.0\times10^5\ \mathrm{m/s}$ goes straight — faster particles bend magnetic-dominant, slower ones electric-dominant.

Worked check (wires). A $0.30\ \mathrm{m}$ wire carrying $5.0\ \mathrm{A}$ perpendicular to $B = 0.20\ \mathrm{T}$ feels $F = BIL = 0.30\ \mathrm{N}$. Two long wires $0.050\ \mathrm{m}$ apart carrying $10\ \mathrm{A}$ each the same way: $F/L = \mu_0 I_1I_2/2\pi r = 4.0\times10^{-4}\ \mathrm{N/m}$, attraction.

Key:

Charged motion and currents: a uniform electric field gives a parabola, a magnetic field a circle at constant kinetic energy with $r = mv/qB$ — the path radius measures $q/m$ — and crossed fields select $v = E/B$; a wire carries $F = BIL\sin\theta$, and parallel wires attract when their currents run the same way with $F/L = \mu_0I_1I_2/2\pi r$.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• An excited gas produces separate coloured lines rather than every wavelength. The pattern is evidence for discrete atomic energy differences.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Atoms have a small positive nucleus and electrons. Emission and absorption spectra arise from transitions between discrete energy levels. A photon energy equals the level difference and obeys E = hf.
• emission spectrum 发射光谱: Wavelength pattern of radiation emitted by a source; energy level 能级: An allowed energy state in a model.

Choose evidence that can test it

• An emitted photon corresponds to a transition to a lower energy level. Absorption requires a compatible energy difference. Rutherford scattering supported a small dense nucleus, but that experiment alone did not establish the complete quantum model.
• Read a labelled energy-level diagram before calculating. Keep joules and electronvolts distinct and use the given constants. Compare attributed spectra at a common wavelength scale and avoid looking at unsafe light sources.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: energy levels are −6.0 and −2.0 eV. A downward transition releases 4.0 eV = 6.4×10⁻¹⁹ J using 1 eV = 1.6×10⁻¹⁹ J. With h = 6.4×10⁻³⁴ J s for this rounded exercise, frequency is 1.0×10¹⁵ Hz.

Example:

Levels are −7.0 and −2.0 eV. Find the energy emitted in the downward transition. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Negative bound-state energies are relative to a chosen zero; they do not mean a negative photon energy is emitted. A larger downward energy difference gives higher frequency.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A negative bound-state energy means an emitted photon has negative energy. This claim is false: Negative bound-state energies are relative to a chosen zero; they do not mean a negative photon energy is emitted. A larger downward energy difference gives higher frequency.

The Geiger–Marsden–Rutherford experiment fired alpha particles at thin gold foil. Most passed straight through — the atom is mostly empty space. A few deflected through large angles, and rare ones bounced almost straight back — impossible unless nearly all the positive charge and mass sit in a tiny, dense nucleus. The observation forced the plum-pudding model out and the nuclear model in.

Atoms are written in nuclear notation $^{\,A}_{Z}X$: $A$ the nucleon number (protons + neutrons), $Z$ the proton number, $X$ the chemical symbol.

Electrons in an atom occupy discrete energy levels. A photon is emitted when an electron drops to a lower level, absorbed when one climbs:

with $E$ the energy difference between the two levels. The photon frequencies an element can emit or absorb are therefore a fingerprint — its emission and absorption spectra — and each line corresponds to one specific transition. Matching a spectrum against known patterns reveals chemical composition, including in stars whose light no probe will ever reach.

Worked check (a transition). An atom has levels at $-1.5\ \mathrm{eV}$ and $-3.4\ \mathrm{eV}$. A downward jump releases $E = 1.9\ \mathrm{eV} = 3.0\times10^{-19}\ \mathrm{J}$, so $f = E/h = (3.0\times10^{-19})/(6.6\times10^{-34}) = 4.6\times10^{14}\ \mathrm{Hz}$ — red light. This pair of levels absorbs the same frequency on the upward transition; other allowed level pairs can absorb other frequencies: the line pair in emission and absorption match.

Key:

The nuclear atom: alpha scattering revealed a tiny dense nucleus; $^{A}_{Z}X$ counts nucleons and protons; electrons sit in discrete levels, emitting or absorbing photons with $E = hf$ set by the level gap — and the line pattern of those transitions identifies each element, near or astronomically far.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A detector records counts even when the classroom source is removed. Background must be considered before attributing every count to the source.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Radioactive decay is spontaneous and random for an individual nucleus. Half-life 半衰期 describes the time for half the undecayed nuclei in a large population to decay, or for background-corrected activity to halve.
• half-life: Time for activity or undecayed population to halve; background radiation 本底辐射: Radiation measured apart from the investigated source.

Choose evidence that can test it

• Subtract background counts measured over the same time interval. Distinguish irradiation from contamination. Shielding, distance and reduced exposure time can reduce risk under a school-controlled procedure.
• Use teacher-managed sources and the school radiation rules. Record count duration and repeat background measurements. Do not extrapolate a half-life from one nucleus or from uncorrected readings.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: measured count rate falls from 100 to 55 counts per minute; background is 10. Source rates are 90 and 45, so one half-life has elapsed. Without subtraction, 55/100 would obscure the exact half in this example.

Example:

An initial source count is 160 per minute. Find it after three half-lives. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• An irradiated object is not automatically radioactive. A half-life does not predict the exact decay time of one nucleus.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Half-life predicts the exact decay time of an individual nucleus. This claim is false: An irradiated object is not automatically radioactive. A half-life does not predict the exact decay time of one nucleus.

Isotopes share proton number but differ in neutron count. A nucleus weighs less than its parts: the mass defect $\Delta m$ is the missing mass, stored as binding energy through

(masses may come in kg, MeV/c² or atomic mass units, $1\ u = 931.5\ \mathrm{MeV/c^2}$). The binding energy per nucleon curve climbs steeply to a peak near $A = 56$ (iron) and falls gently afterward: fusing light nuclei or fissioning heavy ones both release energy — the curve's interpretation is required knowledge.

What holds nuclei together against proton repulsion is the strong nuclear force: attractive, felt only between nucleons, and short-range — it fades beyond a couple of femtometres. Radioactive decay is random and spontaneous: no external trigger, no way to say which nucleus goes next.

Each decay mode changes the nucleus its own way — write the equations conserving $A$ and $Z$:

• $\alpha$: $^{A}_{Z}X \to\ ^{A-4}_{Z-2}Y + ^{4}_{2}\alpha$ (helium nucleus)
• $\beta^-$: $^{A}_{Z}X \to\ ^{A}_{Z+1}Y + ^{0}_{-1}e + \bar{\nu}$ (neutron $\to$ proton, electron, antineutrino)
• $\beta^+$: $^{A}_{Z}X \to\ ^{A}_{Z-1}Y + ^{0}_{+1}e + \nu$ (proton $\to$ neutron, positron, neutrino)
• $\gamma$: no change to $A$ or $Z$ — a photon carries away excess energy.

The three radiations differ in penetration and ionizing power: alpha (paper stops it; most ionizing), beta (a few mm of aluminium), gamma (centimetres of lead; least ionizing). The activity $A$ (becquerels) counts decays per second; a detector reads a count rate from which background radiation must be subtracted. After each half-life half the remaining nuclei decay, so after $n$ half-lives $N = N_0/2^n$.

Worked check (decay equations). Radium: $^{226}_{88}Ra \to ^{222}_{86}Rn + ^{4}_{2}\alpha$. Carbon: $^{14}_{6}C \to ^{14}_{7}N + ^{0}_{-1}e + \bar{\nu}$ — both rows conserve $A$ and $Z$ exactly.

Worked check (binding energy). Using atomic masses consistently, two hydrogen-1 atoms and two neutrons total $4.03298\ u$; a helium-4 atom weighs $4.00260\ u$. The electrons cancel, so the nuclear mass defect is $0.03038\ u$ and binding energy is $0.03038\times931.5=28.3\ \mathrm{MeV}$, or $7.07\ \mathrm{MeV}$ per nucleon.

Worked check (integer half-lives). A sample reads 5200 counts/min against a background of 200. Source alone: 5000. After three half-lives the source contributes $5000/8 = 625$, and the detector reads $625 + 200 = 825$ counts/min.

Key:

Radioactivity: mass defect binds the nucleus ($E = \Delta mc^2$, peak binding per nucleon at iron); the short-range strong force beats proton repulsion; decay is random and spontaneous with $\alpha$, $\beta^\pm$ and $\gamma$ equations conserving $A$ and $Z$; penetration falls and ionizing power rises in the order $\gamma$, $\beta$, $\alpha$; activity halves each half-life once background is removed.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A neutron can trigger a heavy nucleus to split, releasing energy and further neutrons. Whether the process becomes a chain reaction 链式反应 depends on what happens to those neutrons.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• In nuclear fission a heavy nucleus splits into lighter nuclei. Proton number and nucleon number balance in a nuclear equation. The total rest mass can decrease; the difference corresponds to released energy through E = Δmc².
• fission: Splitting of a heavy nucleus into lighter nuclei; chain reaction: A sequence in which products initiate further events.

Choose evidence that can test it

• Separate conservation of nucleon number from conservation of total energy. Released neutrons may initiate further fissions, escape or be absorbed. A controlled reactor and an uncontrolled chain reaction have different neutron-management conditions.
• Balance a supplied nuclear equation, use the specified mass data and identify the system. Analyse models or published reactor data; this is not a school attempt to produce fission or handle reactor materials.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: U-235 absorbs one neutron, forming Ba-141, Kr-92 and three neutrons. Nucleon balance: 235+1 = 141+92+3 = 236. Proton balance: 92 = 56+36. A mass decrease of 2.0×10⁻²⁸ kg corresponds to 1.8×10⁻¹¹ J using c=3.0×10⁸ m/s.

Example:

A U-235 nucleus plus one neutron produces fragments with mass numbers 141 and 92. How many neutrons balance the equation? Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Mass number is not a precise mass in kilograms. Balanced integer labels alone do not calculate the released energy. Radioactive decay and neutron-induced fission are not interchangeable descriptions.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A balanced nucleon-number equation alone gives the energy release in joules. This claim is false: Mass number is not a precise mass in kilograms. Balanced integer labels alone do not calculate the released energy. Radioactive decay and neutron-induced fission are not interchangeable descriptions.

Fission splits a heavy nucleus into two medium fragments plus neutrons, releasing energy — spontaneously for a few nuclides, or induced when a neutron strikes a nucleus such as uranium-235:

Conservation holds exactly: $A$ at $236 = 141 + 92 + 3$, $Z$ at $92 = 56 + 36$. The energy is binding energy set free — the fragments sit higher on the binding-energy-per-nucleon curve than the uranium did (about $8.5$ against $7.6\ \mathrm{MeV}$ per nucleon), so roughly $235 \times 0.9 \approx 210\ \mathrm{MeV}$ per fission; direct mass bookkeeping gives the same answer.

The emitted neutrons are the engine of the chain reaction: each fission can trigger more, sustained when on average at least one product neutron causes another fission. A reactor tames it:

• Moderator (water) slows fast neutrons to thermal speeds — the speeds at which U-235 fissions readily.
• Control rods (boron, cadmium) absorb neutrons, raised or lowered to hold the reaction critical, neither growing nor dying.
• Heat exchanger carries energy from the core to water, making steam for the turbines.
• Shielding (steel and thick concrete) contains radiation from the core.

The fission products are neutron-rich and radioactive, with half-lives from seconds to millennia. They are cooled in water pools for years, then sealed — the long-term answer being deep geological repositories, whose safety over tens of thousands of years is the central consideration of nuclear waste management.

Worked check (energy per fission and per kilogram). A mass difference of $\Delta m = 0.19\ u$ per fission gives $E = 0.19 \times 931.5 = 177\ \mathrm{MeV} \approx 2.8\times10^{-11}\ \mathrm{J}$. One kilogram of U-235 holds $(1000/235) \times 6.0\times10^{23} = 2.6\times10^{24}$ nuclei, so complete fission releases about $2.6\times10^{24} \times 2.8\times10^{-11} = 7\times10^{13}\ \mathrm{J}$ in the idealized complete-fission calculation.

Key:

Fission: neutron-induced splits conserve $A$ and $Z$ and release binding energy (~200 MeV per event) because medium fragments bind tighter than uranium; emitted neutrons chain the reaction, tamed by moderator, control rods, heat exchanger and shielding; the neutron-rich products are intensely radioactive and demand cooled storage and long-term geological isolation.

Supported SL focus. First assessment 2025; full Physics guide acquired (84 PDF pages). Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• Two stars can have the same apparent brightness while having different luminosities. Distance changes the flux 通量 reaching an observer.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Luminosity 光度 is total power emitted; flux is power received per area. A stellar spectrum carries information about surface temperature and composition. Fusion transfers energy as light nuclei combine.
• luminosity: Total emitted power; flux: Power received per unit area.

Choose evidence that can test it

• For isotropic emission without absorption, flux follows an inverse-square relationship with distance. Observed brightness alone therefore cannot establish luminosity.
• Keep distance units consistent, identify which quantities are intrinsic to the star, and distinguish observational evidence from a model of stellar evolution. Do not confuse a red giant stage with every possible final remnant.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: distance doubles while luminosity remains fixed. Use F proportional to 1/d². Flux ratio = 1/2² = 1/4. A flux of 12 units becomes 3 units. This assumes no change in absorption or source output.

Example:

A source gives flux 18 units. Find flux at three times the distance. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• The Sun is not expected to become a supernova. Redshift can support cosmological expansion; it does not mean every nearby object must move away from every observer.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Apparent brightness alone always determines a star luminosity. This claim is false: The Sun is not expected to become a supernova. Redshift can support cosmological expansion; it does not mean every nearby object must move away from every observer.

A star is a balance: outward radiation pressure from the fusion-driven core against inward gravitational compression. When fusion falters, gravity wins and the star evolves.

Fusion — light nuclei merging — needs both extreme temperature (nuclei must be thrown at each other fast enough to overcome their electric repulsion) and extreme density (collisions must be frequent enough to sustain the rate). In stellar cores hydrogen fuses to helium; the mass difference is the starlight:

Stellar mass decides the story: a Sun-like star swells to a red giant, sheds its outer layers and settles as a white dwarf; a much heavier star burns fast, swells to a supergiant, and detonates as a supernova, leaving a neutron star or black hole. The elements themselves are fusion's ash — up to iron in cores, heavier atoms scattered by supernovae.

The Hertzsprung–Russell diagram maps luminosity (vertical) against surface temperature (horizontal, hot on the left). Its main regions: the main sequence running diagonally from hot-bright to cool-dim; red giants and supergiants at the top right (bright but cool, so enormous); white dwarfs at the bottom left (hot but dim, so tiny); the instability strip between. Lines of constant radius run diagonally — radius fixes the L–T combination through

so a star's radius follows from its luminosity and temperature. Temperature and composition are read from the stellar spectrum (its colour and lines).

Distances by parallax: as Earth orbits, nearby stars shift against the background by angle $p$:

with $1\ \mathrm{pc} = 3.26\ \mathrm{ly} = 2.1\times10^5\ \mathrm{AU}$.

Worked check (parallax). A star shifts by $0.25$ arcsec: $d = 1/0.25 = 4.0$ parsecs $= 4.0 \times 3.26 = 13$ light years.

Worked check (a stellar radius). A star has 100 times the Sun's luminosity at the same surface temperature: $L \propto R^2T^4$, so $R = \sqrt{100} = 10$ solar radii — a giant.

Key:

Stars and fusion: radiation pressure balances gravity while fusion lasts; fusion needs extreme temperature (Coulomb barrier) and density, turning 0.0287 u per helium nucleus into 26.7 MeV of light; mass sets evolution — red giant to white dwarf, or supergiant to supernova and neutron star or black hole; the HR diagram's main sequence, giants, supergiants, white dwarfs and instability strip map stars by luminosity and temperature with constant-radius lines from $L = 4\pi R^2\sigma T^4$; parallax measures distance as $d(\mathrm{pc}) = 1/p(\mathrm{arcsec})$.

支援付きSL焦点。第一次評価 2025；物理学ガイド全内容を取得済み（84 PDFページ）。残りのガイド、評価、実習要件は記録された保有状態を維持する。

前提条件：指定された数量と単位を読み取り、算術を使用し、以下のモデル条件に従う。各授業は独立した転移前に独自の方法を発展させる。

これらは元々または明示的に架空の教学例であり、実際の測定値や完了した評価対象学習者調査ではない。

この観察を説明するのは何か？

• ほぼすべてのデータ点を通る直線は有用だが、その傾きには不確かさが残る可能性がある。グラフは、測定範囲内でのモデルの妥当性を示す証拠となる。
• まず予測を立てる。比較したい量や特徴を明記し、2つの説明を区別できる証拠は何かを決定せよ。

モデルを構築する

• 偶然誤差により反復読み取り値が異なる。系統誤差は結果を一方向にずらす。絶対不確かさは測定値と同じ単位を持つ；相対不確かさや百分率不確かさは、不確かさを測定値と比較するものである。
• 不確かさ：測定結果に対する定量化された制限；系誤差：一貫した測定バイアス。

モデルを検証できる証拠を選ぶ

• 積または商に対しては、分率不確かさを加算することが最大不確かさの一般的な近似法である。差に対しては絶対不確かさを加算する。ゼロでない切片はオフセットや不完全なモデルを示唆することがある。
• 軸に単位を表示し、適切なスケールを選択する。妥当な場合は不確かさバーを描き、各点を結ぶのではなく最小二乗近似直線（ベストフィットライン）を描く。課程方法で求められた場合、最も急な傾きと最も緩やかな傾きの可能性のある値を推定する。

既知の値から始める

• 既知の値と単位を明記する。このケースに適応する仮定を持つ関係式を選び、代入前に変形する。
• 既知：長さ = 50.0 mm、不確かさ = 1.0 mm。百分率不確かさ = 絶対不確かさ／値 ×100 = 1.0/50.0×100 = 2.0%。2つの独立して測定された量の商で、最大百分率不確かさが2%と3%の場合、合計された最大推計値は5%である。

例：

40 cmの読み取りには1 cmの絶対不確かさがある。百分率不確かさを求めよ。同じ手順に従う：既知量 → モデル → 関係式 → 代入 → 単位と解釈。

結論とその限界を確認する

• 反復読み取りは平均値の偶然不確かさを減少させるが、ゼロ誤差を自動的に除去するわけではない。測定値の精度を超える桁数の小数点以下は表示しないこと。
• 元の観察に戻り、結果が何を裏付けているか、どのような条件を前提としているかを説明し、対立する説明を検証する方法を一つ示す。

警告：

測定を繰り返すことで校正オフセットが常に消除される。この主張は誤りである：反復読み取りは平均値の偶然不確かさを減少させるが、ゼロ誤差を自動的に除去するわけではない。測定値の精度を超える桁数の小数点以下は表示しないこと。

重要：

不確かさ、傾き、モデル検証：積または商に対しては、分率不確かさを加算することが最大不確かさの一般的な近似法である。差に対しては絶対不確かさを加算する。ゼロでない切片はオフセットや不完全なモデルを示唆することがある。