IB and OCR Physics: Past Paper Question Analysis | IB 与 OCR 物理：历年真题解析

📚 IB and OCR Physics: Past Paper Question Analysis | IB 与 OCR 物理：历年真题解析

Past papers are the single most effective revision resource for IB and OCR Physics students. They reveal how examiners consistently test core concepts, expose common pitfalls, and train you to manage time under pressure. This article walks through typical past paper questions from mechanics, thermal physics, waves, electromagnetism, and nuclear physics, offering step‑by‑step solutions, key command term guidance, and practical advice to sharpen your exam technique.

历年真题是IB与OCR物理学生最高效的复习资料。它们揭示了考官如何反复考察核心概念，暴露了常见失分点，并训练你在压力下管理时间。本文精选力学、热学、波动、电磁学及核物理的典型真题，提供逐步解析、关键指令术语指导以及实用建议，磨砺你的应试技巧。

1. Introduction to Past Paper Analysis | 真题解析概述

Analysing past papers allows you to identify frequently assessed topics. In IB Physics, Topic 2 (Mechanics), Topic 4 (Waves), and Topic 5 (Electricity and magnetism) often carry heavy weightings. OCR A Level Physics emphasises Module 3 (Forces and motion), Module 4 (Electrons, waves, and photons), and Module 6 (Particles and medical physics). Recognising these patterns helps you prioritise revision.

分析真题能帮你识别高频考点。在IB物理中，专题2（力学）、专题4（波动）和专题5（电磁学）常占较大分值。OCR A Level物理则侧重模块3（力与运动）、模块4（电子、波与光子）以及模块6（粒子与医学物理）。认清这些规律有助于你优先复习。

Both syllabuses feature multiple‑choice and structured questions. IB Standard Level has Paper 1 (multiple‑choice), Paper 2 (short‑answer and extended response), and Paper 3 (data‑based and option questions). OCR A Level includes Modelling Physics, Exploring Physics, and Unified Physics papers. Practising these formats under timed conditions builds confidence and reduces exam anxiety.

两大课程都包含选择题与结构化应用题。IB标准水平设有试卷一（选择题）、试卷二（简答与拓展题）和试卷三（数据题与选修专题）。OCR A Level则包括建模物理、探索物理和综合物理试卷。通过限时训练这些题型，可以建立信心并减少考试焦虑。

2. Key Command Terms in IB and OCR Physics | IB 与 OCR 物理中的关键指令术语

‘State’ means give a short, definitive answer without explanation. For example, ‘State the value of the acceleration due to gravity on Earth.’ A typical response: ‘9.81 m s⁻².’ Marks are awarded for precision, not reasoning.

“State”要求给出简短明确的答案，无需解释。例如“State the value of the acceleration due to gravity on Earth.”典型回答为“9.81 m s⁻²”，评分看准确度而非推理过程。

‘Describe’ asks you to recall facts, properties, or a sequence of events. In a waves question, ‘Describe the Doppler effect’ should mention frequency shift due to relative motion between source and observer. Avoid lengthy explanations; stick to what is asked.

“Describe”要求你回顾事实、性质或事件顺序。在波动问题中，“Describe the Doppler effect”应提及源与观察者相对运动导致的频率偏移。避免罗嗦的解释，紧扣问题要求。

‘Explain’ requires using scientific principles to give reasons. If asked ‘Explain why a satellite remains in orbit,’ you must link gravitational force, centripetal acceleration, and velocity. OCR often marks the quality of written communication, so coherent steps matter.

“Explain”要求运用科学原理解释原因。若被问到“Explain why a satellite remains in orbit”，必须联系引力、向心加速度与速度。OCR常评估表达的逻辑性，因此清晰的步骤很重要。

‘Calculate’ and ‘Determine’ signal numerical work. Show the formula, substitute values, and give a final answer with correct units and significant figures. IB expects final answers to 2 or 3 significant figures unless stated otherwise. OCR likewise expects appropriate significant figures, often matching given data.

“Calculate”和“Determine”指示数值计算。要写出公式、代入数值并给出带有正确单位和有效数字的最终答案。IB一般要求2或3位有效数字，除非有特殊说明；OCR同样要求有效数字与已知数据匹配。

‘Sketch’ means draw a rough graph or diagram showing key features (intercepts, shape, asymptotes) but not a precisely plotted graph. For instance, ‘Sketch the velocity‑time graph for a bouncing ball’ should highlight sudden changes in direction and the decreasing peak velocities.

“Sketch”要求绘制显示关键特征（截距、形状、渐近线）的示意图，无需精确标度。例如“Sketch the velocity‑time graph for a bouncing ball”应突出方向突变与逐渐减小的速度峰值。

3. Mechanics: Projectile Motion Questions | 力学：抛体运动真题

A classic IB/OCR question: ‘A ball is kicked from ground level with a speed of 22 m s⁻¹ at an angle of 35° to the horizontal. Calculate (a) the time of flight, (b) the maximum height, and (c) the horizontal range.’ Always resolve velocity into horizontal and vertical components first: u_x = u cosθ, u_y = u sinθ.

一道经典真题：’A ball is kicked from ground level with a speed of 22 m s⁻¹ at an angle of 35° to the horizontal. Calculate (a) the time of flight, (b) the maximum height, and (c) the horizontal range.’ 务必先将速度分解为水平与竖直分量：u_x = u cosθ, u_y = u sinθ。

u_x = 22 cos 35° ≈ 18.0 m s⁻¹, u_y = 22 sin 35° ≈ 12.6 m s⁻¹

Time of flight to return to ground level is t = 2 u_y / g. Substituting: t = 2 × 12.6 / 9.81 ≈ 2.57 s. Many lose marks by forgetting the factor of 2, calculating only the time to the peak.

返回地面时的飞行时间 t = 2 u_y / g。代入得 t = 2 × 12.6 / 9.81 ≈ 2.57 s。许多学生因忘记系数2只计算了升至最高点的时间而失分。

Maximum height: h = u_y² / (2g) = (12.6)² / (2 × 9.81) ≈ 8.10 m. Horizontal range: R = u_x × t = 18.0 × 2.57 ≈ 46.3 m. Or use formula R = u² sin(2θ) / g directly. Always check that the answer is physically reasonable and includes units.

最大高度：h = u_y² / (2g) = (12.6)² / (2 × 9.81) ≈ 8.10 m。水平射程：R = u_x × t = 18.0 × 2.57 ≈ 46.3 m。也可直接用公式 R = u² sin(2θ) / g。务必检查答案的物理合理性与单位。

OCR questions might further ask: ‘Explain why the horizontal velocity remains constant.’ This tests understanding that air resistance is negligible and there is no horizontal force, hence no horizontal acceleration.

OCR试题可能进一步问：“Explain why the horizontal velocity remains constant.”这考查对忽略空气阻力且无水平力故无水平加速度的理解。

4. Thermal Physics: Ideal Gas Law Applications | 热学：理想气体定律应用

An IB past paper question: ‘A sealed container of fixed volume 0.030 m³ contains an ideal gas at 2.0 × 10⁵ Pa and 290 K. Calculate the number of moles of gas.’ Use pV = nRT, so n = pV / (RT). R = 8.31 J K⁻¹ mol⁻¹. n = (2.0 × 10⁵ × 0.030) / (8.31 × 290) ≈ 2.49 mol.

一道IB真题：’A sealed container of fixed volume 0.030 m³ contains an ideal gas at 2.0 × 10⁵ Pa and 290 K. Calculate the number of moles of gas.’ 用 pV = nRT，得 n = pV / (RT)。R = 8.31 J K⁻¹ mol⁻¹。n = (2.0 × 10⁵ × 0.030) / (8.31 × 290) ≈ 2.49 mol。

The follow‑up often varies temperature: ‘The gas is heated to 350 K. Find the new pressure.’ Since volume and n are constant, p₁/T₁ = p₂/T₂, so p₂ = p₁ T₂ / T₁ = 2.0×10⁵ × 350 / 290 ≈ 2.41×10⁵ Pa.

后续常涉及温度变化：“The gas is heated to 350 K. Find the new pressure.”由于体积与n恒定，p₁/T₁ = p₂/T₂，得 p₂ = p₁ T₂ / T₁ = 2.0×10⁵ × 350 / 290 ≈ 2.41×10⁵ Pa。

OCR expects you to convert degrees Celsius to Kelvin correctly. A typical mistake is using 27°C as 27 K; remember T(K) = θ(°C) + 273. Also, be careful with units: pressure in Pa, volume in m³. If given in cm³, convert to m³ by multiplying by 10⁻⁶.

OCR要求正确地将摄氏度换算为开尔文。常见错误是把27°C当成27 K；牢记 T(K) = θ(°C) + 273。同时注意单位：压强用Pa，体积用m³。若题目给的是cm³，须乘以10⁻⁶化为m³。

IB often links this to kinetic theory: ‘Explain the increase in pressure in terms of molecular motion.’ You should discuss increased average kinetic energy, higher collision frequency, and greater momentum change per collision with the container walls.

IB常将之与分子动理论联系：“Explain the increase in pressure in terms of molecular motion.”需讨论平均动能增大、碰撞频率升高以及每次与器壁碰撞的动量变化增大。

5. Waves: Double-Slit Interference | 波动：双缝干涉

A typical OCR question: ‘In a Young’s double‑slit experiment, laser light of wavelength 650 nm passes through slits separated by 0.40 mm. The screen is 2.5 m away. Calculate the fringe spacing.’ Use Δx = λD / d. Convert all units to metres: λ = 650 × 10⁻⁹ m, d = 0.40 × 10⁻³ m, D = 2.5 m.

典型OCR题目：“In a Young’s double‑slit experiment, laser light of wavelength 650 nm passes through slits separated by 0.40 mm. The screen is 2.5 m away. Calculate the fringe spacing.” 运用 Δx = λD / d。全部单位化成米：λ = 650 × 10⁻⁹ m, d = 0.40 × 10⁻³ m, D = 2.5 m。

Δx = (650 × 10⁻⁹ × 2.5) / (0.40 × 10⁻³) ≈ 4.06 × 10⁻³ m (≈ 4.1 mm)

IB questions often extend this to the concept of coherence: ‘State why the two slits must act as coherent sources.’ Answer: they must have a constant phase difference and the same frequency, achievable by using a single laser source.

IB题目常延伸至相干性概念：“State why the two slits must act as coherent sources.” 答案：它们必须有恒定的相位差和相同的频率，这可通过使用单一激光源实现。

Another common query: ‘Describe how the pattern changes if white light is used.’ The central fringe remains white, but higher‑order fringes become spectra with violet on the inner edge and red on the outer edge because Δx depends on λ.

另一常见提问：“Describe how the pattern changes if white light is used.” 中央条纹保持白色，而高级次条纹变成光谱，内侧紫、外侧红，因为Δx依赖于λ。

6. Electricity and Magnetism: Circuit Analysis | 电磁学：电路分析

A typical IB Paper 2 circuit: ‘A 12 V battery with negligible internal resistance is connected to a 6.0 Ω resistor in series with a parallel combination of a 10 Ω resistor and a 15 Ω resistor. Calculate (a) the total resistance, (b) the current drawn from the battery, and (c) the power dissipated in the 10 Ω resistor.’

典型IB试卷二电路题：“A 12 V battery with negligible internal resistance is connected to a 6.0 Ω resistor in series with a parallel combination of a 10 Ω resistor and a 15 Ω resistor. Calculate (a) the total resistance, (b) the current drawn from the battery, and (c) the power dissipated in the 10 Ω resistor.”

Parallel resistance: 1/R_par = 1/10 + 1/15 = 5/30 = 1/6, so R_par = 6.0 Ω. Total resistance R_total = 6.0 + 6.0 = 12 Ω. Battery current I = V / R_total = 12 / 12 = 1.0 A.

并联电阻：1/R_par = 1/10 + 1/15 = 5/30 = 1/6，得 R_par = 6.0 Ω。总电阻 R_total = 6.0 + 6.0 = 12 Ω。电池输出电流 I = V / R_total = 12 / 12 = 1.0 A。

Voltage across parallel block: V_par = I × R_par = 1.0 × 6.0 = 6.0 V. Current through 10 Ω resistor: I₁₀ = V_par / 10 = 0.60 A. Power: P = I²R = (0.60)² × 10 = 3.6 W. Alternatively, P = V²/R = 6.0²/10 = 3.6 W.

并联部分两端电压：V_par = I × R_par = 1.0 × 6.0 = 6.0 V。通过10 Ω电阻的电流：I₁₀ = V_par / 10 = 0.60 A。功率：P = I²R = (0.60)² × 10 = 3.6 W。也可用 P = V²/R = 6.0²/10 = 3.6 W。

OCR requires you to handle internal resistance explicitly. For a cell of emf 1.5 V and internal resistance 0.50 Ω connected to a 4.0 Ω load, terminal p.d. = emf − Ir, where I = emf / (R+r) = 1.5 / (4.0+0.5) = 0.333 A, so terminal p.d. = 1.5 − (0.333×0.5) = 1.33 V.

OCR要求明确处理内阻。对于一个电动势1.5 V、内阻0.50 Ω的电池连接4.0 Ω负载，端电压 = 电动势 − Ir，其中 I = 电动势 / (R+r) = 1.5 / (4.0+0.5) = 0.333 A，故端电压 = 1.5 − (0.333×0.5) = 1.33 V。

7. Circular Motion and Gravitation | 圆周运动与引力

An OCR question on satellites: ‘A satellite orbits Earth at an altitude of 400 km. Calculate its orbital speed. Earth’s radius R_E = 6.4 × 10⁶ m, mass M_E = 6.0 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N m² kg⁻².’ Orbital radius r = R_E + 400 × 10³ m = 6.8 × 10⁶ m. Centripetal force provided by gravity: GMm/r² = mv²/r ⇒ v = √(GM/r).

一道OCR卫星题：“A satellite orbits Earth at an altitude of 400 km. Calculate its orbital speed. Earth’s radius R_E = 6.4 × 10⁶ m, mass M_E = 6.0 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N m² kg⁻².” 轨道半径 r = R_E + 400 × 10³ m = 6.8 × 10⁶ m。向心力由引力提供：GMm/r² = mv²/r ⇒ v = √(GM/r)。

v = √[(6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (6.8 × 10⁶)] ≈ √(5.88 × 10⁷) ≈ 7.67 × 10³ m s⁻¹

IB often asks for the period: T = 2πr / v = 2π × 6.8×10⁶ / 7.67×10³ ≈ 5570 s (about 93 minutes). Students must be comfortable switching between linear and angular velocity. Missed units or powers of ten are a major cause of errors.

IB常要求计算周期：T = 2πr / v = 2π × 6.8×10⁶ / 7.67×10³ ≈ 5570 s（约93分钟）。学生必须熟练转换线速度与角速度。漏写单位或十的幂次是错误的主要原因。

In explaining weightlessness in orbit, both boards expect you to clarify that astronauts are not truly weightless; gravity still acts but is used entirely to provide centripetal force, so the normal contact force is zero, giving the sensation of weightlessness.

在解释轨道中的失重时，两大考试局都期望你阐明宇航员并非真正失重；引力依然存在，但全部提供为向心力，因此法向接触力为零，从而产生了失重感。

8. Quantum and Nuclear Physics: Radioactive Decay | 量子与核物理：放射性衰变

A popular IB nuclear question: ‘A pure sample of a radioactive isotope has an initial activity of 2400 Bq. After 4.0 hours, the activity drops to 300 Bq. Calculate the decay constant λ and the half‑life.’ Use A = A₀ e^(–λt). Divide both sides by A₀: 300/2400 = e^(–λ×4.0) ⇒ 1/8 = e^(–4λ).

一道流行的IB核物理题：“A pure sample of a radioactive isotope has an initial activity of 2400 Bq. After 4.0 hours, the activity drops to 300 Bq. Calculate the decay constant λ and the half‑life.” 运用 A = A₀ e^(–λt)。两边除以A₀：300/2400 = e^(–λ×4.0) ⇒ 1/8 = e^(–4λ)。

Take natural logs: ln(1/8) = –4λ ⇒ λ = –ln(1/8) / 4. ln(1/8) = –ln8 = –2.079, so λ = 2.079 / 4 = 0.520 h⁻¹. Half‑life T₁/₂ = ln2 / λ ≈ 0.693 / 0.520 ≈ 1.33 h (or 1 h 20 min).

取自然对数：ln(1/8) = –4λ ⇒ λ = –ln(1/8) / 4。ln(1/8) = –ln8 = –2.079，故 λ = 2.079 / 4 = 0.520 h⁻¹。半衰期 T₁/₂ = ln2 / λ ≈ 0.693 / 0.520 ≈ 1.33 h（或1小时20分钟）。

OCR frequently uses applications such as carbon‑dating or medical tracers. Example: ‘Explain why a source with a half‑life of a few hours is suitable for medical imaging.’ It balances enough activity for detection with minimising patient radiation dose.

OCR常考查碳定年或医用示踪剂等应用。例如：“Explain why a source with a half‑life of a few hours is suitable for medical imaging.” 它平衡了足够用于检测的活度与最小化患者辐射剂量。

Another typical question is on background radiation: ‘Describe how you would correct a count rate measurement for background radiation.’ You measure the count rate without the source, then subtract it from the measured count rate with the source.

另一典型题涉及本底辐射：“Describe how you would correct a count rate measurement for background radiation.” 测量不含放射源时的计数率，然后从有源时的计数率中减去该值。

9. Energy and Power Calculations | 能量与功率计算

An OCR question scenario: ‘An electric motor lifts a mass of 500 kg vertically at a steady speed of 2.0 m s⁻¹. Calculate the minimum power output of the motor.’ Power = force × velocity. Force = weight = mg = 500 × 9.81 = 4905 N. P = 4905 × 2.0 = 9810 W ≈ 9.8 kW.

一道OCR情境题：“An electric motor lifts a mass of 500 kg vertically at a steady speed of 2.0 m s⁻¹. Calculate the minimum power output of the motor.” 功率 = 力 × 速度。力 = 重力 = mg = 500 × 9.81 = 4905 N。P = 4905 × 2.0 = 9810 W ≈ 9.8 kW。

If the motor is only 70% efficient, the electrical input power needed = output power / efficiency = 9.8 kW / 0.70 ≈ 14 kW. Students often forget efficiency or invert the ratio.

若电动机效率只有70%，所需电输入功率 = 输出功率 / 效率 = 9.8 kW / 0.70 ≈ 14 kW。学生常忘记效率或颠倒比值。

IB may link this to gravitational potential energy: ‘How much energy is used to lift the mass through 12 m?’ E = mgh = 500 × 9.81 × 12 = 58 860 J. Then calculate time taken using speed: t = d/v = 12/2 = 6 s, so average power = E/t = 58 860/6 = 9 810 W, consistent. Understanding the equivalence reinforces concepts.

IB可能联系重力势能：“How much energy is used to lift the mass through 12 m?” E = mgh = 500 × 9.81 × 12 = 58 860

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IB and OCR Physics: Past Paper Question Analysis | IB 与 OCR 物理：历年真题解析