📚 IB Physics Past Paper Questions: Practice and Detailed Solutions | IB 物理真题演练与解析
Welcome to this intensive revision resource, where we work through typical IB Physics past paper questions and provide step-by-step solutions. By engaging with these examples across key topics, you will sharpen your problem‑solving skills, learn to avoid common pitfalls, and build confidence for your final exams.
欢迎使用这本强化复习资源，我们将一起演练典型的IB物理真题并给出分步解析。通过这些涵盖核心知识点的题目练习，你将提升解题能力，学会避开常见陷阱，为最终考试建立信心。
1. General Approach to Past Paper Questions | 真题演练概述
Always begin by reading the question carefully, underlining command terms such as ‘calculate’, ‘explain’, or ‘determine’. Identify the given data and convert all units to SI where necessary.
始终从仔细读题开始，在“计算”、“解释”、“确定”等指令性词语下划线。找出已知数据，必要时将所有单位转换为国际单位制。
Draw a clear, labelled diagram whenever possible; it helps visualise forces, fields, or wavefronts. Write down the relevant equations from the data booklet and assign symbols to each quantity before substituting numbers.
尽可能画出清晰、带标注的示意图，这有助于将力、场或波阵面形象化。从公式册中写出相关方程，给每个物理量分配符号，再代入数值。
After obtaining a numerical answer, check if it is reasonable and whether the correct number of significant figures has been used. Finally, re‑read the question to confirm that you have answered exactly what was asked.
得到数值答案后，检查其是否合理，以及有效数字位数是否正确。最后，重新读题，确认你已准确回答了题目所问。
2. Mechanics – Projectile Motion | 力学 – 抛体运动
Question: A ball is kicked from ground level with an initial speed of 22.0 m·s⁻¹ at an angle of 35.0° above the horizontal. Air resistance is negligible. Calculate (a) the time of flight, (b) the maximum height reached, and (c) the horizontal range.
题目： 一颗足球从地面以初速度22.0 m·s⁻¹、仰角35.0°踢出，空气阻力可忽略。计算 (a) 飞行时间，(b) 达到的最大高度，以及 (c) 水平射程。
Solution (a): Resolve the initial velocity: uₓ = 22.0 cos 35.0° ≈ 18.0 m·s⁻¹, uᵧ = 22.0 sin 35.0° ≈ 12.6 m·s⁻¹. Use vertical motion: when the ball returns to ground, displacement sᵧ = 0. Using s = uᵧ t + ½ a t² with a = –9.81 m·s⁻² gives 0 = 12.6 t – 4.905 t². Solving, t = 2.57 s (time of flight).
解析 (a)： 分解初速度：uₓ = 22.0 cos 35.0° ≈ 18.0 m·s⁻¹，uᵧ = 22.0 sin 35.0° ≈ 12.6 m·s⁻¹。分析竖直运动：球回到地面时，位移 sᵧ = 0。用 s = uᵧ t + ½ a t²，取 a = –9.81 m·s⁻² 得 0 = 12.6 t – 4.905 t²。解得 t = 2.57 s（飞行时间）。
Solution (b): At maximum height, vᵧ = 0. Use vᵧ² = uᵧ² + 2a sᵧ: 0 = (12.6)² + 2(–9.81) H, giving H = 8.10 m.
解析 (b)： 在最高点，vᵧ = 0。利用 vᵧ² = uᵧ² + 2a sᵧ：0 = (12.6)² + 2(–9.81) H，得 H = 8.10 m。
Solution (c): Horizontal motion is uniform: range R = uₓ × t_total = 18.0 × 2.57 = 46.3 m.
解析 (c)： 水平方向做匀速运动：射程 R = uₓ × 飞行总时间 = 18.0 × 2.57 = 46.3 m。
3. Thermal Physics – Ideal Gas Law | 热学 – 理想气体定律
Question: A sealed cylinder contains 0.12 mol of an ideal gas at a pressure of 1.8 × 10⁵ Pa and temperature 320 K. (a) Calculate the volume of the gas. (b) The gas is heated at constant volume until its pressure becomes 2.4 × 10⁵ Pa. Determine the new temperature.
题目： 密封气缸内装有0.12 mol理想气体，压强为1.8×10⁵ Pa，温度为320 K。(a) 计算气体体积。(b) 气体在体积不变的条件下加热，直到压强变为2.4×10⁵ Pa。求此时的新温度。
Solution (a): Apply pV = nRT, with R = 8.31 J·K⁻¹·mol⁻¹. V = nRT / p = (0.12 × 8.31 × 320) / (1.8×10⁵) = 1.77×10⁻³ m³ (or about 1.77 L).
解析 (a)： 应用 pV = nRT，其中 R = 8.31 J·K⁻¹·mol⁻¹。V = nRT / p = (0.12 × 8.31 × 320) / (1.8×10⁵) = 1.77×10⁻³ m³（约1.77 L）。
Solution (b): For a fixed mass at constant volume, p/T = constant. Thus p₁/T₁ = p₂/T₂. T₂ = T₁ × (p₂/p₁) = 320 × (2.4×10⁵ / 1.8×10⁵) = 427 K (or about 154°C).
解析 (b)： 对于质量一定、体积不变的气体，p/T 为常量。因此 p₁/T₁ = p₂/T₂。T₂ = T₁ × (p₂/p₁) = 320 × (2.4×10⁵ / 1.8×10⁵) = 427 K（约154°C）。
4. Waves – Standing Waves on a String | 波 – 弦上的驻波
Question: A guitar string of length 0.65 m is fixed at both ends. The speed of waves on the string is 410 m·s⁻¹. (a) Find the fundamental frequency. (b) Calculate the frequency of the third harmonic.
题目： 一根吉他弦长0.65 m，两端固定。弦上波速为410 m·s⁻¹。(a) 求基频。(b) 计算第三谐波的频率。
Solution (a): For the fundamental, wavelength λ₁ = 2L = 2 × 0.65 = 1.30 m. Using v = fλ, f₁ = v / λ₁ = 410 / 1.30 = 315 Hz (to 3 significant figures).
解析 (a)： 基频对应的波长 λ₁ = 2L = 2 × 0.65 = 1.30 m。由 v = fλ 得 f₁ = v / λ₁ = 410 / 1.30 = 315 Hz（保留三位有效数字）。
Solution (b): The harmonics follow fₙ = n f₁. For the third harmonic, n = 3, so f₃ = 3 × 315 = 945 Hz. Alternatively, λ₃ = 2L/3 = 0.433 m, f₃ = 410 / 0.433 ≈ 947 Hz (consistent).
解析 (b)： 谐频满足 fₙ = n f₁。对于第三谐波 n=3，故 f₃ = 3 × 315 = 945 Hz。或由 λ₃ = 2L/3 = 0.433 m 得 f₃ = 410 / 0.433 ≈ 947 Hz（一致）。
5. Electricity and Magnetism – Circuit Analysis | 电磁学 – 电路分析
Question: A battery of emf 9.0 V and internal resistance 1.2 Ω is connected to an external resistor of 5.8 Ω. Calculate (a) the current in the circuit, (b) the terminal potential difference, and (c) the power dissipated in the external resistor.
题目： 一个电动势为9.0 V、内阻为1.2 Ω的电池与一个5.8 Ω的外部电阻相连。计算 (a) 电路中的电流，(b) 路端电压，以及 (c) 外部电阻消耗的功率。
Solution (a): Total resistance R_total = r + R = 1.2 + 5.8 = 7.0 Ω. Current I = ε / R_total = 9.0 / 7.0 = 1.286 A (≈ 1.29 A).
解析 (a)： 总电阻 R_total = r + R = 1.2 + 5.8 = 7.0 Ω。电流 I = ε / R_total = 9.0 / 7.0 = 1.286 A（≈ 1.29 A）。
Solution (b): Terminal p.d. V = ε – I r = 9.0 – (1.286×1.2) = 9.0 – 1.54 = 7.46 V. Alternatively, V = I R = 1.286 × 5.8 = 7.46 V.
解析 (b)： 路端电压 V = ε – I r = 9.0 – (1.286×1.2) = 9.0 – 1.54 = 7.46 V。也可用 V = I R = 1.286 × 5.8 = 7.46 V 求得。
Solution (c): Power P = I² R = (1.286)² × 5.8 = 9.6 W. Or P = V I = 7.46 × 1.286 = 9.6 W.
解析 (c)： 功率 P = I² R = (1.286)² × 5.8 = 9.6 W。或 P = V I = 7.46 × 1.286 = 9.6 W。
6. Nuclear Physics – Radioactive Decay | 核物理 – 放射性衰变
Question: A sample of radioactive iodine‑131 has a half‑life of 8.0 days. At a certain instant, the sample contains 5.0 × 10¹² nuclei. (a) Calculate the decay constant λ in s⁻¹. (b) How many nuclei remain after 24 days?
题目： 一个放射性碘‑131样品的半衰期为8.0天。在某一时刻，样品含有5.0×10¹²个原子核。(a) 计算衰变常量 λ，单位取 s⁻¹。(b) 24天后还剩多少个原子核？
Solution (a): First convert half‑life to seconds: T½ = 8.0 × 24 × 3600 = 6.912 × 10⁵ s. λ = ln 2 / T½ = 0.693 / (6.912×10⁵) = 1.00 × 10⁻⁶ s⁻¹.
解析 (a)： 先将半衰期换算为秒：T½ = 8.0 × 24 × 3600 = 6.912×10⁵ s。λ = ln 2 / T½ = 0.693 / (6.912×10⁵) = 1.00×10⁻⁶ s⁻¹。
Solution (b): Number of half‑lives in 24 days: n = 24 / 8 = 3. Remaining fraction = (½)³ = ⅛. Nuclei left N = 5.0×10¹² × ⅛ = 6.25×10¹¹. Using exponential form: N = N₀ e^(–λt), t = 24×3600×24 s, gives the same result.
解析 (b)： 24天含有的半衰期个数：n = 24 / 8 = 3。剩余比例 = (½)³ = ⅛。剩余原子核数 N = 5.0×10¹² × ⅛ = 6.25×10¹¹。用指数形式 N = N₀ e^(–λt)，代入 t = 24×3600×24 s 也得到相同结果。
7. Relativity – Time Dilation (HL) | 相对论 – 时间膨胀（HL）
Question: A spaceship travels away from Earth at a constant speed of 0.80c. An astronaut on board measures a time interval of 2.0 years between two events. What is the time interval measured by an observer on Earth?
题目： 一艘宇宙飞船以0.80c的恒定速度飞离地球。船上宇航员测得两个事件的时间间隔为2.0年。地球上的观察者测得的时间间隔是多少？
Solution: This is a time dilation scenario: the proper time Δt₀ = 2.0 years. Lorentz factor γ = 1 / √(1 – v²/c²) = 1 / √(1 – 0.80²) = 1 / √(0.36) = 1 / 0.6 = 1.667. Observed time Δt = γ Δt₀ = 1.667 × 2.0 = 3.33 years.
解析： 这是时间膨胀情景：本征时间 Δt₀ = 2.0 年。洛伦兹因子 γ = 1 / √(1 – v²/c²) = 1 / √(1 – 0.80²) = 1 / √(0.36) = 1 / 0.6 = 1.667。测得的时间 Δt = γ Δt₀ = 1.667 × 2.0 = 3.33 年。
Thus, the Earth‑based observer records a longer time interval, in agreement with the concept that moving clocks run slow.
因此，地球观察者记录的时间间隔更长，这与“运动时钟变慢”的概念一致。
8. Option Topic – Wave Phenomena (Doppler Effect) | 选项课题 – 波动现象（多普勒效应）
Question: A police car siren emits a sound of frequency 800 Hz. The car moves at 30 m·s⁻¹ directly towards a stationary observer. Take the speed of sound in air as 340 m·s⁻¹. Calculate the frequency heard by the observer.
题目： 一辆警车鸣笛发出800 Hz的声音，同时以30 m·s⁻¹的速度径直驶向一位静止的观察者。设空气中声速为340 m·s⁻¹。计算观察者听到的频率。
Solution: For a source moving towards a stationary observer, the observed frequency f ‘ = f × v / (v – v_s), where v_s is the speed of the source. f ‘ = 800 × 340 / (340 – 30) = 800 × 340 / 310 ≈ 877 Hz.
解析： 声源向静止观察者运动时，观测频率 f ‘ = f × v / (v – v_s)，其中 v_s 为声源速度。f ‘ = 800 × 340 / (340 – 30) = 800 × 340 / 310 ≈ 877 Hz。
The frequency is shifted higher, which is the familiar rise in pitch as a siren approaches.
频率升高，这正是警车接近时音调变高的现象。
9. Data Analysis and Error Calculation | 数据分析与误差计算
Question: In an experiment to determine the resistance of a wire, the following data were obtained. Plot a graph of potential difference V (V) against current I (A) and use it to find the resistance. Estimate the absolute uncertainty in the resistance if the uncertainty in each V reading is ±0.05 V and in each I reading is ±0.01 A.
Data: I = 0.20, 0.40, 0.60, 0.80, 1.00 A; V = 0.95, 1.90, 2.85, 3.80, 4.75 V.
题目： 在测定导线电阻的实验中，获得以下数据。作电压 V (V) 对电流 I (A) 的图像，并用其求电阻。若每个 V 读数的绝对不确定度为±0.05 V，每个 I 读数的为±0.01 A，估算电阻的绝对不确定度。
数据：I = 0.20, 0.40, 0.60, 0.80, 1.00 A；V = 0.95, 1.90, 2.85, 3.80, 4.75 V。
Solution: Plot the points: they lie very close to a straight line through the origin. The gradient R = ΔV / ΔI. Using the first and last points: R = (4.75 – 0.95) / (1.00 – 0.20) = 3.80 / 0.80 = 4.75 Ω. More accurate slope from a best‑fit line remains about 4.75 Ω.
解析： 描点：它们几乎落在一条过原点的直线上。斜率 R = ΔV / ΔI。用首尾点：R = (4.75 – 0.95) / (1.00 – 0.20) = 3.80 / 0.80 = 4.75 Ω。最佳拟合线的斜率更准确，但仍约为4.75 Ω。
For uncertainty: the largest and smallest possible slopes give R_max and R_min. Taking extreme points with error bars, steepest line: R_max ≈ (4.80 – 0.90) / (0.80) ≈ 4.875 Ω; shallowest line: R_min ≈ (4.70 – 1.00) / (0.80) ≈ 4.625 Ω. Uncertainty ΔR = (R_max – R_min)/2 ≈ 0.125 Ω, so R = 4.75 ± 0.13 Ω.
不确定度：由可能的最大和最小斜率可得 R_max 和 R_min。考虑误差棒极值点，最陡线：R_max ≈ (4.80 – 0.90) / (0.80) ≈ 4.875 Ω；最缓线：R_min ≈ (4.70 – 1.00) / (0.80) ≈ 4.625 Ω。不确定度 ΔR = (R_max – R_min)/2 ≈ 0.125 Ω，因此 R = 4.75 ± 0.13 Ω。
10. Common Mistakes and Tips | 常见错误与应考技巧
Mistake 1 – Ignoring direction: When using vector equations such as v² = u² + 2as, ensure that acceleration a is assigned the correct sign (negative for deceleration). Similarly, in momentum conservation, define a positive direction before writing equations.
错误1 – 忽略方向： 使用矢量方程如 v² = u² + 2as 时，务必给加速度 a 赋予正确符号（减速时取负）。同理，动量守恒时，先规定正方向再列方程。
Mistake 2 – Incorrect significant figures: IB examiners are strict about significant figures. Carry full calculator precision during intermediate steps and round only the final answer to the same number of significant figures as the least precise given data.
错误2 – 有效数字错误： IB考官对有效数字要求严格。中间步骤保留计算器全部精度，仅在最终答案处按已知数据中最少有效数字位数进行四舍五入。
Tip: Always annotate graphs clearly. In Paper 2 and Paper 3, draw large diagrams, label axes with quantity and unit, and show error bars where necessary. A well‑drawn graph can earn generous marks even if some calculations contain slips.
应考技巧： 始终清晰标注图表。试卷二和试卷三中，画大尺寸示意图，用物理量和单位标注轴，必要时画出误差棒。即使部分计算有小错，一幅精美的图表也能获得可观分数。
Time management: Allocate roughly 1 minute per mark. If stuck on a part, move on and return later; a partially correct method may still receive method points.
时间管理： 大致按每分钟1分的节奏分配时间。若某一部分卡住，先跳过往后做，稍后回头；部分正确的方法仍可获得方法分。
Published by TutorHao | IB Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply