📚 IGCSE OCR Physics Calculation Practice | IGCSE OCR 物理计算题专项训练
Calculation questions form a significant portion of the IGCSE OCR Physics exam. Mastering numerical problems requires not only recalling equations but also selecting the right formula, converting units correctly, and presenting solutions step by step. This article provides targeted practice across core topics, with worked examples, common pitfalls and strategy tips to help you tackle calculation questions with confidence.
计算题在 IGCSE OCR 物理考试中占有很大比重。掌握数值计算不仅需要记住公式，还要能选择正确的方程、准确换算单位，并一步步展示解题过程。本文针对核心主题提供专项训练，包含范例、常见错误和策略提示，帮助你自信地应对计算题。
1. Kinematics – Speed, Velocity and Acceleration | 运动学——速度、速率与加速度
Use the equation v = s / t for average speed, and a = (v – u) / t for acceleration. Remember that velocity is a vector; direction matters. Always check if units are consistent, for example convert km to m and hours to seconds when necessary.
使用公式 v = s / t 求平均速率，用 a = (v – u) / t 求加速度。注意速度是矢量，方向很重要。始终检查单位是否一致，比如必要时将千米换算成米，小时换算成秒。
A car accelerates uniformly from rest to 25 m/s in 10 s. (a) Calculate its acceleration. (b) How far does it travel during this time? Use s = (u + v)t / 2 for displacement. Solution: a = (25 – 0) / 10 = 2.5 m/s². s = (0 + 25) × 10 / 2 = 125 m.
一辆汽车从静止匀加速到 25 m/s，用时 10 s。(a) 计算加速度。(b) 这段时间内行驶了多远？使用位移公式 s = (u + v)t / 2。解答：a = (25 – 0) / 10 = 2.5 m/s²。s = (0 + 25) × 10 / 2 = 125 m。
2. Motion Graphs and Area Under Curve | 运动图像与曲线下面积
Velocity–time graphs provide a visual way to determine acceleration (gradient) and distance travelled (area under the graph). For non-uniform motion, counting squares or using geometry for standard shapes may be required.
速度–时间图像能直观地求加速度（斜率）和行驶距离（图像下方面积）。对于非匀变速运动，可能需要数格或利用标准图形的几何形状求面积。
A velocity–time graph shows a straight line from (0, 0) to (8 s, 20 m/s). The distance is the area of the triangle: ½ × base × height = ½ × 8 × 20 = 80 m. The acceleration is gradient = (20 – 0) / 8 = 2.5 m/s².
一条速度–时间图显示从 (0, 0) 到 (8 s, 20 m/s) 的直线。距离等于三角形面积：½ × 底 × 高 = ½ × 8 × 20 = 80 m。加速度等于斜率 = (20 – 0) / 8 = 2.5 m/s²。
3. Forces and Newton’s Second Law | 力与牛顿第二定律
F = m × a is fundamental. Resultant force is the vector sum of all forces acting on an object. When multiple forces are present, resolve them or find the net force before applying the equation. Also link with weight W = m × g.
F = m × a 是基础公式。合力是作用在物体上所有力的矢量和。当存在多个力时，先分解或求净力再应用公式。也要结合重力公式 W = m × g。
A 1200 kg car experiences a driving force of 3000 N and a resistive force of 600 N. The resultant force = 3000 – 600 = 2400 N. Acceleration a = F / m = 2400 / 1200 = 2.0 m/s².
一辆 1200 kg 的汽车受到驱动力 3000 N 和阻力 600 N。合力 = 3000 – 600 = 2400 N。加速度 a = F / m = 2400 / 1200 = 2.0 m/s²。
4. Momentum and Impulse | 动量与冲量
Momentum p = m × v. The principle of conservation of momentum applies to collisions and explosions. Impulse = change in momentum = F × t. Remember to assign positive and negative signs for direction in one-dimensional problems.
动量 p = m × v。动量守恒定律适用于碰撞和爆炸。冲量 = 动量变化 = F × t。记得在一维问题中为方向分配正负号。
Ball A (0.5 kg) moves at 4 m/s right and strikes stationary Ball B (0.3 kg). After collision, A moves at 1 m/s right. Find velocity of B. Conserve momentum: 0.5×4 + 0.3×0 = 0.5×1 + 0.3×v → 2.0 = 0.5 + 0.3v → v = 5 m/s to the right.
球 A（0.5 kg）以 4 m/s 向右运动，撞击静止的球 B（0.3 kg）。碰撞后 A 以 1 m/s 向右运动。求 B 的速度。动量守恒：0.5×4 + 0.3×0 = 0.5×1 + 0.3×v → 2.0 = 0.5 + 0.3v → v = 5 m/s 向右。
5. Work, Energy and Power | 功、能量与功率
Work done W = F × d (when force and distance are parallel). Kinetic energy KE = ½ m v². Gravitational potential energy GPE = m g h. Power P = W / t or P = E / t. Efficiency = useful output / total input.
做的功 W = F × d（当力与距离平行时）。动能 KE = ½ m v²。重力势能 GPE = m g h。功率 P = W / t 或 P = E / t。效率 = 有用输出 / 总输入。
A 50 kg student climbs 3 m high stairs in 4 s. GPE gained = 50 × 10 × 3 = 1500 J (taking g = 10 m/s²). Power = 1500 / 4 = 375 W. If the student’s body only works at 25% efficiency, total energy expended = 1500 / 0.25 = 6000 J.
一名 50 kg 的学生在 4 s 内爬上 3 m 高的楼梯。获得的 GPE = 50 × 10 × 3 = 1500 J（取 g = 10 m/s²）。功率 = 1500 / 4 = 375 W。若该学生身体的效率仅为 25%，则总消耗能量 = 1500 / 0.25 = 6000 J。
6. Density and Pressure | 密度与压强
Density ρ = m / V. Pressure p = F / A. For liquid pressure, p = h ρ g. Pay attention to units: volume in m³, area in m². Often need to rearrange formulas to find missing quantities.
密度 ρ = m / V。压强 p = F / A。对于液体压强，p = h ρ g。注意单位：体积用 m³，面积用 m²。常常需要变换公式来求未知量。
A cube of side 0.1 m has mass 0.8 kg. Density = 0.8 / (0.1)³ = 0.8 / 0.001 = 800 kg/m³. The pressure exerted on a table if it rests on one face: area = 0.01 m², force = weight = 8 N, so p = 8 / 0.01 = 800 Pa.
一个边长为 0.1 m 的立方体，质量为 0.8 kg。密度 = 0.8 / (0.1)³ = 0.8 / 0.001 = 800 kg/m³。如果它一个面放在桌面上，压强：面积 = 0.01 m²，压力 = 重量 = 8 N，故 p = 8 / 0.01 = 800 Pa。
7. Thermal Physics – Specific Heat Capacity and Latent Heat | 热学——比热容与潜热
Energy transferred Q = m c Δθ for temperature change. For change of state, Q = m L. Use c for specific heat capacity, L for specific latent heat (fusion or vaporisation). Take care with units of mass (kg) and temperature (°C or K).
温度变化时的能量转移 Q = m c Δθ。物态变化时 Q = m L。c 是比热容，L 是比潜热（熔化或汽化）。注意质量（kg）和温度（°C 或 K）的单位。
How much energy is needed to melt 2 kg of ice at 0 °C? L_f (water) = 334 000 J/kg. Q = 2 × 334 000 = 668 000 J. To then raise the water temperature to 20 °C, Q = 2 × 4200 × 20 = 168 000 J. Total = 836 000 J.
将 2 kg 0 °C 的冰熔化需要多少能量？水的熔化潜热 L_f = 334 000 J/kg。Q = 2 × 334 000 = 668 000 J。随后将水加热到 20 °C，Q = 2 × 4200 × 20 = 168 000 J。总计 = 836 000 J。
8. Wave Equation and Lens Formula | 波动方程与透镜公式
Wave speed v = f λ. Frequency f = 1 / T. For lenses, the thin lens equation 1/f = 1/u + 1/v, with sign conventions. Magnification m = v / u. Real-is-positive convention is common in OCR; check specification.
波速 v = f λ。频率 f = 1 / T。透镜方面，薄透镜公式 1/f = 1/u + 1/v，注意符号规则。放大率 m = v / u。OCR 通常使用实物为正的符号约定，请核对大纲。
A wave has frequency 500 Hz and wavelength 0.68 m. Speed v = 500 × 0.68 = 340 m/s. For a converging lens, an object at u = 30 cm, focal length f = 20 cm: 1/20 = 1/30 + 1/v → 1/v = 1/20 – 1/30 = 1/60 → v = 60 cm. Image is real, 60 cm from lens.
一列波的频率为 500 Hz，波长为 0.68 m。波速 v = 500 × 0.68 = 340 m/s。对于会聚透镜，物距 u = 30 cm，焦距 f = 20 cm：1/20 = 1/30 + 1/v → 1/v = 1/20 – 1/30 = 1/60 → v = 60 cm。成实像，距透镜 60 cm。
9. Electricity – Ohm’s Law, Resistance and Circuits | 电学——欧姆定律、电阻与电路
V = I × R. For resistors in series, R_total = R1 + R2 + … In parallel, 1/R_total = 1/R1 + 1/R2 + … Power P = I V = I² R = V² / R. Also use Q = I t for charge. Always label units clearly: A, V, Ω, W, C.
V = I × R。串联电阻：R_total = R1 + R2 + … 并联电阻：1/R_total = 1/R1 + 1/R2 + … 功率 P = I V = I² R = V² / R。电量 Q = I t。始终清晰标注单位：A、V、Ω、W、C。
Two 10 Ω resistors in parallel: 1/R = 1/10 + 1/10 = 2/10 → R = 5 Ω. Total current from a 12 V supply: I = V / R = 12 / 5 = 2.4 A. Power dissipated by one resistor: each gets half the total current? Actually in parallel voltage across each is 12 V, so I per resistor = 12/10 = 1.2 A, P = 1.2 × 12 = 14.4 W or P = V²/R = 144/10 = 14.4 W.
两个 10 Ω 电阻并联：1/R = 1/10 + 1/10 = 2/10 → R = 5 Ω。12 V 电源提供的总电流：I = V / R = 12 / 5 = 2.4 A。一个电阻消耗的功率：每个电阻上电压均为 12 V，因此每个分支电流 = 12/10 = 1.2 A，P = 1.2 × 12 = 14.4 W，或用 P = V²/R = 144/10 = 14.4 W。
10. Mains Electricity and Transformers | 市电与变压器
Power transmission: P = I V. For an ideal transformer, V_p / V_s = N_p / N_s and I_p V_p = I_s V_s (assuming 100% efficiency). Use the turns ratio to find unknown voltages or currents. Be aware of step-up and step-down applications.
电力传输：P = I V。理想变压器满足 V_p / V_s = N_p / N_s，且 I_p V_p = I_s V_s（假设 100% 效率）。利用匝数比求未知电压或电流。要清楚升压和降压的应用。
A transformer has 5000 turns on the primary and 250 turns on the secondary. Primary voltage is 230 V. Secondary voltage V_s = (N_s / N_p) × V_p = (250/5000) × 230 = 11.5 V. If the secondary current is 2 A, primary current (assuming 100% efficiency) I_p = (V_s I_s) / V_p = (11.5 × 2) / 230 = 0.1 A.
一个变压器初级线圈 5000 匝，次级 250 匝。初级电压 230 V。次级电压 V_s = (N_s / N_p) × V_p = (250/5000) × 230 = 11.5 V。若次级电流为 2 A，则初级电流（假设效率 100%） I_p = (V_s I_s) / V_p = (11.5 × 2) / 230 = 0.1 A。
11. Nuclear Physics – Half-life and Activity | 核物理——半衰期与活度
Activity A = – dN/dt, measured in becquerels (Bq). The half-life is the time for half the nuclei to decay. You may need to calculate remaining nuclei after n half-lives: N = N₀ (1/2)^n. Sometimes count rate is used as a proxy for activity.
活度 A = – dN/dt，单位为贝克勒尔（Bq）。半衰期是半数原子核发生衰变所需的时间。你可能需要计算 n 个半衰期后剩余的原子核数量：N = N₀ (1/2)^n。有时用计数率近似表示活度。
A sample has initial activity 800 Bq. After 24 minutes, activity drops to 100 Bq. How many half-lives? 800 → 400 → 200 → 100 = 3 half-lives. Thus half-life = 24 / 3 = 8 minutes. After 32 minutes (4 half-lives), activity = 800 × (1/2)^4 = 800 × 1/16 = 50 Bq.
某样品初始活度为 800 Bq。24 分钟后活度降至 100 Bq。经过多少个半衰期？800 → 400 → 200 → 100 = 3 个半衰期。因此半衰期 = 24 / 3 = 8 分钟。32 分钟（4 个半衰期）后，活度 = 800 × (1/2)^4 = 800 × 1/16 = 50 Bq。
12. General Calculation Strategies and Common Errors | 通用计算策略与常见错误
Always show the formula, substitution, and final answer with correct units. Write down known quantities and convert to SI (e.g., cm to m, g to kg). Check if the question asks for standard form or significant figures. Watch out for: using wrong formula, forgetting to square or take square roots, misplacing decimal points, and direction signs in vector problems.
始终写出公式、代入过程和最终答案及正确单位。写下已知量并转换为国际单位制（如 cm 转 m，g 转 kg）。检查题目是否要求用科学记数法或有效数字。常见错误包括：用错公式，忘记平方或开平方根，小数点错位，以及在矢量问题中遗漏方向符号。
Practice by redoing examples without looking at solutions, timing yourself. On exam day, underline the values in the question and label them. If stuck, write down the relevant formula – you may earn marks. Finally, ask: is the answer reasonable? A car probably doesn’t travel 1000 m in 2 seconds!
练习时先不看答案重做例题并计时。考试时，划出题目中的数值并标注。如果卡住了，写下可能相关的公式——或许能拿到分数。最后，问自己：答案合理吗？一辆车不可能在 2 秒内行驶 1000 米！
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