📚 IGCSE OCR Science: Calculations Masterclass | IGCSE OCR 科学：计算题专项训练
This masterclass targets the most common numerical question types in the IGCSE OCR Science specifications (Physics, Chemistry, and Biology). By reviewing core formulas, rearrangements, unit conversions, and applying them to worked examples, you will build confidence and accuracy. The emphasis is on systematic problem-solving using the ‘Given, Formula, Substitute, Solve’ method.
本文针对 IGCSE OCR 科学（物理、化学、生物）中最常见的计算题型进行专项训练。通过回顾核心公式、公式变形、单位换算，并结合典型例题的应用，你将逐步建立信心并提高准确率。全文突出系统化的解题方法——’已知、公式、代入、求解’四步法。
1. Speed, Distance and Time | 速度、距离与时间
The average speed of an object is the distance travelled divided by the time taken. Always check units: if distance is in metres (m) and time in seconds (s), speed will be in metres per second (m/s).
物体的平均速度等于通过的距离除以所用时间。务必检查单位：如果距离的单位是米（m），时间的单位是秒（s），那么速度的单位就是米每秒（m/s）。
v = s ÷ t
Worked example: A cyclist travels 450 m in 30 s. Calculate her average speed. Given: s = 450 m, t = 30 s. Formula: v = s / t = 450 / 30 = 15 m/s.
例题：一名自行车手在 30 秒内骑行了 450 米。计算她的平均速度。已知：s = 450 m，t = 30 s。公式：v = s / t = 450 / 30 = 15 m/s。
Rearranging is essential: to find distance, use s = v × t; to find time, use t = s / v. When speed is given in km/h, convert to m/s by dividing by 3.6 (since 1 km/h = 1000 m / 3600 s).
公式变形必不可少：求距离用 s = v × t；求时间用 t = s / v。当速度以 km/h 给定时，需除以 3.6 换算成 m/s（因为 1 km/h = 1000 m / 3600 s）。
2. Density, Mass and Volume | 密度、质量与体积
Density (ρ) is mass per unit volume. The standard units are kilogram per cubic metre (kg/m³) or gram per cubic centimetre (g/cm³). Remember that 1 g/cm³ = 1000 kg/m³.
密度（ρ）是单位体积的质量。标准单位是千克每立方米（kg/m³）或克每立方厘米（g/cm³）。记住 1 g/cm³ = 1000 kg/m³。
ρ = m ÷ V
If a block of metal has a mass of 200 g and a volume of 40 cm³, its density is ρ = 200 / 40 = 5 g/cm³. For irregular objects, volume is often measured by water displacement in a measuring cylinder.
如果一块金属的质量为 200 克，体积为 40 立方厘米，其密度为 ρ = 200 / 40 = 5 g/cm³。对于不规则形状的物体，通常使用量筒通过排水法测量体积。
You must be able to rearrange the formula: m = ρ × V and V = m / ρ. Always ensure that mass and volume units are consistent before substituting.
你必须能够进行公式变形：m = ρ × V 和 V = m / ρ。代入值前务必保证质量和体积单位一致。
3. Acceleration and Equations of Motion | 加速度与运动方程
Acceleration (a) is the rate of change of velocity. If the velocity changes by Δv over time t, acceleration is given by:
加速度（a）是速度的变化率。如果速度在时间 t 内变化了 Δv，加速度可用下式计算：
a = (v – u) ÷ t
where u = initial velocity, v = final velocity. This equation assumes constant acceleration. Example: A car accelerates from 5 m/s to 20 m/s in 3 s. a = (20 – 5) / 3 = 5 m/s².
其中 u = 初速度，v = 末速度。此公式假设加速度恒定。例题：一辆汽车在 3 秒内从 5 m/s 加速到 20 m/s，加速度 a = (20 – 5) / 3 = 5 m/s²。
For motion under constant acceleration, the equation v² = u² + 2as is also used, where s is displacement. Always state the direction when acceleration is negative (deceleration).
对于匀加速运动，也会用到公式 v² = u² + 2as，其中 s 为位移。当加速度为负值（减速）时，要说明方向。
4. Force, Mass and Newton’s Second Law | 力、质量与牛顿第二定律
The resultant force acting on an object equals the product of its mass and acceleration. This is written as:
作用在物体上的合外力等于其质量与加速度的乘积。写作：
F = m × a
Force is measured in newtons (N), mass in kilograms (kg), and acceleration in m/s². A force of 1 N gives a mass of 1 kg an acceleration of 1 m/s².
力的单位是牛顿（N），质量的单位是千克（kg），加速度的单位是 m/s²。1 N 的力能使 1 kg 的质量产生 1 m/s² 的加速度。
Example: A trolley of mass 0.5 kg is pulled with a resultant force of 2 N. Its acceleration is a = F / m = 2 / 0.5 = 4 m/s². Remember to consider all forces when determining the resultant force.
例题：一辆质量为 0.5 kg 的手推车受到 2 N 的合外力，其加速度 a = F / m = 2 / 0.5 = 4 m/s²。在确定合外力时，务必考虑所有作用力。
5. Momentum | 动量
Momentum (p) is the product of mass and velocity. It is a vector quantity, so direction matters.
动量（p）是质量与速度的乘积。它是矢量，因此方向至关重要。
p = m × v
Momentum is conserved in collisions and explosions, provided no external forces act. In a collision: total momentum before = total momentum after. If one object is stationary before, the calculation simplifies.
在没有外力作用的情况下，碰撞和爆炸过程中动量守恒。碰撞中：碰撞前总动量 = 碰撞后总动量。若碰撞前有一个物体静止，计算会简化。
Example: A 2 kg trolley moving at 3 m/s hits a stationary 1 kg trolley. They stick together. Total momentum before = (2 × 3) + (1 × 0) = 6 kg m/s. After collision, combined mass = 3 kg, so velocity v = 6 / 3 = 2 m/s.
例题：一辆 2 kg 的手推车以 3 m/s 的速度撞上一辆静止的 1 kg 手推车，它们粘在一起。碰撞前总动量 = (2 × 3) + (1 × 0) = 6 kg m/s。碰撞后总质量 = 3 kg，因此速度 v = 6 / 3 = 2 m/s。
6. Work, Energy and Power | 功、能量与功率
Work done (W) is the energy transferred when a force moves an object. It is calculated as:
功（W）是力使物体移动时所转移的能量。计算公式为：
W = F × d
where F is the force in newtons, d is the distance moved in the direction of the force in metres. Work is measured in joules (J). Gravitational potential energy gained: GPE = m × g × h (g = 10 N/kg on Earth for IGCSE).
其中 F 表示力（单位：牛顿），d 表示沿力方向移动的距离（单位：米）。功的单位是焦耳（J）。增加的重力势能：GPE = m × g × h（IGCSE 中 g 地球取 10 N/kg）。
Kinetic energy: KE = ½ × m × v². Power (P) is the rate of doing work or transferring energy: P = W / t or P = E / t, with the unit watt (W) = J/s.
动能：KE = ½ × m × v²。功率（P）表示做功或能量转换的快慢：P = W / t 或 P = E / t，单位是瓦特（W）= J/s。
Example: A crane lifts a 200 kg load through 12 m. Work done = mgh = 200 × 10 × 12 = 24 000 J. If it takes 8 s, power = 24 000 / 8 = 3000 W.
例题：一台起重机将 200 kg 的重物吊起 12 m。做功 = mgh = 200 × 10 × 12 = 24 000 J。若用时 8 秒，功率 = 24 000 / 8 = 3000 W。
7. Current, Voltage and Resistance (Ohm’s Law) | 电流、电压与电阻（欧姆定律）
For a resistor at constant temperature, the current (I) flowing through it is directly proportional to the potential difference (V) across it. The relationship is:
对于温度恒定的电阻器，流经它的电流（I）与它两端的电势差（V）成正比。关系式为：
V = I × R
where V is in volts (V), I in amperes (A), R in ohms (Ω). Rearrangements: I = V / R and R = V / I. Use this for series and parallel circuits carefully – remember to find total resistance first in series (R_total = R₁ + R₂ + …).
其中 V 的单位是伏特（V），I 的单位是安培（A），R 的单位是欧姆（Ω）。变形公式：I = V / R 和 R = V / I。应用在串联和并联电路中时要仔细——串联电路先求总电阻（R_total = R₁ + R₂ + …）。
Example: A lamp with a resistance of 12 Ω carries a current of 0.5 A. The pd across it is V = I × R = 0.5 × 12 = 6 V.
例题：某灯泡的电阻为 12 Ω，通过的电流为 0.5 A。它两端的电势差 V = I × R = 0.5 × 12 = 6 V。
8. Electrical Power and Energy Transfer | 电功率与电能转移
Electrical power can be calculated using three related equations. The most common is:
电功率可以用三个相关的公式计算。最常用的是：
P = I × V
Other forms: P = I² × R and P = V² / R. Energy transferred (E) is power multiplied by time: E = P × t. Energy is measured in joules (J), time in seconds (s). In domestic use, energy is often expressed in kilowatt-hours (kWh).
其他形式：P = I² × R 和 P = V² / R。转移的电能（E）等于功率乘以时间：E = P × t。电能的单位是焦耳（J），时间单位是秒（s）。在家庭用电中，电能常以千瓦时（kWh）表示。
Example: A 230 V heater draws a current of 4 A. Its power is P = 230 × 4 = 920 W. In 10 minutes (600 s), energy = 920 × 600 = 552 000 J (or 0.153 kWh).
例题：一个 230 V 的加热器通过 4 A 的电流。它的功率 P = 230 × 4 = 920 W。在 10 分钟（600 秒）内，消耗电能 = 920 × 600 = 552 000 J（即 0.153 kWh）。
9. Moles and Molar Mass (Chemistry) | 摩尔与摩尔质量（化学）
The amount of substance is measured in moles. The number of moles (n) is found from mass and molar mass (Mᵣ) using:
物质的数量以摩尔衡量。摩尔数（n）可通过质量和摩尔质量（Mᵣ）求得：
n = m ÷ M
where m is mass in grams, M is molar mass in g/mol. Molar mass is the relative formula mass in grams. Example: Calculate moles in 8 g of NaOH (Mᵣ = 40). n = 8 / 40 = 0.2 mol.
其中 m 为质量（单位：克），M 为摩尔质量（单位：g/mol）。摩尔质量即以克为单位的相对式量。例题：计算 8 g NaOH（Mᵣ = 40）的摩尔数，n = 8 / 40 = 0.2 mol。
You can also find the mass of a product from the mole ratio in a balanced equation. Always work in moles first, then convert to mass.
你还可以根据配平方程式中的摩尔比求出产物的质量。务必先用摩尔计算，再换算成质量。
10. Concentration and Volume of Solutions (Chemistry) | 溶液的浓度与体积（化学）
Concentration (c) is the amount of solute dissolved in a given volume of solution. It is often expressed in mol/dm³ (molarity) or g/dm³.
浓度（c）表示溶解在一定体积溶液中的溶质的量，通常以 mol/dm³（摩尔浓度）或 g/dm³ 表示。
c = n ÷ V
where n is moles of solute, V is volume of solution in dm³. Rearranged: n = c × V. Example: 0.5 mol of NaCl is dissolved in 2 dm³ of water. Concentration = 0.5 / 2 = 0.25 mol/dm³.
其中 n 为溶质的摩尔数，V 为溶液体积（单位：dm³）。变形：n = c × V。例题：将 0.5 mol 的 NaCl 溶于 2 dm³ 水中，浓度 = 0.5 / 2 = 0.25 mol/dm³。
Conversions: 1 dm³ = 1000 cm³. So if volume is given in cm³, divide by 1000 to get dm³. Always check the units carefully.
换算关系：1 dm³ = 1000 cm³。因此若体积以 cm³ 给出，需除以 1000 得到 dm³。务必仔细检查单位。
11. Magnification (Biology) | 放大倍率（生物）
In microscopy and biological drawings, magnification (M) tells you how much larger an image is compared to the actual specimen. The formula is:
在显微镜操作和生物绘图中，放大倍率（M）表示图像相比实际标本放大了多少。公式为：
M = I ÷ A
where I = image size, A = actual size. Both must be in the same unit, but image size is often in millimetres (mm) and actual size in micrometres (µm). Recall: 1 mm = 1000 µm.
其中 I = 图像尺寸，A = 实际尺寸。二者须使用相同单位，但图像尺寸通常以毫米（mm）表示，实际尺寸以微米（µm）表示。记住：1 mm = 1000 µm。
Example: A cell image measures 20 mm. The actual cell is 0.02 mm long. Magnification = 20 / 0.02 = ×1000. Alternatively, if actual size is 50 µm and image size is 5 mm (5000 µm), M = 5000 / 50 = ×100.
例题：一个细胞的图像长 20 mm，实际细胞长 0.02 mm。放大倍率 = 20 / 0.02 = ×1000。或者，若实际尺寸 50 µm，图像尺寸 5 mm（5000 µm），M = 5000 / 50 = ×100。
You may need to rearrange to find actual size: A = I / M. Always convert sizes to the same unit before dividing.
有时需要变形公式求实际尺寸：A = I / M。务必在相除之前将尺寸换算为统一单位。
12. General Tips for Calculation Questions | 计算题通用技巧
- Show all working: OCR examiners award marks for correct substitution and rearrangement, even if the final answer is wrong.
- 写出完整步骤：OCR 考官对正确的代入和变形式给予步骤分，哪怕最终答案错了。
- Check units: Convert all quantities to SI base units (m, kg, s, A, etc.) unless specified otherwise. Watch out for prefixes like k, m, μ.
- 检查单位：除非另有规定，将所有物理量换算成 SI 基本单位（m、kg、s、A 等）。注意 k、m、μ 等词头。
- Use standard form: For very large or very small numbers, use powers of ten to avoid errors (e.g., 0.005 A = 5 × 10⁻³ A).
- 使用标准形式：对非常大或非常小的数，用 10 的幂避免错误（如 0.005 A = 5 × 10⁻³ A）。
- Check significant figures: Give final answers to the same number of significant figures as the least precise data in the question, usually 2 or 3.
- 注意有效数字：最终答案的有效数字位数应等于题目中精度最低的数据的有效数字位数，通常为 2 或 3 位。
- Know your formula sheet: OCR provides some formulas, but not all. Memorise the ones not on the sheet, and practise rearrangement.
- 熟记公式表：OCR 会提供部分公式，但并非全部。记住那些不在公式表上的公式，并练习变形。
Practice daily with past paper questions, and always time yourself. Start with the easier one-step calculations before progressing to multi-step problems.
每天用真题练习并计时。先做简单的一步计算题，再挑战多步计算题。
Published by TutorHao | IGCSE OCR Science Calculations | 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