📚 IGCSE Edexcel Physics: Calculation Question Drills | IGCSE Edexcel 物理：计算题专项训练
Calculation questions form the backbone of IGCSE Edexcel Physics papers, and mastering them requires more than just knowing the formulas. It demands a systematic approach: extracting data, converting units, selecting the right equation, rearranging it correctly, and plugging in numbers with confidence. This guide is designed to walk you through each of these skills with targeted drills, worked examples and common pitfalls, so you can tackle any numerical problem calmly in the exam.
计算题是 IGCSE Edexcel 物理试卷的核心，掌握计算题需要的不仅仅是记住公式，更需要一套系统的方法：提取数据、转换单位、选择正确的方程、正确变形，然后自信地代入数值。本指南将通过针对性的训练、示例解析和常见错误分析，带你逐一攻克这些技能，让你能够在考试中从容应对任何数值问题。
1. Understanding the Question and Data | 理解题目与数据
Before you reach for your calculator, underline or highlight every numerical value given in the question, together with its unit. Also identify the quantity you are asked to find and its expected unit. This simple habit prevents you from rushing into a formula with the wrong numbers or missing hidden information such as ‘at rest’ meaning initial velocity u = 0 m/s.
在你拿起计算器之前，先划出题目中给出的每一个数值及其单位，同时明确题目要求你求的是什么物理量及其应有的单位。这个简单的习惯可以防止你匆忙代入错误的数据，或忽略隐藏信息，比如“静止”意味着初速度 u = 0 m/s。
Many IGCSE questions embed data in diagrams or graphs. Always transfer these numbers onto your working space. If a velocity-time graph shows a final velocity of 15 m/s and a time of 6 s, write v = 15 m/s, t = 6 s before deciding which SUVAT equation to use. Re-reading the question after you have an answer is also vital – does your result make sense? For example, a car’s acceleration is unlikely to be 150 m/s².
许多 IGCSE 题目会将数据隐藏在示意图或图表中。务必把这些数字誊写到你的草稿区。如果一张速度-时间图显示末速度为 15 m/s、时间为 6 s，那么在选择 SUVAT 方程之前，先写下 v = 15 m/s、t = 6 s。得出答案后再回读一遍题目也至关重要——你的结果合理吗？例如，一辆汽车的加速度不太可能是 150 m/s²。
2. Formula Recall and Selection | 公式记忆与选择
Edexcel provides a formula sheet, but you must know which equation corresponds to which context. The key is to match the variables you have with the variables in the formula. If you know initial velocity, acceleration and time, and need final velocity, v = u + at is your immediate choice. The table below lists essential formulas you will use repeatedly.
Edexcel 会提供公式表，但你必须清楚哪个方程对应哪种情境。关键是将你手头已知的变量与公式中的变量匹配起来。如果你知道初速度、加速度和时间，需要求末速度，那么 v = u + at 就是你的首选。下表列出了你会反复使用的基本公式。
| Quantity | Formula | Typical units |
|---|---|---|
| Velocity | v = s / t | m/s |
| Acceleration | a = (v – u) / t | m/s² |
| SUVAT – without s | v = u + at | – |
| SUVAT – without v | s = ut + ½at² | – |
| SUVAT – without t | v² = u² + 2as | – |
| SUVAT – average speed | s = (u+v)t / 2 | – |
| Force | F = m × a | N |
| Weight | W = m × g | N |
| Kinetic energy | Eₖ = ½mv² | J |
| Gravitational potential energy | Eₚ = mgh | J |
| Work done | W = F × d | J |
| Power | P = W / t or P = E / t | W |
| Pressure | p = F / A | Pa |
| Density | ρ = m / V | kg/m³ |
| Moment | M = F × d | Nm |
| Wave speed | v = f × λ | m/s |
| Ohm’s law | V = I × R | V |
| Electrical power | P = I × V or P = I²R | W |
| Energy transferred | E = I × V × t | J |
| Efficiency | η = (useful output / total input) × 100% | % |
Practise writing these formulas from memory every day, and say aloud what each symbol stands for. In the exam, if you are unsure, check the units given: a formula yielding m/s when you expect m/s² is probably wrong.
每天凭记忆默写这些公式，并口头说出每个符号代表什么。考试中如果你不确定，可以检查单位：一个得到 m/s 的公式在你预期 m/s² 时就很可能是错的。
3. Unit Conversion Mastery | 单位换算精通
IGCSE marks are often lost because students forget to convert grams to kilograms, centimetres to metres, or minutes to seconds. Always convert to SI base units before substituting into a formula: mass in kg, length in m, time in s, force in N, energy in J. Write down the conversion factor explicitly, e.g., 250 g = 0.25 kg.
学生在 IGCSE 考试中常因忘记将克换算为千克、厘米换算为米或分钟换算为秒而失分。务必在代入公式前换算成 SI 基本单位：质量用 kg、长度用 m、时间用 s、力用 N、能量用 J。要明确写下换算系数，例如 250 g = 0.25 kg。
For area and volume, the conversion requires squaring or cubing the linear factor. 1 cm² = (0.01 m)² = 1 × 10⁻⁴ m², and 1 cm³ = 1 × 10⁻⁶ m³. When a question gives pressure in N/cm², convert it to N/m² (Pa) by multiplying by 10 000. Always double-check powers of ten.
面积和体积的换算需要对长度系数进行平方或立方运算。1 cm² = (0.01 m)² = 1 × 10⁻⁴ m²，1 cm³ = 1 × 10⁻⁶ m³。如果题目给出的压强单位是 N/cm²，要将其转换为 Pa (N/m²) 就需要乘以 10 000。务必反复检查数量级。
4. Rearranging Equations | 方程变形
Many calculation questions require you to rearrange a formula before substituting numbers. Practise the algebra step by step: to make ‘a’ the subject of v² = u² + 2as, first subtract u² from both sides, then divide by 2s. Write every intermediate step – this reduces careless errors and shows the examiner your method even if the final answer is slightly off.
许多计算题要求你先对方程进行变形，再代入数值。一步一步地练习代数操作：要将 v² = u² + 2as 改写成以 a 为主语的表达式，首先两边减去 u²，再除以 2s。写出每一步中间过程，这能减少粗心错误，而且即使最终答案有些偏差，也能向考官展示你的解题思路。
A common trap is mishandling squares and square roots. After rearranging to v² = u² + 2as, you must take the square root of the entire right-hand side, not just the last term. Use brackets on your calculator: sqrt(u² + 2as). Similarly, when finding v from kinetic energy, v = √(2Eₖ / m). Practise these rearrangements with simple numbers to build confidence.
一个常见的陷阱是错误处理平方和平方根。变形得到 v² = u² + 2as 后，你必须对整个右侧开平方，而不是只对最后一项。在计算器上使用括号：sqrt(u² + 2as)。类似地，由动能求速度时，v = √(2Eₖ / m)。用简单的数字练习这些变形，以建立信心。
5. Motion Calculations (SUVAT) | 运动学计算 (SUVAT)
SUVAT equations only apply when acceleration is constant. Begin by listing the five quantities: s (displacement), u (initial velocity), v (final velocity), a (acceleration), t (time). Identify which three you know and which one you need. The equation that omits the unknown you don’t need will give the quickest route.
SUVAT 方程仅适用于加速度恒定的情形。首先列出五个物理量：s（位移）、u（初速度）、v（末速度）、a（加速度）和 t（时间）。确定你知道哪三个，需要求哪一个。那个不包含你不需要的未知量的方程会给出最快捷的计算路径。
Drill example: A cyclist accelerates uniformly from 2 m/s to 8 m/s over a distance of 30 m. Find the acceleration. Known: u = 2, v = 8, s = 30. Unknown: a. Use v² = u² + 2as → 8² = 2² + 2a×30 → 64 = 4 + 60a → 60a = 60 → a = 1.0 m/s². Notice how working with symbols first reduces numerical clutter.
训练示例：一名自行车骑手从 2 m/s 匀加速到 8 m/s，行驶距离为 30 m。求加速度。已知：u = 2，v = 8，s = 30。未知：a。使用 v² = u² + 2as → 8² = 2² + 2a×30 → 64 = 4 + 60a → 60a = 60 → a = 1.0 m/s²。注意先使用符号运算可以减少数字杂乱。
For vertical motion under gravity, a = 10 m/s² (downwards). Be careful with sign conventions: if you take upwards as positive, then a = -10 m/s². An object thrown upwards with u = +15 m/s will have v = 0 at its highest point.
对于重力作用下的竖直运动，a = 10 m/s²（向下）。注意正负号规定：若取向上为正，则 a = -10 m/s²。以 u = +15 m/s 向上抛出的物体，在最高点处 v = 0。
6. Forces and Newton’s Laws Calculations | 力与牛顿定律计算
Resultant force problems often combine F = ma with weight W = mg. Always draw a free-body diagram to show all forces acting on the object. For a lift accelerating upwards, the tension T in the cable must overcome both the weight and provide the net force: T – mg = ma, so T = m(g + a).
合力问题常将 F = ma 与重力 W = mg 结合起来。一定要画出受力示意图，标出作用在物体上的所有力。对于向上加速的电梯，缆绳张力 T 既要克服重力，又要提供净力：T – mg = ma，因此 T = m(g + a)。
Worked drill: A 1200 kg car experiences a driving force of 4500 N and a resistive force of 900 N. Calculate its acceleration. Resultant force = 4500 – 900 = 3600 N. Using F = ma → 3600 = 1200 × a → a = 3.0 m/s². Always subtract opposing forces first to find the net force.
训练示例：一辆 1200 kg 的汽车受到 4500 N 的驱动力和 900 N 的阻力。计算其加速度。合力 = 4500 – 900 = 3600 N。利用 F = ma → 3600 = 1200 × a → a = 3.0 m/s²。务必先减去反向力求净力。
7. Energy, Work and Power | 能量、功与功率
Energy calculations often involve converting between kinetic energy (Eₖ = ½mv²) and gravitational potential energy (Eₚ = mgh). In an ideal system with no friction, total mechanical energy is conserved: ½mv² + mgh = constant. Use this to find the speed of a falling object or the height reached by a projectile.
能量计算常涉及动能 (Eₖ = ½mv²) 和重力势能 (Eₚ = mgh) 之间的转换。在无摩擦的理想系统中，总机械能守恒：½mv² + mgh = 常数。利用这一点可以求出下落物体的速度，或抛射体达到的高度。
When work is done against a force, W = F × d, where d is the distance moved in the direction of the force. Power is the rate of doing work, P = W / t. A 60 W lamp running for 5 minutes transfers E = P × t = 60 × (5×60) = 18 000 J. Never forget to convert time to seconds.
当克服某个力做功时，W = F × d，其中 d 是在力的方向上移动的距离。功率是做功的速率，P = W / t。一盏 60 W 的灯工作 5 分钟，传递的能量为 E = P × t = 60 × (5×60) = 18 000 J。永远不要忘记将时间换算成秒。
8. Density, Pressure and Moments | 密度、压强与力矩
Density ρ = m / V requires volume in m³. For a regular solid, calculate volume from lengths; for an irregular object, the question may give the volume directly or describe a displacement method. Example: a block of mass 0.5 kg has dimensions 10 cm × 5 cm × 4 cm. Volume = 0.10 × 0.05 × 0.04 = 2 × 10⁻⁴ m³. Density = 0.5 / (2×10⁻⁴) = 2500 kg/m³.
密度 ρ = m / V 需要体积以 m³ 为单位。对于规则固体，可通过长度计算体积；对于不规则物体，题目可能直接给出体积或描述排水法。示例：一个质量为 0.5 kg 的物块尺寸为 10 cm × 5 cm × 4 cm。体积 = 0.10 × 0.05 × 0.04 = 2 × 10⁻⁴ m³。密度 = 0.5 / (2×10⁻⁴) = 2500 kg/m³。
Pressure in a fluid at depth h is p = ρgh, where ρ is the liquid density and g = 10 N/kg. The total pressure at that depth is atmospheric pressure plus ρgh. Moment calculations rely on the principle of moments: sum of clockwise moments = sum of anticlockwise moments about a pivot. Remember that moment = force × perpendicular distance from pivot.
在深度为 h 的流体中，压强为 p = ρgh，其中 ρ 是液体密度，g = 10 N/kg。该深度处的总压强为大气压加上 ρgh。力矩计算依据力矩原理：对一个支点而言，顺时针力矩之和等于逆时针力矩之和。记住力矩 = 力 × 力的作用线到支点的垂直距离。
9. Waves: Speed, Frequency and Wavelength | 波：波速、频率与波长
The wave equation v = f λ is straightforward, but watch out for unit prefixes: kHz must become Hz, MHz → 10⁶ Hz, cm or mm must become metres. A wave with frequency 250 kHz and wavelength 0.5 cm has f = 250 000 Hz, λ = 0.005 m, so v = 250 000 × 0.005 = 1250 m/s.
波动方程 v = f λ 很简单，但要当心单位前缀：kHz 必须转换为 Hz，MHz → 10⁶ Hz，cm 或 mm 必须转换为米。一列频率为 250 kHz、波长为 0.5 cm 的波，f = 250 000 Hz，λ = 0.005 m，因此 v = 250 000 × 0.005 = 1250 m/s。
When waves move from deep to shallow water, frequency remains constant, speed decreases, and wavelength decreases. Use the same equation to find the new wavelength: λ’ = v’ / f. If speed halves, wavelength halves. Always identify which quantity stays the same before calculating.
当波从深水区进入浅水区时，频率保持不变，波速减小，波长减小。用同一个公式求新波长：λ’ = v’ / f。如果波速减半，波长也减半。在计算之前，一定要先判明哪个物理量保持不变。
10. Electricity: Ohm’s Law and Power | 电学：欧姆定律与电功率
Ohm’s law V = IR is the starting point for most circuit calculations. For series circuits, current is the same everywhere, total resistance R_total = R₁ + R₂ + …, and supply voltage is shared. For parallel circuits, voltage across each branch is the same, total current is the sum of branch currents, and combined resistance is found using 1/R_total = 1/R₁ + 1/R₂.
欧姆定律 V = IR 是大多数电路计算的起点。对于串联电路，各处电流相同，总电阻 R_total = R₁ + R₂ + …，电源电压被分配。对于并联电路，各支路两端电压相同，总电流等于各支路电流之和，总电阻通过 1/R_total = 1/R₁ + 1/R₂ 计算。
Electrical power can be calculated as P = IV, P = I²R, or P = V²/R. Choose the form that uses known quantities directly. A 12 V, 24 W lamp draws current I = P/V = 24/12 = 2 A, and its resistance is R = V/I = 12/2 = 6 Ω. Energy transferred over 10 minutes is E = P × t = 24 × 600 = 14 400 J.
电功率可以通过 P = IV、P = I²R 或 P = V²/R 来计算。选择能直接利用已知量的公式形式。一盏 12 V、24 W 的灯，其工作电流 I = P/V = 24/12 = 2 A，电阻 R = V/I = 12/2 = 6 Ω。10 分钟内转移的能量为 E = P × t = 24 × 600 = 14 400 J。
11. Efficiency and Sankey Diagrams | 效率与桑基图
Efficiency = (useful energy output / total energy input) × 100 %. This can be applied to devices, energy transfers, or power. An electric motor lifting a 4 kg mass through 5 m does useful work = mgh = 4 × 10 × 5 = 200 J. If the motor draws 500 J of electrical energy, efficiency = (200 / 500) × 100 = 40%.
效率 = （有用的能量输出 / 总能量输入）× 100%。这可以应用于设备、能量转换或功率。一台电动机将 4 kg 的重物提升 5 m，所做的有用功 = mgh = 4 × 10 × 5 = 200 J。若电动机消耗了 500 J 的电能，则效率 = (200 / 500) × 100 = 40%。
Sankey diagrams visualise energy flow; the width of the arrows is proportional to the amount of energy. To calculate wasted energy, subtract useful output from total input. A gas boiler with 80% efficiency wastes 20% of the chemical energy as heat to the surroundings. Exam questions may ask you to calculate the wasted power or energy from given values.
桑基图将能量流可视化；箭头的宽度与能量的多少成正比。要计算浪费的能量，只需用总输入减去有用输出。一台效率为 80% 的燃气锅炉会将 20% 的化学能作为热量浪费到环境中。考题可能会要求你根据给定数值计算浪费的功率或能量。
12. Common Pitfalls and Exam Tips | 常见错误与考试技巧
Pitfall 1: Using the wrong mass – distinguish between mass (kg) and weight (N). In F = ma, it is mass, not weight. Pitfall 2: Forgetting to square the velocity in kinetic energy. ½ × m × v² is not the same as ½ × m × v × 2. Pitfall 3: Ignoring direction in momentum or velocity; sign errors can flip your answer.
常见错误一：混淆质量
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