IGCSE CIE Physics: Intensive Calculation Practice | IGCSE CIE 物理：计算题专项训练

📚 IGCSE CIE Physics: Intensive Calculation Practice | IGCSE CIE 物理：计算题专项训练

Calculation questions are the backbone of the IGCSE CIE Physics exam. They test your ability to select the correct formula, handle units, and present logical working. This targeted practice guide revisits the most frequently assessed calculation topics, walks through worked examples, and highlights common pitfalls. Mastering these skills will boost both your confidence and your grade.

计算题是 IGCSE CIE 物理考试的核心组成部分，考查你选用正确公式、处理单位和展示解题逻辑的能力。本专项训练指南将重温最高频的计算题型，通过范例讲解和易错点提醒，帮助你扎实掌握答题技巧，提升应考信心和得分。

1. Key Equations and Units Review | 关键公式与单位回顾

Before attempting any calculation, ensure you can instantly recall these fundamental equations and their SI units. A quick mental checklist of symbols and units prevents costly errors.

开始计算前，请确保能立即想起下列基本公式及其国际单位。快速核对符号与单位可以避免无谓的失分。

Formula	Symbols & Units
average speed = total distance / total time	v = s / t (m/s, m, s)
acceleration = change in velocity / time taken	a = (v – u) / t (m/s², m/s, s)
force = mass × acceleration	F = m a (N, kg, m/s²)
weight = mass × gravitational field strength	W = m g (N, kg, N/kg)
momentum = mass × velocity	p = m v (kg m/s, kg, m/s)
kinetic energy = ½ × mass × speed²	KE = ½ m v² (J, kg, (m/s)²)
gravitational potential energy = mass × g × height	GPE = m g h (J, kg, N/kg, m)
work done = force × distance moved in direction of force	W = F d (J, N, m)
power = work done / time taken = energy transferred / time	P = W / t (W, J, s)
pressure = force / area	p = F / A (Pa, N, m²)
liquid pressure = density × g × depth	p = ρ g h (Pa, kg/m³, N/kg, m)
density = mass / volume	ρ = m / V (kg/m³, kg, m³)
wave speed = frequency × wavelength	v = f λ (m/s, Hz, m)
refractive index = sin i / sin r	n = sin i / sin r (no unit)
critical angle: sin c = 1 / n	sin c = 1 / n (degrees)
Ohm’s law: voltage = current × resistance	V = I R (V, A, Ω)
electrical power = current × voltage	P = I V (W, A, V)
electrical energy = power × time	E = P t (J, W, s)
thermal energy = mass × specific heat capacity × temperature change	Q = m c Δθ (J, kg, J/(kg °C), °C)
latent heat = mass × specific latent heat	Q = m L (J, kg, J/kg)

Always convert quantities to SI base or derived units before substituting. For example, change km to m, g to kg, cm² to m², and minutes to seconds. Keeping units inside the calculation helps you catch mistakes.

务必在代入公式前将所有物理量转换为国际基本或导出单位。例如将千米换成米，克换成千克，平方厘米换成平方米，分钟换成秒。计算过程中保留单位可以帮你发现错误。

2. Kinematics: Speed, Velocity and Acceleration | 运动学：速率、速度与加速度

Kinematics problems usually involve uniform acceleration. Memorise the ‘suvat’ equations and identify which quantities are given. Always state the positive direction.

运动学问题大多涉及匀加速运动。牢记 suvat 方程组，并识别题目给出了哪些量。解题前先明确正方向。

Example: A cyclist accelerates uniformly from rest at 1.5 m/s² for 20 seconds. Calculate the final velocity and the distance travelled.

例题：一名骑行者由静止开始以 1.5 m/s² 的加速度匀加速行驶 20 秒。求未速度和走过的距离。

Step 1: Write down knowns: initial velocity u = 0 m/s, acceleration a = 1.5 m/s², time t = 20 s.

步骤一：列出已知量：初速度 u = 0 m/s，加速度 a = 1.5 m/s²，时间 t = 20 s。

Step 2: Select v = u + a t. Substitute: v = 0 + (1.5 × 20) = 30 m/s.

步骤二：选用 v = u + a t 代入：v = 0 + 1.5 × 20 = 30 m/s。

Step 3: For distance, use s = u t + ½ a t² = 0 + 0.5 × 1.5 × 400 = 300 m. Alternatively, use average speed: v_av = (u+v)/2 = 15 m/s, so s = 15 × 20 = 300 m.

步骤三：求距离，用 s = u t + ½ a t² = 0 + 0.5 × 1.5 × 400 = 300 m。或利用平均速度：v_av = (u+v)/2 = 15 m/s，s = 15 × 20 = 300 m。

Common mistake: using average speed formula for non-uniform acceleration. It only works when acceleration is constant.

常见错误：在非匀加速情况下使用平均速度公式。该公式仅适用于加速度恒定。

3. Forces and Motion | 力与运动

Newton’s second law governs most force calculations. Always begin with a free-body diagram if multiple forces act. Remember that resultant force = mass × acceleration.

牛顿第二定律支配大部分力的计算。若存在多个力，先画受力分析图。牢记合力 = 质量 × 加速度。

Example: A 1200 kg car experiences a driving force of 5000 N and a total resistive force of 1400 N. Find the acceleration.

例题：一辆 1200 kg 的汽车受到 5000 N 的驱动力和 1400 N 的总阻力。求加速度。

Resultant force = 5000 N – 1400 N = 3600 N. Then a = F / m = 3600 / 1200 = 3.0 m/s².

合力 = 5000 N – 1400 N = 3600 N。a = F / m = 3600 / 1200 = 3.0 m/s²。

For momentum, use p = m v. Momentum is a vector; assign positive and negative directions in collision problems. Impulse = change in momentum = F t.

动量用 p = m v。动量是矢量，碰撞问题中需规定正负方向。冲量 = 动量变化量 = F t。

Hooke’s Law: F = k x where x is extension in metres. Watch out for units: if k is given in N/cm, convert to N/m.

胡克定律：F = k x，其中 x 为伸长量（米）。注意单位：若 k 以 N/cm 给出，须换算为 N/m。

Weight: W = m g. On Earth g = 9.8 N/kg, but the exam often uses 10 N/kg unless stated otherwise. Always check the front of the paper.

重量：W = m g。地球上 g = 9.8 N/kg，但试题常采用 10 N/kg，除非另有说明。务必查看试卷首页的数值。

4. Energy, Work and Power | 能量、功与功率

Energy calculations frequently combine kinetic energy and gravitational potential energy. Remember to square the speed in KE = ½ m v² — a single missed square loses many marks.

能量计算常结合动能与重力势能。记住动能 KE = ½ m v² 中的速度必须平方——漏掉平方会失掉大量分数。

Example: A ball of mass 2.0 kg is dropped from a height of 5.0 m. Ignoring air resistance, calculate its speed just before hitting the ground.

例题：质量为 2.0 kg 的球从 5.0 m 高处落下。忽略空气阻力，求落地前的速率。

Use energy conservation: loss in GPE = gain in KE. m g h = ½ m v². Cancel m: g h = ½ v². So v = √(2 g h) = √(2 × 10 × 5) = √100 = 10 m/s.

利用能量守恒：减少的重力势能 = 增加的动能。m g h = ½ m v²。消去 m，g h = ½ v²，所以 v = √(2 g h) = √(2 × 10 × 5) = √100 = 10 m/s。

Work done: W = F d. The distance must be in the same direction as the force. Power: P = W / t. Efficiency = useful output / total input, often expressed as a percentage.

功：W = F d，距离须与力同向。功率：P = W / t。效率 = 有用输出 / 总输入，常用百分数表示。

For electrical power, P = I V and P = I² R. These are useful when analysing energy transfers in circuits.

电功率公式 P = I V 和 P = I² R 在分析电路能量转移时十分有用。

5. Pressure and Density | 压强与密度

Pressure questions often require unit conversion for area (cm² to m²). 1 m² = 10 000 cm², so 1 cm² = 1 × 10⁻⁴ m². Missing this conversion is one of the most frequent errors.

压强题常需换算面积单位（cm² 换为 m²）。1 m² = 10 000 cm²，故 1 cm² = 1 × 10⁻⁴ m²。漏掉换算是最常见错误之一。

For liquid pressure, p = ρ g h. Depth h is measured vertically from the liquid surface. The shape of the container does not affect the pressure at a given depth.

液体压强 p = ρ g h，深度 h 从液面垂直向下测量。给定深度处的压强与容器形状无关。

Density: ρ = m / V. When measuring volume, 1 litre = 1000 cm³ = 0.001 m³. Always match mass and volume units to get density in kg/m³ or g/cm³.

密度：ρ = m / V。体积测量中 1 升 = 1000 cm³ = 0.001 m³。质量与体积单位要对应，使密度单位为 kg/m³ 或 g/cm³。

Example: A brick of dimensions 20 cm × 10 cm × 5 cm has a mass of 3 kg. Calculate the maximum pressure it can exert on a table.

例题：一块尺寸为 20 cm × 10 cm × 5 cm 的砖质量为 3 kg。求其放在桌面上时能产生的最大压强。

Weight = m g = 3 × 10 = 30 N. Minimum area = smallest face = 0.10 m × 0.05 m = 0.005 m² (convert cm to m). Maximum p = F / A_min = 30 / 0.005 = 6000 Pa.

重量 = m g = 30 N。最小接触面积 = 最小的面 = 0.10 m × 0.05 m = 0.005 m²。最大压强 p = F / A_min = 30 / 0.005 = 6000 Pa。

6. Waves: The Wave Equation and Refraction | 波动：波动方程与折射

Wave speed, frequency and wavelength are linked by v = f λ. Frequency can be found from the period: f = 1 / T. Be comfortable rearranging the equation for any variable.

波速、频率与波长的关系为 v = f λ。频率可由周期求出：f = 1 / T。应熟练针对任一变量整理公式。

Example: A water wave has a wavelength of 0.4 m and a frequency of 5 Hz. Calculate its speed. If the frequency doubles while the speed remains the same, what happens to the wavelength?

例题：水波波长为 0.4 m，频率为 5 Hz。计算其波速。若波速不变而频率加倍，波长如何变化？

v = f λ = 5 × 0.4 = 2.0 m/s. Doubling frequency gives λ = v / (2f) = 2.0 / 10 = 0.2 m, so wavelength halves.

v = 5 × 0.4 = 2.0 m/s。频率加倍后，λ = 2.0 / 10 = 0.2 m，波长减半。

Refractive index n = sin i / sin r. The angles are always measured from the normal. For critical angle, sin c = 1 / n. Make sure your calculator is in degree mode.

折射率 n = sin i / sin r，角度始终从法线量起。临界角公式 sin c = 1 / n。确保计算器处于角度模式。

When light travels from a denser to a less dense medium at an angle greater than the critical angle, total internal reflection occurs.

当光从光密介质射向光疏介质且入射角大于临界角时，发生全内反射。

7. Electricity: Ohm’s Law, Power and Resistance | 电学：欧姆定律、功率与电阻

Ohm’s law V = I R applies to ohmic conductors at constant temperature. For series circuits: current is the same everywhere; total resistance R_total = R₁ + R₂ + …; supply voltage is shared.

欧姆定律 V = I R 适用于恒温下的欧姆导体。串联电路中电流处处相等；总电阻 R_total = R₁ + R₂ + …；电源电压按电阻分配。

For parallel circuits: voltage across each branch equals the supply voltage; total current is the sum of branch currents; total resistance is given by 1/R_total = 1/R₁ + 1/R₂ + … This often confuses candidates, so practise rearrangement.

并联电路中各支路电压等于电源电压；总电流为各支路电流之和；总电阻满足 1/R_total = 1/R₁ + 1/R₂ + … 考生常混淆这点，务必多练习变形。

Example: Two resistors 6 Ω and 3 Ω are connected in parallel across a 12 V battery. Calculate the current through each resistor and the total current from the battery.

例题：6 Ω 和 3 Ω 的电阻并联在 12 V 电池上。求通过每个电阻的电流和电池总电流。

For 6 Ω: I = V / R = 12 / 6 = 2 A. For 3 Ω: I = 12 / 3 = 4 A. Total current = 2 + 4 = 6 A. Total resistance = V / I_total = 12 / 6 = 2 Ω, which matches 1/(1/6 + 1/3).

6 Ω 电阻：I = 12/6 = 2 A；3 Ω 电阻：I = 12/3 = 4 A。总电流 = 6 A，总电阻 = 12/6 = 2 Ω，与并联公式一致。

Electrical energy E = P t, and one kilowatt-hour (kWh) is the energy used by a 1000 W device for 1 hour. Convert watts to kilowatts and time to hours when calculating cost.

电能 E = P t，1 千瓦时（kWh）是 1000 W 电器工作 1 小时消耗的能量。计算电费时将瓦转千瓦、时间转小时。

8. Thermal Physics: Specific Heat and Latent Heat | 热物理：比热容与潜热

When a substance changes temperature, use Q = m c Δθ. The symbol Δθ (or ΔT) represents the temperature change. Use consistent units: m in kg, c in J/(kg °C), Δθ in °C, Q in J.

温度变化时使用 Q = m c Δθ，Δθ 代表温度变化。须统一单位：质量用 kg，比热容用 J/(kg °C)，Δθ 用 °C，热量用 J。

During a change of state, temperature remains constant. The energy required is Q = m L, where L is the specific latent heat (fusion or vaporisation). Do not use Δθ in this formula.

物态变化时温度保持不变，所需热量 Q = m L，L 为比潜热（熔化或汽化）。该公式中不得出现 Δθ。

Example: How much energy is needed to melt 0.50 kg of ice at 0 °C? Specific latent heat of fusion of ice = 334

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IGCSE CIE Physics: Intensive Calculation Practice | IGCSE CIE 物理：计算题专项训练