📚 IGCSE CCEA Physics: Calculation Questions Practice | IGCSE CCEA 物理：计算题专项训练
Welcome to this intensive revision guide focused on calculation questions for the CCEA IGCSE Physics specification. Mastering these numerical problems is essential for achieving a high grade, as they test not only your knowledge of formulae but also your ability to convert units, rearrange equations, and present final answers with appropriate significant figures. This article covers the most frequently examined calculation topics, providing step‑by‑step approaches and worked examples to build your confidence. Follow along, practise with the embedded questions, and remember that consistent application of the correct method will lead to success in the examination.
欢迎来到这本针对 CCEA IGCSE 物理考试计算题的强化复习指南。掌握这些数字问题对于获得高分至关重要，因为它们不仅考查你对公式的记忆，还考查单位换算、方程变形以及用合适有效数字给出最终答案的能力。本文涵盖了最常考的计算主题，提供分步骤的方法和例题解析，帮助你建立信心。跟着练习，记住始终使用正确的方法就会在考试中取得成功。
1. Speed and Acceleration | 速度与加速度
Speed, distance and time are linked by the equation v = s / t, where v is speed, s is distance and t is time. When an object moves at constant acceleration, we use a = (v – u) / t, with u being initial velocity, v final velocity and a acceleration. Always ensure that units are consistent: distance in metres (m), time in seconds (s), speed in m/s, and acceleration in m/s². In some exam questions you will need to convert km to m or minutes to seconds before substituting values.
速度、距离和时间由公式 v = s / t 联系，其中 v 是速度，s 是距离，t 是时间。当物体匀加速运动时，我们使用 a = (v – u) / t，u 是初速度，v 是末速度，a 是加速度。务必保持单位一致：距离用米 (m)，时间用秒 (s)，速度用 m/s，加速度用 m/s²。在一些考题中，你需要先将千米换算为米，或将分钟换算为秒，然后再代入数值。
Worked example: A sprinter accelerates from rest to 10 m/s in 4 seconds. Calculate the acceleration and the distance covered during this time. For acceleration: a = (10 – 0) / 4 = 2.5 m/s². To find distance, use the equation s = (u + v) × t / 2: s = (0 + 10) × 4 / 2 = 20 m. Notice how we choose the appropriate equation for the data given.
例题：一名短跑运动员从静止开始加速，4 秒内达到 10 m/s。计算加速度以及这段时间内跑过的距离。加速度：a = (10 – 0) / 4 = 2.5 m/s²。求距离时，使用公式 s = (u + v) × t / 2：s = (0 + 10) × 4 / 2 = 20 m。注意针对所给数据选择合适的公式。
2. Force and Newton’s Second Law | 力与牛顿第二定律
Newton’s second law states that the resultant force acting on an object is equal to the product of its mass and acceleration: F = m × a. Force is measured in newtons (N), mass in kilograms (kg), and acceleration in m/s². The weight of an object is a specific force caused by gravity: W = m × g, where g is the gravitational field strength (on Earth, approximately 9.8 N/kg, often taken as 10 N/kg in CCEA questions). Free‑body diagrams help to resolve forces and find the resultant before applying the formula.
牛顿第二定律指出，作用在物体上的合力等于其质量与加速度的乘积：F = m × a。力的单位是牛顿 (N)，质量是千克 (kg)，加速度是 m/s²。物体的重量是由重力引起的一种特殊的力：W = m × g，其中 g 是引力场强度（在地球上约为 9.8 N/kg，CCEA 考试中常取 10 N/kg）。受力示意图有助于分解力并先求出合力，再应用公式。
Often you will combine the two equations. For example, a 2 kg mass hangs on a rope. The tension T in the rope must balance the weight: T – W = m × a. If the mass is accelerating upward at 1.5 m/s², then T – (2 × 10) = 2 × 1.5, so T = 20 + 3 = 23 N. Always define the positive direction clearly in your working.
常常需要将两个方程结合。例如，一个 2 kg 的质量悬挂在绳子上。绳子张力 T 必须与重量平衡：T – W = m × a。如果该质量以 1.5 m/s² 向上加速，那么 T – (2 × 10) = 2 × 1.5，所以 T = 20 + 3 = 23 N。解题时始终要清晰地规定正方向。
3. Work, Energy and Power | 功、能与功率
Work done is defined as the force multiplied by the distance moved in the direction of the force: W = F × d, with work in joules (J). Energy can exist in different forms, and two important mechanical forms are kinetic energy, Eₖ = ½ × m × v², and gravitational potential energy, Eₚ = m × g × h. Power is the rate of doing work or transferring energy: P = W / t, measured in watts (W). When a question asks for the power of a machine, you often need to calculate the work done first and then divide by the time taken.
功定义为力乘以在力的方向上移动的距离：W = F × d，功的单位是焦耳 (J)。能量可以以不同形式存在，两种重要的机械能是动能 Eₖ = ½ × m × v² 和重力势能 Eₚ = m × g × h。功率是做功或转换能量的速率：P = W / t，单位是瓦特 (W)。当题目要求计算机器的功率时，你通常需要先算出所做的功，再除以所用的时间。
In a typical problem, a 0.5 kg ball is dropped from a height of 4 m. Calculate its speed just before hitting the ground, assuming no air resistance. Using conservation of energy: loss in Eₚ = gain in Eₖ. m × g × h = ½ × m × v², so v² = 2 × g × h = 2 × 10 × 4 = 80, therefore v = √80 ≈ 8.94 m/s. Notice that the mass cancels, so any object dropped from the same height reaches the same speed.
在典型问题中，一个 0.5 kg 的小球从 4 m 高处落下。假设无空气阻力，计算其刚好撞击地面前的速度。利用能量守恒：减少的重力势能 = 增加的动能。m × g × h = ½ × m × v²，所以 v² = 2 × g × h = 2 × 10 × 4 = 80，因此 v = √80 ≈ 8.94 m/s。注意质量被消去，因此任何物体从相同高度落下都会达到相同的速度。
4. Density and Pressure | 密度与压强
Density is mass per unit volume: ρ = m / V. The SI unit is kg/m³, but you may also encounter g/cm³. Remember that 1 g/cm³ = 1000 kg/m³. For regular solids, volume can be found from geometry; for irregular objects, use the displacement method. Pressure is force per unit area: P = F / A, with pascals (Pa) equal to N/m². In fluids, pressure increases with depth and can be calculated as P = ρ × g × h, where h is the depth below the surface.
密度是单位体积的质量：ρ = m / V。国际单位是 kg/m³，但你也会遇到 g/cm³。记住 1 g/cm³ = 1000 kg/m³。对于规则固体，体积可以通过几何方法求得；对于不规则物体，使用排水法。压强是单位面积上的力：P = F / A，单位帕斯卡 (Pa) 等于 N/m²。在流体中，压强随深度增加，可以用 P = ρ × g × h 计算，其中 h 是表面以下的深度。
A common examination task involves a rectangular block of dimensions 0.2 m × 0.1 m × 0.05 m and mass 2 kg resting on a table. Calculate the maximum and minimum pressure it can exert. Area of the smallest face = 0.1 × 0.05 = 0.005 m², so maximum pressure = F / A = (2 × 10) / 0.005 = 4000 Pa. For minimum pressure, use the largest face area = 0.2 × 0.1 = 0.02 m², giving P = 20 / 0.02 = 1000 Pa.
一种常见的考试题涉及一个尺寸为 0.2 m × 0.1 m × 0.05 m、质量为 2 kg 的长方体放在桌子上。计算它能产生的最大和最小压强。最小面的面积 = 0.1 × 0.05 = 0.005 m²，因此最大压强 = F / A = (2 × 10) / 0.005 = 4000 Pa。对于最小压强，使用最大面面积 = 0.2 × 0.1 = 0.02 m²，得出 P = 20 / 0.02 = 1000 Pa。
5. Current, Voltage and Resistance in Circuits | 电路中的电流、电压与电阻
Ohm’s law is the foundation of circuit calculations: V = I × R, where V is potential difference in volts (V), I is current in amperes (A), and R is resistance in ohms (Ω). For components connected in series, the current is the same everywhere, the total voltage is shared, and the total resistance is the sum of individual resistances: Rtotal = R₁ + R₂ + …. In parallel circuits, the voltage across each branch is the same, the total current is the sum of branch currents, and the total resistance is found from 1 / Rtotal = 1 / R₁ + 1 / R₂ + ….
欧姆定律是电路计算的基础：V = I × R，其中 V 是电势差（伏特 V），I 是电流（安培 A），R 是电阻（欧姆 Ω）。对于串联的元件，电流处处相同，总电压被分配，总电阻等于各电阻之和：Rtotal = R₁ + R₂ + …。在并联电路中，各支路两端电压相同，总电流等于各支路电流之和，总电阻由 1 / Rtotal = 1 / R₁ + 1 / R₂ + … 求得。
Consider a 6 Ω and a 3 Ω resistor connected in parallel, and this combination is connected in series with a 4 Ω resistor across a 12 V battery. First, find the parallel resistance: 1 / Rparallel = 1/6 + 1/3 = 1/2, so Rparallel = 2 Ω. Total circuit resistance = 2 + 4 = 6 Ω. Then total current from battery: Itotal = Vtotal / Rtotal = 12 / 6 = 2 A. The voltage across the parallel block = Itotal × Rparallel = 2 × 2 = 4 V. Therefore the current through the 6 Ω resistor = 4 V / 6 Ω ≈ 0.67 A. Breaking the problem into steps prevents confusion.
考虑一个 6 Ω 和一个 3 Ω 的电阻并联，然后这个组合与一个 4 Ω 电阻串联，接在 12 V 电池上。首先，求并联电阻：1 / Rparallel = 1/6 + 1/3 = 1/2，所以 Rparallel = 2 Ω。电路总电阻 = 2 + 4 = 6 Ω。然后电池提供的总电流：Itotal = Vtotal / Rtotal = 12 / 6 = 2 A。并联块两端的电压 = Itotal × Rparallel = 2 × 2 = 4 V。因此流过 6 Ω 电阻的电流 = 4 V / 6 Ω ≈ 0.67 A。将问题拆分成步骤可以避免混淆。
6. Electrical Power and Energy | 电功率与电能
Electrical power can be expressed in three useful forms: P = I × V, P = I² × R, and P = V² / R. The appropriate version depends on the given quantities. Energy transferred in an electrical device is E = P × t, where t is time in seconds, giving energy in joules. In domestic contexts, kilowatt‑hours (kW h) may be used: energy (kW h) = power (kW) × time (h). One kW h equals 3.6 × 10⁶ J. Exam questions often require you to combine these with cost calculations.
电功率可以用三种有用的形式表达：P = I × V、P = I² × R 和 P = V² / R。选择哪一种取决于已知量。电器中转移的电能为 E = P × t，其中 t 是时间（秒），能量单位为焦耳。在家庭用电中，可以用千瓦时 (kW h)：电能 (kW h) = 功率 (kW) × 时间 (h)。1 kW h 等于 3.6 × 10⁶ J。考试题常要求你结合这些公式计算电费。
A worked example: a 230 V heater has a resistance of 50 Ω. Determine the power rating and the cost of running it for 3 hours if each kW h costs 15 pence. Using P = V² / R: P = (230)² / 50 = 52900 / 50 = 1058 W = 1.058 kW. Energy used in 3 h = 1.058 kW × 3 h = 3.174 kW h. Cost = 3.174 × 15 p = 47.61 p. Always check whether to give answers in pence or pounds.
例题：一个 230 V 的加热器具有 50 Ω 的电阻。计算其额定功率，以及如果每 kW h 电费为 15 便士，运行 3 小时的成本。使用 P = V² / R：P = (230)² / 50 = 52900 / 50 = 1058 W = 1.058 kW。3 小时使用的电能 = 1.058 kW × 3 h = 3.174 kW h。费用 = 3.174 × 15 便士 = 47.61 便士。始终注意答案应该用便士还是英镑给出。
7. Wave Speed, Frequency and Wavelength | 波速、频率与波长
All waves obey the wave equation: v = f × λ, where v is wave speed (m/s), f is frequency (Hz), and λ is wavelength (m). This relationship is fundamental for both transverse and longitudinal waves, including sound, light and water waves. In ripple tank experiments or electromagnetic spectrum questions, you may be given any two of the three quantities and asked to find the third. Do not forget that period T = 1 / f, and the wave equation can also be written as v = λ / T.
所有波都遵循波动方程：v = f × λ，其中 v 是波速 (m/s)，f 是频率 (Hz)，λ 是波长 (m)。这一关系对横波和纵波都基本适用，包括声波、光波和水波。在波纹槽实验或电磁波谱题目中，通常会给出三个量中的任意两个，要求你求出第三个。不要忘记周期 T = 1 / f，波动方程也可写作 v = λ / T。
A radio station transmits waves with frequency 100 MHz and wavelength 3 m. The speed v = (100 × 10⁶ Hz) × 3 m = 3 × 10⁸ m/s, which confirms that electromagnetic waves travel at the speed of light in a vacuum. If a water wave has a speed of 25 cm/s and a wavelength of 5 cm, its frequency is f = v / λ = 25 / 5 = 5 Hz. Always express speed and wavelength in the same length unit before dividing.
某广播电台发射频率为 100 MHz、波长为 3 m 的波。波速 v = (100 × 10⁶ Hz) × 3 m = 3 × 10⁸ m/s，这证实了电磁波在真空中以光速传播。如果一个水波波速为 25 cm/s、波长为 5 cm，其频率 f = v / λ = 25 / 5 = 5 Hz。在做除法之前，始终将波速和波长表示为相同的长度单位。
8. Specific Heat Capacity | 比热容
When an object is heated, the temperature rise depends on its mass, the material’s specific heat capacity and the energy supplied. The equation is Q = m × c × Δθ, where Q is heat energy (J), m is mass (kg), c is specific heat capacity (J/(kg °C)), and Δθ is temperature change (°C or K). Water has a high specific heat capacity of about 4200 J/(kg °C), which is a common value in numerical problems. Rearranging to find any unknown is a key skill.
当物体受热时，温度升高的程度取决于其质量、材料的比热容以及提供的能量。方程是 Q = m × c × Δθ，其中 Q 是热能 (J)，m 是质量 (kg)，c 是比热容 (J/(kg °C))，Δθ 是温度变化 (°C 或 K)。水的比热容较大，约为 4200 J/(kg °C)，这是数值题中的常见数值。变换公式以求出任何一个未知量是一项关键技能。
Example: An electric heater supplies 10 000 J of energy to a 0.5 kg aluminium block (c = 900 J/(kg °C)). Find the temperature rise. Rearranging: Δθ = Q / (m × c) = 10000 / (0.5 × 900) = 10000 / 450 ≈ 22.2 °C. Some questions combine this with electrical power: if the heater operates at 50 W, the time taken can be found from Q = P × t. Then t = 10000 / 50 = 200 s. Always show the substitution step clearly.
例题：一个电加热器给一个 0.5 kg 的铝块（c = 900 J/(kg °C)）提供 10 000 J 的能量。求温度升高多少。变形公式：Δθ = Q / (m × c) = 10000 / (0.5 × 900) = 10000 / 450 ≈ 22.2 °C。有些题目会结合电功率来考：如果加热器功率为 50 W，那么所需时间可由 Q = P × t 求出。则 t = 10000 / 50 = 200 s。始终清晰地展示代入步骤。
9. Radioactive Decay and Half‑life | 放射性衰变与半衰期
The half‑life of a radioactive isotope is the time taken for half the nuclei in a sample to decay, or for the count rate to fall to half its original value. You can calculate the remaining mass or activity after a given number of half‑lives using: remaining amount = original amount × (½)ⁿ, where n is the number of half‑lives elapsed (n = total time / half‑life). Alternatively, sketch a decay curve and read values from the graph.
放射性同位素的半衰期是指样品中一半的原子核发生衰变所需的时间，或者计数率下降到初始值一半所需的时间。你可以使用以下方法计算经过一定数量半衰期后的剩余质量或活度：剩余量 = 初始量 × (½)ⁿ，其中 n 是经历的半衰期个数（n = 总时间 / 半衰期）。或者，可以画一条衰变曲线并从图中读取数值。
A sample of iodine‑131 has a half‑life of 8 days and an initial mass of 40 mg. Calculate the mass remaining after 24 days. Number of half‑lives, n = 24 / 8 = 3. Remaining mass = 40 mg × (½)³ = 40 × 1/8 = 5 mg. If a question asks for the time to decay to a certain fraction, rearrange: 40 × (½)ⁿ = 2.5 → (½)ⁿ = 1/16 → n = 4, so time = 4 × 8 = 32 days. This method is quick and reliable.
一个碘‑131 样品的半衰期为 8 天，初始质量为 40 mg。计算 24 天后的剩余质量。半衰期个数 n = 24 / 8 = 3。剩余质量 = 40 mg × (½)³ = 40 × 1/8 = 5 mg。如果题目要求计算衰变到某一分数所需的时间，则进行变形：40 × (½)ⁿ = 2.5 → (½)ⁿ = 1/16 → n = 4，所以时间 = 4 × 8 = 32 天。这种方法快捷而可靠。
10. Efficiency | 效率
Efficiency measures how well a device converts input energy into useful output energy. It can be expressed as a decimal or a percentage: Efficiency = (useful output energy or power / total input energy or power) × 100%. No real machine is 100% efficient due to energy losses, often as heat due to friction or electrical resistance. When calculating efficiency, you must identify the ‘useful’ component correctly from the description.
效率衡量一个设备将输入能量转化为有用输出能量的程度。它可以用小数或百分比表示：效率 = (有用的输出能量或功率 / 总的输入能量或功率) × 100%。由于存在能量损失（通常为摩擦或电阻发热），没有真实的机器能达到 100% 的效率。在计算效率时，你必须根据描述正确识别出“有用”的部分。
For instance, an electric motor lifts a 20 N weight through a height of 3 m, doing useful work = 20 × 3 = 60 J. The motor draws 100 J of electrical energy. Efficiency = (60 / 100) × 100% = 60%. The remaining 40 J is wasted as heat and sound. In a power station context, overall efficiency is the product of individual efficiencies. Practising these multi‑step calculations is vital for CCEA exams, where such questions often combine different topics.
例如，一台电动机将一个重 20 N 的物体提升了 3 m，所做的有用功 = 20 × 3 = 60 J。电动机消耗了 100 J 的电能。效率 = (60 / 100) × 100% = 60%。剩下的 40 J 以热和声的形式浪费。在发电站的情景中，总效率是各环节效率的乘积。对于 CCEA 考试来说，练习这些多步骤计算至关重要，因为这类题目常常跨不同主题。
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