📚 GCSE Edexcel Physics: Tackling Calculations | GCSE Edexcel 物理：计算题专项训练
Success in GCSE Edexcel Physics depends heavily on your ability to handle numerical problems with confidence. This article breaks down the essential calculation topics, from kinematics to nuclear decay, providing clear methods, example solutions, and key exam tips. Each section is presented in both English and Chinese to support bilingual learners, ensuring you can follow the logic no matter which language you prefer. By working through these targeted drills, you will master unit conversions, formula selection, and multi-step reasoning required for top marks.
想在 GCSE Edexcel 物理考试中取得好成绩，熟练处理计算题至关重要。本文拆解了从运动学到核衰变等核心计算专题，提供清晰的解题方法、例题详解和关键的考试技巧。每个小节均以中英双语配对呈现，无论你更熟悉哪种语言，都能跟上逻辑脉络。通过这一系列针对性训练，你将牢固掌握单位换算、公式选择以及高分所需的跨步骤推理能力。
1. Formula Recall and Unit Conversions | 公式记忆与单位换算
Before tackling any calculation, you must be able to recall the relevant equation and convert all quantities into standard SI units. Edexcel provides a formula sheet, but knowing the equations by heart saves time. Common unit traps include grams instead of kilograms, centimetres instead of metres, and minutes instead of seconds.
在开始任何计算之前，你必须能迅速回忆相关公式，并将所有物理量换算为国际单位制 (SI)。Edexcel 考试会提供公式表，但熟记公式可以争取更多时间。常见的单位陷阱包括误用克 (g) 代替千克 (kg)、厘米 (cm) 代替米 (m)、分钟 (min) 代替秒 (s)。
Always write down the given values with units and convert them first. For example, a mass of 250 g becomes 0.25 kg, a distance of 3.2 km becomes 3200 m, and a time of 4 minutes becomes 240 s. Check that your final answer has the correct unit and a sensible number of significant figures (usually 2 or 3).
始终先写下已知物理量及其单位，然后进行换算。例如，250 g 的质量应写作 0.25 kg，3.2 km 的距离应换为 3200 m，4 min 的时间应换为 240 s。最后要检查答案的单位是否正确、有效数字是否合理（通常保留 2 至 3 位）。
If an equation contains a constant such as g = 9.8 m/s² or the speed of light c = 3.0 × 10⁸ m/s, use the value given in the question. When no value is specified, use the standard value from your data sheet.
如果公式中含有常量，例如重力加速度 g = 9.8 m/s² 或光速 c = 3.0 × 10⁸ m/s，请使用题目中给出的数值。若题目未提供，则采用公式表上的标准值。
2. Kinematics Calculations | 运动学计算
The SUVAT equations describe uniform acceleration in a straight line. The five equations are:
v = u + at
s = ut + ½at²
v² = u² + 2as
s = (u+v)/2 × t
s = vt – ½at²
Remember that u is initial velocity, v is final velocity, a is acceleration, t is time, and s is displacement. Always identify which quantities you know and which one you need to find, then pick the equation that contains only them.
匀速直线运动由 SUVAT 方程组描述。五个公式分别为：v = u + at；s = ut + ½at²；v² = u² + 2as；s = (u+v)/2 × t；s = vt – ½at²。请记住 u 为初速度，v 为末速度，a 为加速度，t 为时间，s 为位移。先明确已知量和待求量，然后选择只包含这些量对应的公式。
Example: A car accelerates from rest at 3.0 m/s² for 8.0 s. Find the distance travelled.
Known: u = 0, a = 3.0 m/s², t = 8.0 s. Unknown: s.
Use s = ut + ½at² = 0 + ½×3.0×(8.0)² = 96 m.
例题：一辆汽车从静止开始以 3.0 m/s² 的加速度加速 8.0 s，求行驶的距离。已知 u = 0，a = 3.0 m/s²，t = 8.0 s，未知量为 s。选用 s = ut + ½at²，代入得 0 + 0.5×3.0×64 = 96 m。
Example: A ball is thrown vertically upwards with 15 m/s. How high does it go? (Take g = 10 m/s²)
At the highest point v = 0, a = –10 m/s², u = 15 m/s. Use v² = u² + 2as → 0 = 15² + 2×(–10)×s → s = 11.25 m.
例题：一球以 15 m/s 的速度竖直上抛，求最大高度（取 g = 10 m/s²）。在最高点 v = 0，a = –10 m/s²，u = 15 m/s。用 v² = u² + 2as → 0 = 225 – 20s → s = 11.25 m。
3. Newton’s Second Law (F = ma) | 牛顿第二定律 (F = ma)
Resultant force = mass × acceleration. F = m a. The unit of force is the newton (N), equivalent to kg·m/s². When multiple forces act on an object, you must first calculate the resultant force. For example, a 5 kg block is pushed with 20 N to the right while friction opposes with 5 N; the resultant force is 15 N to the right, giving a = F/m = 15/5 = 3.0 m/s².
合力 = 质量 × 加速度，F = m a。力的单位是牛顿 (N)，相当于 kg·m/s²。当物体受多个力作用时，需先求出合力。例如，一块 5 kg 的木块受到向右 20 N 的推力，摩擦力向左为 5 N，则合力向右为 15 N，加速度 a = F/m = 15/5 = 3.0 m/s²。
Weight is a specific force due to gravity: W = m g, where g is the gravitational field strength (9.8 N/kg on Earth). Do not confuse mass and weight.
重量是由于重力产生的力：W = m g，其中 g 为引力场强度（地球表面约为 9.8 N/kg）。注意区分质量和重量。
Example: An 800 kg elevator accelerates upwards at 1.2 m/s². The tension T in the cable must overcome weight and provide acceleration. T – mg = ma → T = m(g + a) = 800×(9.8 + 1.2) = 800×11 = 8800 N.
例题：一部 800 kg 的电梯以 1.2 m/s² 的加速度上升，求缆绳拉力 T。T 既要平衡重力，又要产生加速度：T – mg = ma → T = m(g + a) = 800×(9.8+1.2) = 8800 N。
4. Momentum and Conservation | 动量与动量守恒
Momentum p = m v. In a closed system, total momentum before a collision or explosion equals total momentum after, provided no external forces act. Remember momentum is a vector; assign positive and negative directions.
动量 p = m v。在封闭系统中，只要没有外力作用，碰撞或爆炸前后的总动量守恒。动量是矢量，需规定正负方向。
Example: A 2 kg trolley moving at 3 m/s to the right collides and sticks to a stationary 1 kg trolley. Total momentum before = 2×3 + 1×0 = 6 kg·m/s. After collision, combined mass = 3 kg, so v = p/m = 6/3 = 2 m/s.
例题：一辆 2 kg 的小车以 3 m/s 向右运动，与静止的 1 kg 小车碰撞后粘在一起。碰前总动量 = 2×3 + 1×0 = 6 kg·m/s。碰后总质量 3 kg，速度 v = 6/3 = 2 m/s。
In explosions, total momentum before is zero, so pieces fly apart with equal and opposite momentum.
在爆炸问题中，初始总动量为零，因此碎片以等大反向的动量飞散。
5. Work, Energy, and Efficiency | 功、能量及效率
Work done = force × distance moved in the direction of the force: W = F d. Kinetic energy: KE = ½ m v². Gravitational potential energy: GPE = m g h. The principle of conservation of energy states that energy cannot be created or destroyed, only transferred. In many problems, GPE lost = KE gained, or work done against friction dissipates energy.
功 = 力 × 沿力方向移动的距离：W = F d。动能：KE = ½ m v²。重力势能：GPE = m g h。能量守恒定律指出，能量不会凭空产生或消失，只会转移。在许多问题中，重力势能的减少量等于动能的增加量，或者克服摩擦力做功消耗能量。
Example: A 500 kg roller coaster car drops from a height of 20 m. Assuming no friction, find its speed at the bottom. GPE lost = m g h = 500×9.8×20 = 98 000 J. This equals KE gained: ½ m v² = 98 000 → v = √(2×98 000/500) = √392 ≈ 19.8 m/s.
例题：一辆 500 kg 的过山车从 20 m 高处下落，不计摩擦，求底部速度。减少的重力势能 = 500×9.8×20 = 98 000 J。这部分转化为动能：½×500×v² = 98 000 → v = √(2×98 000/500) ≈ 19.8 m/s。
Efficiency = (useful energy output / total energy input) × 100%. For example, a motor lifts a weight doing 200 J of useful work but consumes 500 J of electrical energy. Efficiency = (200/500)×100 = 40%.
效率 = (有用输出能量 / 总输入能量) × 100%。例如，一个电动机将重物提起做 200 J 有用功，但消耗了 500 J 电能，效率为 (200/500)×100 = 40%。
6. Power Calculations | 功率计算
Power is the rate of doing work: P = E / t or P = W / t. The unit is the watt (W), where 1 W = 1 J/s. Another useful form is P = F v for an object moving at constant speed against a force.
功率是做功的速率：P = E / t 或 P = W / t。单位是瓦特 (W)，1 W = 1 J/s。针对匀速运动的物体，若克服恒定外力，还可用 P = F v。
Example: A crane lifts a 200 kg load through 15 m in 10 s. Find the minimum power output. Work done = m g h = 200×9.8×15 = 29 400 J. Power = 29 400 / 10 = 2940 W.
例题：一台起重机在 10 s 内将 200 kg 的重物提升 15 m，求最小输出功率。做功 = 200×9.8×15 = 29 400 J，功率 = 29 400 / 10 = 2940 W。
When using P = F v, ensure the speed is in m/s. A car exerts a driving force of 500 N at 20 m/s; its power output = 500 × 20 = 10 000 W = 10 kW.
使用 P = F v 时，速度单位须为 m/s。一辆汽车在 20 m/s 时产生 500 N 的驱动力，输出功率 = 500 × 20 = 10 000 W = 10 kW。
7. Density and Pressure | 密度与压强
Density ρ = mass / volume: ρ = m / V. Units are kg/m³. To measure density of an irregular solid, use a displacement can and a balance. Pressure p = F / A, measured in pascals (Pa) where 1 Pa = 1 N/m². Pressure in a liquid column: p = h ρ g.
密度 ρ = 质量 / 体积：ρ = m / V，单位为 kg/m³。测量不规则固体的密度时，可用溢水罐和天平。压强 p = F / A，单位为帕斯卡 (Pa)，1 Pa = 1 N/m²。液体柱产生的压强：p = h ρ g。
Example: A cube of side 0.02 m has a mass of 0.12 kg. Its volume = (0.02)³ = 8×10⁻⁶ m³. Density = 0.12 / 8×10⁻⁶ = 15 000 kg/m³.
例题：一个边长为 0.02 m 的立方体质量为 0.12 kg。体积 = (0.02)³ = 8×10⁻⁶ m³，密度 = 0.12 / 8×10⁻⁶ = 15 000 kg/m³。
For pressure: a force of 40 N acts on an area of 0.5 m², so p = 40/0.5 = 80 Pa. In liquid pressure, a swimming pool depth of 2 m filled with water (ρ = 1000 kg/m³) gives p = 2 × 1000 × 9.8 = 19 600 Pa.
压强方面：40 N 的力作用在 0.5 m² 的面积上，p = 40/0.5 = 80 Pa。液体压强：泳池水深 2 m，水密度 1000 kg/m³，p = 2 × 1000 × 9.8 = 19 600 Pa。
8. Wave Speed: v = f λ | 波速公式：v = f λ
The wave equation links speed v (m/s), frequency f (Hz), and wavelength λ (m): v = f λ. This applies to all waves, including sound, water waves, and electromagnetic waves. Remember that speed of electromagnetic waves in a vacuum is 3.0 × 10⁸ m/s.
波速公式将波速 v (m/s)、频率 f (Hz) 和波长 λ (m) 联系起来：v = f λ。该公式适用于所有波，包括声波、水波和电磁波。要记住真空中电磁波的速率是 3.0 × 10⁸ m/s。
Example: A radio wave has a frequency of 100 MHz. What is its wavelength? f = 100 × 10⁶ Hz, v = 3.0 × 10⁸ m/s. λ = v / f = 3.0×10⁸ / 1.0×10⁸ = 3.0 m.
例题：某无线电波频率为 100 MHz，求其波长。f = 100 × 10⁶ Hz，v = 3.0 × 10⁸ m/s，λ = v / f = 3 m。
You might be given a ripple tank measurement: 20 waves pass a point in 5 seconds, and the wavelength is 0.1 m. Frequency f = number/time = 20/5 = 4 Hz. Wave speed = 4 × 0.1 = 0.4 m/s.
有时题目给出水波实验数据：5 秒内通过某点 20 个完整波，波长为 0.1 m。频率 f = 波数 / 时间 = 20/5 = 4 Hz，波速 = 4 × 0.1 = 0.4 m/s。
9. Circuit Calculations: Ohm’s Law and Electrical Power | 电路计算：欧姆定律与电功率
Ohm’s law: V = I R. Electrical power: P = I V and P = I² R. Energy transferred: E = P t or E = I V t. Charge: Q = I t. Always ensure time is in seconds when calculating energy in joules.
欧姆定律：V = I R。电功率：P = I V 和 P = I² R。能量转移：E = P t 或 E = I V t。电荷量：Q = I t。计算电能（焦耳）时务必将时间转换为秒。
Example: A 12 Ω resistor carries a current of 2.0 A. Voltage = I R = 24 V. Power = I V = 2.0 × 24 = 48 W. Energy used in 30 s = 48 × 30 = 1440 J.
例题：一个 12 Ω 的电阻上流过 2.0 A 的电流，电压 V = 2.0 × 12 = 24 V。功率 = 2.0 × 24 = 48 W，30 s 内消耗的能量 = 48 × 30 = 1440 J。
For combinations: in series, total resistance R_total = R₁ + R₂ + …; current is the same. In parallel, 1/R_total = 1/R₁ + 1/R₂; voltage across each branch is the same. Practice calculating total resistance and then using Ohm’s law to find current.
串并联组合：串联时总电阻 R_total = R₁ + R₂ + …，电流处处相等。并联时 1/R_total = 1/R₁ + 1/R₂，各支路电压相等。需练习先求出总电阻，再利用欧姆定律求总电流。
10. Transformer Equation | 变压器公式
The transformer equation relates voltages and turns: Vₚ / Vₛ = Nₚ / Nₛ. For an ideal transformer, input power ≈ output power, so Vₚ Iₚ = Vₛ Iₛ. Step-up transformers have more secondary turns; step-down have fewer. Always confirm whether you are dealing with primary or secondary quantities.
变压器公式联系了电压与匝数：Vₚ / Vₛ = Nₚ / Nₛ。对理想变压器，输入功率 ≈ 输出功率，因此 Vₚ Iₚ = Vₛ Iₛ。升压变压器次级匝数更多，降压变压器次级匝数更少。务必先分清初级与次级量。
Example: A transformer has 200 primary turns and 50 secondary turns. The primary voltage is 240 V. Find the secondary voltage. Vₛ = Vₚ × (Nₛ / Nₚ) = 240 × (50/200) = 60 V. It is a step-down transformer.
例题：某变压器初级 200 匝，次级 50 匝，初级电压 240 V，求次级电压。Vₛ = 240 × (50/200) = 60 V，这是一个降压变压器。
If the primary current is 0.5 A, the secondary current can be found from Vₚ Iₚ = Vₛ Iₛ → 240 × 0.5 = 60 × Iₛ → Iₛ = 2.0 A. Notice that lower voltage gives higher current (step-down increases current).
若初级电流为 0.5 A，次级电流可由功率相等求出：240 × 0.5 = 60 × Iₛ → Iₛ = 2.0 A。可见降压变压器会升高电流。
11. Radioactive Decay and Half‑life | 放射性衰变与半衰期
The half‑life T½ is the time taken for half the unstable nuclei in a sample to decay, or for the count rate to halve. Calculations often involve repeated halving: after n half‑lives, the remaining activity = initial activity × (½)ⁿ.
半衰期 T½ 是指样本中不稳定原子核的数量（或计数率）减半所需的时间。计算时常采用反复减半的思路：经过 n 个半衰期后，剩余活度 = 初始活度 × (½)ⁿ。
Example: A sample has an initial count rate of 600 counts/min and a half‑life of 2 hours. After 6 hours (i.e. 3 half‑lives), count rate = 600 × (½)³ = 600 × 1/8 = 75 counts/min.
例题：一样品初始计数率为 600 次/min，半衰期为 2 小时。6 小时后（即 3 个半衰期），计数率 = 600 × (½)³ = 75 次/min。
You may need to find the number of half‑lives from a graph of activity vs time. Read the time for the activity to drop from 100% to 50%; that is one half‑life. Use the same value to predict when activity drops to 25%, etc.
有时需从活度-时间图中读取半衰期。找出活度从 100% 降至 50% 所用的时间，即为一个半衰期。再以此推算下降到 25% 的时刻。
12. Exam Tips for Calculations | 计算题考试技巧
Always show your working in a logical order: write the equation, substitute numbers with units, and then write the final answer with the correct unit. Even if the final answer is wrong, a clear method earns most marks. Check whether the question asks for standard form or a specific number of significant figures.
始终按逻辑顺序展示计算步骤：先写公式，再代入带单位的数值，最后写出带有正确单位的计算结果。即使最后答案有误，清晰的解题过程也能获得大部分分数。注意题目是否要求用标准形式或指定的有效数字位数。
If a calculation has multiple steps, break it down. For example, to find the speed of a pendulum bob at the lowest point, first calculate the vertical drop to determine GPE lost, then equate to KE. Write each step as a new line so the examiner can follow your reasoning.
如果计算涉及多步推理，要逐层拆解。例如求摆锤在最低点的速率，先计算下落高度以得出失去的重力势能，再令其等于动能。每步另起一行，使考官清楚你的思路。
Finally, double-check unit consistency. Convert minutes to seconds, cm to m, and g to kg before substituting. A quick dimensional check (does the unit of the answer match the quantity?) can catch silly mistakes.
最后要复查单位一致性。代入公式前，务必将分钟转为秒、厘米转为米、克转为千克。快速检查量纲（答案的单位是否与所求量一致）能有效避免低级错误。
With consistent practice using these techniques, you will approach Edexcel Physics calculation questions methodically and maximize your marks.
坚持用上述方法练习，你就能有条不紊地应对 Edexcel 物理计算题，最大限度地拿下分数。
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