📚 Calculation Practice for IGCSE WJEC Physics | IGCSE WJEC 物理计算题专项训练
Mastering calculations is essential for success in IGCSE WJEC Physics. This revision guide walks you through the most common types of numerical problems, from basic unit conversions to multi-step energy and electricity questions. Each section provides the key formula, a worked example, and step-by-step solution reasoning so you can build confidence and tackle any calculation question in the exam.
掌握计算题对于在 IGCSE WJEC 物理考试中取得成功至关重要。本复习指南带你梳理最常见的计算题型，从基本的单位换算到多步骤的能量和电学问题。每个小节都提供了关键公式、解题示例和逐步推理过程，帮助你建立信心，轻松应对考试中的任何计算题。
1. Understanding Units and Converting | 理解单位与换算
ur = 3600 s, and 1 tonne = 1000 kg. Writing units next to every number helps avoid errors.
许多计算错误源于忘记将单位换算成国际基本单位。在使用公式前，请检查距离是否为米（m）、时间为秒（s）、质量为千克（kg）、力为牛顿（N）。常见换算包括：1 km = 1000 m，1 min = 60 s，1 h = 3600 s，1 t = 1000 kg。在每个数字旁边写上单位有助于避免出错。
Example: A cyclist travels 7.2 km in 12 minutes. Convert both quantities to SI units.
7.2 km = 7.2 × 1000 = 7200 m; 12 min = 12 × 60 = 720 s.
示例： 一名骑行者用 12 分钟行驶了 7.2 km。请将两个物理量换算为国际单位。
7.2 km = 7.2 × 1000 = 7200 m；12 min = 12 × 60 = 720 s。
2. Speed, Distance and Time | 速度、距离和时间
The relationship between speed, distance and time is given by the formula:
速度、距离和时间之间的关系由以下公式给出：
v = d / t
where v is average speed in m/s, d is distance in m, and t that the speed is constant or that the question asks for average speed. If the motion has different stages, calculate the total distance and total time separately.
其中 v 是平均速度（m/s），d 是距离（m），t 是时间（s）。你可以将公式变形为 d = v × t 或 t = d / v。始终确认速度是恒定的，或者题目要求的是平均速度。如果运动包含不同阶段，应分别计算总距离和总时间。
Worked example: A train covers 1500 m in 50 s at constant speed. Find its speed.
v = d / t = 1500 / 50 = 30 m/s.
he same speed, total time = total distance / speed = (1500 + 900) / 30 = 2400 / 30 = 80 s.
解题示例： 一列火车以恒定速度在 50 s 内行驶了 1500 m。求它的速度。
v = d / t = 1500 / 50 = 30 m/s。
如果它以相同速度又行驶了 900 m，总时间 = 总距离 / 速度 = (1500 + 900) / 30 = 2400 / 30 = 80 s。
3. Acceleration | 加速度
Acceleration tells us how quickly velocity changes. The key equation for uniform acceleration is:
加速度描述速度变化的快慢。匀加速运动的关键方程是：
a = (v - u) / t
where a is acceleration in m/s², v is final velocity, u is initial velocity, and t is time taken. Deceleration produces a negative value of a n v² = u² + 2 a d, which links final velocity, initial velocity, acceleration and distance d. Rearranging these formulae correctly is vital.
其中 a 是加速度（m/s²），v 是末速度，u 是初速度，t 是所用时间。减速会产生负的 a 值。你可能还需要用到 v² = u² + 2 a d 这个方程，它将末速度、初速度、加速度和距离 d 联系起来。正确变形这些公式至关重要。
Worked example: A car accelerates from 5 m/s to 25 m/s in 4 s. Calculate its acceleration.
a = (v - u) / t = (25 - 5) / 4 = 20 / 4 = 5 m/s².
ce of 40 m, find the deceleration using v² = u² + 2 a d. Here u = 25 m/s, v = 0, d = 40 m.
0 = 25² + 2 × a × 40 → 0 = 625 + 80 a → a = -625 / 80 = -7.81 m/s² (deceleration).
解题示例： 一辆汽车在 4 s 内从 5 m/s 加速到 25 m/s。计算它的加速度。
a = (v - u) / t = (25 - 5) / 4 = 20 / 4 = 5 m/s²。
如果汽车随后刹车并在 40 m 的距离内停下，用 v² = u² + 2 a d 求减速度。这里 u = 25 m/s，v = 0，d = 40 m。
0 = 25² + 2 × a × 40 → 0 = 625 + 80 a → a = -625 / 80 = -7.81 m/s²（减速度）。
4. Forces and Newton's Second Law | 力与牛顿第二定律
Newton's second law relates force, mass and acceleration:
牛顿第二定律将力、质量和加速度联系在一起：
F = m × a
where F is the resultant force in newtons (N), m is mass in kg, and a ed). Always identify the resultant force when multiple forces act. Draw a free-body diagram if necessary.
其中 F 是合力，单位牛顿（N），m 是质量（kg），a 是加速度（m/s²）。物体的重力是一个特例：W = m × g，g = 9.8 m/s²（或按题目要求取 10 m/s²）。当存在多个力时，一定要找出合力。必要时画出受力示意图。
Worked example: A 1200 kg car experiences a driving force of 4000 N and a resistive force of 1000 N. Find the acceleration.
Resultant force = 4000 - 1000 = 3000 N.
a = F / m = 3000 / 1200 = 2.5 m/s².
What is the weight of the car? W = m g = 1200 × 9.8 = 11760 N.
解题示例： 一辆 1200 kg 的汽车受到 4000 N 的驱动力和 1000 N 的阻力。求加速度。
合力 = 4000 - 1000 = 3000 N。
a = F / m = 3000 / 1200 = 2.5 m/s²。
这辆车的重力是多少？W = m g = 1200 × 9.8 = 11760 N。
5. Work, Energy and Power | 功、能与功率
Work done is energy transferred when a force moves an object:
功是力使物体移动时所传递的能量：
W = F × d
Gravitational potential energy (GPE) and kinetic energy (KE) are given by:
重力势能（GPE）和动能（KE）由以下公式给出：
GPE = m g h KE = ½ m v²
Power is the rate of doing work:
功率是做功的快慢：
P = W / t P = E / t
ioned. When a falling object converts GPE to KE, we often set m g h = ½ m v² and cancel m.
在能量守恒问题中，除非提及摩擦，否则假设没有能量损失。当落体将重力势能转化为动能时，我们常设 m g h = ½ m v²，并消去 m。
Worked example: d from a height of 10 m. Calculate its speed just before hitting the ground (g = 10 m/s²).
GPE at top = m g h = 2 × 10 × 10 = 200 J.
KE at bottom = ½ m v² = 200 J → ½ × 2 × v² = 200 → v² = 200 → v = √200 ≈ 14.1 m/s.
解题示例： 一个 2 kg 的球从 10 m 高处落下。求它刚撞击地面前的速度（g = 10 m/s²）。
顶部的 GPE = m g h = 2 × 10 × 10 = 200 J。
底部的 KE = ½ m v² = 200 J → ½ × 2 × v² = 200 → v² = 200 → v = √200 ≈ 14.1 m/s。
6. Density and Pressure | 密度与压强
Density relates mass and volume:
密度将质量和体积联系起来：
ρ = m / V
where ρ is density in kg/m³, m is mass in kg, and V is volume in m³. For regular solids, calculate volume using geometry; for irregular objects, use water displacement. Pressure is force per unit area:
其中 ρ 是密度（kg/m³），m 是质量（kg），V 是体积（m³）。对于规则固体，利用几何计算体积；对于不规则物体，使用排水法测量。压强是单位面积上的压力：
p = F / A
Liquid pressure depends on depth and density:
液体压强取决于深度和密度：
p = ρ g h
Remember to use consistent units: 1 cm² = 1 × 10⁻⁴ m².
注意使用一致的单位：1 cm² = 1 × 10⁻⁴ m²。
Worked example: A cube of side 2 cm has a mass of 16 g. Find its density in kg/m³.
Volume = 2 cm × 2 cm × 2 cm = 8 cm³ = 8 × 10⁻⁶ m³. Mass = 16 g = 0.016 kg.
ρ = m / V = 0.016 / 8×10⁻⁶ = 2000 kg/m³.
s on a table, pressure = F / A = (0.016 × 9.8) / (0.02 × 0.02) = 0.1568 / 0.0004 = 392 Pa.
解题示例： 一个边长为 2 cm 的立方体质量为 16 g。求它的密度，以 kg/m³ 表示。
体积 = 2 cm × 2 cm × 2 cm = 8 cm³ = 8 × 10⁻⁶ m³。质量 = 16 g = 0.016 kg。
ρ = m / V = 0.016 / 8×10⁻⁶ = 2000 kg/m³。
如果该立方体静置在桌面上，压强 = F / A = (0.016 × 9.8) / (0.02 × 0.02) = 0.1568 / 0.0004 = 392 Pa。
7. Electrical Circuits: Ohm's Law and Power | 电路：欧姆定律与电功率
Ohm's law links voltage, current and resistance:
欧姆定律将电压、电流和电阻联系起来：
V = I × R
Electrical power can be expressed in three useful forms:
电功率可以用三种有用的形式表示：
P = I V P = I² R P = V² / R
circuit is given by E = P t, and charge by Q = I t. When combining resistors in series, Rtotal = R₁ + R₂. In parallel, use 1/Rtotal each quantity: current in amperes (A), voltage in volts (V), resistance in ohms (Ω), power in watts (W).
电路中的能量传递用 E = P t 表示，电荷量用 Q = I t 表示。当电阻串联时，Rtotal = R₁ + R₂。并联时，使用 1/Rtotal = 1/R₁ + 1/R₂。始终标明每个物理量的单位：电流为安培（A），电压为伏特（V），电阻为欧姆（Ω），功率为瓦特（W）。
Worked example: A 12 V battery powers a lamp with a resistance of 6 Ω. Find the current and the power dissipated.
I = V / R = 12 / 6 = 2 A.
P = I V = 2 × 12 = 24 W, or P = V² / R = (12²) / 6 = 144 / 6 = 24 W.
How much energy does the lamp use in 5 minutes? t = 5 × 60 = 300 s; E = P t = 24 × 300 = 7200 J.
解题示例： 一个 12 V 的电池为一个 6 Ω 的灯泡供电。求电流大小和消耗的功率。
I = V / R = 12 / 6 = 2 A。
P = I V = 2 × 12 = 24 W，或 P = V² / R = (12²) / 6 = 144 / 6 = 24 W。
灯泡在 5 分钟内用了多少能量？t = 5 × 60 = 300 s；E = P t = 24 × 300 = 7200 J。
8. Wave Speed Equation | 波速方程
All waves obey the wave equation:
所有波都遵循波动方程：
v = f × λ
where v is wave speed in m/s, f is frequency in hertz (Hz), and λ (lambda) is wavelength in metres. You may also need the relationship between period T and frequency:
其中 v 是波速（m/s），f 是频率（Hz），λ（拉姆达）是波长（m）。你可能还需要周期 T 与频率的关系：
T = 1 / f
travels at about 340 m/s in air; for electromagnetic waves in a vacuum, v = 3.0 × 10⁸ m/s.
当题目涉及回声或声纳时，记住波传播到物体并返回，因此距离是单程距离的两倍。声音在空气中的传播速度约为 340 m/s；对于真空中的电磁波，v = 3.0 × 10⁸ m/s。
Worked example: A water wave has a wavelength of 0.5 m and a frequency of 4 Hz. Calculate its speed.
v = f λ = 4 × 0.5 = 2 m/s.
If the period is T, then T = 1 / f = 1 / 4 = 0.25 s.
解题示例： 一列水波的波长为 0.5 m，频率为 4 Hz。计算它的波速。
v = f λ = 4 × 0.5 = 2 m/s。
如果周期为 T，则 T = 1 / f = 1 / 4 = 0.25 s。
9. Efficiency Calculations | 效率计算
Efficiency measures how much of the input energy is converted into useful output. The formula is:
效率衡量有多少输入能量转化为有用的输出。公式为：
Efficiency = (Useful energy output / Total energy input) × 100%
tput / total power input). Efficiency has no units and is always less than or equal to 100% for real systems. Remember to multiply by 100 to express it as a percentage.
它也可以用功率表示：效率 = （有用输出功率 / 总输入功率）。效率没有单位，对于实际系统总是小于或等于 100%。记住乘以 100 以百分数表示。
Worked example: ork of GPE. The motor draws 200 J of electrical energy. Find the efficiency (g = 10 m/s²).
Useful work = m g h = 5 × 10 × 2 = 100 J.
Efficiency = (100 / 200) × 100% = 50%.
or runs for 4 s, total input power = 200 / 4 = 50 W; useful output power = 100 / 4 = 25 W.
解题示例： 一台电动机将一个 5 kg 的重物提升 2 m，做了重力势能的有用功。电动机消耗了 200 J 的电能。求效率（g = 10 m/s²）。
有用功 = m g h = 5 × 10 × 2 = 100 J。
效率 = (100 / 200) × 100% = 50%。
如果电动机运行了 4 s，总输入功率 = 200 / 4 = 50 W；有用输出功率 = 100 / 4 = 25 W。
10. Mixed Calculation Practice and Tips | 混合计算练习与技巧
In the exam, you may face multi-step problems that combine several concepts. Approach them systematically: first, list the quantities you know and what you need to find. Second, identify the relevant equation(s) and convert all units to SI. Third, carry out the calculation step by step, writing down each substituted value. Fourth, check that your answer has the correct unit and is sensible in magnitude.
考试中你可能会遇到融合多个概念的混合计算题。解题时要系统化：首先，列出已知量和待求量。其次，确定相关方程并将所有单位换算成国际单位。第三，逐步进行计算，并写下每个代入的数值。第四，检查答案的单位是否正确，数值是否合理。
Example of a mixed problem: A 0.8 kg toy car accelerates from rest to 4 m/s in 2 s on a smooth surface. Find (i) the acceleration, (ii) the force needed, (iii) the kinetic energy gained, and (iv) the power of the motor.
(i) a = (v - u)/t = (4 - 0)/2 = 2 m/s².
(ii) F = m a = 0.8 × 2 = 1.6 N.
(iii) KE = ½ m v² = ½ × 0.8 × 4² = 0.4 × 16 = 6.4 J.
(iv) Power = work done / time = KE gained / time = 6.4 / 2 = 3.2 W.
混合题示例： 一辆 0.8 kg 的玩具车在光滑表面上从静止开始，在 2 s 内加速到 4 m/s。求 (i) 加速度，(ii) 所需的力，(iii) 获得的动能，(iv) 电动机的功率。
(i) a = (v - u)/t = (4 - 0)/2 = 2 m/s²。
(ii) F = m a = 0.8 × 2 = 1.6 N。
(iii) KE = ½ m v² = ½ × 0.8 × 4² = 0.4 × 16 = 6.4 J。
(iv) 功率 = 做功 / 时间 = 获得的动能 / 时间 = 6.4 / 2 = 3.2 W。
Common pitfalls include forgetting to square the velocity in KE calculations, mixing up mass and weight, and using non-SI units. Practice regularly with past-paper questions, and always keep a formula list handy until you have memorised the essential equations.
常见错误包括在动能计算中忘记将速度平方、混淆质量与重力，以及使用非国际单位。通过真题定期练习，并随时准备一份公式表，直到你记住所有重要方程。
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