📚 IGCSE WJEC Physics: Typical Example Problems Explained | IGCSE WJEC 物理：典型例题详解
Mastering IGCSE WJEC Physics requires more than just memorising facts — you need to solve typical exam‑style problems with confidence. This article presents eight carefully selected worked examples that cover the core topics of the syllabus, from motion and forces to electricity, waves and thermal physics. Each problem is broken down step by step, linking essential concepts, formula application and unit handling. Whether you are revising for a mock exam or preparing for the final paper, working through these examples will sharpen your problem‑solving skills and deepen your understanding of the physical principles tested by WJEC.
要掌握 IGCSE WJEC 物理，不仅需要记忆事实，还需要自信地解决典型的考试风格的问题。本文精选了八个涵盖课程核心主题的详细解析例题，从运动与力到电学、波动和热物理。每个问题都逐步分解，结合关键概念、公式应用和单位处理。无论你是在准备模拟考试还是备战最终试卷，通过这些例题将提升你的解题能力，加深你对 WJEC 测试的物理原理的理解。
1. Motion with Constant Acceleration | 匀加速运动
A car accelerates uniformly from rest and reaches a velocity of 12 m/s in 3.0 seconds. Calculate its acceleration and the distance travelled during this time.
一辆汽车从静止开始匀加速，在 3.0 秒内达到 12 m/s 的速度。计算它的加速度以及在这段时间内行驶的距离。
Step 1: Identify the given quantities. Initial velocity u = 0 m/s, final velocity v = 12 m/s, time t = 3.0 s. Acceleration a = (v – u) / t = (12 – 0) / 3.0 = 4.0 m/s².
步骤 1：识别已知量。初速度 u = 0 m/s，末速度 v = 12 m/s，时间 t = 3.0 s。加速度 a = (v – u) / t = (12 – 0) / 3.0 = 4.0 m/s²。
Step 2: Choose the appropriate equation for distance. Since acceleration is constant, use s = ut + ½at². Substituting: s = 0 × 3.0 + ½ × 4.0 × (3.0)² = 0 + 2.0 × 9.0 = 18 m.
步骤 2：选择合适的距离公式。由于加速度恒定，使用 s = ut + ½at²。代入：s = 0 × 3.0 + ½ × 4.0 × (3.0)² = 0 + 2.0 × 9.0 = 18 m。
The acceleration is 4.0 m/s² and the car travels 18 metres. Always check that units are consistent and that your final answers carry the correct unit symbols.
加速度为 4.0 m/s²，汽车行驶了 18 米。始终检查单位是否一致，最终答案是否带有正确的单位符号。
2. Conservation of Momentum | 动量守恒
A 1200 kg car travelling at 15 m/s collides with a stationary 800 kg car. The two cars stick together. Find the velocity of the combined wreckage immediately after the collision.
一辆 1200 kg 的汽车以 15 m/s 的速度行驶，与一辆静止的 800 kg 汽车相撞。两车粘在一起。求碰撞后瞬间组合残骸的速度。
Momentum is conserved in isolated systems. Total momentum before collision = total momentum after. Before: p = (1200 kg × 15 m/s) + (800 kg × 0) = 18000 kg m/s.
在孤立系统中动量守恒。碰撞前总动量 = 碰撞后总动量。碰撞前：p = (1200 kg × 15 m/s) + (800 kg × 0) = 18000 kg m/s。
After the collision, the combined mass is (1200 + 800) = 2000 kg. Let the common final velocity be v. Then 2000 × v = 18000, so v = 18000 / 2000 = 9.0 m/s.
碰撞后，总质量为 (1200 + 800) = 2000 kg。设共同末速度为 v。则 2000 × v = 18000，所以 v = 18000 / 2000 = 9.0 m/s。
The wreckage moves at 9.0 m/s in the original direction of the moving car. Note that momentum is a vector; here all motion is along a straight line, so direction is given by the sign.
残骸以 9.0 m/s 的速度沿初始运动方向移动。注意动量是矢量；这里所有运动都在一条直线上，因此方向由符号给出。
3. Kinetic Energy and Work Done | 动能与做功
A force of 50 N pushes a 10 kg box over a distance of 4.0 m along a frictionless horizontal surface, starting from rest. Calculate the final kinetic energy of the box and its final speed.
一个 50 N 的力将一个 10 kg 的箱子在无摩擦水平面上从静止推过 4.0 m 的距离。计算箱子的末动能和末速度。
The work done by the force is W = F × d = 50 N × 4.0 m = 200 J. Since there is no friction, all this work goes into the kinetic energy of the box. Therefore the final kinetic energy Eₖ = 200 J.
力所做的功为 W = F × d = 50 N × 4.0 m = 200 J。由于没有摩擦，所有功都转化为箱子的动能。因此末动能 Eₖ = 200 J。
Kinetic energy is given by Eₖ = ½mv². Rearranging: v = √(2Eₖ / m) = √(2 × 200 / 10) = √40 ≈ 6.32 m/s.
动能由 Eₖ = ½mv² 给出。变形得：v = √(2Eₖ / m) = √(2 × 200 / 10) = √40 ≈ 6.32 m/s。
The box reaches a speed of about 6.32 m/s. This problem illustrates the work–energy principle: the net work done on an object equals its change in kinetic energy.
箱子达到了约 6.32 m/s 的速度。这道题说明了功-能原理：对物体做的净功等于其动能的变化量。
4. Density and Pressure in a Liquid | 密度与液体压强
A rectangular block of metal measures 2.0 cm × 3.0 cm × 5.0 cm and has a mass of 240 g. Calculate (a) the density of the metal in g/cm³ and kg/m³, and (b) the pressure exerted by the block when it rests on its largest face on a table.
一块金属长方体尺寸为 2.0 cm × 3.0 cm × 5.0 cm，质量为 240 g。计算 (a) 金属的密度（以 g/cm³ 和 kg/m³ 表示），以及 (b) 当它以最大面平放在桌上时施加的压强。
Volume = 2.0 × 3.0 × 5.0 = 30 cm³. Density = mass / volume = 240 g / 30 cm³ = 8.0 g/cm³. In kg/m³: 8.0 g/cm³ = 8.0 × 1000 = 8000 kg/m³.
体积 = 2.0 × 3.0 × 5.0 = 30 cm³。密度 = 质量 / 体积 = 240 g / 30 cm³ = 8.0 g/cm³。换算为 kg/m³：8.0 g/cm³ = 8.0 × 1000 = 8000 kg/m³。
The largest face has area 3.0 cm × 5.0 cm = 15 cm² = 15 × 10⁻⁴ m² = 1.5 × 10⁻³ m². Weight = mg = 0.24 kg × 9.8 N/kg = 2.352 N. Pressure = force / area = 2.352 N / (1.5 × 10⁻³ m²) ≈ 1570 Pa (or 1.57 kPa).
最大面的面积为 3.0 cm × 5.0 cm = 15 cm² = 15 × 10⁻⁴ m² = 1.5 × 10⁻³ m²。重量 = mg = 0.24 kg × 9.8 N/kg = 2.352 N。压强 = 力 / 面积 = 2.352 N / (1.5 × 10⁻³ m²) ≈ 1570 Pa（或 1.57 kPa）。
Always convert to SI units before calculating pressure. The density conversion factor is 1000: 1 g/cm³ = 1000 kg/m³.
在计算压强之前，务必转换为国际单位制（SI）。密度换算因子是 1000：1 g/cm³ = 1000 kg/m³。
5. Ohm’s Law and Series Circuits | 欧姆定律与串联电路
A 12 V battery is connected to two resistors in series: R₁ = 4.0 Ω and R₂ = 8.0 Ω. Calculate (a) the total resistance, (b) the current in the circuit, and (c) the p.d. across each resistor.
一个 12 V 的电池与两个串联电阻连接：R₁ = 4.0 Ω，R₂ = 8.0 Ω。计算 (a) 总电阻，(b) 电路中的电流，(c) 每个电阻两端的电压。
For series resistors, total resistance R_total = R₁ + R₂ = 4.0 + 8.0 = 12.0 Ω. Using Ohm’s law, the total current I = V / R_total = 12 V / 12.0 Ω = 1.0 A.
对于串联电阻，总电阻 R_total = R₁ + R₂ = 4.0 + 8.0 = 12.0 Ω。使用欧姆定律，总电流 I = V / R_total = 12 V / 12.0 Ω = 1.0 A。
The current is the same through all components in series. Thus p.d. across R₁: V₁ = I × R₁ = 1.0 A × 4.0 Ω = 4.0 V. Across R₂: V₂ = I × R₂ = 8.0 V. Note that V₁ + V₂ = 12 V, consistent with Kirchhoff’s voltage law.
串联电路中所有元件的电流相同。因此 R₁ 两端的电压：V₁ = I × R₁ = 1.0 A × 4.0 Ω = 4.0 V。R₂ 两端：V₂ = I × R₂ = 8.0 V。注意 V₁ + V₂ = 12 V，符合基尔霍夫电压定律。
This problem highlights the key characteristics of a series circuit: the resistance adds up, the current is the same everywhere, and the supply voltage is shared across the components.
这道题突出了串联电路的关键特征：电阻相加，电流处处相同，电源电压在元件之间分配。
6. Wave Speed, Frequency and Wavelength | 波速、频率与波长
A water wave travels at 2.4 m/s. Its wavelength is measured to be 1.8 m. Calculate the frequency of the wave and the time period of one oscillation.
一个水波以 2.4 m/s 的速度传播。测得波长为 1.8 m。计算该波的频率和一次振荡的时间周期。
The wave equation is v = f × λ, where v is wave speed, f is frequency and λ is wavelength. Rearranging: f = v / λ = 2.4 m/s / 1.8 m = 1.33 Hz (to 3 significant figures).
波动方程为 v = f × λ，其中 v 是波速，f 是频率，λ 是波长。变形得：f = v / λ = 2.4 m/s / 1.8 m = 1.33 Hz（保留三位有效数字）。
The time period T is the reciprocal of frequency: T = 1 / f = 1 / 1.333… = 0.75 s. Alternatively, T = λ / v = 1.8 / 2.4 = 0.75 s.
时间周期 T 是频率的倒数：T = 1 / f = 1 / 1.333… = 0.75 s。或者，T = λ / v = 1.8 / 2.4 = 0.75 s。
Make sure to use consistent units (metres and seconds) and be comfortable with the relationship between period and frequency. In WJEC exams, wave‑type identification (transverse or longitudinal) often accompanies such calculations.
确保使用一致的单位（米和秒），并熟悉周期与频率之间的关系。在 WJEC 考试中，此类计算常伴随波的类型识别（横波或纵波）。
7. Specific Heat Capacity | 比热容
An electric kettle contains 800 g of water. A heater supplies 6000 J of energy and the temperature of the water rises from 25 °C to 30 °C. Assuming no heat losses, calculate the specific heat capacity of water.
一个电热水壶装有 800 g 水。加热器提供 6000 J 的能量，水温从 25 °C 上升到 30 °C。假设没有热量损失，计算水的比热容。
We use the equation ΔE = m × c × Δθ, where ΔE is the energy supplied, m is the mass, c is the specific heat capacity, and Δθ is the temperature change.
我们使用方程 ΔE = m × c × Δθ，其中 ΔE 是提供的能量，m 是质量，c 是比热容，Δθ 是温度变化。
Mass = 800 g = 0.800 kg. Temperature rise Δθ = 30 – 25 = 5 °C (or 5 K). Rearranging: c = ΔE / (m × Δθ) = 6000 J / (0.800 kg × 5 K) = 6000 / 4.0 = 1500 J/(kg°C).
质量 = 800 g = 0.800 kg。温度升高 Δθ = 30 – 25 = 5 °C（或 5 K）。变形得：c = ΔE / (m × Δθ) = 6000 J / (0.800 kg × 5 K) = 6000 / 4.0 = 1500 J/(kg°C)。
The calculated value is 1500 J/(kg°C), which is lower than the accepted figure for water (4180 J/(kg°C)) — the question uses a simplified model. In reality, some energy heats the kettle and is lost to the surroundings; this experiment typically yields around 4200 J/(kg°C) when corrected.
计算值为 1500 J/(kg°C)，低于水的公认值（4180 J/(kg°C)）——该题使用了一个简化模型。现实中，部分能量加热了水壶并散失到环境中；经过修正，该实验通常得出约 4200 J/(kg°C) 的结果。
This example reminds you to convert mass to kilograms and to note that a temperature difference of 1 °C equals 1 K. In WJEC, you may be asked to suggest improvements to the experiment to reduce heat losses.
这个例子提醒你将质量转换为千克，并注意 1 °C 的温差等于 1 K。在 WJEC 中，你可能会被要求提出减少热量损失的实验改进建议。
8. The Motor Effect and Fleming’s Left‑Hand Rule | 电动机效应与弗莱明左手定则
A straight wire of length 0.20 m carries a current of 3.0 A perpendicular to a uniform magnetic field of flux density 0.80 T. Calculate the magnitude of the force acting on the wire. State the direction of the force if the current flows from north to south and the magnetic field is directed from east to west.
一根长 0.20 m 的直导线载有 3.0 A 的电流，且与密度为 0.80 T 的均匀磁场垂直。计算作用在导线上的力的大小。如果电流从北向南，磁场从东向西，说明力的方向。
The force on a current‑carrying conductor in a magnetic field is given by F = B × I × L (since the wire is perpendicular to the field, sin 90° = 1). Substituting: F = 0.80 T × 3.0 A × 0.20 m = 0.48 N.
磁场中对载流导体的力由 F = B × I × L 给出（因为导线与磁场垂直，sin 90° = 1）。代入：F = 0.80 T × 3.0 A × 0.20 m = 0.48 N。
To find the direction, use Fleming’s left‑hand rule: thumb = force, first finger = magnetic field (west), second finger = current (south). If the field points east‑to‑west, point first finger west. Point second finger south (direction of current). The thumb then points downwards. So the force acts vertically downwards.
要确定方向，使用弗莱明左手定则：拇指 = 力，食指 = 磁场（西），中指 = 电流（南）。如果磁场从东向西，食指指向西。中指指向南（电流方向）。拇指指向下方。因此力垂直向下作用。
The magnitude of the force is 0.48 N, and it acts downwards. For WJEC, you must be able to apply Fleming’s left‑hand rule for motors and explain how it relates to the direction of force, field and current.
力的大小为 0.48 N，方向向下。对于 WJEC，你必须能够应用弗莱明左手定则来处理电动机，并解释其与力、磁场和电流方向的关系。
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