📚 Year 10 WJEC Physics: Unit Test Mock Paper Analysis | Year 10 WJEC 物理：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test paper for Year 10 WJEC Physics, covering key topics such as electricity, energy, waves and the electromagnetic spectrum. Each section presents a typical exam-style question, explains the underlying principles step by step, and highlights common pitfalls.
本文详细解析一套针对 Year 10 WJEC 物理的单元测试模拟卷，涵盖电学、能量、波和电磁波谱等核心主题。每个小节展示一道典型考题，逐步讲解基本原理，并指出常见错误。
1. Ohm’s Law and Current Calculation | 欧姆定律与电流计算
A circuit is set up with a 12 Ω resistor connected to a 6 V battery. Calculate the current flowing through the resistor.
一个电路由 12 Ω 的电阻和 6 V 的电池连接而成。计算流过该电阻的电流。
Ohm’s law states: V = I × R, where V is voltage (V), I is current (A), and R is resistance (Ω). Rearranging gives I = V ÷ R = 6 V ÷ 12 Ω = 0.5 A. The current is 0.5 amperes.
欧姆定律为：V = I × R，其中 V 是电压（伏特），I 是电流（安培），R 是电阻（欧姆）。移项得 I = V ÷ R = 6 V ÷ 12 Ω = 0.5 A。电流是 0.5 安培。
Always ensure that the resistance is measured in ohms and the voltage in volts. Many WJEC questions ask for units; forgetting to include ‘A’ can lose a mark.
务必确保电阻单位为欧姆，电压单位为伏特。WJEC 常考单位，遗漏 ‘A’ 会失分。
2. Resistors in Series: Total Resistance | 串联电阻：总电阻
Two resistors of 5 Ω and 15 Ω are connected in series. What is the total resistance of the circuit?
两个分别为 5 Ω 和 15 Ω 的电阻串联。电路的总电阻是多少？
In a series circuit, the total resistance is the sum of individual resistances: Rtotal = R₁ + R₂ = 5 Ω + 15 Ω = 20 Ω.
在串联电路中，总电阻为各个电阻之和：Rtotal = R₁ + R₂ = 5 Ω + 15 Ω = 20 Ω。
If a third resistor were added in series, the total resistance would increase further. This is because the current has to pass through more resistive components, making it harder for charge to flow.
如果再加一个电阻串联，总电阻将进一步增大。这是因为电流需要经过更多电阻元件，阻碍电荷流动。
3. Energy Efficiency of Electrical Appliances | 电器能效
A lamp converts 100 J of electrical energy into 10 J of light and 90 J of heat. What is its efficiency as a percentage?
一盏灯将 100 焦耳的电能转化为 10 焦耳的光能和 90 焦耳的热能。其效率的百分比是多少？
Efficiency = (useful energy output ÷ total energy input) × 100% = (10 J ÷ 100 J) × 100% = 10%.
效率 =（有用能量输出 ÷ 总能量输入）× 100% =（10 J ÷ 100 J）× 100% = 10%。
An efficiency can never exceed 100% due to the conservation of energy. WJEC often includes a follow-up question about how to improve efficiency, e.g. using LED bulbs that waste less energy as heat.
由于能量守恒，效率永远不可能超过 100%。WJEC 经常会追问如何提高效率，例如使用浪费较少热能的 LED 灯泡。
4. Wave Speed, Frequency and Wavelength | 波速、频率与波长
A water wave has a frequency of 8 Hz and a wavelength of 0.5 m. Calculate the wave speed.
一个水波频率为 8 Hz，波长为 0.5 m。计算波速。
The wave equation is v = f × λ, where v is speed (m/s), f is frequency (Hz), and λ is wavelength (m). Substituting gives v = 8 Hz × 0.5 m = 4 m/s.
波动方程为 v = f × λ，其中 v 为波速（米/秒），f 为频率（赫兹），λ 为波长（米）。代入得 v = 8 Hz × 0.5 m = 4 m/s。
v = f λ
In examinations, ensure you can rearrange the equation to find frequency (f = v ÷ λ) or wavelength (λ = v ÷ f). This applies to both transverse and longitudinal waves.
考试中要掌握公式变形，求频率时 f = v ÷ λ，求波长时 λ = v ÷ f。这适用于横波和纵波。
5. Electromagnetic Spectrum: Uses and Risks | 电磁波谱：用途与风险
Which type of electromagnetic wave is used by a television remote control? State one risk associated with overexposure to this type of wave.
电视遥控器使用的是哪种电磁波？指出过度暴露于这种波的一个风险。
The remote control uses infrared radiation. Prolonged exposure to intense infrared can cause skin burns.
遥控器使用红外线。长时间暴露在强红外线下会导致皮肤灼伤。
The electromagnetic spectrum from low to high frequency is: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. WJEC requires you to know uses, e.g. radio waves for communication, microwaves for cooking, and X-rays for medical imaging. Risks include ionising damage from UV, X-rays and gamma rays.
电磁波谱从低频率到高频率依次为：无线电波、微波、红外线、可见光、紫外线、X 射线、伽马射线。WJEC 要求了解各波段的用途，如无线电波用于通信，微波用于烹饪，X 射线用于医学成像。风险方面，紫外线、X 射线和伽马射线具有电离危害。
6. Specific Heat Capacity Calculations | 比热容计算
A 1.5 kg aluminium block is heated using an electric heater. Its temperature rises from 22 °C to 40 °C. The specific heat capacity of aluminium is 900 J/kg°C. Calculate the energy transferred to the block.
一个 1.5 kg 的铝块用电加热器加热，温度从 22 °C 升至 40 °C。铝的比热容为 900 J/kg°C。计算传递给铝块的能量。
Temperature change Δθ = 40 °C – 22 °C = 18 °C. Energy E = m c Δθ = 1.5 kg × 900 J/kg°C × 18 °C = 24,300 J.
温度变化 Δθ = 40 °C – 22 °C = 18 °C。能量 E = m c Δθ = 1.5 kg × 900 J/kg°C × 18 °C = 24 300 J。
E = m c Δθ
The large unit of J/kg°C means that a lot of energy is needed to raise aluminium’s temperature. Students often mix up Δθ with the final temperature; always subtract initial from final temperature.
比热容单位为 J/kg°C 意味着升高铝的温度需要大量能量。学生经常混淆 Δθ 与最终温度；一定要用最终温度减去初始温度。
7. Kinetic Energy and Motion | 动能与运动
A cyclist and bicycle have a combined mass of 80 kg and are moving at a speed of 5 m/s. Calculate their kinetic energy.
自行车和人总质量为 80 kg，以 5 m/s 的速度运动。计算他们的动能。
Kinetic energy KE = ½ m v² = ½ × 80 kg × (5 m/s)² = 40 × 25 = 1000 J.
动能 KE = ½ m v² = ½ × 80 kg × (5 m/s)² = 40 × 25 = 1000 J。
If the speed doubles (10 m/s), KE becomes ½ × 80 × 100 = 4000 J – four times greater. This concept is frequently tested in WJEC multiple-choice questions.
如果速度加倍（10 m/s），动能变为 ½ × 80 × 100 = 4000 J——增加了四倍。这是 WJEC 选择题的常见考点。
8. Electrical Power and Energy Transferred | 电功率与能量转移
A kettle has a power rating of 2200 W. It is switched on for 2 minutes. Calculate the energy transferred in joules.
一个水壶的额定功率为 2200 W，开启 2 分钟。计算转移的能量（单位：焦耳）。
First convert time to seconds: 2 min = 120 s. Energy E = P × t = 2200 W × 120 s = 264,000 J.
首先将时间换算为秒：2 min
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