Year 11 Eduqas Physics: Unit Test Mock Paper Walkthrough | Year 11 Eduqas 物理：单元测试模拟卷解析

📚 Year 11 Eduqas Physics: Unit Test Mock Paper Walkthrough | Year 11 Eduqas 物理：单元测试模拟卷解析

Mock papers are an essential tool for Year 11 students preparing for the Eduqas GCSE Physics examinations. This walkthrough takes you through a representative unit test covering core topics such as energy, waves, electricity, forces, radioactivity, and space. Each question is analysed with detailed solutions, highlighting key concepts, common mistakes, and exam technique. The aim is to boost your confidence and consolidate your understanding so that you can apply your knowledge effectively under timed conditions.

模拟试卷是 Year 11 学生备考 Eduqas GCSE 物理的重要资源。本解析带你完成一份涵盖能量、波动、电学、力学、放射性和宇宙物理等核心主题的典型单元测试卷。每道题目都提供详细解答，突出关键概念、常见错误和应试技巧，旨在增强自信、巩固理解，帮助你在限时条件下高效运用知识。

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1. Kinetic and Gravitational Potential Energy | 动能与重力势能

Gravitational potential energy (GPE) is stored due to height in a gravitational field. The equation is:

重力势能（GPE）来自物体在引力场中的高度。其公式为：

GPE = m g h

where m is mass (kg), g is gravitational field strength (9.8 N/kg on Earth, sometimes approximated to 10 N/kg), and h is vertical height (m).

式中 m 为质量（kg），g 为引力场强度（地球表面为 9.8 N/kg，有时近似取 10 N/kg），h 为垂直高度（m）。

For the crane lifting a 500 kg block through 12 m: GPE = 500 × 9.8 × 12 = 58 800 J (or 58.8 kJ). Always check which g value the question expects.

起重机将 500 kg 的重物提升 12 m：GPE = 500 × 9.8 × 12 = 58 800 J（即 58.8 kJ）。务必按题目给出的 g 值计算。

If the object falls freely, GPE converts into kinetic energy (KE). Setting KE = GPE gives:

若物体自由下落，重力势能转化为动能（KE）。令 KE = GPE 可得：

½ m v² = m g h

Mass cancels, so impact speed v = √(2 g h) = √(2 × 9.8 × 12) ≈ 15.3 m/s. The result is independent of mass.

质量 m 可约去，因此落地速度 v = √(2 g h) = √(2 × 9.8 × 12) ≈ 15.3 m/s。该结果与物体的质量无关。

Common mistake: forgetting to take the square root after solving v² = 2gh. Always complete the final step and include units.

常见错误：根据 v² = 2gh 求出 v² 后忘记开平方。务必完成最后一步，并标明单位。

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2. Series and Parallel Circuits | 串联与并联电路

In a series circuit, total resistance R_total = R₁ + R₂ + … . Current is the same everywhere, and the sum of p.d.s equals the battery voltage.

在串联电路中，总电阻 R_总 = R₁ + R₂ + … 。电流处处相等，各元件两端的电势差之和等于电源电压。

For the 12 V battery with 4 Ω and 6 Ω in series: R_total = 4 + 6 = 10 Ω. Current I = V / R = 12 / 10 = 1.2 A.

给定 12 V 电源，串联 4 Ω 和 6 Ω 电阻：总电阻 = 4 + 6 = 10 Ω。电流 I = V / R = 12 / 10 = 1.2 A。

The p.d. across the 6 Ω resistor is V₆ = I × R = 1.2 × 6 = 7.2 V. The remaining 4.8 V drops across the 4 Ω resistor.

6 Ω 电阻两端的电势差 V₆ = I × R = 1.2 × 6 = 7.2 V。剩余 4.8 V 落在 4 Ω 电阻上。

If another 6 Ω is added in parallel with the existing 6 Ω, their combined resistance becomes R_parallel = (6 × 6) / (6 + 6) = 3 Ω. The total circuit resistance drops to 4 + 3 = 7 Ω, increasing total current.

若再并联一个 6 Ω 电阻，其并联等效电阻为 (6×6)/(6+6)=3 Ω。此时总电阻降为 4 + 3 = 7 Ω，总电流增大。

Remember: adding parallel branches always reduces total resistance and raises current drawn from the supply.

注意：增加并联支路一定减小总电阻，并增大电源输出的电流。

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3. Wave Calculations and Applications | 波的计算与应用

The wave speed equation links speed v, frequency f, and wavelength λ:

波速公式关联速度 v、频率 f 和波长 λ：

v = f λ

For a 50 Hz wave with λ = 6.8 m: v = 50 × 6.8 = 340 m/s (typical speed of sound in air). Period T = 1/f = 1/50 = 0.02 s.

对于 50 Hz、波长 6.8 m 的波：v = 50 × 6.8 = 340 m/s（典型的声速）。周期 T = 1/f = 1/50 = 0.02 s。

When the wave enters shallow water and slows to 25 m/s, the frequency remains constant. The new wavelength becomes λ’ = v’ / f = 25 / 50 = 0.5 m.

当波进入浅水区，速度降至 25 m/s，频率保持不变。新的波长 λ’ = v’ / f = 25 / 50 = 0.5 m。

A key exam point: frequency never changes when a wave passes from one medium to another; wave speed and wavelength adjust proportionally.

重要考点：波从一种介质进入另一种介质时频率不变；波速与波长同比例变化。

Always show the substitution step and check that units match — m/s, Hz (s⁻¹), and m.

务必展示代入步骤，并检查单位是否匹配—— m/s、Hz（s⁻¹）和 m。

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4. Radioactive Decay and Half-Life | 放射性衰变与半衰期

Half-life is the time taken for half the unstable nuclei in a sample to decay, or for the activity/count rate to halve.

半衰期指样品中一半的不稳定原子核发生衰变，或活度/计数率减半所需要的时间。

Carbon-14 has a half-life of 5730 years. Starting with 80 mg, after 17 190 years three half-lives have passed (17 190 ÷ 5730 = 3). Remaining mass: 80 → 40 → 20 → 10 mg.

碳-14 的半衰期为 5730 年。初始 80 mg，经过 17 190 年即三个半衰期（17 190÷5730=3）。剩余质量：80→40→20→10 mg。

A decay graph showing activity against time will display a smooth curve that drops by half every 5730 years. Avoid common errors such as assuming linear decrease.

画活度-时间衰变曲线时，应绘出一条每 5730 年减半的光滑曲线。常见错误是误认为放射性活度呈线性减少。

Nuclear equations must balance mass and atomic numbers. For alpha decay, ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He. Both mass (238=234+4) and charge (92=90+2) are conserved.

核方程需满足质量数与电荷数守恒。α 衰变实例：²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He。质量数（238=234+4）与电荷数（92=90+2）均守恒。

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5. Newton’s Second Law and Weight | 牛顿第二定律与重量

Newton’s second law states: resultant force = mass × acceleration:

牛顿第二定律：合力 = 质量 × 加速度：

F = m a

For a 1200 kg car with a resultant force of 3000 N, acceleration a = F / m = 3000 / 1200 = 2.5 m/s².

一辆 1200 kg 的汽车受到 3000 N 的合力，加速度 a = F / m = 3000 / 1200 = 2.5 m/s²。

Weight is the force of gravity on a mass: W

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