📚 Year 11 OCR Physics: Unit Test Mock Paper Walkthrough | Year 11 OCR 物理：单元测试模拟卷解析
This mock paper walkthrough is designed to help Year 11 students tackle typical OCR Physics unit test questions. By working through key examples from energy, forces, electricity, the particle model, radioactivity, waves, magnetism and practical skills, you will sharpen your problem-solving techniques and avoid common pitfalls. Use this guide alongside your revision notes to build confidence before the real exam.
这份模拟卷解析旨在帮助 11 年级学生应对典型的 OCR 物理单元测试题目。通过逐题分析涵盖能量、力、电学、粒子模型、放射性、波动、磁学以及实验技能的关键例题，你将提高解题技巧并避开常见误区。请将本指南与你的复习笔记结合使用，在实际考试前树立信心。
1. Overview of the Mock Paper and Key Topics | 模拟卷概览及关键主题
This mock paper reflects the style and demand of an OCR Gateway Physics unit test. Questions span P1 Matter, P2 Forces, P3 Electricity, P4 Magnetism, P5 Waves, P6 Radioactivity and practical data handling. The paper balances calculations, short-answer explanations and graph work, testing both your knowledge and application skills. Marks are typically allocated for correct units, significant figures and clear working out.
这份模拟卷反映了 OCR Gateway 物理单元测试的风格与难度。试题涵盖了 P1 物质、P2 力、P3 电学、P4 磁学、P5 波、P6 放射性以及实验数据处理。试卷均衡地考查了计算、简答题和图表处理，既检验知识掌握，也考查应用能力。通常，正确单位、有效数字和清晰的解题步骤都能获得相应分数。
2. Energy Changes and Efficiency | 能量变化与效率
tational field strength g = 10 N/kg. The electrical energy supplied to the motor is 150 J.
一台电机以恒定速度将 5 kg 的重物竖直向上提升 2 m。取重力场强 g = 10 N/kg。供给电机的电能为 150 J。
W = F d
The work done against gravity equals the weight (force) multiplied by the vertical distance. First calculate the weight: Wgravity = m g = 5 × 10 = 50 N. Then work done = 50 N × 2 m = 100 J.
克服重力所做的功等于重力（力）乘以竖直距离。先计算重量：W重力 = m g = 5 × 10 = 50 N。因此做功 = 50 N × 2 m = 100 J。
η = (useful output / total input) × 100%
Efficiency = (100 J / 150 J) × 100% ≈ 66.7% to friction and sound. Always check that the useful output does not exceed the total input – efficiency can never be greater than 100%.
效率 = (100 J / 150 J) × 100% ≈ 66.7%。剩下的 50 J 大部分因摩擦和声音以热能形式耗散。务必检查有用输出不会超过总输入——效率绝不可能大于 100%。
3. Forces and Motion: Newton's Second Law | 力与运动：牛顿第二定律
A resultant force of 6 N acts on a trolley of mass 2 kg. Calculate its acceleration and describe the motion shown on a velocity–time graph if the force is removed after 4 seconds.
一个 6 N 的合力作用在一辆质量为 2 kg 的小车上。计算其加速度，并描述如果 4 秒后撤去该力，速度–时间图像将如何显示其运动。
F = m a
removed, the resultant force becomes zero and the trolley continues at a constant speed in a straight line (Newton's first law). The v–t graph would show a straight line sloping up for 4 s, then a horizontal line.
移项可得 a = F / m = 6 N / 2 kg = 3 m/s²。小车以 3 m/s² 加速 4 秒，达到速度 v = u + a t = 0 + 3 × 4 = 12 m/s。撤去力后，合力变为零，小车将保持匀速直线运动（牛顿第一定律）。对应的 v–t 图将显示前 4 秒为向上倾斜的直线，之后为水平线。
4. Electrical Circuits: Ohm's Law and Resistance | 电路：欧姆定律与电阻
A 12 V battery is connected across a series circuit containing a 4 Ω and a 6 Ω resistor. Calculate the total resistance, the current flowing, and the potential difference across the 6 Ω resistor.
一个 12 V 电池连接到一个由 4 Ω 和 6 Ω 电阻串联组成的电路上。计算总电阻、电路中的电流以及 6 Ω 电阻两端的电压。
Rtotal = R₁ + R₂ (series)
Rtotal = 4 Ω + 6 Ω = 10 Ω. Using Ohm' e remaining 4.8 V is dropped across the 4 Ω resistor. Always check that the sum of the individual p.d.s equals the supply voltage.
Rtotal = 4 Ω + 6 Ω = 10 Ω。根据欧姆定律，I = V / R = 12 V / 10 Ω = 1.2 A。相同的电流流过两个电阻。6 Ω 电阻两端的电压：V = I × R = 1.2 A × 6 Ω = 7.2 V。剩下的 4.8 V 加在 4 Ω 电阻上。务必验证各电阻电压之和等于电源电压。
5. Density and the Particle Model | 密度与粒子模型
s density in g/cm³ and state whether it will float in water (density 1 g/cm³). Explain your reasoning in terms of the particle arrangement.
一个长方体块的尺寸为 5 cm × 4 cm × 3 cm，质量为 540 g。求其密度（以 g/cm³ 为单位），并判断它是否会浮在水面（水的密度为 1 g/cm³）。用粒子排列的知识解释你的推理。
ρ = m / V
0 g/cm³ is much greater than 1 g/cm³, the block will sink. The particles in the solid block are packed much more closely than water particles, giving a much larger mass per unit volume. An object sinks if its density is greater than that of the fluid it is placed in.
体积 V = 5 cm × 4 cm × 3 cm = 60 cm³。密度 ρ = 540 g / 60 cm³ = 9.0 g/cm³。由于 9.0 g/cm³ 远大于 1 g/cm³，该块会下沉。固体块中的粒子比水粒子排列紧密得多，因此单位体积的质量更大。如果物体的密度大于它所置流体密度，物体就会下沉。
6. Specific Heat Capacity Calculation | 比热容计算
An electric heater rated at 50 W is used to heat 0.5 kg of water for 1260 seconds. The temperature of the water rises from 20 °C to 50 °C. Calculate the specific heat capacity of water using these results. Assume no energy losses to the surroundings.
一个额定功率为 50 W 的电热器用来加热 0.5 kg 的水，加热时间为 1260 秒。水温从 20 °C 升到 50 °C。用这些数据计算水的比热容。假设没有能量损失到周围环境。
E = m c Δθ
es c = E / (m Δθ) = 63 000 J / (0.5 kg × 30 °C) = 4200 J/(kg °C). This matches the accepted value for water. Remember that specific heat capacity is the energy required to raise 1 kg of a substance by 1 °C.
先求提供的热量：E = P t = 50 W × 1260 s = 63 000 J。温度变化 Δθ = 50 °C − 20 °C = 30 °C。移项公式可得 c = E / (m Δθ) = 63 000 J / (0.5 kg × 30 °C) = 4200 J/(kg °C)。这一结果与公认的水的比热容一致。记住比热容是指 1 kg 物质升高 1 °C 所需的能量。
7. Radioactive Decay and Half-Life | 放射性衰变与半衰期
A sample of a radioactive isotope has an initial count rate of 800 counts per minute. After 9 days, the count rate falls to 100 counts per minute. Determine the half-life of the isotope and explain what half-life means for an individual nucleus.
一种放射性同位素样本初始计数率为每分钟 800 次。9 天后，计数率降至每分钟 100 次。求该同位素的半衰期，并解释半衰期对单个原子核意味着什么。
count rate halves each half‑life
800 → 400 (1 half‑life) → 200 (2 half‑lives) → 100 (3 half‑lives). Three half‑lives took 9 days, so one half‑life = 9 days / 3 = 3 days. The half‑ roperty; we cannot predict exactly when a single nucleus will decay, only that it has a 50% chance of decaying within one half‑life.
800 → 400（1 个半衰期）→ 200（2 个半衰期）→ 100（3 个半衰期）。3 个半衰期用时 9 天，因此半衰期 = 9 天 / 3 = 3 天。半衰期是样本中一半不稳定原子核发生衰变所需的时间。这是一个统计性质；我们无法准确预测单个原子核何时衰变，只能知道它在一个半衰期内有 50% 的衰变几率。
8. Wave Properties: Speed, Frequency, Wavelength | 波的性质：波速、频率、波长
A water wave has a frequency of 5 Hz and a wavelength of 0.2 m. Calculate the wave speed. Then explain how the wavelength would change if the same wave entered shallower water where its speed decreased to 0.8 m/s while its frequency remained constant.
一个水波的频率为 5 Hz，波长为 0.2 m。计算波速。然后解释如果该波进入更浅的水域，波速降至 0.8 m/s 而频率保持不变，其波长将如何变化。
v = f λ
sing v = f λ, λ = v / f = 0.8 m/s ÷ 5 Hz = 0.16 m. The wavelength decreases. This explains why waves appear to 'bunch up' near a shore.
Published by TutorHao | Year 11 Physics Revision Series | aleveler.com
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