📚 GCSE OCR Physics: Unit Test Paper | GCSE OCR 物理：单元测试卷
This unit test paper is designed to help you revise essential topics from the GCSE OCR Physics specification. Each question targets a key concept, from energy stores to radioactivity, and includes detailed worked solutions. Use this resource to test your understanding, practise calculations, and build confidence before your exam.
本单元测试卷旨在帮助你复习 GCSE OCR 物理考试大纲中的核心主题。每个问题都针对一个关键概念，从能量储存到放射性，并配有详细的解题步骤。利用这份资料检验你的理解、练习计算，并在考试前建立信心。
1. Energy Stores and Transfers | 能量储存与转移
Question: A student lifts a 5.0 kg barbell from the floor to a height of 1.8 m. Calculate the increase in gravitational potential energy (g = 9.8 N/kg).
问题：一名学生将 5.0 kg 的杠铃从地面举至 1.8 m 高处。计算重力势能的增加量 (g = 9.8 N/kg)。
Solution: The gravitational potential energy store increases by Ep = mgh. Substitute the values: Ep = 5.0 × 9.8 × 1.8 = 88.2 J. The energy has been transferred mechanically from the chemical store of the student’s muscles.
解答：重力势能储存的增加量 Ep = mgh。代入数值：Ep = 5.0 × 9.8 × 1.8 = 88.2 J。该能量通过机械方式从学生肌肉的化学能储存转移而来。
2. Circuit Calculations | 电路计算
Question: A filament lamp has a potential difference of 12 V across it and a current of 0.25 A flowing through it. Calculate the resistance of the lamp and the power dissipated.
问题：一个白炽灯两端的电势差为 12 V，流过的电流为 0.25 A。计算灯的电阻和耗散功率。
Solution: Using Ohm’s law, R = V / I = 12 / 0.25 = 48 Ω. Power is given by P = I V = 0.25 × 12 = 3.0 W. Note that the resistance of a filament lamp increases as it gets hotter, but under these steady conditions it is 48 Ω.
解答：根据欧姆定律，R = V / I = 12 / 0.25 = 48 Ω。功率 P = I V = 0.25 × 12 = 3.0 W。注意白炽灯的电阻随温度升高而增大，但在该稳定条件下电阻为 48 Ω。
3. Newton’s Laws and Acceleration | 牛顿定律与加速度
Question: A 1200 kg car accelerates from rest to 20 m/s in 8.0 s. Calculate the resultant force acting on the car, assuming constant acceleration.
问题：一辆 1200 kg 的汽车从静止加速到 20 m/s，用时 8.0 s。假设加速度恒定，计算作用在汽车上的合力。
Solution: First find acceleration: a = (v − u) / t = (20 − 0) / 8.0 = 2.5 m/s². Then apply Newton’s second law: F = m a = 1200 × 2.5 = 3000 N. The resultant force is 3000 N in the direction of motion.
解答：先求加速度：a = (v − u) / t = (20 − 0) / 8.0 = 2.5 m/s²。再应用牛顿第二定律：F = m a = 1200 × 2.5 = 3000 N。合力为 3000 N，方向与运动方向相同。
4. Wave Properties | 波的性质
Question: A water wave has a frequency of 4.0 Hz and a wavelength of 0.75 m. Calculate the wave speed. The wave then passes into shallower water where its speed decreases to 2.4 m/s. Calculate the new wavelength, assuming the frequency remains the same.
问题：一个水波的频率为 4.0 Hz，波长为 0.75 m。计算波速。该波随后进入较浅水域，波速降低至 2.4 m/s。假设频率不变，计算新的波长。
Solution: Wave speed v = f λ = 4.0 × 0.75 = 3.0 m/s. In shallower water, v’ = f λ’, so λ’ = v’ / f = 2.4 / 4.0 = 0.60 m. The wavelength shortens as the wave slows down.
解答：波速 v = f λ = 4.0 × 0.75 = 3.0 m/s。在浅水中，v’ = f λ’，因此 λ’ = v’ / f = 2.4 / 4.0 = 0.60 m。波速减小时波长变短。
5. Radioactive Decay | 放射性衰变
Question: A sample of radioactive isotope has a half-life of 15 hours. Initially the sample contains 800 million unstable nuclei. Estimate the number of unstable nuclei remaining after 45 hours.
问题：一种放射性同位素样品的半衰期为 15 小时。最初样品含有 8 亿个不稳定原子核。估计 45 小时后剩余的不稳定原子核数量。
Solution: 45 hours is three half-lives (15 h × 3). After each half-life the number of nuclei halves. After 1 half-life: 400 million; after 2: 200 million; after 3: 100 million. So 100 million nuclei remain. The count rate would also fall to one-eighth of the original.
解答：45 小时是三个半衰期 (15 h × 3)。每经过一个半衰期，原子核数目减半。1 个半衰期后：4 亿；2 个后：2 亿；3 个后：1 亿。因此剩余 1 亿个原子核。计数率也会降至原来的八分之一。
6. Specific Heat Capacity | 比热容
Question: An aluminium block of mass 0.80 kg is heated by an electric heater that supplies 24 kJ of energy. The temperature of the block rises from 20 °C to 45 °C. Calculate the specific heat capacity of aluminium.
问题：一个质量为 0.80 kg 的铝块被电热器加热，电热器提供了 24 kJ 的能量。铝块的温度从 20 °C 升至 45 °C。计算铝的比热容。
Solution: Use ΔE = m c Δθ. Rearranging: c = ΔE / (m Δθ) = 24000 / (0.80 × 25) = 24000 / 20 = 1200 J/(kg °C). This is close to the standard value of 900 J/(kg °C); the difference may be due to heat losses.
解答：应用 ΔE = m c Δθ。变形得：c = ΔE / (m Δθ) = 24000 / (0.80 × 25) = 24000 / 20 = 1200 J/(kg °C)。该值接近标准值 900 J/(kg °C)；差异可能是由于热量损失造成的。
7. Electromagnetic Spectrum | 电磁波谱
Question: Sort the following electromagnetic waves in order of increasing wavelength: ultraviolet, radio, visible, X-ray. State which of these waves has the highest frequency and which transfers the most energy per photon.
问题：将以下电磁波按波长递增的顺序排列：紫外线、无线电波、可见光、X射线。指出哪一种波的频率最高，哪一种波每个光子传递的能量最大。
Solution: Order of increasing wavelength: X-ray, ultraviolet, visible, radio. Frequency and photon energy decrease as wavelength increases. Therefore, X-rays have the highest frequency and transfer the most energy per photon. Radio waves have the longest wavelength and the lowest photon energy.
解答：波长递增的顺序：X射线、紫外线、可见光、无线电波。频率和光子能量随波长增大而减小。因此，X射线频率最高，光子能量最大。无线电波波长最长，光子能量最低。
8. Forces and Elasticity | 力与弹性
Question: A spring has a spring constant of 50 N/m. A force of 7.5 N is applied to stretch the spring. Calculate the extension of the spring. Assume the spring does not exceed its limit of proportionality.
问题：一根弹簧的劲度系数为 50 N/m。施加 7.5 N 的力来拉伸弹簧。计算弹簧的伸长量。假设弹簧未超过其比例极限。
Solution: Hooke’s law: F = k x, so x = F / k = 7.5 / 50 = 0.15 m (or 15 cm). The elastic potential energy stored in the stretched spring would be Ee = ½ k x² = 0.5 × 50 × (0.15)² = 0.5625 J.
解答：胡克定律：F = k x，因此 x = F / k = 7.5 / 50 = 0.15 m（即 15 cm）。拉伸弹簧储存的弹性势能为 Ee = ½ k x² = 0.5 × 50 × (0.15)² = 0.5625 J。
9. Density Practical | 密度实验
Question: A student measures the mass of an irregular stone as 156 g. She then places it into a measuring cylinder containing 80 cm³ of water; the water level rises to 138 cm³. Calculate the density of the stone in g/cm³ and in kg/m³.
问题：一名学生测得一块不规则石头的质量为 156 g。然后她将石头放入装有 80 cm³ 水的量筒中；水面升至 138 cm³。计算该石头的密度，分别以 g/cm³ 和 kg/m³ 表示。
Solution: Volume of stone = 138 − 80 = 58 cm³. Density ρ = mass / volume = 156 / 58 = 2.689… ≈ 2.7 g/cm³. To convert to kg/m³: 1 g/cm³ = 1000 kg/m³, so ρ ≈ 2700 kg/m³. This density suggests the stone might be granite.
解答：石头体积 = 138 − 80 = 58 cm³。密度 ρ = 质量 / 体积 = 156 / 58 = 2.689… ≈ 2.7 g/cm³。换算为 kg/m³：1 g/cm³ = 1000 kg/m³，因此 ρ ≈ 2700 kg/m³。此密度表明该石头可能是花岗岩。
10. Mains Electricity and Power | 家庭用电与功率
Question: An electric kettle is connected to the UK mains supply (230 V) and is rated at 2.2 kW. Calculate the current drawn by the kettle and the energy transferred if it is used for 3.0 minutes. Give the energy in joules and in kilowatt-hours.
问题：一个电热水壶接入英国市电 (230 V)，额定功率为 2.2 kW。计算水壶的工作电流，以及使用 3.0 分钟转移的能量。分别以焦耳和千瓦时为单位表示。
Solution: Current I = P / V = 2200 / 230 = 9.565… ≈ 9.6 A. Time in seconds: t = 3.0 × 60 = 180 s. Energy E = P t = 2200 × 180 = 396 000 J (or 396 kJ). In kilowatt-hours: 2.2 kW × (3/60) h = 0.11 kWh. The energy heats the water, increasing its internal store.
解答：电流 I = P / V = 2200 / 230 = 9.565… ≈ 9.6 A。时间以秒计：t = 3.0 × 60 = 180 s。能量 E = P t = 2200 × 180 = 396 000 J（即 396 kJ）。以千瓦时为单位：2.2 kW × (3/60) h = 0.11 kWh。这些能量加热了水，增加了水的内能储存。
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