📚 Year 10 OCR Physics: Mock Unit Test Walkthrough | Year 10 OCR 物理：单元测试模拟卷解析
A mock unit test is one of the best ways to identify gaps in your understanding before the real exam. This walkthrough breaks down typical OCR Year 10 Physics questions, showing you how to approach each topic systematically and avoid common mistakes.
单元模拟卷是考前查漏补缺的最佳方式之一。本文逐题解析典型的 OCR Year 10 物理试题，教你如何系统性地处理各个知识点，并避开常见错误。
1. Understanding the Mock Test Structure | 理解模拟卷结构
The OCR Year 10 Physics mock paper usually contains a mix of multiple-choice, short-answer and 4–6 mark extended writing questions. All questions are based on the Gateway Science Physics A specification, covering topics from energy and matter to electricity and forces.
OCR Year 10 物理模拟卷通常包含选择题、简答题和 4–6 分的扩展写作题。所有题目均基于 Gateway Science 物理 A 大纲，覆盖从能量、物质到电学和力学的知识点。
You are expected to recall key equations, convert units, and explain scientific ideas clearly using correct terminology. Marks are awarded not only for the final answer but also for showing your working and logical structure.
考试要求你背诵关键公式、转换单位，并用准确术语清晰地解释科学概念。得分点不仅在于最终答案，也包括展示解题步骤和逻辑结构。
2. Energy Stores and Transfers | 能量储存与转移
A ball is dropped from rest. Describe the energy transfers that take place until it hits the ground. As the ball falls, energy is transferred mechanically from the gravitational potential energy store to the kinetic energy store of the ball. Some energy is also transferred mechanically to the thermal energy store of the surroundings due to air resistance.
一个球从静止释放。描述球下落到地面的能量转移过程。球下落时，能量通过机械做功从重力势能储存转移到球的动能储存。由于空气阻力，部分能量也通过机械做功转移到周围环境的热能储存。
When the ball hits the ground, energy is transferred by mechanical work (deformation) to the thermal energy store of the ball, the ground and the surroundings, and also as sound waves. The total energy remains conserved throughout.
当球撞击地面时，能量通过机械做功（形变）转移到球、地面和周围环境的热能储存，也以声波形式传递。整个过程中总能量守恒。
Common mistake: Some students say energy is 'lost' or 'used up'. Always use the precise language of energy stores and transfers. Energy is never destroyed.
常见错误：部分学生说能量被“丢失”或“用尽”。务必使用能量储存和转移的精确语言。能量永不消失。
3. Efficiency Calculations | 效率计算
rcentage. Efficiency = (useful output energy transfer / total input energy transfer) × 100%.
一台电动机转换了 250 J 的电能，做了 160 J 的有用机械功。计算效率，并以百分比给出答案。效率 = （有用输出能量转移 ÷ 总输入能量转移）× 100%。
Efficiency = 160 / 250 = 0.64, so 64% d final percentage. The remaining 90 J is dissipated as thermal energy to the surroundings – this is the wasteful energy transfer.
效率 = 160 ÷ 250 = 0.64，即 64%。务必展示公式、代入和最终百分比。其余 90 J 作为热能散逸到周围环境中——这是浪费的能量转移。
Efficiency = (160 ÷ 250) × 100% = 64%
If the motor were 100% efficient, it would convert all input energy into useful work, but that is impossible in practice. Low-efficiency devices waste more energy and cost more to run.
若电动机效率为 100%，则所有输入能量都将转化为有用功，但现实中不可能实现。低效率设备浪费更多能源，运行成本也更高。
4. Series and Parallel Circuits | 串联与并联电路
In a series circuit, the current is the same through all components. The total resistance is the sum of individual resistances. The potential difference from the battery is shared between components.
在串联电路中，通过所有元件的电流相同。总电阻等于各电阻之和。电池提供的电势差在元件之间分配。
In a parallel circuit, the potential difference across each branch is the same. The total current from the battery splits at junctions. The total resistance is less than the smallest individual resistance because there are multiple paths for charge to flow.
在并联电路中，各支路两端的电势差相等。电池提供的总电流在分支点分流。总电阻小于最小的单个电阻，因为电荷有多条通路。
A typical 6-mark question might ask you to compare the brightness of bulbs in series and parallel circuits. Bulbs in parallel shine brighter than those in series when connected to the same battery, because each gets the full battery voltage and receives a larger current.
典型的 6 分题可能会要求你比较串联和并联电路中灯泡的亮度。连接到同一电池时，并联灯泡比串联灯泡更亮，因为每个灯泡获得全部电池电压，并通过较大电流。
5. Ohm's Law and Resistance | 欧姆定律与电阻
across it is 6.0 V. Calculate the resistance. Using V = I × R, we rearrange to R = V / I.
一个定值电阻两端电势差为 6.0 V 时，通过的电流为 2.5 A。计算电阻。由 V = I × R 得 R = V / I。
R = 6.0 V ÷ 2.5 A = 2.4 Ω
Resistance is measured in ohms (Ω). The slope of a voltage–current graph for an ohmic conductor is constant, indicating resistance remains constant at constant temperature. If the wire gets hot, the resistance increases because ions vibrate more, making it harder for electrons to flow.
电阻的单位是欧姆（Ω）。欧姆导体的电压–电流图像斜率恒定，表明在恒温下电阻保持不变。如果导线变热，电阻会增大，因为离子振动加剧，阻碍电子流动。
Remember: the component must be ohmic (e.g., fixed resistor at constant temperature) for the straight-line graph to apply. Diodes and filament lamps are non-ohmic.
记住：只有欧姆元件（如恒温下的定值电阻）才能得到直线图像。二极管和灯丝是非欧姆元件。
6. Density and States of Matter | 密度与物质状态
2.0 × 2.0 = 8.0 cm³. Density = 16.0 g / 8.0 cm³ = 2.0 g/cm³. This density is close to aluminium.
密度 = 质量 ÷ 体积。一个边长为 2.0 cm 的规则立方体质量为 16.0 g。计算其密度并说明可能是何种材料。体积 = 2.0 × 2.0 × 2.0 = 8.0 cm³。密度 = 16.0 g / 8.0 cm³ = 2.0 g/cm³。此密度接近铝。
The particle arrangement explains density differences. Solids have particles close together in a regular pattern, so they have high density. Gases have particles far apart, so density is very low. When a substance changes state, mass stays the same but volume changes, altering density.
粒子排列方式可解释密度差异。固体粒子紧密且排列规则，因此密度高。气体粒子间距大，密度非常低。物质状态变化时，质量不变但体积改变，从而改变密度。
Determining density of an irregular solid: measure mass using a balance; measure volume by displacement of water in a measuring cylinder. Ensure the object is fully submerged and read the meniscus at eye level.
测量不规则固体的密度：用天平测质量；用量筒排水法测体积。确保物体完全浸没，并平视读取弯月面。
7. Specific Heat Capacity | 比热容
Specific heat capacity is the energy required to raise the temperature of 1 kg of a substance by 1 °C. A heater supplies 12 000 J to a 2.0 kg block of metal, raising its temperature from 20 °C to 35 °C. Calculate the specific heat capacity.
比热容是使 1 kg 物质温度升高 1 °C 所需的能量。一个加热器向 2.0 kg 的金属块提供 12 000 J 能量，使其温度从 20 °C 升至 35 °C。计算此金属的比热容。
Using ΔE = m × c × Δθ, c = ΔE / (m × Δθ) = 12 000 / (2.0 × 15) = 12 000 / 30 = 400 J/kg°C.
温差 Δθ = 35 − 20 = 15 °C。用 ΔE = m × c × Δθ 得 c = ΔE ÷ (m × Δθ) = 12 000 ÷ (2.0 × 15) = 12 000 ÷ 30 = 400 J/kg°C。
c = ΔE / (m × Δθ) = 400 J/kg°C
Substances with high specific heat capacity, like water, can store large amounts of energy with a small temperature rise. This is why water is used in central heating systems and why coastal areas have milder climates.
比热容大的物质（如水）可在温度变化不大的情况下储存大量能量。这就是水用于中央供暖系统以及沿海地区气候温和的原因。
8. Forces and Motion Graphs | 力与运动图像
A car accelerates from rest and then maintains a constant speed. Sketch a distance–time graph and a velocity–time graph. The distance–time graph for acceleration is a curve with increasing gradient; for constant speed, it is a straight line. The velocity–time graph shows a straight line sloping upward during acceleration, then a horizontal line at constant velocity.
一辆车从静止加速，然后保持匀速。画出距离–时间图像和速度–时间图像。加速阶段的距离–时间图像是一条斜率递增的曲线；匀速段是一条直线。速度–时间图像中，加速段为向上倾斜的直线，匀速段为水平线。
The area under a velocity–time graph represents displacement. The gradient of a velocity–time graph gives acceleration. If the acceleration section has a gradient of 5 m/s², the velocity increases by 5 m/s each second.
速度–时间图像下的面积表示位移。速度–时间图像的斜率表示加速度。若加速段斜率为 5 m/s²，则速度每秒增加 5 m/s。
When forces on an object are balanced, the velocity remains constant (or stays at zero). Unbalanced forces cause acceleration in the direction of the resultant force. This is Newton's first law.
当物体受力平衡时，速度保持不变（或保持零）。不平衡力会在合力方向上产生加速度。这就是牛顿第一定律。
9. Work Done and Power | 做功与功率
e. Weight = m × g = 200 × 10 = 2 000 N. Work done = 2 000 N × 8.0 m = 16 000 J (or 16 kJ).
做功 = 力 × 沿力方向移动的距离。一台起重机将 200 kg 的重物垂直提升 8.0 m。计算做功。重力 = m × g = 200 × 10 = 2 000 N。做功 = 2 000 N × 8.0 m = 16 000 J（或 16 kJ）。
ne takes 4.0 s to lift the load, its power output is 16 000 J / 4.0 s = 4 000 W or 4.0 kW.
功率是做功的快慢：P = W ÷ t。若起重机用时 4.0 s 提升重物，则输出功率为 16 000 J ÷ 4.0 s = 4 000 W 即 4.0 kW。
One watt is one joule per second. In questions about climbing stairs, weight (force) is multiplied by vertical height to find work done. Always check for consistent units: force in newtons, distance in metres, giving joules.
一瓦特等于每秒一焦耳。在爬楼梯问题中，重力（力）乘以垂直高度即可求做功。务必检查单位统一：力用牛顿，距离用米，得出焦耳。
10. National Grid and Transformers | 国家电网与变压器
The National Grid distributes electricity from power stations to consumers. Step-up transformers increase the voltage to hundreds of thousands of volts, reducing current for the same power. This minimises energy losses (as heat) in transmission cables, because Ploss = I² × R.
国家电网将电力从发电站输送到用户。升压变压器将电压升至数十万伏，在同等功率下降低电流。这减少了输电线中的能量损失（以热的形式），因为 Pₗₒₛₛ = I² × R。
Step-down transformers then reduce the voltage to a safe level (e.g., 230 V) for homes. Transformers only work with alternating current (a.c.). A transformer has a primary coil and a secondary coil wound on a laminated iron core.
随后，降压变压器将电压降至安全级别（如 230 V）供家庭使用。变压器仅适用于交流电（a.c.）。变压器由绕在叠片铁芯上的初级线圈和次级线圈组成。
Vp / Vs = Np / Ns
Where Vp, Vs are primary and secondary voltages, and Np, Ns are the numbers of turns. If a transformer has 500 turns on the primary and 50 on the secondary, and the primary voltage is 230 V, Vs = 230 × (50 / 500) = 23 V.
其中 Vp、Vs 为初级和次级电压，Np、Ns 为匝数。若变压器初级 500 匝、次级 50 匝，初级电压为 230 V，则 Vs = 230 ×（50 ÷ 500）= 23 V。
11. Radioactive Decay | 放射性衰变
An unstable nucleus emits alpha, beta or gamma radiation to become more stable. Alpha particles are helium nuclei (2 protons + 2 neutrons), highly ionising but stopped by paper. Beta particles are fast electrons (from neutron turning into proton), moderately ionising and stopped by a few mm of aluminium. Gamma rays are electromagnetic waves, weakly ionising but require thick lead or concrete for shielding.
不稳定的原子核会放射 α、β 或 γ 射线以变得更稳定。α 粒子是氦核（2 个质子 + 2 个中子），电离能力强，但可被纸张阻挡。β 粒子是高速电子（中子转变为质子时产生），电离能力中等，几毫米铝片可阻挡。γ 射线是电磁波，电离能力弱，但需要厚铅板或混凝土屏蔽。
urs, after 6 hours the activity falls to 400 counts/s (two half-lives: 1 600 → 800 → 400).
半衰期是计数率（或原子核数量）减半所需的时间。若某样品初始计数率为 1 600 次/秒，半衰期为 3 小时，则 6 小时后活度降为 400 次/秒（经历两次半衰期：1 600 → 800 → 400）。
Radioactive decay is random, but you can use the half-life to predict changes over time. Background radiation must be subtracted from readings in experiments.
放射性衰变是随机的，但你可利用半衰期预测随时间的变化。实验中必须从读数中扣除背景辐射。
12. Tips for Exam Success | 考试成功技巧
Always read the question carefully, looking for command words like 'describe', 'explain', 'calculate' and 'compare'. Underline key numbers and units. Show all steps in calculations – even if your final answer is wrong, you can still earn method marks.
仔细阅读题目，留意如“描述”、“解释”、“计算”和“比较”等指令词。划出关键数字和单位。计算时写出所有步骤——即使最终答案错误，你仍能获得方法分。
Use scientific vocabulary accurately. For example, 'energy is transferred to the thermal store' not 'heat is produced'. When plotting graphs, choose sensible scales, label axes with quantity and unit, and draw a line of best fit.
准确使用科学词汇。例如，“能量转移到热能储存”而非“产生热”。绘制图像时，选择合适的坐标尺度，标注轴名和单位，并画出最佳拟合线。
Time management is crucial. Allocate roughly one minute per mark. For longer written questions, plan a few bullet points in the margin before writing. Check your answers if you have time at the end, especially unit conversions and significant figures.
时间管理至关重要。大约每分钟得一分。对于较长的写作题，在答题前先在页边列出要点。如有时间，最后检查答案，尤其是单位换算和有效数字。
Published by TutorHao | Physics Revision Series | aleveler.com
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