📚 Year 10 OCR Physics Unit Test Mock Paper Walkthrough | Year 10 OCR 物理单元测试模拟卷解析
This mock paper walkthrough covers the core topics typically assessed in the Year 10 OCR Physics unit test, including energy stores and transfers, electricity, particle model of matter, atomic structure, and required practical skills. Each question is broken down with step‑by‑step reasoning, key formulas, and common pitfalls. Use this guide to sharpen your problem‑solving technique and boost your confidence before the real test.
本份模拟卷解析涵盖了 Year 10 OCR 物理单元测试中经常出现的核心主题，包括能量储存与转移、电学、粒子物质模型、原子结构以及必修实验技能。每道题目都配有逐步推理、关键公式和常见易错点。请利用本指南打磨解题技巧，在正式考试前提升信心。
1. Question 1 – Energy Stores and Transfers | 能量储存与转移
d of the ball just before it hits the ground. Ignore air resistance and take g = 9.8 N/kg.
一个质量为 0.2 kg 的球从 5.0 m 高度落下。计算球刚好撞击地面前的速率。忽略空气阻力，取 g = 9.8 N/kg。
gy (KE) because no external work is done against air resistance. The GPE lost equals the KE gained. Write: mgh = ½mv². Cancel mass m from both sides, then v² = 2gh. Substituting the numbers gives v² = 2 × 9.8 × 5.0 = 98, so v = √98 ≈ 9.9 m/s. Always check that the mass cancels – a common mistake is to carry the mass into the kinetic energy expression prematurely.
这是一道经典的能量守恒问题。球下落时，重力势能完全转化为动能，因为没有空气阻力做负功。损失的重力势能等于获得的动能。列出方程：mgh = ½mv²。两边消去质量 m，得到 v² = 2gh。代入数字得 v² = 2 × 9.8 × 5.0 = 98，因此 v = √98 ≈ 9.9 m/s。请务必检查质量是否被消去——常见的错误是过早将质量代入动能表达式。
2. Question 2 – Specific Heat Capacity Calculation | 比热容计算
An electric kettle contains 0.50 kg of water at 20 °C. It is switched on until the water reaches boiling point at 100 °C. The specific heat capacity of water is 4200 J/(kg°C). Calculate the energy transferred to the water.
一个电热水壶装有 0.50 kg、20 °C 的水，通电加热至沸点 100 °C。水的比热容为 4200 J/(kg°C)。计算传递给水的能量。
e temperature of a substance is given by ΔE = m c Δθ. Here the temperature change Δθ = 100 – additional latent heat would be required, but the question only asks for the energy to reach 100 °C. Watch out for unit conversions – temperature change in Celsius is equal to the change in kelvin.
提升物质温度所需的能量由公式 ΔE = m c Δθ 给出。此处温度变化 Δθ = 100 – 20 = 80 °C。代入数值：ΔE = 0.50 × 4200 × 80 = 168 000 J，即 168 kJ。该公式仅在没有状态变化时有效。如果继续沸腾，还需要额外的潜热，但此题只要求到达 100 °C 所需能量。注意单位换算——摄氏度的温度变化与开尔文的温度变化数值相等。
3. Question 3 – Ohm’s Law in a Fixed Resistor | 定值电阻的欧姆定律
A filament lamp is connected to a variable power supply. The table shows the p.d. across the lamp and the current through it. Plot a graph of p.d. against current. Describe how the resistance changes as the p.d. increases and explain why this happens.
一个灯丝灯泡连接到可调电源上。表格给出了灯泡两端的电压和通过它的电流。绘制电压-电流图。描述随着电压升高电阻如何变化，并解释原因。
The current– te, so the graph curves. Resistance is given by R = V/I. Since the graph bends towards the voltage axis, V/I increases, meaning resistance increases. This happens because the metal filament gets hotter, and the metal ions vibrate more vigorously, making it harder for free electrons to flow through. This is a classic required practical – ph has labelled axes with units and a smooth curve of best fit, ignoring anomalous points.
灯丝灯泡的电流-电压特性是非线性的。随着电压升高，电流也增大，但增大的速率减慢，因此图像弯曲。电阻由 R = V/I 给出。由于图像向电压轴弯曲，V/I 增大，意味着电阻增大。其原因是金属灯丝温度升高，金属离子振动加剧，使得自由电子更难通过。这是一个典型的必修实验——请确保你的坐标轴标有物理量和单位，并画出光滑的最佳拟合曲线，忽略异常点。
4. Question 4 – Density of a Regular Solid | 规则固体的密度
minium in g/cm³ and state whether it is likely to float in water (density of water = 1.0 g/cm³).
一名学生测得一个矩形铝块的质量为 270 g。该铝块的尺寸为 10.0 cm × 5.0 cm × 2.0 cm。计算铝的密度（以 g/cm³ 为单位），并判断它在水中是否可能浮起（水的密度 = 1.0 g/cm³）。
e aluminium’s density is greater than that of water, the block will sink. Many students invert the density formula or forget to calculate volume correctly. Also remember that for an object to float, its density must be less than the fluid’s density. Gases and liquids are treated the same way, although measuring volumes may differ.
密度 ρ = 质量 / 体积。首先计算体积：V = 10.0 × 5.0 × 2.0 = 100 cm³。然后 ρ = 270 g / 100 cm³ = 2.7 g/cm³。因为铝的密度大于水的密度，铝块会下沉。许多学生会弄反密度公式或忘记正确计算体积。还要记住，物体浮起的条件是它的密度小于流体的密度。气体和液体测量方法类似，只是体积测量可能不同。
5. Question 5 – Internal Energy and Change of State | 内能与物态变化
Explain why the temperature of a substance remains constant while it is melting, even though energy is still being transferred to it.
解释为什么物质在熔化过程中温度保持不变，即使它仍在接受能量传递。
During melting, the energy supplied does not increase the kinetic energy of the particles, but rather breaks the intermolecular bonds holding the particles in fixed positions. This increases the potential energy of the particles while their average kinetic energy stays the same. Since temperature is a measure of the average kinetic energy of the particles, the temperature remains constant. This is called the latent heat of fusion. The same principle applies to boiling (latent heat of vaporisation). A common misconception is to think that energy input always raises temperature; here it changes the state without a temperature change.
在熔化过程中，供给的能量并没有增加粒子的动能，而是用于打破将粒子固定在确定位置上的分子间作用力。这使得粒子的势能增加，而平均动能保持不变。由于温度是粒子平均动能的度量，因此温度保持恒定。这称为熔化潜热。同样的原理也适用于沸腾（汽化潜热）。常见的误解是认为输入能量总会升高温度；在此处，能量改变了物态而温度不变。
6. Question 6 – Types of Nuclear Radiation | 核辐射的类型
Complete the table by giving the range in air and ionising power for alpha, beta, and gamma radiations. Then state which type of radiation can be stopped by a sheet of paper.
请填写表格，给出 α、β 和 γ 辐射在空气中的射程以及电离能力。然后说明哪种辐射能被一张纸阻挡。
Alpha particles have a range of only a few centimetres in air and are highly ionising because they have a large mass and a charge of +2. Beta particles can travel about a metre in air and are moderately ionising. Gamma rays travel several metres in air (or virtually unlimited, as their intensity decreases with distance) and are weakly ionising; they are electromagnetic waves. Alpha radiation is stopped by a sheet of paper or skin. Beta requires a few millimetres of aluminium. Gamma needs several centimetres of lead or thick concrete. Remember the pattern: the more ionising the radiation, the less penetrating it is.
α 粒子在空气中的射程只有几厘米，且电离能力很强，因为它们质量大且带 +2 电荷。β 粒子可在空气中行进约一米，电离能力中等。γ 射线在空气中可行进数米（事实上射程几乎无限，但强度随距离减弱），电离能力很弱；它们是电磁波。α 辐射能被一张纸或皮肤阻挡。β 辐射需要几毫米的铝片阻挡。γ 辐射需要几厘米的铅或厚混凝土。记住规律：电离能力越强的辐射，穿透能力越弱。
7. Question 7 – Series and Parallel Resistance | 串联与并联电阻
Two resistors, 10 Ω and 15 Ω, are connected first in series and then in parallel across a 6 V battery. Calculate the total resistance in each case and determine which arrangement draws the larger current from the battery.
将两个电阻（10 Ω 和 15 Ω）先后串联和并联在 6 V 电池两端。分别计算每种情况的总电阻，并判断哪种接法从电池取用更大的电流。
. Current I = 6 / 6 = 1.0 A. The parallel arrangement has a lower total resistance and therefore draws a larger current. Always add fractions carefully for parallel resistors – tly. In parallel, total resistance is always less than the smallest individual resistance.
串联：Rₜₒₜₐₗ = R₁ + R₂ = 10 + 15 = 25 Ω。电流 I = V / Rₜₒₜₐₗ = 6 / 25 = 0.24 A。并联：1/Rₜₒₜₐₗ = 1/10 + 1/15 = 3/30 + 2/30 = 5/30，因此 Rₜₒₜₐₗ = 30/5 = 6 Ω。电流 I = 6 / 6 = 1.0 A。并联接法总电阻更小，因此取用的电流更大。计算并联电阻时要仔细相加倒数——常见的错误是直接将电阻值相加。并联时总电阻总是小于最小的单个电阻。
8. Question 8 – Electrical Power and Energy | 电功率与电能
An electric heater operates at 230 V and draws a current of 4.0 A. Calculate the power of the heater and the energy transferred in 5 minutes. Give your answer in joules and in kilowatt‑hours.
一个电暖器在 230 V 电压下工作，通过的电流为 4.0 A。计算暖器的功率以及在 5 分钟内传递的能量。答案用焦耳和千瓦时表示。
E = 276 000 / 3 600 000 = 0.0767 kWh (about 0.077 kWh). Alternatively, power 0.92 kW × (5/60) h = 0.0767 kWh. Pay careful attention to time units: power in watts and time in seconds give joules; power in kilowatts and time in hours give kilowatt‑hours.
功率 P = I V = 4.0 × 230 = 920 W（920 J/s）。时间 t = 5 × 60 = 300 s。能量 E = P × t = 920 × 300 = 276 000 J。换算为千瓦时：1 kWh = 3.6 × 10⁶ J（3 600 000 J）。因此 E = 276 000 / 3 600 000 = 0.0767 kWh（约 0.077 kWh）。也可用 0.92 kW ×（5/60）h = 0.0767 kWh。务必注意时间单位：功率用瓦特、时间用秒得到焦耳；功率用千瓦、时间用小时得到千瓦时。
9. Question 9 – Half‑life from a Decay Graph | 从衰变图求半衰期
The decay curve of a radioactive isotope shows that the count rate falls from 800 counts per minute to 200 counts per minute in 6.0 hours. Determine the half‑life of the isotope and sketch the remaining count rate after 12 hours.
某种放射性同位素的衰变曲线显示，计数率在 6.0 小时内从每分钟 800 次下降到 200 次。确定该同位素的半衰期，并画出 12 小时后的剩余计数率。
The count rate has halved twice: 800 → 400 (one half‑life) → 200 (second half‑life). Thus two half‑lives have passed in 6.0 hours, so one half‑life = 6.0 / 2 = 3.0 hours. After 12 hours (which is 12 / 3 = 4 half‑ s to think that the sample has halved from 800 to 200 in one step, which would give a half‑ nt the number of halvings carefully. The graph should be a smooth exponential decay curve.
计数率经历了两次减半：800 → 400（一个半衰期）→ 200（第二个半衰期）。因此 6.0 小时内经过了两个半衰期，一个半衰期 = 6.0 / 2 = 3.0 小时。12 小时后（即 12 / 3 = 4 个半衰期），计数率将减半四次：800 → 400 → 200 → 100 → 50 次/分钟。常见的错误是认为从 800 到 200 只经历了一次减半，从而得出半衰期为 6 小时。务必仔细数清减半次数。绘制的图表应是一条光滑的指数衰减曲线。
10. Question 10 – Required Practical: Measuring Density of an Irregular Object | 必修实验：测量不规则物体的密度
Describe a method to determine the density of an irregularly shaped stone using a displacement can (Eureka can) and a balance. Include the measurements needed and the formula used.
描述一种使用排水罐（溢流罐）和天平测定一块不规则形状石头密度的方法。包括所需测量的量及使用的公式。
r under the spout. Carefully lower the stone into the can on a thin thread. The displaced water runs into the cylinder; measure its volume, V, which equals the volume of the stone. Density is then ρ = m / V. The stone must be completely submerged and not touch the sides, and no air bubbles should be trapped. This method works for any insoluble, non‑ very large objects, a directly measured displacement in a measuring cylinder may be used.
首先，用天平测量石头的质量，记录为 m。将溢流罐注满水直到水从溢流嘴滴下，然后将量筒放在溢流嘴下方。用细线小心地将石头放入罐中。排开的水流入量筒；测量其体积 V，该体积即石头的体积。密度则由 ρ = m / V 计算得出。石头必须完全浸没且不触碰罐壁，不能残留气泡。此方法适用于任何不溶于水且不吸水的固体。对于非常大的物体，可以用量筒直接测量排水体积。
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