📚 Deep Analysis of Past Exam Papers for Eduqas Year 10 Physics | Eduqas 10年级物理历年真题深度解析
Studying past exam papers is one of the most effective ways to prepare for your Eduqas GCSE Physics assessments. By analysing recurring question types, mark schemes and common pitfalls, you can develop a clear strategy for answering both calculation-based and descriptive questions. This article breaks down typical exam questions topic by topic, showing you exactly how to approach them and how to avoid the mistakes that cost marks.
研读历年真题是备战Eduqas GCSE物理考试最有效的方法之一。通过分析反复出现的题型、评分方案和常见失分点，你可以形成清晰的答题策略，从容应对计算题和描述题。本文按主题逐一拆解典型真题，向你展示正确的解题思路，并帮你避开那些容易丢分的陷阱。
1. Energy Stores and Transfers | 能量储存与转换
A classic exam question asks you to describe the energy transfers for a falling object, such as a dropped ball. At the start, the ball has gravitational potential energy (GPE) due to its height. As it falls, GPE is converted to kinetic energy (KE). Upon hitting the ground, some KE is stored as elastic potential energy during the brief compression, and the rest is transferred to the thermal energy store of the ball, floor and surroundings. The ball then rebounds as elastic potential energy converts back to KE and finally to GPE, but it never reaches the original height because energy has been dissipated.
典型的考题要求你描述一个下落物体（例如掉落的球）的能量转移过程。开始时，球因高度而具有重力势能。下落过程中，重力势能转化为动能。球撞击地面时，部分动能在短暂的压缩过程中储存为弹性势能，其余的能量转移到球、地面和周围环境的热能储存中。随后球反弹，弹性势能再转化为动能，最后变为重力势能，但由于能量已经散失，球无法达到原来的高度。
When marking these questions, examiners look for precise use of 'store' and 'transfer' terminology. Avoid loose phrases like 'energy is lost' – always specify the store it is transferred to. Diagrams such as energy flow charts or Sankey diagrams can support your answer, but clear written descriptions are essential for full marks.
阅卷时，考官看重你能否准确使用“储存”和“转移”这两个术语。避免使用“能量损失”这类模糊说法——一定要指明能量转移到了哪个储存。能量流程图或桑基图可以为答案增色，但清晰书面描述才是获得满分的关键。
ceives 200 W of electrical power and the useful mechanical power is shown as a 120 W wide arrow. The wasted thermal power is the difference: 200 W − 120 W = 80 W. Efficiency = (useful power / total power) × 100 = (120/200) × 100 = 60%.
一种常见的数据分析变式是给出桑基图，要求你计算有用功率输出或浪费的能量。例如，一台电动机接收200 W的电功率，图中显示有用机械功率的箭头宽度对应120 W。浪费的热功率就是差值：200 W − 120 W = 80 W。效率 = (有用功率 / 总功率) × 100 = (120/200) × 100 = 60%。
2. Efficiency and Sankey Diagrams | 效率与桑基图
Efficiency calculations appear almost every year. The standard equation is:
Efficiency = (Useful energy output / Total energy input) × 100%
效率计算几乎每年都考。标准公式为：
效率 = (有用能量输出 / 总能量输入) × 100%
Beware of unit conversions: energy may be given in kilojoules (kJ) while power is in watts (J/s). Always convert to joules before substituting. A common error is using the wasted energy value instead of the useful output; always identify the 'useful' transfer intended by the device.
注意单位换算：能量可能以千焦（kJ）给出，而功率以瓦特（J/s）表示。代入公式前一定要换算为焦耳。常见的错误是使用浪费的能量值而不是有用输出；务必明确该设备预期的“有用”转移是什么。
Sankey diagram questions test your ability to draw arrows to scale. The width of each arrow is proportional to the amount of energy. If the total input is 100 J and the wasted thermal energy is 40 J, the useful arrow width should represent 60 J. Examiners expect arrows drawn with a ruler and correctly labelled with the energy store and value.
桑基图题目考察你按比例画箭头的能力。每个箭头的宽度与能量大小成正比。如果总输入为100 J，浪费的热能为40 J，那么有用箭头的宽度应代表60 J。考官要求用尺子绘制箭头，并正确标注能量储存和数值。
3. Electrical Circuits and Ohm's Law | 电路与欧姆定律
Ohm' h is a straight line through the origin, indicating that resistance is constant. You may be asked to calculate resistance at a point: R = V / I. If V = 6.0 V and I = 0.4 A, then R = 6.0 / 0.4 = 15 Ω.
欧姆定律类题目经常给出定值电阻或白炽灯的电流-电压图线。对于定值电阻，图线是一条通过原点的直线，表明电阻恒定。你可能会被要求计算某一点的电阻：R = V / I。若V = 6.0 V，I = 0.4 A，则R = 6.0 / 0.4 = 15 Ω。
For a filament lamp, the line curves because the resistance increases as the temperature of the filament rises. In an exam, you must explain that as current increases, the metal ions vibrate more, causing more collisions with electrons, so resistance increases. This is a high-mark explanation question.
对于白炽灯，图线是弯曲的，因为灯丝温度升高导致电阻增大。考试中你必须解释：随着电流增大，金属离子振动加剧，与电子的碰撞增多，因此电阻增大。这是一道分值较高的解释题。
ion keeping the current constant or low to avoid heating, measuring the wire length accurately with a metre rule, and using the formula R = V/I. Repeating readings and plotting a graph of resistance against length to show proportionality are key marking points.
一道典型的实验题会要求你描述探究导线长度如何影响电阻的实验。你要提到保持电流恒定或尽可能小以避免发热，用米尺精确测量导线长度，并使用公式R = V/I。重复读数、画出电阻-长度图线以展示正比关系，这些都是重要的得分点。
4. Series and Parallel Circuits | 串联与并联电路
Past papers frequently test the differences between series and parallel circuits using identical components. In a series circuit, the current is the same everywhere, but the potential difference is shared. If two identical lamps are in series with a 12 V battery, each lamp receives 6 V. If one lamp breaks, the circuit is open and both go out.
历年真题经常用相同元件对比串联和并联电路。串联电路中电流处处相等，但电压被分担。如果两个相同灯泡串联在12 V电池上，每个灯泡得到6 V。若一个灯泡断路，整个电路断开，两个灯泡都熄灭。
and total current equals the sum of branch currents. If a 12 V battery is connected to two parallel branches each with a 6 Ω resistor, the current in each branch is 12 / 6 = 2 A, and the total current from the battery is 2 + 2 = 4 A. A common calculation error is adding resistances as if in series. For parallel, 1/Rtotal = 1/R₁ + 1/R₂.
在并联电路中，每条支路都获得电池的全部电压，因此并联的灯泡更亮。电流在节点处分流，总电流等于各支路电流之和。若一个12 V电池连接两条支路，每条支路有一个6 Ω的电阻，则每条支路电流为12 / 6 = 2 A，电池提供的总电流为2 + 2 = 4 A。常见的计算错误是按串联方式将电阻直接相加。并联时应该用1/Rtotal = 1/R₁ + 1/R₂。
1 / Rtotal = 1 / R₁ + 1 / R₂
1 / Rtotal = 1 / R₁ + 1 / R₂
Examiners reward the correct rearrangement: Rtotal sistors only. Be prepared to use this formula and then apply V = IR to find total current.
考官认可正确的变形：对于两个电阻，Rtotal = (R₁ × R₂) / (R₁ + R₂)。你要做好准备使用该公式，然后应用V = IR求总电流。
5. Particle Model and Specific Heat Capacity | 粒子模型与比热容
tions link energy, mass and temperature change: ΔE = m c Δθ. A typical numerical question: 'How much energy is needed to heat 2.5 kg of water from 20 °C to 80 °C? (c = 4200 J/kg°C)' 5 × 4200 × 60 = 630,000 J (or 630 kJ). Always state the unit, as missing units lose marks.
比热容题目将能量、质量和温度变化联系起来：ΔE = m c Δθ。典型的数值题：“将2.5 kg水从20 °C加热到80 °C需要多少能量？（c = 4200 J/kg°C）”。解：Δθ = 80 − 20 = 60 °C；ΔE = 2.5 × 4200 × 60 = 630,000 J（即630 kJ）。务必注明单位，遗漏单位会丢分。
Practical questions on specific heat capacity often ask you to describe how to reduce heat loss to the surroundings, such as using a lid and insulation. The biggest source of error is energy being lost to the air, which makes the measured temperature rise smaller than expected, leading to a calculated c that is higher than the true value. You must be able to explain how this systematic error affects the result.
比热容的实验题常要求你描述如何减少热量散失到周围环境，例如加上盖子和使用保温材料。最大的误差来源是能量散失到空气中，这使得测得的温升低于预期，从而导致计算出的比热容值高于真实值。你必须能够解释这种系统误差如何影响结果。
ΔE = m c Δθ
ΔE = m c Δθ
6. Radioactive Decay and Half-Life | 放射性衰变与半衰期
Half-life questions usually present a decay graph of activity against time. You might be asked to determine half-life by finding the time taken for the count rate to drop from, say, 800 counts per second to 400. If this takes 3 hours, the half-life is 3 hours. A common pitfall is reading the time from the x-axis for one 'half' of the original activity rather than half of any value; always verify by checking a second half-life.
半衰期题目通常会给出活度随时间变化的衰变曲线图。你可能会被要求通过找出计数率从例如800次/秒降至400次/秒所需的时间来求半衰期。如果这需要3小时，那么半衰期就是3小时。常见误区是读取纵轴上初始活度的一半所对应的时间，而不是从任意数值的一半去验证；一定要通过检查第二个半衰期来确认。
Explaining half-life in terms of the random nature of decay is a key 6-mark question. You should state that radioactive decay is random and cannot be predicted for a single nucleus, but with a large number of nuclei the average time for half to decay is constant. The half-life is the time for the number of undecayed nuclei to halve. Uses include carbon dating and determining the thickness of paper in industry.
从衰变的随机性解释半衰期是一道典型的高分题（6分）。你应该说明放射性衰变是随机的，无法预测单个原子核何时衰变，但对于大量原子核，半数衰变所需的平均时间是恒定的。半衰期就是未衰变原子核数目减半所需的时间。其应用包括碳年代测定和工业中纸张厚度的测量。
ion remaining is (1/2)³ = 1/8. So if an initial mass is 80 g, after 3 half-lives the remaining radioactive mass is 80 / 8 = 10 g. Remember that the mass of the material may not change dramatically because decay products remain.
涉及半衰期的计算经常要求你求出经过若干个半衰期后剩余的比例。经过3个半衰期，剩余比例为 (1/2)³ = 1/8。因此，若初始质量为80 g，则3个半衰期后剩余的放射性质量为80 / 8 = 10 g。但要注意材料的总质量可能变化不大，因为衰变产物仍留在其中。
7. Forces and Motion Graphs | 力与运动图像
aight line from 0 m/s to 20 m/s over 5 seconds, acceleration = (20 − 0) / 5 = 4 m/s². The area under the graph gives distance travelled. For a triangle shape during acceleration, area = ½ × base × height = ½ × 5 × 20 = 50 m.
速度-时间图像是Eduqas考试的必考内容。你必须能够根据斜率计算加速度：a = (v − u) / t。如果图像显示从0 m/s到20 m/s的直线，时间为5秒，则加速度 = (20 − 0) / 5 = 4 m/s²。图线下的面积表示行驶的距离。对于加速过程中的三角形区域，面积 = ½ × 底 × 高 = ½ × 5 × 20 = 50 m。
Distance-time graph interpretation: a steep straight line means constant high speed. A horizontal line means the object is stationary. A curved line with increasing steepness indicates acceleration. Typical exam questions ask you to describe the motion in stages. Use precise wording: 'constant speed', 'accelerating', 'stationary', 'decelerating'. Avoid vague terms like 'moving fast'.
距离-时间图像解读：一条陡直的直线表示匀速高速运动。水平线表示物体静止。一条越来越陡的曲线表示加速。典型的考题要求你分阶段描述运动状态。要使用精确的措辞：“匀速”、“加速”、“静止”、“减速”。避免使用“快速移动”等模糊说法。
Newton' e question gives a velocity-time graph, first find acceleration, then apply F = m a. Always check if resistive forces are mentioned; resultant force = driving force − friction.
牛顿第二定律的计算常与运动图像结合：F = m a。若一辆1200 kg的汽车以3 m/s²加速，合力 = 1200 × 3 = 3600 N。当题目给出速度-时间图像时，先求加速度，再应用F = m a。务必检查是否提及阻力；合力 = 驱动力 − 摩擦力。
8. Wave Properties and the Wave Equation | 波的性质与波动方程
of 0.75 m. Calculate the speed. v = 440 × 0.75 = 330 m/s. Be prepared to convert units: kHz to Hz, cm to m. Always state the formula, show substitution, and give the unit.
波动方程v = f λ是高频考点，常以声波或水波为背景。例题：一个声波的频率为440 Hz，波长为0.75 m，求波速。v = 440 × 0.75 = 330 m/s。要做好单位换算的准备：千赫兹换算为赫兹，厘米换算为米。务必写出公式、代入过程和单位。
v = f × λ
v = f × λ
Describing the differences between transverse and longitudinal waves appears in many past papers. For transverse waves (e.g., light, ripples on water), oscillations are perpendicular to the direction of energy transfer. For longitudinal waves (e.g., sound), oscillations are parallel to the direction of energy transfer, forming compressions and rarefactions. Diagrams must be drawn with labels.
描述横波与纵波的区别在许多真题中出现。对于横波（如光、水面涟漪），振动方向与能量传递方向垂直。对于纵波（如声音），振动方向与能量传递方向平行，形成压缩区和稀疏区。作图时需要标注清楚。
The ripple tank required practical: you may be asked to describe measuring wavelength and frequency to calculate wave speed. Place a ruler in the tank to measure the distance of multiple wavelengths, then divide by the number of waves. Count the number of waves passing a point in a set time for frequency. Use a strobe light to 'freeze' the wave pattern if possible. This practical is a classic 6-mark question.
水波槽必做实验：考生可能被要求描述测量波长和频率以计算波速的方法。将尺子放入水槽中测量多个波长的总长度，再除以波的个数。记录一定时间内通过某点的波数求频率。若条件允许，使用频闪仪“冻结”波面。这个实验是经典的6分大题。
9. Electromagnetic Spectrum and Uses | 电磁波谱与应用
Eduqas questions often ask you to order the electromagnetic spectrum by wavelength or frequency, from radio waves (longest wavelength, lowest frequency) to gamma rays (shortest wavelength, highest frequency). A mnemonic such as 'Rabbits Mate In Very Unusual eXpensive Gardens' (Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma) can help memorise the order.
Eduqas考题常要求你按波长或频率排列电磁波谱，从无线电波（波长最长，频率最低）到伽马射线（波长最短，频率最高）。可以用记忆口诀辅助记忆顺序（如：无线电、微波、红外线、可见光、紫外线、X射线、伽马射线）。
Linking each wave type to a specific use is essential. Radio waves: broadcasting and communications. Microwaves: satellite communication and cooking. Infrared: remote controls and thermal imaging. Visible light: fibre optic communications. Ultraviolet: security markings and sterilisation. X-rays: medical imaging. Gamma rays: cancer treatment and sterilising medical equipment. Explanations should reference the wave's ability to penetrate materials or interact with matter.
将每种波与具体用途联系起来非常重要。无线电波：广播与通信。微波：卫星通信与烹饪。红外线：遥控器与热成像。可见光：光纤通信。紫外线：验钞防伪与杀菌。X射线：医学影像。伽马射线：癌症治疗与医疗设备消毒。解释时要提及波的穿透能力或与物质的相互作用。
A typical 3-mark question: 'Explain why ultraviolet radiation can cause skin cancer but infrared cannot.' You must state that UV has a higher frequency (and thus higher energy photons) than IR. UV photons have enough energy to damage DNA in skin cells, leading to mutations, whereas IR photons only cause heating.
典型的3分题：“解释为什么紫外线能导致皮肤癌而红外线不能。”你必须说明紫外线比红外线频率更高（因此光子能量更大）。紫外光子具有足够的能量损伤皮肤细胞中的DNA，导致突变，而红外光子只能引起热效应。
10. Space Physics: Red-Shift and Big Bang | 空间物理：红移与大爆炸
Red-shift is a common analysis and explain topic. When light from distant galaxies is observed, the spectral lines are shifted towards the red end of the spectrum. This indicates that galaxies are moving away from us, providing evidence that the Universe is expanding. For a 6-mark question, structure your answer: observation, explanation (Doppler effect for light), and conclusion (Universe began from a single point – Big Bang).
红移是常见的分析与解释主题。当观察到来自遥远星系的光时，其光谱线向光谱的红端移动。这表明星系正在远离我们，为宇宙膨胀提供了证据。对于6分大题，要这样组织答案：观测现象、解释（光的多普勒效应）及结论（宇宙起源于一个奇点——大爆炸）。
Cosmic microwave background radiation (CMBR) often appears alongside red-shift as the second major piece of evidence for the Big Bang theory. You must explain that CMBR is the leftover thermal radiation from the early hot, dense Universe, now cooled to about 2.7 K, and is detected uniformly in all directions. This uniformity supports the theory that the Universe started from one explosion.
宇宙微波背景辐射（CMBR）经常与红移同时作为大爆炸理论的第二大证据出现。你必须解释CMBR是早期高温致密宇宙遗留下来的热辐射，如今已冷却到约2.7 K，并且在各个方向上均匀地被探测到。这种均匀性支持了宇宙始于一次爆炸的理论。
Be careful to use precise scientific language: avoid saying galaxies 'explode' outward; instead say 'space itself is expanding'. This distinction is frequently tested in multiple-choice and explanation questions to tackle misconceptions.
注意使用精确的科学语言：避免说星系向外“爆炸”，而应说“空间本身正在膨胀”。这一区别常在选择题和解释题中出现，以纠正常见迷思。
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