📚 Year 11 Eduqas Physics: Unit Test Mock Paper Walkthrough | Year 11 Eduqas 物理：单元测试模拟卷解析
Welcome to this comprehensive walkthrough of a typical Year 11 Eduqas Physics unit test mock paper. This article breaks down key question types, provides step-by-step solutions, and highlights common pitfalls. By working through these examples, you will strengthen your understanding of core physics concepts and examination technique.
欢迎阅读这篇全面的 Year 11 Eduqas 物理单元测试模拟卷解析。本文将分解关键题型，提供分步解答，并指出常见易错点。通过这些例题练习，你可以巩固核心物理概念并提升应试技巧。
1. Question 1: Uniform Acceleration | 题目1：匀加速直线运动
A car accelerates uniformly from rest with an acceleration of 3 m/s² for a time of 8 seconds. Calculate the final velocity of the car and the distance it travels during this period.
一辆汽车从静止开始以 3 m/s² 的加速度匀加速运动，持续 8 秒。计算汽车的最终速度以及在这段时间内行驶的距离。
Step 1 – v = u + at. Initial velocity u = 0, a = 3 m/s², t = 8 s. Therefore v = 0 + 3 × 8 = 24 m/s.
第1步 – 使用 v = u + at 计算末速度。初速度 u = 0，a = 3 m/s²，t = 8 s。因此 v = 0 + 3 × 8 = 24 m/s。
Step 2 – 64 = 96 m. Alternatively, average velocity = (0+24)/2 = 12 m/s, distance = 12 × 8 = 96 m.
第2步 – 计算行驶距离。使用 s = ut + ½at²。代入得 s = 0 × 8 + ½ × 3 × 8² = 0 + ½ × 3 × 64 = 96 m。或者用平均速度 = (0+24)/2 = 12 m/s，距离 = 12 × 8 = 96 m。
Common mistake: forgetting to square the time in the distance formula. Always write the units and check they are consistent.
常见错误：在距离公式中忘记对时间取平方。始终写明单位并检查其一致性。
2. Question 2: Energy Conservation – Falling Object | 题目2：能量守恒 – 落体问题
/kg and ignoring air resistance, calculate the kinetic energy of the stone just before it hits the ground.
一块质量为 2 kg 的石块从距地面 45 m 高的悬崖上掉下。取 g = 10 N/kg，忽略空气阻力，计算石块着地前的动能。
Step 1 – (GPE) at the top is converted entirely into kinetic energy (KE) at the bottom. GPE = mgh.
第1步 – 确定能量转化：顶端的重力势能 (GPE) 全部转化为底端的动能 (KE)。GPE = mgh。
Step 2 – = 2 kg, g = 10 N/kg, h = 45 m. GPE = 2 × 10 × 45 = 900 J. Therefore KE at impact = 900 J.
第2步 – 代入：m = 2 kg，g = 10 N/kg，h = 45 m。GPE = 2 × 10 × 45 = 900 J。因此着地时的 KE = 900 J。
If asked for speed: use KE = ½mv², then v = √2KE/m = √2×900/2 = √900 = 30 m/s. Remember to state the direction of energy transfer.
如果要求计算速度：使用 KE = ½mv²，则 v = √2KE/m = √2×900/2 = √900 = 30 m/s。记得说明能量传递的方向。
3. Question 3: Electrical Circuits – Ohm’s Law and Charge | 题目3：电路 – 欧姆定律与电荷
A 6 Ω resistor is connected across a 3 V battery. Determine the current flowing through the resistor and the charge that passes through it in 2 minutes.
一个 6 Ω 的电阻器连接在 3 V 的电池上。计算流经电阻器的电流，以及在 2 分钟内通过它的电荷量。
Step 1 – Use Ohm’s law: V = IR. Rearranging gives I = V / R = 3 V ÷ 6 Ω = 0.5 A.
第1步 – 使用欧姆定律: V = IR。整理得 I = V / R = 3 V ÷ 6 Ω = 0.5 A。
Step 2 – ime t must be in seconds: 2 minutes = 2 × 60 = 120 s. Q = 0.5 A × 120 s = 60 C (coulombs).
第2步 – 电荷量 Q = I × t。时间 t 必须以秒为单位：2 分钟 = 2 × 60 = 120 s。Q = 0.5 A × 120 s = 60 C（库仑）。
Always convert time to seconds. Check the resistor value does not change unless temperature varies; here we assume constant conditions.
始终将时间换算为秒。检查电阻值是否变化，除非温度变化；这里我们假设条件恒定。
4. Question 4: Wave Speed, Frequency and Wavelength | 题目4：波速、频率与波长
A water ripple has a wavelength of 0.5 m and a period of 0.2 seconds. Calculate the frequency of the wave and its speed.
水波纹的波长为 0.5 m，周期为 0.2 秒。计算波的频率和波速。
Step 1 – Frequency f = 1 / T. T = 0.2 s, so f = 1 ÷ 0.2 = 5 Hz.
第1步 – 频率 f = 1 / T。T = 0.2 s，所以 f = 1 ÷ 0.2 = 5 Hz。
Step 2 – = f × λ. λ = 0.5 m, f = 5 Hz. v = 5 × 0.5 = 2.5 m/s. Remember the unit: metres per second.
第2步 – 波速 v = f × λ。λ = 0.5 m，f = 5 Hz。v = 5 × 0.5 = 2.5 m/s。记住单位：米每秒。
Tip: In exam questions, sometimes frequency is given in kHz; convert to Hz (×1000).
提示：在考题中，频率有时以 kHz 给出；需换算成 Hz（×1000）。
5. Question 5: Half-Life and Radioactive Decay | 题目5：半衰期与放射性衰变
The half-life of iodine-131 is 8 days. Initially a sample has an activity of 1600 Bq. After how many days will the activity fall to 200 Bq?
碘-131 的半衰期为 8 天。开始时样品的活度为 1600 Bq。经过多少天后活度会降至 200 Bq？
Step 1 – half-life, activity halves. 1600 → 800 (1 half-life) → 400 (2) → 200 (3). So 3 half-lives.
第1步 – 确定半衰期个数：每经过一个半衰期，活度减半。1600 → 800（1个半衰期）→ 400（2个）→ 200（3个）。所以是 3 个半衰期。
Step 2 – Total time = number of half-lives × half-life duration = 3 × 8 days = 24 days.
第2步 – 总时间 = 半衰期个数 × 半衰期时长 = 3 × 8 天 = 24 天。
You could also use the formula N = N₀(½)t/T. Set N₀=1600, N=200, T=8 d. Solving gives t=24 d. Check your division carefully.
也可用公式 N = N₀(½)t/T。设 N₀=1600，N=200，T=8 天。求解得 t=24 天。仔细核对除法。
6. Question 6: Specific Heat Capacity | 题目6：比热容
0.5 kg of aluminium is heated from 25 °C to 75 °C. The specific heat capacity of aluminium is 900 J/kg°C. Calculate the energy transferred to the aluminium.
0.5 kg 的铝从 25 °C 被加热到 75 °C。铝的比热容为 900 J/kg°C。计算传递到铝的能量。
Step 1 – Temperature change Δθ = 75 °C – 25 °C = 50 °C.
第1步 – 温度变化 Δθ = 75 °C – 25 °C = 50 °C。
Step 2 – kg × 900 J/kg°C × 50 °C. Multiply: 0.5 × 900 = 450, then 450 × 50 = 22 500 J (or 22.5 kJ).
第2步 – 能量 E = m c Δθ = 0.5 kg × 900 J/kg°C × 50 °C。乘法：0.5 × 900 = 450，然后 450 × 50 = 22 500 J（或 22.5 kJ）。
Be careful with the units: the specific heat capacity unit means per kilogram per degree Celsius. Always write the final unit as joules.
注意单位：比热容单位意味着每千克每摄氏度。最后结果的单位务必写成焦耳。
7. Question 7: Newton’s Third Law – Rocket Propulsion | 题目7：牛顿第三定律 – 火箭推进
Explain how a water rocket launches upwards by ejecting water downwards, using Newton’s third law of motion.
请用牛顿第三运动定律解释水火箭通过向下喷水而向上发射的过程。
Newton’s third law states: whenever two objects interact, the forces they exert on each other are equal in size and opposite in direction. The water rocket expels water downwards; this is the action force. The water exerts an equal and opposite reaction force back on the rocket, pushing it upwards.
牛顿第三定律指出：两个物体相互作用时，彼此施加的力大小相等、方向相反。水火箭向下喷水，这是作用力。水同时对火箭施加一个大小相等、方向相反的反作用力，将火箭向上推。
It is important to identify the force pairs: the rocket pushes on the water (downward), the water pushes on the rocket (upward). Even though masses differ, the forces are equal. The rocket accelerates upward because its mass is relatively small and air/water are expelled rapidly.
关键在于确定力对：火箭对水施加向下的力，水对火箭施加向上的力。尽管质量不同，但力的大小相等。火箭向上加速是因为它的质量相对较小，而水被快速喷出。
8. Question 8: Electromagnetic Induction – Experimental Description | 题目8：电磁感应 – 实验描述
Describe an experiment to demonstrate electromagnetic induction. Include the apparatus and what is observed.
描述一个演示电磁感应的实验，包括实验器材和观察到的现象。
Apparatus needed: a coil of wire connected to a sensitive ammeter (or galvanometer), and a bar magnet. Move the bar magnet into the coil – the ammeter needle deflects, showing a current is induced. When the magnet stops moving, the current returns to zero. Pull the magnet out, and the needle deflects in the opposite direction.
所需器材：一个连接灵敏电流计（或检流计）的线圈，以及一块条形磁铁。将磁铁移入线圈，电流计指针发生偏转，表明产生了感应电流。当磁铁停止运动时，电流回到零。将磁铁拉出，指针反向偏转。
The induced current is produced because the magnetic field through the coil is changing. Moving the magnet faster increases the rate of change of magnetic flux, resulting in a larger current. This is electromagnetic induction, discovered by Faraday.
产生感应电流的原因是穿过线圈的磁场在变化。磁铁移动越快，磁通量变化率越大，产生的电流也越大。这就是法拉第发现的电磁感应现象。
9. Question 9: Pressure, Force and Area | 题目9：压强、力与面积
An elephant exerts a weight of 40 000 N on the ground. When standing on one foot, the area of contact is 0.08 m². Calculate the pressure in pascals exerted on the ground by that one foot.
一头大象对地面的重力为 40 000 N。当它单脚站立时，接触面积为 0.08 m²。计算该只脚对地面施加的压强（以帕斯卡为单位）。
Step 1 – Recall pressure formula: P = F / A, where F is force in newtons and A is area in m².
第1步 – 回忆压强公式: P = F / A，其中 F 为力（牛顿），A 为面积（平方米）。
Step 2 – m². P = 40 000 ÷ 0.08 = 500 000 Pa (or 5 × 10⁵ Pa). You can also express this as 500 kPa.
第2步 – 代入：F = 40 000 N，A = 0.08 m²。P = 40 000 ÷ 0.08 = 500 000 Pa（或 5 × 10⁵ Pa）。也可表示为 500 kPa。
If the elephant stands on four feet, the area is four times, so pressure is one quarter. Always check the area is in m² – convert from cm² by dividing by 10 000.
如果大象四脚站立，面积变为四倍，压强则为四分之一。务必检查面积单位为 m²——若为 cm² 则需除以 10 000 转换。
10. Question 10: Redshift and the Expanding Universe | 题目10：红移与宇宙膨胀
Explain how the observation of redshift in light from distant galaxies provides evidence for the Big Bang theory and an expanding universe.
解释观测到遥远星系光谱的红移如何为大爆炸理论和宇宙膨胀提供证据。
When light from distant galaxies is analysed, the characteristic spectral lines are shifted towards the red end of the spectrum. This redshift occurs because the galaxies are moving away from us, stretching the wavelength of the light – an effect of the Doppler shift.
分析来自遥远星系的光时，特征谱线会向光谱的红端移动。这种红移的发生是因为这些星系正在远离我们，光的波长被拉长——这是多普勒效应的一种体现。
The more distant the galaxy, the greater the redshift and hence the faster it is receding. This relationship supports the idea that the universe began from a single point (the Big Bang) and has been expanding ever since. The redshift is not due to galaxies moving through space but due to the expansion of space itself.
星系越远，红移越大，因此其退行速度越快。这种关系支持了宇宙始于一个奇点（大爆炸）并且此后一直在膨胀的观点。红移并非星系在空间中穿行所致，而是空间本身膨胀的结果。
Cosmic microwave background radiation is another key piece of evidence, but redshift data alone strongly indicates the expansion of the universe.
宇宙微波背景辐射是另一项关键证据，但仅红移数据已强有力地表明宇宙在膨胀。
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