📚 Year 11 CCEA Physics: Mock Unit Test Walkthrough | CCEA 物理模拟单元测试解析
This walkthrough analyses a typical CCEA Year 11 Physics unit test paper, covering motion, forces, energy, waves, electricity, atomic physics and more. Each question is broken down step by step with clear reasoning, helping you master the problem-solving skills needed for your actual exam.
本文针对一份典型的 CCEA 11 年级物理单元测试模拟卷进行逐题解析，内容涵盖运动、力、能量、波、电路、原子物理等核心模块。每道题都配有清晰的步骤拆解和思路点拨，帮助你牢固掌握解题方法，自信迎战真实考试。
1. Velocity-Time Graph Analysis | 速度-时间图分析
A racing car accelerates uniformly from rest to 28 m/s in 7.0 s, maintains that top speed for a further 12.0 s, then decelerates uniformly to rest in 5.0 s. The motion is represented on a velocity-time graph.
一辆赛车从静止开始匀加速，7.0 秒内达到 28 m/s，然后保持该速度 12.0 秒，最后匀减速至静止，用时 5.0 秒。整个运动过程体现在速度-时间图上。
ation in the final stage is (0 - 28) m/s ÷ 5.0 s = -5.6 m/s². The total distance travelled is the area under the entire v-t graph.
第一阶段加速度对应图线斜率。a = (v - u) / t，初速度为零，a = 28 m/s ÷ 7.0 s = 4.0 m/s²。第三阶段的减速度为 (0 - 28) m/s ÷ 5.0 s = -5.6 m/s²。总行驶距离等于整个 v-t 图下方的面积。
Area = (½ × 7 × 28) + (12 × 28) + (½ × 5 × 28) = 98 + 336 + 70 = 504 m
Always sketch the graph roughly before calculating areas. Breaking the area into a triangle, rectangle and triangle makes the arithmetic straightforward and reduces errors.
在计算面积之前，最好先大致画出图形。将总面积拆成一个三角形、一个矩形再加一个三角形，计算清晰且不易出错。
2. Resultant Force and Newton's Second Law | 合力和牛顿第二定律
A 900 kg car experiences a forward driving force of 3200 N. Air resistance and friction provide a total backward force of 500 N. Determine the resultant force and the acceleration of the car.
一辆质量为 900 kg 的汽车受到 3200 N 向前驱动力，空气阻力和摩擦共产生 500 N 向后的阻力。求合力及汽车的加速度。
00 kg = 3.0 m/s². The direction of acceleration is always the same as the resultant force.
合力 = 驱动力 - 阻力 = 3200 N - 500 N = 2700 N，方向向前。根据 F = m a，加速度 a = F / m = 2700 N ÷ 900 kg = 3.0 m/s²，方向与合力一致。
se the driving force alone in F = m a. Also check that mass is in kg, not grams or tonnes.
务必先求出净合力，很多同学会错误地直接把驱动力代入 F = m a。同时注意质量单位必须是 kg，不能直接用 g 或 t。
3. Work Done, Gravitational Potential Energy and Power | 做功、重力势能与功率
m vertically upward through a height of 15 m in 24 s at constant speed. Take g = 9.8 N/kg.
一台起重机以恒定速度将 120 kg 的钢梁竖直提升 15 m，用时 24 s，取 g = 9.8 N/kg。
ant. Work done = force × distance = 1176 N × 15 m = 17 640 J. This work increases the beam' gravitational potential energy (GPE). Power = work done ÷ time = 17 640 J ÷ 24 s = 735 W.
钢梁重力 = m g = 120 × 9.8 = 1176 N。匀速提升时拉力等于重力。做功 = 力 × 距离 = 1176 N × 15 m = 17 640 J，这部分功转化为钢梁的重力势能。功率 = 做功 ÷ 时间 = 17 640 J ÷ 24 s = 735 W。
GPE gained = m g h = 120 × 9.8 × 15 = 17 640 J
Exactly matches work done, illustrating the conservation of energy principle. When the question mentions 'constant speed', resultant force is zero and lifting force balances weight.
做功与增加的重力势能完全相等，体现了能量守恒。题目中若出现“恒定速度”，即意味着合力为零，提升力与重力平衡，这一点在受力分析中非常关键。
4. Hooke's Law and Spring Constant | 胡克定律与弹簧常数
A spring of original length 0.20 m extends to 0.26 m when a 6.0 N load is attached. Calculate the spring constant k. Predict the extension for a 9.0 N load, assuming the elastic limit is not exceeded.
一根原长 0.20 m 的弹簧挂上 6.0 N 的负载后伸长至 0.26 m。计算弹簧常数 k，并预测在不超过弹性极限的情况下，挂 9.0 N 负载时的伸长量。
Extension e = 0.26 - 0.20 = 0.06 m. Hooke' N load, e = F / k = 9.0 N ÷ 100 N/m = 0.09 m, giving a new length of 0.20 + 0.09 = 0.29 m.
伸长量 e = 0.26 - 0.20 = 0.06 m。根据 F = k e，得 k = 6.0 N ÷ 0.06 m = 100 N/m。当负载为 9.0 N 时，伸长量 e = F / k = 9.0 N ÷ 100 N/m = 0.09 m，此时弹簧总长为 0.20 + 0.09 = 0.29 m。
Always work with extension (change in length), not the total length, when applying Hooke's Law. Double-check unit conversion: extension must be in metres to obtain k in N/m.
运用胡克定律时一定要用伸长量（长度变化量），而不是弹簧总长。并且长度务必换算成米，以确保 k 的单位是 N/m。
5. Density and Pressure | 密度与压强
ine the pressure exerted on a table when the block rests on its largest face. g = 10 N/kg.
一块矩形铝块质量为 8.1 kg，尺寸为 0.30 m × 0.20 m × 0.050 m。计算其密度，然后求铝块以最大面置于桌面时对桌面产生的压强。取 g = 10 N/kg。
t face area is 0.30 m × 0.20 m = 0.060 m². Pressure p = F / A = 81 N ÷ 0.060 m² = 1350 Pa.
体积 V = 0.30 × 0.20 × 0.050 = 0.0030 m³。密度 ρ = m / V = 8.1 kg ÷ 0.0030 m³ = 2700 kg/m³。重力 W = m g = 8.1 × 10 = 81 N。最大面面积为 0.30 m × 0.20 m = 0.060 m²，压强 p = F / A = 81 N ÷ 0.060 m² = 1350 Pa。
The block exerts minimum pressure on its largest face. If it were placed on its smallest face, the pressure would be much larger, demonstrating how force distribution affects pressure.
以最大面放置时压强最小；若以最小面放置，压强会显著增大，这很好地说明了压力分布对压强的影响。
6. Radioactive Decay and Half-Life Calculations | 放射性衰变与半衰期计算
A sample of radioactive isotope has an initial count rate of 2400 counts per minute, corrected for background radiation. Its half-life is 6.0 hours. Find the count rate after 18 hours and the fraction of the original isotope remaining after one full day.
某放射性同位素样品的初始计数率为 2400 次/分钟（已扣除本底辐射）。其半衰期为 6.0 小时。求 18 小时后的计数率，以及一整日后剩余的原始同位素比例。
. For one day (24 hours), number of half-lives = 24 ÷ 6 = 4. Fraction remaining = (½)⁴ = 1/16.
18 小时经历的半衰期个数为 18 ÷ 6 = 3。每次半衰期计数率减半：2400 → 1200 → 600 → 300 次/分钟，所以 18 小时后为 300 次/分钟。一整天（24 小时）的半衰期个数为 24 ÷ 6 = 4，剩余同位素比例 = (½)⁴ = 1/16。
Always subtract background radiation before halving. If a question doesn't state 'corrected for background', you must subtract the background count rate first, then apply the half-life pattern.
务必先扣除本底辐射再计算半衰。如果题目未说明“已扣本底”，则需要自行减掉本底计数率，再按照半衰规律处理。
7. Nuclear Equations for Alpha and Beta Decay | α 衰变和 β 衰变的核方程
Complete the following nuclear decay equations:
完成下列核衰变方程：
²³⁴₉₀Th → ²³⁰₈₈Ra + ? (alpha decay)
¹⁴₆C → ¹⁴₇N + ? (beta-minus decay)
²³⁴₉₀Th → ²³⁰₈₈Ra + ⁴₂He. For beta-minus decay, a neutron turns into a proton and an electron (beta particle) is emitted. Mass number stays 14, atomic number increases from 6 to 7, so the emitted particle is ⁰₋₁e (or β⁻). The equation: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e.
α 衰变中质量数减 4，原子序数减 2。缺失粒子为 α 粒子 ⁴₂He，完整方程为 ²³⁴₉₀Th → ²³⁰₈₈Ra + ⁴₂He。β⁻ 衰变中，一个中子转变为质子，释放出一个电子（β 粒子）。质量数不变，原子序数从 6 增加到 7，因此发射出的粒子是 ⁰₋₁e。方程为 ¹⁴₆C → ¹⁴₇N + ⁰₋₁e。
In CCEA papers, you may also be asked to recognise gamma emission, which does not change the mass number or atomic number but is often written as ⁰₀γ. Always balance the top numbers (mass) and bottom numbers (atomic charge) on both sides of the arrow.
在 CCEA 试题中，还可能涉及到 γ 辐射，不改变质量和电荷数，常写作 ⁰₀γ。书写核方程时，要始终确保箭头两侧上标（质量数）总和相等，下标（电荷数/原子序数）总和也相等。
8. Kinetic Theory and Gas Pressure | 气体动理论及气压
A fixed mass of gas is sealed in a rigid container. Use the kinetic particle model to explain why the pressure increases when the gas is heated strongly.
一个固定质量的气体密封在刚性容器中。请用粒子动理论模型解释，为什么强烈加热气体会导致压强增大。
When the gas is heated, the average kinetic energy of the particles increases – they move faster. Since the container is rigid, the volume does not change. Faster-moving particles collide with the walls more frequently and with greater force per collision. Both effects increase the average force exerted on unit area, therefore the pressure rises.
加热气体时，粒子平均动能增大，运动速度更快。由于容器是刚性的，体积不变。运动更快的粒子单位时间内撞击器壁的次数更多，且每次撞击的力更大。这两个效果都使得单位面积上受到的平均力增大，因此压强升高。
| Factor / 因素 | Change / 变化 | Effect on Pressure / 对压强的影响 |
|---|---|---|
| Particle speed / 粒子速度 | Increases / 增加 | More frequent & harder collisions / 更频繁、更剧烈的碰撞 |
| Container volume / 容器体积 | Fixed / 不变 | No reduction in collision surface / 碰撞表面积无减小 |
A common exam phrase is 'the force per unit area increases because particles hit the walls harder and more often'. Make sure you link both collision rate and force per collision to the pressure increase.
试卷中常见的表述是“因为粒子对器壁的撞击更剧烈更频繁，所以单位面积受力增大”。答题时要同时提到碰撞频率和单次碰撞力度，才能完整解释压强增大。
9. Moments and Equilibrium | 力矩与平衡
A uniform wooden plank of length 4.0 m and weight 200 N is pivoted at its centre. A 350 N weight is placed 0.80 m from the pivot on the left side. Determine the position on the right side where a 280 N load must be placed to balance the plank horizontally.
一块长 4.0 m、重 200 N 的均匀木板在其中点支起。在支点左侧 0.80 m 处放置 350 N 的重物。若要木板水平平衡，应在右侧距支点多远处放置一个 280 N 的负载？
The plank' ise moment. Anticlockwise moment = 350 N × 0.80 m = 280 N m. Let d be the required distance on the right. Clockwise moment = 280 N × d. So 280 × d = 280, giving d = 1.0 m. Place the 280 N load 1.0 m to the right of the pivot.
木板自重作用在中心即支点处，因此对支点不产生力矩。平衡时顺时针力矩等于逆时针力矩。左侧逆时针力矩 = 350 N × 0.80 m = 280 N m。设右侧所需距离为 d，则顺时针力矩 = 280 N × d。由 280 × d = 280 解得 d = 1.0 m，因此将 280 N 的负载放在支点右侧 1.0 m 处。
When a beam is described as 'uniform' and 'pivoted at its centre', you can ignore its weight in moment calculations. If the pivot were not at the centre, you would have to include the plank's weight acting at its midpoint.
若题干出现“均匀”且“支于中点”，计算力矩时可忽略木板自重。但如果支点不在中点，则必须将木板重量视为作用在其几何中心处进行计算。
10. Specific Heat Capacity and Energy Efficiency | 比热容与能量效率
A kettle is used to heat 1.5 kg of water from 18 °C to 96 °C. The specific heat capacity of water is 4200 J/kg°C. The kettle has a power of 2200 W and takes 250 s to heat the water. Calculate the useful energy output and the efficiency of the kettle.
一个电热水壶将 1.5 kg 的水从 18 °C 加热到 96 °C。水的比热容为 4200 J/kg°C。水壶功率为 2200 W，加热耗时 250 s。计算有用输出能量及水壶的效率。
2200 W × 250 s = 550 000 J. Efficiency = (useful output energy ÷ total input energy) × 100% = (491 400 / 550 000) × 100% ≈ 89.3%.
温度升高 Δθ = 96 - 18 = 78 °C。有用能量 Q = m c Δθ = 1.5 × 4200 × 78 = 491 400 J。总输入电能 = P × t = 2200 W × 250 s = 550 000 J。效率 = (有用输出能量 ÷ 总输入能量) × 100% = (491 400 / 550 000) × 100% ≈ 89.3%。
Efficiency can be expressed as a decimal (0.893) or a percentage. In CCEA papers, you are often expected to round sensibly, e.g. to three significant figures unless the question states otherwise. Also remember that some energy is always lost to the surroundings in practice.
效率可用小数（0.893）或百分数表示。CCEA 试卷中通常要求合理取整，如保留三位有效数字，除非题目另有规定。此外，实际中总有一部分能量散失到环境中，这是常识性采分点。
Published by TutorHao | Physics Revision Series | aleveler.com
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