📚 Year 11 CCEA Physics: Unit Test Mock Paper Walkthrough | CCEA 物理单元测试模拟卷解析
Mock tests are an essential part of preparing for your CCEA Year 11 Physics examination. This walkthrough covers representative questions from mechanics, electricity, waves, radioactivity and more. For each question, we show step-by-step reasoning that mirrors what examiners expect, helping you to avoid common pitfalls and build confidence.
模拟测试是备考 CCEA 11 年级物理考试的重要环节。本解析涵盖力学、电学、波动、放射性等代表性题型，逐步展示解题思路并贴近评分标准，帮助你避开常见陷阱，增强信心。
1. Velocity and Acceleration | 速度与加速度
A car accelerates uniformly from rest to 20 m/s in 5.0 seconds. Calculate the acceleration and the distance travelled during this time.
一辆汽车从静止匀加速到 20 m/s，用时 5.0 秒。计算加速度和行驶距离。
ar starts from rest, u = 0. Rearranging gives a = (v - u) / t = (20 - 0) / 5.0 = 4.0 m/s².
求加速度使用公式 v = u + at。汽车从静止出发，初速度 u = 0。变形得 a = (v - u) / t = (20 - 0) / 5.0 = 4.0 m/s²。
= (0 + 20)/2 × 5.0 = 10 × 5.0 = 50 m. Always include the correct units in your final answer.
位移可用 s = ut + ½at² 计算。代入 u = 0，得 s = 0 × 5.0 + ½ × 4.0 × (5.0)² = 0.5 × 4.0 × 25 = 50 m。也可采用平均速度法：s = (u + v)/2 × t = (0 + 20)/2 × 5.0 = 10 × 5.0 = 50 m。最终答案务必带上正确单位。
2. Newton's Second Law | 牛顿第二定律
A trolley of mass 2.0 kg experiences a resultant force of 6.0 N. Determine its acceleration. What would happen to the acceleration if the mass were doubled while the force remained the same?
一辆质量为 2.0 kg 的小车受到 6.0 N 的合力。求加速度；若力不变而质量加倍，加速度会如何变化？
Newton' hows that acceleration is directly proportional to force and inversely proportional to mass.
牛顿第二定律 F = m × a，因此 a = F / m = 6.0 / 2.0 = 3.0 m/s²。这表明加速度与合力成正比，与质量成反比。
². The acceleration halves, demonstrating that for a constant force, a larger mass results in a smaller acceleration.
若质量加倍为 4.0 kg，力仍为 6.0 N，则 a = 6.0 / 4.0 = 1.5 m/s²。加速度减半，说明在恒力作用下，质量越大加速度越小。
3. Kinetic Energy and Gravitational Potential Energy | 动能与重力势能
ir resistance and taking g = 9.8 m/s², calculate its speed just before it hits the ground.
一块质量为 0.5 kg 的石头从 8.0 m 高处自由下落。忽略空气阻力，取 g = 9.8 m/s²，求它落地前的瞬时速度。
39.2 J. This energy converts entirely into kinetic energy, KE = ½mv², assuming no losses.
先求失去的重力势能：GPE = mgh = 0.5 × 9.8 × 8.0 = 39.2 J。假设无能量损失，这些能量全部转化为动能，KE = ½mv²。
9.2 gives v² = (2 × 39.2) / 0.5 = 156.8, so v = √156.8 ≈ 12.5 m/s. A quicker method is v = √2gh = √2 × 9.8 × 8.0 = √156.8, yielding the same result.
令 ½mv² = 39.2，得 v² = (2 × 39.2) / 0.5 = 156.8，v = √156.8 ≈ 12.5 m/s。更快的推导是 v = √2gh = √2 × 9.8 × 8.0 = √156.8，结果一致。
4. Ohm's Law and Series Circuits | 欧姆定律与串联电路
Two resistors of 4.0 Ω and 6.0 Ω are connected in series to a 5.0 V battery. Calculate the total current flowing and the potential difference across the 6.0 Ω resistor.
两个电阻分别为 4.0 Ω 和 6.0 Ω，串联后接在 5.0 V 的电池上。计算电路中的总电流以及 6.0 Ω 电阻两端的电压。
In a series circuit, the total resistance Rtotal = R₁ + R₂ = 4.0 + 6.0 = 10.0 Ω. Using Ohm's law, I = V / Rtotal = 5.0 / 10.0 = 0.50 A. The current is the same through all components in series.
串联电路中，总电阻 Rtotal = R₁ + R₂ = 4.0 + 6.0 = 10.0 Ω。根据欧姆定律，I = V / Rtotal = 5.0 / 10.0 = 0.50 A。串联电路中各处电流相等。
0 V. The remaining 2.0 V drops across the 4.0 Ω resistor, adding up to the supply voltage.
6.0 Ω 电阻两端的电压 V = I × R = 0.50 × 6.0 = 3.0 V。其余 2.0 V 降落在 4.0 Ω 电阻上，总电压等于 5.0 V 电源电压。
5. Wave Speed, Frequency and Wavelength | 波速、频率与波长
A tuning fork produces sound waves with a frequency of 256 Hz. The wavelength is measured as 1.34 m. Calculate the speed of sound in air. Predict the new wavelength if the frequency is doubled but the wave speed stays the same.
一个音叉发出频率为 256 Hz 的声波，测得波长为 1.34 m。计算空气中的声速；若频率加倍而波速不变，波长将变为多少？
343 m/s (approximately). This is close to the known speed of sound in air at room temperature.
波速、频率和波长满足 v = f × λ。代入数据得 v = 256 × 1.34 = 343 m/s（约）。这与室温下空气中声速的已知值接近。
he wavelength is halved, which illustrates the inverse relationship between frequency and wavelength for a constant wave speed.
如果频率加倍到 512 Hz，波速仍为 343 m/s，则波长 λ = v / f = 343 / 512 = 0.67 m。波长减半，说明波速恒定时频率与波长成反比。
6. Radioactive Decay and Half-Life | 放射性衰变与半衰期
A radioactive isotope has a half-life of 2.0 hours. At the start of an experiment, the count rate is 1600 counts per minute. Determine the count rate after 8.0 hours.
某放射性同位素的半衰期为 2.0 小时。实验开始时计数率为 1600 次/分。求 8.0 小时后的计数率。
time / half-life = 8.0 / 2.0 = 4 half-lives. Each half-life reduces the activity by half.
先求经过的半衰期个数：总时间/半衰期 = 8.0 / 2.0 = 4 个半衰期。每经过一个半衰期，活度减半。
1/16 = 100 counts per minute. This tells us the sample is still detectable but much weaker.
经过 1 个半衰期：1600 × ½ = 800；2 个：400；3 个：200；4 个：100 次/分。用指数表示，计数率 = 初始计数率 × (½)⁴ = 1600 × 1/16 = 100 次/分。可见样品虽然仍可探测，但已大为减弱。
7. Specific Heat Capacity | 比热容
A 0.50 kg aluminium block is heated by an electrical heater that supplies 9000 J of energy. The temperature rises from 20 °C to 30 °C. Calculate the specific heat capacity of aluminium. Explain why a large block of the same material would have a smaller temperature rise for the same energy input.
一个 0.50 kg 的铝块被电热器加热，提供了 9000 J 的能量，温度从 20 °C 升至 30 °C。计算铝的比热容，并解释为何质量更大的同种材料在相同能量输入下温升更小。
5.0 = 1800 J/(kg°C). This value matches the expected specific heat capacity of aluminium.
热量公式为 ΔE = m × c × Δθ，Δθ 为温度变化。变形得 c = ΔE / (m × Δθ) = 9000 / (0.50 × 10) = 9000 / 5.0 = 1800 J/(kg°C)，符合铝的典型比热容值。
nt. For instance, a 1.0 kg block would only heat by 5 °C: Δθ = 9000 / (1.0 × 1800) = 5 °C.
质量越大，需要分配能量的粒子越多。相同 9000 J 能量下，m 增大使 Δθ 减小（c 不变）。例如 1.0 kg 的铝块升温仅为 5 °C：Δθ = 9000 / (1.0 × 1800) = 5 °C。
8. Hooke's Law and Spring Constant | 胡克定律与弹簧常数
nstant k. Describe what you would do in the lab to check if the elastic limit has been exceeded.
一根弹簧在悬挂 0.80 kg 的重物后伸长了 0.12 m。取 g = 9.8 m/s²，求弹簧常数 k。描述在实验室如何检测是否超过弹性限度。
The weight provides the stretching force: F = mg = 0.80 × 9.8 = 7.84 N. According to Hooke' elastic limit). Therefore, k = F / x = 7.84 / 0.12 = 65.3 N/m (to 3 significant figures).
重物产生的拉力 F = mg = 0.80 × 9.8 = 7.84 N。根据胡克定律 F = k × x（在弹性限度内），k = F / x = 7.84 / 0.12 = 65.3 N/m（保留三位有效数字）。
To test the elastic limit, add weights incrementally, recording the extension each time. Plot a graph of force against extension. A straight line through the origin indicates Hooke's law is obeyed. If the graph begins to curve, the elastic limit has been exceeded and permanent deformation occurs.
验证弹性限度可逐步增加重物并记录伸长量。作拉力-伸长图，过原点的直线段说明遵循胡克定律。若图线开始弯曲，说明已超弹性限度，弹簧将发生永久形变。
9. Electrical Power and Energy | 电功率与电能
A light bulb is labelled "6.0 V, 12 W". It is connected to a 6.0 V power supply. Calculate the current through the bulb and the charge that flows in 10 minutes.
一个灯泡标有“6.0 V, 12 W”，连接到 6.0 V 电源。计算通过灯泡的电流及 10 分钟内流过的电荷量。
× V. Rearranging, I = P / V = 12 / 6.0 = 2.0 A. The bulb draws 2.0 A at its rated voltage.
电功率、电压和电流满足 P = I × V。变形得 I = P / V = 12 / 6.0 = 2.0 A。灯泡在额定电压下工作电流为 2.0 A。
C. This large amount of charge represents the flow of many electrons through the filament.
电荷量 Q = I × t。10 分钟 = 10 × 60 = 600 s，故 Q = 2.0 × 600 = 1200 C。这个巨大的电荷量意味着大量电子流过灯丝。
10. Motor Effect and Fleming's Left-Hand Rule | 电动机效应与弗莱明左手定则
A straight wire of length 0.15 m carries a current of 4.0 A and is placed perpendicular to a uniform magnetic field of 0.25 T. Calculate the force exerted on the wire. Use Fleming's left-hand rule to determine the direction of the force if the current flows east and the magnetic field points north.
一根长 0.15 m 的直导线通有 4.0 A 电流，垂直于 0.25 T 的匀强磁场放置。计算导线所受的力。若电流向东、磁场向北，用弗莱明左手定则判断力的方向。
.25 × 4.0 × 0.15 = 0.15 N. Ensure that the length is in metres and the field is in teslas.
垂直磁场中的通电导线受力 F = B × I × L。代入数据得 F = 0.25 × 4.0 × 0.15 = 0.15 N。务必用米为长度单位、特斯拉为磁感应强度单位。
Fleming's left-hand rule: hold the thumb, forefinger and second finger mutually at right angles. Point the Forefinger in the direction of the magnetic Field (north), the seCond finger in the direction of the Current (east); the thuMb then points in the direction of the force. In this case, the force is upwards.
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