📚 Year 11 Cambridge Physics Mock Unit Test Analysis | 剑桥物理单元测试模拟卷解析
This article walks you through a typical Year 11 Cambridge IGCSE Physics unit test mock paper, highlighting key concepts, examiner expectations, and common pitfalls. Each section breaks down a classic exam-style question drawn from mechanics, waves, electricity, and thermal physics, providing detailed step-by-step solutions and revision tips. Use this as a self-check or a structured revision guide to boost your confidence before the real test.
本文将带你逐题解析一份典型的 Year 11 剑桥 IGCSE 物理单元测试模拟卷，突出核心概念、得分要点与常见错误。每一节聚焦一道力学、波、电学或热学的高频题型，给出详细的阶梯式解答与复习建议。你可以把它当作自测工具或系统复习指南，在真正考试前巩固信心。
1. Kinematics: Interpreting a Velocity-Time Graph | 运动学：解读速度-时间图
A classic structured question asks: a car accelerates uniformly from rest to 20 m/s in 10 s, maintains a steady 20 m/s for 20 s, then decelerates uniformly to rest in a further 10 s. Parts (a)-(c) target graph sketching, acceleration, and distance travelled.
一道经典的简答题这样描述：一辆汽车由静止开始匀加速，10 s 后达到 20 m/s，然后匀速行驶 20 s，最后用 10 s 匀减速到静止。题目设问通常包括作图、加速度计算和总路程求解。
The velocity-time graph is a trapezoid: a rising straight line from (0,0) to (10,20), a horizontal line from (10,20) to (30,20), and a falling straight line from (30,20) to (40,0). Mark axes clearly and label key values.
速度-时间图是一个梯形：从 (0,0) 到 (10,20) 的上升直线，从 (10,20) 到 (30,20) 的水平线，以及从 (30,20) 到 (40,0) 的下降直线。作图时要清晰标注坐标轴和关键数值。
Acceleration during the first stage is calculated using a = (v - u) / t. With u = 0, v = 20 m/s, t = 10 s, we get:
第一阶段加速度用 a = (v - u) / t 求得。初速度 u = 0，末速度 v = 20 m/s，时间 t = 10 s，可得：
a = (20 - 0) / 10 = 2.0 m/s²
Distance is found from the area under the graph. Many students forget that area of a velocity-time graph equals displacement (and distance in straight-line motion). The total area is the sum of two triangles and a rectangle:
距离由图像下方面积求得。很多学生忘记了速度-时间图的面积等于位移（在直线运动中等于路程）。总面积是两个三角形加一个矩形的面积之和：
Distance = ½ × 10 × 20 + 20 × 20 + ½ × 10 × 20 = 100 + 400 + 100 = 600 m
A common mistake is using the constant acceleration equations for the entire journey: these only apply per stage. Also, check that you are not confusing distance with displacement when direction reverses; here the motion is in one direction, so they are equal.
常见错误是对全程统一使用匀变速公式，这些公式只能分阶段使用。此外，当运动方向改变时要区分距离与位移；本题始终沿同一方向运动，两者相等。
2. Forces and Resultant Force Calculation | 力与合力的计算
A typical force problem presents a box of mass 10 kg being pulled along a rough horizontal surface by a rope at 30° to the horizontal with a tension of 50 N. The frictional force opposing motion is 20 N. You are required to find the resultant horizontal force.
典型的力学习题：一个质量为 10 kg 的箱子放在粗糙水平面上，一根绳子与水平面成 30° 角，对箱子施加 50 N 的拉力，摩擦力为 20 N。要求计算水平方向合力。
tal component does the pulling: F_horizontal = T cosθ = 50 × cos30°. Using cos30° ≈ 0.866:
首先将拉力分解为水平和竖直分量。水平分量提供牵引：F_horizontal = T cosθ = 50 × cos30°。cos30° ≈ 0.866，所以：
F_horizontal = 50 × 0.866 = 43.3 N
3.3 N (forward). Many candidates mistakenly use the 50 N directly without resolving, or forget to subtract friction.
水平方向合力 = 拉力水平分量 - 摩擦力 = 43.3 N - 20 N = 23.3 N（向前）。很多考生直接使用 50 N 而没有分解，或是忘记减去摩擦力。
In questions where the box might lift, you would also consider the vertical component (T sinθ) to check if it reduces the normal contact force, but here the box remains on the ground.
如果题目涉及箱子是否被提离地面，还需要分析竖直分量对支持力的影响，但本题箱子始终在地面上。
3. Newton's Second Law Application | 牛顿第二定律的应用
Building on the previous result, if the box has a mass of 10 kg, use Newton's second law to determine its acceleration. The net force acting is 23.3 N.
接上题，如果箱子的质量为 10 kg，利用牛顿第二定律求加速度。作用于箱子的净力为 23.3 N。
F = m a ⇒ a = F / m = 23.3 / 10 = 2.33 m/s²
The acceleration is in the direction of the resultant force. Always state the direction (forward/to the right). A common slip is to use the pulling force or the frictional force alone instead of the resultant. Remember: only the unbalanced force causes acceleration.
加速度方向与合力方向一致。务必说明方向（向前/向右）。常犯的错误是单独使用拉力或摩擦力，而忘了用合力。记住：只有未平衡的力才会产生加速度。
4. Energy Transfers in a Pendulum | 单摆中的能量转换
st point and released. Find the maximum speed of the bob. Take g = 10 N/kg for simplicity.
能量守恒题型偏爱单摆。一个质量为 0.2 kg 的摆球被提高到最低点上方 0.10 m 后释放，求摆球的最大速度。取 g = 10 N/kg 以简化计算。
At the highest point, the bob has gravitational potential energy (GPE) and zero kinetic energy (KE). At the lowest point, all GPE has been converted to KE.
在最高点，摆球具有重力势能（GPE），动能为零。在最低点，所有重力势能都转化为动能。
GPE = mgh = 0.2 × 10 × 0.10 = 0.20 J
KE = ½ m v² → 0.20 = ½ × 0.2 × v² → v² = 2.0 → v ≈ 1.41 m/s
equired; always check the paper. Also, remember that energy is a scalar and direction of velocity is not needed for maximum speed calculation.
注意：有时 g 需要取 9.8 m/s²，务必看试卷要求。另外，能量是标量，计算最大速率时不需要考虑速度方向。
5. Wave Properties and the Wave Equation | 波的性质与波动方程
A typical wave question provides frequency and wavelength. For instance, a water wave has a frequency of 50 Hz and a wavelength of 0.68 m. Calculate the wave speed and the time period.
波动题常给出频率和波长。例如，一个水波的频率为 50 Hz，波长为 0.68 m。请计算波速和周期。
The wave equation links speed (v), frequency (f), and wavelength (λ):
波动方程将波速 (v)、频率 (f) 和波长 (λ) 联系起来：
v = f λ = 50 × 0.68 = 34 m/s
Time period T is the reciprocal of frequency:
周期 T 是频率的倒数：
T = 1 / f = 1 / 50 = 0.02 s
Many marks are lost by confusing period and frequency, or using incorrect units. Always give v in m/s, f in Hz, λ in m, and T in s.
很多同学因混淆周期和频率或单位使用错误而失分。波速的单位始终是 m/s，频率是 Hz，波长是 m，周期是 s。
6. Reflection and Refraction of Light | 光的反射与折射
dex n = 1.5. The angle of incidence is 40°. Find the angle of refraction inside the glass.
有关角度计算的折射题频繁出现。例如：一束光从空气射入折射率 n = 1.5 的玻璃块，入射角为 40°。求玻璃内的折射角。
Snell's law: n₁ sinθ₁ = n₂ sinθ₂. For air, n₁ = 1.00, sin40° ≈ 0.643. So:
斯涅尔定律：n₁ sinθ₁ = n₂ sinθ₂。空气 n₁ = 1.00，sin40° ≈ 0.643。因此：
1.00 × 0.643 = 1.5 × sin r ⇒ sin r = 0.643 / 1.5 ≈ 0.4287 ⇒ r ≈ 25.4°
Candidates often reverse the indices or forget to use sin. Also, when the ray goes from glass to air, check for total internal reflection if the angle exceeds the critical angle.
考生经常颠倒折射率或忘记使用正弦函数。另外，当光线从玻璃进入空气时，如果入射角超过临界角，要检验是否发生全反射。
7. Electrical Circuits and Ohm's Law | 电路与欧姆定律
A straightforward circuit question: a 10 Ω resistor is connected across a 6 V battery of negligible internal resistance. Find the current in the circuit and the power dissipated by the resistor.
简单电路题：一个 10 Ω 电阻器连接在 6 V 电源上（内阻忽略不计）。求电路中的电流和电阻器消耗的功率。
Ohm' = V / R = 6 / 10 = 0.6 A. The direction of current is conventional (positive to negative).
欧姆定律：I = V / R = 6 / 10 = 0.6 A。电流方向使用惯例（由正极流向负极）。
Power can be calculated in three ways: P = VI, P = I²R, or P = V²/R. Using P = VI:
功率可以用三种方式计算：P = VI、P = I²R 或 P = V²/R。利用 P = VI 得：
P = 6 × 0.6 = 3.6 W
A typical mistake is applying the power formula to the wrong component or using total circuit values when the question asks for a single resistor. Also, if internal resistance is given, you must find the terminal voltage first.
常见错误是将功率公式用在错误元件上，或在只问单个电阻时使用了整个电路的总值。若题目给出电源内阻，则需先求出端电压。
8. Series and Parallel Resistors | 串联与并联电阻
More complex circuits test your ability to combine resistances. A circuit has a 4 Ω and a 12 Ω resistor connected in parallel, and this combination is connected in series with an 8 Ω resistor. Calculate the total resistance of the circuit.
更复杂的电路考查电阻的组合能力。一个电路包含一个 4 Ω 和一个 12 Ω 的电阻并联，然后整体与一个 8 Ω 电阻串联。求电路的总电阻。
For parallel combination: 1/R_parallel = 1/R₁ + 1/R₂ = 1/4 + 1/12 = (3+1)/12 = 4/12 ⇒ 1/R_parallel = 1/3 ⇒ R_parallel = 3 Ω.
并联部分：1/R_parallel = 1/R₁ + 1/R₂ = 1/4 + 1/12 = (3+1)/12 = 4/12，倒数得 R_parallel = 3 Ω。
resistors, the combined resistance is always smaller than the smallest individual resistor; here 3 Ω is less than both 4 Ω and 12 Ω, which serves as a quick sanity check.
总电阻 = R_parallel + 8 Ω = 3 + 8 = 11 Ω。记住：并联电阻的等效电阻总小于最小的单个电阻；此处 3 Ω 小于 4 Ω 和 12 Ω，可用作快速验算。
9. Radioactive Decay and Half-life | 放射性衰变与半衰期
A classic half-life problem: a radioactive sample has an initial activity of 800 Bq and a half-life of 2 hours. What is its activity after 6 hours?
经典的半衰期问题：一个放射性样品的初始活度为 800 Bq，半衰期为 2 小时。求 6 小时后的活度。
Determine the number of half-lives: 6 hours / 2 hours per half-life = 3 half-lives.
确定半衰期个数：6 小时 ÷ 2 小时/个 = 3 个半衰期。
f-life halves the activity: 800 → 400 → 200 → 100 Bq. So after 6 hours, activity = 100 Bq.
每个半衰期使活度减半：800 → 400 → 200 → 100 Bq。因此 6 小时后活度为 100 Bq。
Common errors include adding the half-lives instead of dividing or misreading the time interval. Always check if time given is exactly a multiple of the half-life; if not, you may need an exponential calculation, though at IGCSE it's usually whole multiples.
常见错误是用加法处理半衰期而不是减半，或读错时间间隔。始终检查题目给出的时间是否为半衰期的整倍数，若非整倍数则可能需指数计算，不过 IGCSE 阶段通常设计为整倍数。
10. Specific Heat Capacity Calculation | 比热容计算
Thermal physics questions routinely ask for energy needed to heat a substance. Example: 0.5 kg of water is heated from 20 °C to 80 °C. The specific heat capacity of water is 4200 J/(kg °C). Calculate the energy transferred to the water.
热学题目经常要求计算加热物质所需的能量。例如：将 0.5 kg 水从 20 °C 加热到 80 °C，水的比热容为 4200 J/(kg °C)。求水吸收的能量。
Use the formula ΔE = m c Δθ, where Δθ is the temperature change (80 - 20 = 60 °C).
使用公式 ΔE = m c Δθ，其中 Δθ 是温度变化量 (80 - 20 = 60 °C)。
ΔE = 0.5 × 4200 × 60 = 126,000 J (or 126 kJ)
Always express the answer in joules unless asked otherwise. Some students mistakenly use the final temperature instead of the temperature change, or forget to convert kg to g (but here c is in J/(kg °C), so mass must be in kg).
题目无特殊要求时答案应以焦耳表示。有些学生误用末温代替温度变化，或忘记质量单位需与比热容单位匹配（此处 c 采用 J/(kg °C)，因此质量只能用 kg）。
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