📚 Year 12 WJEC Physics: Unit Test Mock Paper Walkthrough | Year 12 WJEC 物理：单元测试模拟卷解析
Mock exams are a vital part of AS Physics revision, and this walkthrough tackles a full WJEC-style unit test covering Mechanics, Materials, Waves, Electricity and Quantum phenomena. We will work through selected questions, highlighting key formulas, common pitfalls and the level of detail required by WJEC mark schemes.
模拟考试是 AS 物理复习的关键环节，本文解析一套完整的 WJEC 风格单元测试卷，涵盖力学、材料、波、电学和量子现象。我们将逐步解答典型题目，强调关键公式、常见错误以及 WJEC 评分标准所要求的答题深度。
1. Kinematics: Uniform Acceleration | 运动学：匀加速运动
A classic question presents a cyclist accelerating uniformly from rest with a = 1.8 m/s² over 12 s. You are asked to find the final velocity and the distance covered.
经典题目给出一个自行车手从静止开始以 a = 1.8 m/s² 匀加速 12 s，要求计算末速度和行驶的距离。
Start by listing knowns: u = 0, a = 1.8 m/s², t = 12 s. The fastest route to velocity is v = u + at.
首先列出已知量：u = 0, a = 1.8 m/s², t = 12 s。求速度的最快途径是使用 v = u + at。
v = 0 + (1.8)(12) = 21.6 m/s
For displacement, use s = ut + ½at², which simplifies to s = ½ × 1.8 × (12)² = 129.6 m.
计算位移时，使用 s = ut + ½at²，化简为 s = ½ × 1.8 × (12)² = 129.6 m。
WJEC marks are awarded for selecting the correct SUVAT equation, substituting values with units and giving the answer to an appropriate number of significant figures (here 3 s.f.).
WJEC 的评分标准包括选择正确的 SUVAT 方程、代入带单位的数值并给出合理有效数字（此处为 3 位有效数字）。
- Always write the equation first; examiners penalise ‘number soup’. —— 务必先写出公式；考官会扣“数字大杂烩”的分数。
- Watch for signs: if the object decelerates, a becomes negative. —— 注意正负号：若物体减速，a 取负值。
2. Projectile Motion | 抛体运动
WJEC often asks about a ball kicked horizontally from a cliff. Assume initial horizontal velocity ux = 8.0 m/s, vertical drop height = 20 m. Find time of flight and horizontal range.
WJEC 经常考查从悬崖水平踢出的球。设水平初速度 ux = 8.0 m/s，竖直下落高度 = 20 m，求飞行时间和水平射程。
Resolve motion independently. Vertically, uy = 0, ay = 9.81 m/s², sy = 20 m (downward). Use sy = uyt + ½ayt².
独立分解运动。竖直方向：uy = 0，ay = 9.81 m/s²，sy = 20 m（向下）。使用 sy = uyt + ½ayt²。
20 = 0 + ½ × 9.81 × t² ⇒ t = √(40/9.81) ≈ 2.02 s
Horizontally acceleration is zero, so range = ux × t = 8.0 × 2.02 ≈ 16.2 m.
水平方向加速度为零，因此射程 = ux × t = 8.0 × 2.02 ≈ 16.2 m。
Many candidates incorrectly mix horizontal and vertical components. Draw a clear diagram and treat the two directions separately to avoid losing marks.
许多考生错误地混合了水平和竖直分量。画清晰的示意图并分开处理两个方向，可避免失分。
3. Newton’s Laws and Free-Body Diagrams | 牛顿定律与受力图
A mock question shows a 5.0 kg block pulled along a rough surface by a force of 25 N at 30° above the horizontal. Coefficient of kinetic friction μₖ = 0.20. Determine the acceleration.
模拟题展示一个 5.0 kg 的物块在粗糙水平面上受到与水平成 30° 的 25 N 拉力，动摩擦系数 μₖ = 0.20。求加速度。
Resolve the pulling force: Fx = 25 cos30° ≈ 21.65 N, Fy = 25 sin30° = 12.5 N. The normal reaction N = mg − Fy = (5.0×9.81) − 12.5 = 36.55 N.
分解拉力：Fx = 25 cos30° ≈ 21.65 N，Fy = 25 sin30° = 12.5 N。支持力 N = mg − Fy = (5.0×9.81) − 12.5 = 36.55 N。
Friction f = μₖN = 0.20 × 36.55 = 7.31 N. Net horizontal force = 21.65 − 7.31 = 14.34 N. Acceleration a = Fnet/m = 14.34/5.0 ≈ 2.87 m/s².
摩擦力 f = μₖN = 0.20 × 36.55 = 7.31 N。水平合力 = 21.65 − 7.31 = 14.34 N。加速度 a = Fnet/m = 14.34/5.0 ≈ 2.87 m/s²。
Always draw a free-body diagram showing weight, normal reaction, applied force and friction. WJEC expects you to state Newton’s second law before substituting values.
务必画出受力图，标出重量、支持力、拉力和摩擦力。WJEC 期待你在代入数值前先陈述牛顿第二定律。
4. Work, Energy and Power | 功、能与功率
A crane lifts a 200 kg load vertically at a constant speed of 0.50 m/s. Calculate the power output of the motor and the work done in lifting the load through 15 m.
起重机以 0.50 m/s 的恒定速度竖直提升 200 kg 的重物。计算电机输出功率和将重物提升 15 m 所做的功。
At constant speed, tension T = mg = 200 × 9.81 = 1962 N. Power = force × velocity = 1962 × 0.50 = 981 W (≈ 0.98 kW).
匀速时，拉力 T = mg = 200 × 9.81 = 1962 N。功率 = 力 × 速度 = 1962 × 0.50 = 981 W（≈ 0.98 kW）。
Work done = force × distance moved in direction of force = 1962 × 15 = 29 430 J (≈ 29.4 kJ). Alternatively, work equals the gain in gravitational potential energy mgh = 200 × 9.81 × 15, giving the same value.
做功 = 力 × 沿力方向的位移 = 1962 × 15 = 29 430 J（≈ 29.4 kJ）。也可由重力势能增加量 mgh = 200 × 9.81 × 15 获得，数值相同。
Be careful to convert units correctly; power is often expected in watts or kilowatts, and work in joules.
注意正确换算单位；功率通常用瓦特或千瓦，功用焦耳。
5. Conservation of Momentum | 动量守恒
A typical question: a 3.0 kg trolley moving at 4.0 m/s collides with a stationary 2.0 kg trolley, and they stick together. Find the common velocity after the collision.
典型题目：一辆 3.0 kg 的小车以 4.0 m/s 的速度撞上一辆静止的 2.0 kg 小车并粘在一起，求碰后的共同速度。
Total momentum before = m₁u₁ + m₂u₂ = 3.0×4.0 + 2.0×0 = 12 kg m/s. After collision, total mass = 5.0 kg, so velocity v = 12 / 5.0 = 2.4 m/s.
碰撞前总动量 = m₁u₁ + m₂u₂ = 3.0×4.0 + 2.0×0 = 12 kg m/s。碰后总质量 = 5.0 kg，因此速度 v = 12 / 5.0 = 2.4 m/s。
Always specify that momentum is conserved in the absence of external forces. WJEC mark schemes reward a statement of the conservation law as well as the correct vector direction.
务必说明在没有外力时动量守恒。WJEC 评分标准奖励写出守恒定律并指明正确的矢量方向。
If the collision is elastic, kinetic energy is also conserved; check that the total KE before equals the total KE after. This mock question involves a perfectly inelastic collision because they stick together.
若是弹性碰撞，动能也守恒；需验证碰前总动能等于碰后总动能。因两车粘在一起，本题属于完全非弹性碰撞。
6. Stress, Strain and Young Modulus | 应力、应变与杨氏模量
A steel wire of diameter 0.80 mm and original length 2.50 m stretches by 4.2 mm under a load of 45 N. Determine the Young modulus of the material.
一根直径 0.80 mm、原长 2.50 m 的钢丝在 45 N 载荷下伸长 4.2 mm，求材料的杨氏模量。
Cross-sectional area A = π(d/2)² = π(0.40×10⁻³)² ≈ 5.027×10⁻⁷ m². Stress σ = F/A = 45 / 5.027×10⁻⁷ ≈ 8.95×10⁷ Pa.
横截面积 A = π(d/2)² = π(0.40×10⁻³)² ≈ 5.027×10⁻⁷ m²。应力 σ = F/A = 45 / 5.027×10⁻⁷ ≈ 8.95×10⁷ Pa。
Strain ε = ΔL/L = 4.2×10⁻³ / 2.50 = 1.68×10⁻³. Young modulus E = σ/ε = 8.95×10⁷ / 1.68×10⁻³ ≈ 5.33×10¹⁰ Pa (≈ 53 GPa).
应变 ε = ΔL/L = 4.2×10⁻³ / 2.50 = 1.68×10⁻³。杨氏模量 E = σ/ε = 8.95×10⁷ / 1.68×10⁻³ ≈ 5.33×10¹⁰ Pa（≈ 53 GPa）。
Examiners expect unit conversions to metres before calculation. Also be ready to interpret a stress–strain graph and identify the elastic limit, yield point and breaking stress.
考官要求在计算前将单位换算为米。同时需准备解读应力–应变图，识别弹性极限、屈服点和断裂应力。
7. Wave Speed and Standing Waves | 波速与驻波
A string fixed at both ends vibrates in its second harmonic at 120 Hz. The length of the string is 0.90 m. Calculate the wave speed and the fundamental frequency.
一根两端固定的弦以 120 Hz 的第二谐频振动，弦长 0.90 m。计算波速和基频。
For the second harmonic (first overtone), the string has one full wavelength: λ = L = 0.90 m. Wave speed v = fλ = 120 × 0.90 = 108 m/s.
对于第二谐频（第一泛音），弦上有一个完整波长：λ = L = 0.90 m。波速 v = fλ = 120 × 0.90 = 108 m/s。
Fundamental frequency f₁ = v/(2L) = 108 / (2×0.90) = 60 Hz. Alternatively, f₁ = f₂/2 = 60 Hz, since frequency is proportional to harmonic number.
基频 f₁ = v/(2L) = 108 / (2×0.90) = 60 Hz。也可通过 f₁ = f₂/2 = 60 Hz 计算，因为频率与谐波数成正比。
Always draw the standing wave pattern to confirm nodes and antinodes. For the nth harmonic the length L = n(λ/2).
务必画出驻波图以确认波节和波腹。对于第 n 次谐波，有 L = n(λ/2)。
8. Refraction and Total Internal Reflection | 折射与全内反射
Light travels from glass (n = 1.52) into air. Find the critical angle and explain what happens if the angle of incidence is 50°.
光从玻璃（n = 1.52）射向空气，求临界角并说明当入射角为 50° 时会发生什么。
Critical angle θc = sin⁻¹(n₂/n₁) = sin⁻¹(1.00/1.52). So θc = sin⁻¹(0.6579) ≈ 41.1°.
临界角 θc = sin⁻¹(n₂/n₁) = sin⁻¹(1.00/1.52)，即 θc ≈ 41.1°。
If the angle of incidence (50°) is greater than the critical angle, total internal reflection occurs; all light is reflected back into the glass. No refraction into air takes place.
若入射角（50°）大于临界角，则发生全内反射；全部光线反射回玻璃中，没有折射到空气中。
In a WJEC written answer you must use the phrase ‘total internal reflection’ and mention that the angle of incidence exceeds the critical angle. A clear ray diagram earns additional marks.
在 WJEC 笔试答案中必须使用“全内反射”这一术语，并说明入射角大于临界角。清晰的光路图可获得额外分数。
9. Photoelectric Effect | 光电效应
Zinc plate has a work function Φ = 4.3 eV. Ultraviolet light of frequency 1.2×10¹⁵ Hz is incident on it. Determine whether electrons are emitted and, if so, find their maximum kinetic energy.
锌板的逸出功 Φ = 4.3 eV。频率为 1.2×10¹⁵ Hz 的紫外光照射其上。判断是否会发射电子，若会则求最大动能。
Photon energy E = hf = (6.63×10⁻³⁴) × (1.2×10¹⁵) = 7.956×10⁻¹⁹ J. Convert to eV: divide by 1.6×10⁻¹⁹ → E ≈ 4.97 eV.
光子能量 E =
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