📚 Year 13 WJEC Physics: Mock Unit Test Analysis & Solutions | Year 13 WJEC 物理：单元测试模拟卷解析
Welcome to this extended walkthrough of a Year 13 WJEC Physics mock unit test. The paper covers the core A2 topics examined in the WJEC Physics specification: simple harmonic motion, circular motion, capacitors, electric fields, radioactivity, ideal gases, electromagnetic induction, energy in SHM, magnetic fields and gravitational fields. Every worked example below is written in the order a marker expects to see it - state the equation, substitute the values, then give the answer with a unit and a sensible number of significant figures. Use the analysis sections and the error table at the end to turn each mock into a revision plan rather than just a score.
欢迎阅读这份 Year 13 WJEC 物理单元模拟卷的完整解析。本卷覆盖 WJEC 物理考纲中的 A2 核心主题：简谐运动、圆周运动、电容器、电场、放射性、理想气体、电磁感应、简谐运动中的能量、磁场与引力场。以下每个例题都按照阅卷人期望的书写顺序呈现——先写方程，再代入数值，最后给出带单位、有效数字合理的答案。请利用文末的试卷分析与易错点表格，把一次模考变成一份复习计划，而不仅仅是一个分数。
1. Simple Harmonic Motion - Maximum Velocity and Acceleration | 简谐运动：最大速度与最大加速度
A 0.500 kg mass attached to a spring oscillates with an amplitude of 0.080 m and a frequency of 2.00 Hz. Determine the maximum velocity and the maximum acceleration of the mass.
一个 0.500 kg 的物体连接在弹簧上，以振幅 0.080 m、频率 2.00 Hz 振动。求物体的最大速度和最大加速度。
Step 1 is to convert frequency into angular frequency, because every SHM energy and velocity equation in the specification is written in terms of ω, not f.
第一步是把频率换算为角频率，因为考纲中所有简谐运动的速度和能量方程都是用 ω 而不是 f 写出的。
ω = 2πf = 2π × 2.00 = 4π ≈ 12.57 rad s⁻¹
resultant force (and therefore acceleration) is zero. The defining equation is vₘₐₓ = ωA.
最大速度出现在物体经过平衡位置时，此时合力（因而加速度）为零。其定义式为 vₘₐₓ = ωA。
vₘₐₓ = ωA = 4π × 0.080 ≈ 1.01 m s⁻¹
s greatest, so aₘₐₓ = ω²A and it is always directed back towards the equilibrium position.
最大加速度出现在运动的两端，此时位移最大，因此 aₘₐₓ = ω²A，且方向始终指向平衡位置。
aₘₐₓ = ω²A = (4π)² × 0.080 ≈ 12.6 m s⁻²
Examiner points worth remembering:
值得记住的得分要点：
- Amplitude must be in metres before substitution; 8.0 cm entered as 8.0 is the single most common lost mark.
- The calculator must be in radian mode whenever a sine or cosine of ωt is evaluated.
- The mass of 0.500 kg is not needed for vₘₐₓ or aₘₐₓ, but it is needed if the question also asks for the total energy or the spring constant.
- 代入前振幅必须化为米；把 8.0 cm 直接写成 8.0 是最常见的失分点。
- 只要对 ωt 取正弦或余弦，计算器就必须处于弧度模式。
- 0.500 kg 的质量对求 vₘₐₓ、aₘₐₓ 无用，但若题目还要求总能量或劲度系数，就必须用到它。
2. Circular Motion - Centripetal Force and Banking Angle | 圆周运动：向心力与倾斜角
A car of mass 1200 kg moves around a circular bend of radius 50 m at a constant speed of 15 m/s. Calculate the centripetal force and the ideal angle of banking assuming no friction.
一辆质量为 1200 kg 的汽车以 15 m/s 的恒定速率通过半径为 50 m 的圆形弯道。请计算向心力，以及无摩擦时的理想倾斜角。
force directed towards the centre of the circle, and its magnitude is given by F = mv²/r.
向心力是指向圆心的合力，其大小由 F = mv²/r 给出。
F = mv²/r = (1200 × 15²) / 50 = 5400 N
On a frictionless banked track this force is supplied entirely by the horizontal component of the normal reaction, which leads to the standard banking result.
在无摩擦的倾斜弯道上，该力完全由支持力的水平分量提供，由此得到标准的倾斜角公式。
tan θ = v² / (rg) = 15² / (50 × 9.81) = 225 / 490.5 ≈ 0.459
θ ≈ 24.6°
Notice that the mass cancels in the banking equation, so the ideal angle is independent of the mass of the vehicle. That is exactly why the same road camber works for a heavy lorry and a small car.
注意，在倾斜角方程中质量被约去，因此理想倾斜角与车辆质量无关。这正是同一路面横向坡度对重型卡车和小轿车同样有效的原因。
- If the question states a coefficient of friction, banking only reduces the friction needed; you must then resolve vertically and horizontally and include the friction term.
- Speed must be in m s⁻¹, radius in metres, and g taken as 9.81 m s⁻² unless told otherwise.
- 若题目给出摩擦系数，倾斜只起减小所需摩擦力的作用；此时必须竖直与水平方向分别列方程，并加入摩擦项。
- 速率单位必须是 m s⁻¹、半径必须是米；除特别说明外 g 取 9.81 m s⁻²。
3. Capacitors - Charge, Energy Stored and Time Constant | 电容器：电荷量、储能与时间常数
A 470 μF capacitor is charged to 12.0 V and then discharged through a 10 kΩ resistor. Calculate the charge stored, the energy stored, the time constant, and the voltage across the capacitor after 5.0 s.
一个 470 μF 的电容器被充电到 12.0 V，然后通过一个 10 kΩ 的电阻放电。求储存的电荷量、储存的能量、时间常数，以及 5.0 s 后电容器两端的电压。
llows the definition of capacitance, Q = CV, with capacitance converted into farads first.
电荷量由电容的定义式 Q = CV 给出，先把电容换算成法拉。
Q = CV = 470 × 10⁻⁶ × 12.0 = 5.64 × 10⁻³ C = 5.64 mC
W = ½CV², and this is the energy that will be dissipated in the resistor during discharge.
已充电电容器储存的能量为 W = ½CV²，这就是放电过程中将在电阻上耗散的能量。
W = ½CV² = ½ × 470 × 10⁻⁶ × 12.0² ≈ 3.38 × 10⁻² J
s units of seconds and equals the time for the charge to fall to 1/e of its initial value.
RC 电路的时间常数为 τ = RC，单位为秒，其数值等于电荷量降到初始值 1/e 所需的时间。
τ = RC = 10 × 10³ × 470 × 10⁻⁶ = 4.7 s
For exponential discharge the voltage obeys V = V₀e−t/RC, so after 5.0 s:
指数放电时电压遵循 V = V₀e−t/RC，因此 5.0 s 后：
V = 12.0 × e−5.0/4.7 ≈ 12.0 × 0.345 ≈ 4.14 V
- = RC ln(V₀/V), so the capacitor reaches half voltage in 0.693τ ≈ 3.26 s - a useful check.
- The graph of ln V against time for a discharging capacitor is a straight line of gradient −1/RC.
- 对指数式变形可得 t = RC ln(V₀/V)，因此电容器在 0.693τ ≈ 3.26 s 时降到一半电压——这是一个很实用的检验方法。
- 放电电容器的 ln V 对时间图像是一条直线，斜率为 −1/RC。
4. Electric Fields - Uniform Field Between Parallel Plates | 电场：平行板间的匀强电场
Two parallel plates separated by 5.0 cm have a potential difference of 250 V between them. Calculate the electric field strength, the force on an electron placed between the plates, and the acceleration of that electron.
两块相距 5.0 cm 的平行板之间存在 250 V 的电势差。求板间的电场强度、置于板间电子所受的力，以及该电子的加速度。
For a uniform field the strength is the potential gradient, E = V/d, with d in metres.
匀强电场的场强等于电势梯度 E = V/d，其中 d 以米为单位。
E = V/d = 250 / 0.050 = 5000 V m⁻¹ = 5.0 kV m⁻¹
n a charged particle in a uniform field is F = qE, where q is the magnitude of the charge.
带电粒子在匀强电场中所受的力为 F = qE，其中 q 为电荷量的大小。
F = qE = 1.6 × 10⁻¹⁹ × 5000 = 8.0 × 10⁻¹⁶ N
Newton's second law then gives the acceleration, using the electron rest mass 9.11 × 10⁻³¹ kg.
再由牛顿第二定律求加速度，电子静质量取 9.11 × 10⁻³¹ kg。
a = F/m = 8.0 × 10⁻¹⁶ / 9.11 × 10⁻³¹ ≈ 8.78 × 10¹⁴ m s⁻²
full potential difference is W = qV = 1.6 × 10⁻¹⁹ × 250 = 4.0 × 10⁻¹⁷ J, which is 250 eV.
同一物理过程还有第二条路径，即用能量而非力来处理：电荷跨越整个电势差时所做的功为 W = qV = 1.6 × 10⁻¹⁹ × 250 = 4.0 × 10⁻¹⁷ J，即 250 eV。
- The force is directed from the positive plate to the negative plate for a positive charge, and opposite to that for an electron.
- Field lines are parallel and equally spaced, which is what defines the field as uniform.
- Electric potential V and field strength E have different units and are never interchangeable: E is measured in V m⁻¹ or N C⁻¹.
- 对正电荷来说，力的方向由正极板指向负极板；对电子而言方向相反。
- 电场线平行且等间距，这正是电场为匀强电场的标志。
- 电势 V 与场强 E 单位不同、绝不可混用：E 的单位是 V m⁻¹ 或 N C⁻¹。
5. Radioactivity - Decay Constant, Activity and Half-Life | 放射性：衰变常数、活度与半衰期
ulate the decay constant in s⁻¹ and estimate the number of undecayed nuclei in the source.
一份钴-60 样品的半衰期为 5.27 年，测得活度为 3.7 × 10⁷ Bq。请计算以 s⁻¹ 为单位的衰变常数，并估算源中未衰变原子核的数量。
ponential and spontaneous, so the decay constant is related to half-life by λ = ln 2 / t½.
放射性衰变是自发的指数过程，因此衰变常数与半衰期的关系为 λ = ln 2 / t½。
t½ = 5.27 × 3.15 × 10⁷ ≈ 1.66 × 10⁸ s
λ = ln 2 / t½ = 0.
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