Year 13 WJEC Physics: Unit Test Mock Paper Walkthrough | Year 13 WJEC 物理单元测试模拟卷解析

📚 Year 13 WJEC Physics: Unit Test Mock Paper Walkthrough | Year 13 WJEC 物理单元测试模拟卷解析

This walkthrough tackles a full mock paper designed for Year 13 WJEC Physics, covering core topics from Units 3 and 4 such as simple harmonic motion, gravitational and electric fields, capacitors, magnetism, nuclear decay, and astrophysics. Each question is broken down step by step so that you can identify common pitfalls, reinforce key equations, and build confidence for all structured questions.

本文是对一套专为 WJEC 十三年级物理设计的高仿真模拟卷的逐题解析，覆盖单元 3 和单元 4 的核心内容——简谐运动、引力场与电场、电容器、磁学、核衰变和天体物理。每道题目都做了分步拆解，帮助你发现常见失分点，巩固关键方程，建立面对所有结构题的信心。

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1. Simple Harmonic Motion – Pendulum and Spring | 简谐运动——摆与弹簧

A simple pendulum of length 0.75 m swings with small amplitude. (a) Calculate the period T. (b) A mass‑spring system has force constant k = 50 N m⁻¹ and mass m = 0.20 kg. It is released from rest at displacement 4.0 cm. Determine the angular frequency ω and the displacement at t = 0.50 s. Take g = 9.81 m s⁻².

一摆长 0.75 m 的单摆以微小振幅摆动。(a) 计算周期 T。(b) 一弹簧振子的劲度系数 k = 50 N m⁻¹，质量 m = 0.20 kg，从静止释放时位移为 4.0 cm。求角频率 ω 及 t = 0.50 s 时的位移。取 g = 9.81 m s⁻²。

For the pendulum, the period depends only on length and g. The formula T = 2π√(L/g) gives T = 2π√(0.75/9.81) ≈ 2π√0.07645 ≈ 2π × 0.2765 = 1.74 s.

对单摆而言，周期只取决于摆长和 g。由 T = 2π√(L/g) 得 T = 2π√(0.75/9.81) ≈ 2π√0.07645 ≈ 2π × 0.2765 = 1.74 s。

For the mass‑spring system, angular frequency ω = √(k/m) = √(50/0.20) = √250 ≈ 15.8 rad s⁻¹. The motion is described by x = A cos(ωt) because it starts at maximum displacement. Substituting A = 4.0 cm and t = 0.50 s gives x = 4.0 cos(15.8 × 0.50) = 4.0 cos(7.90 rad). Reducing the angle: 7.90 rad is equivalent to 452.6°; after removing 360° we get 92.6°, whose cosine is approximately -0.0454. Hence x ≈ 4.0 × (-0.0454) = -0.18 cm, meaning the mass is 0.18 cm on the opposite side of the equilibrium position.

弹簧振子的角频率 ω = √(k/m) = √(50/0.20) = √250 ≈ 15.8 rad s⁻¹。因从最大位移开始运动，位移方程写作 x = A cos(ωt)。代入 A = 4.0 cm，t = 0.50 s 得 x = 4.0 cos(15.8 × 0.50) = 4.0 cos(7.90 rad)。化简角度：7.90 rad 相当于 452.6°，减去 360° 后为 92.6°，cos 92.6° ≈ -0.0454。因此 x ≈ 4.0 × (-0.0454) = -0.18 cm，即物体在平衡位置另一侧 0.18 cm 处。

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2. Gravitational Fields and Potential | 引力场与引力势

At a point 2.0 Earth radii from the centre of the Earth, calculate the gravitational potential V and the escape velocity from that point. Use the surface value g = 9.81 N kg⁻¹ and Earth’s radius R = 6.4 × 10⁶ m.

在距离地心 2.0 倍地球半径的一点处，计算引力势 V 和从该点逃逸所需的速度。已知地表 g = 9.81 N kg⁻¹，地球半径 R = 6.4 × 10⁶ m。

We first find GM from gR². Since g = GM/R², we have GM = 9.81 × (6.4 × 10⁶)² = 4.02 × 10¹⁴ m³ s⁻². Gravitational potential at distance r is V = –GM/r. With r = 2R = 1.28 × 10⁷ m, V = –4.02 × 10¹⁴ / (1.28 × 10⁷) ≈ –3.14 × 10⁷ J kg⁻¹.

先用 gR² 求出 GM。由 g = GM/R² 得 GM = 9.81 × (6.4 × 10⁶)² = 4.02 × 10¹⁴ m³ s⁻²。在距离 r 处的引力势 V = –GM/r，将 r = 2R = 1.28 × 10⁷ m 代入，V = –4.02 × 10¹⁴ / (1.28 × 10⁷) ≈ –3.14 × 10⁷ J kg⁻¹。

Escape velocity from that location is v = √(2GM/r). Substituting r = 2R yields v = √(GM/R) = √(gR) = √(9.81 × 6.4 × 10⁶) = √(6.278 × 10⁷) ≈ 7.92 × 10³ m s⁻¹. Notice how the escape velocity at 2R is exactly 1/√2 times the surface escape velocity (11.2 km s⁻¹).

该点的逃逸速度 v = √(2GM/r)。代入 r = 2R 后得到 v = √(GM/R) = √(gR) = √(9.81 × 6.4 × 10⁶) = √(6.278 × 10⁷) ≈ 7.92 × 10³ m s⁻¹。注意 2R 处的逃逸速度正好是表面逃逸速度 (11.2 km s⁻¹) 的 1/√2 倍。

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3. Electric Fields – Two Point Charges | 电场——两个点电荷

Two point charges, +2.0 μC and –4.0 μC, are placed 0.30 m apart in air. Determine the resultant electric field at the midpoint between them.

两个点电荷 +2.0 μC 和 –4.0 μC 在空气中相距 0.30 m。求连线中点处的合电场强度。

Set the +2.0 μC charge at x = 0 and the –4.0 μC charge at x = 0.30 m. The midpoint is at x = 0.15 m, equidistant (0.15 m) from each charge. The field due to the positive charge points away from it, i.e. to the right. Using E = kQ/r² with k = 8.99 × 10⁹, E₁ = (8.99 × 10⁹ × 2.0 × 10⁻⁶) / (0.15)² = 7.99 × 10⁵ N C⁻¹ to the right.

将 +2.0 μC 电荷置于 x = 0，–4.0 μC 电荷置于 x = 0.30 m。中点 x = 0.15 m 到两电荷距离均为 0.15 m。正电荷的电场方向背离电荷，即向右。由 E = kQ/r²，k = 8.99 × 10⁹，E₁ = (8.99 × 10⁹ × 2.0 × 10⁻⁶) / (0.15)² = 7.99 × 10⁵ N C⁻¹ 向右。

The field caused by the negative charge points towards it. At the midpoint, this direction is also to the right because the negative charge is on the right and attracts a positive test charge. E₂ = (8.99 × 10⁹ × 4.0 × 10⁻⁶) / (0.15)² = 1.598 × 10⁶ N C⁻¹ to the right. The resultant field E = E₁ + E₂ = 2.40 × 10⁶ N C⁻¹ to the right.

负电荷的电场方向指向自身。在中点处，因负电荷位于右侧，对正检验电荷的吸引力也朝右。E₂ = (8.99 × 10⁹ × 4.0 × 10⁻⁶) / (0.15)² = 1.598 × 10⁶ N C⁻¹ 向右。合电场 E = E₁ + E₂ = 2.40 × 10⁶ N C⁻¹，方向向右。

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4. Capacitor Discharge – Time Constant | 电容器放电——时间常数

A 10 μF capacitor is charged to 12 V and then discharged through a 500 kΩ resistor. Calculate the time constant τ of the circuit and the voltage across the capacitor after 5.0 s.

一只 10 μF 的电容器被充电至 12 V，然后通过 500 kΩ 的电阻放电。求电路的时间常数 τ 及 5.0 s 后电容器两端的电压。

The time constant is simply τ = RC = 500 × 10³ Ω × 10 × 10⁻⁶ F = 5.0 s. During discharge, the voltage obeys V = V₀ e^(–t/τ). With V₀ = 12 V and t = τ = 5.0 s, V = 12 × e⁻¹ = 12 × 0.3679 ≈ 4.42 V.

时间常数 τ = RC = 500 × 10³ Ω × 10 × 10⁻⁶ F = 5.0 s。放电过程中电压遵循 V = V₀ e^(–t/τ)。将 V₀ = 12 V，t = τ = 5.0 s 代入，V = 12 × e⁻¹ = 12 × 0.3679 ≈ 4.42 V。

You could also be asked to find the current or the charge remaining. The current at 5.0 s would be I = (V₀/R) e⁻¹ = (12/500×10³) × 0.368 ≈ 8.8 μA, and the charge Q = CV = 10 × 10⁻⁶ × 4.42 = 4.42 × 10⁻⁵ C.

你也可能被要求求电流或剩余电荷。5.0 s 时的电流 I = (V₀/R) e⁻¹ = (12/500×10³) × 0.368 ≈ 8.8 μA，电荷 Q = CV = 10 × 10⁻⁶ × 4.42 = 4.42 × 10⁻⁵ C。

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5. Magnetic Force on a Moving Charge | 运动电荷所受的磁力

An electron enters a uniform magnetic field of flux density 0.50 T at a speed of 2.0 × 10⁷ m s⁻¹, moving perpendicular to the field. Calculate the magnetic force on the electron and the radius of its circular path.

一电子以 2.0 × 10⁷ m s⁻¹ 的速度垂直于均匀磁场进入磁场，磁通密度为 0.50 T。计算电子所受磁力及其圆周运动半径。

The force is given by F = Bev, where e = 1.60 × 10⁻¹⁹ C. F = 0.50 × 1.60 × 10⁻¹⁹ × 2.0 × 10⁷ = 1.6 × 10⁻¹² N. This force provides the centripetal force mv²/r, so r = mv/(Be).

磁力公式为 F = Bev，其中 e = 1.60 × 10⁻¹⁹ C。F = 0.50 × 1.60 × 10⁻¹⁹ × 2.0 × 10⁷ = 1.6 × 10⁻¹² N。该力提供向心力 mv²/r，因此 r = mv/(Be)。

Substituting the electron mass m = 9.11 × 10⁻³¹ kg: r = (9.11 × 10⁻³¹ × 2.0 × 10⁷) / (1.60 × 10⁻¹⁹ × 0.50) = (1.822 × 10⁻²³) / (8.0 × 10⁻²⁰) ≈ 2.28 × 10⁻⁴ m, or about 0.23 mm. This small radius reflects the tiny mass of the electron; the same field would produce a much larger radius for a heavier particle.

代入电子质量 m = 9.11 × 10⁻³¹ kg：r = (9.11 × 10⁻³¹ × 2.0 × 10⁷) / (1.60 × 10⁻¹⁹ × 0.50) = (1.822 × 10⁻²³)

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