Year 13 CIE Physics Unit Test Mock Exam Analysis | Year 13 CIE 物理单元测试模拟卷解析

📚 Year 13 CIE Physics Unit Test Mock Exam Analysis | Year 13 CIE 物理单元测试模拟卷解析

Mock exams are a vital tool for reinforcing core concepts and honing exam technique in Year 13 CIE Physics. This article breaks down a typical unit test paper, covering key topics from circular motion to nuclear physics. Each section presents a challenging question followed by a step-by-step solution, helping you master the reasoning behind the answers.

模拟考试是巩固 Year 13 CIE 物理核心概念和磨练应试技巧的重要手段。本文剖析一份典型的单元测试卷，涵盖从圆周运动到核物理的关键主题。每一小节呈现一道具有挑战性的题目，并给出逐步解析，帮助你掌握解题背后的逻辑。

1. Circular Motion: Conical Pendulum | 圆周运动：圆锥摆

A conical pendulum consists of a mass of 0.50 kg tied to a light string of length 1.2 m. The mass moves in a horizontal circle with the string at 30° to the vertical. Calculate the period of revolution. (Take g = 9.81 m s⁻²)

一个圆锥摆由质量为 0.50 kg 的小球拴在一根长 1.2 m 的轻绳上构成。小球在水平面内做圆周运动，绳与竖直方向夹角为 30°。计算旋转周期。（取 g = 9.81 m·s⁻²）

Resolve forces radially and vertically. The tension T provides the centripetal force via its horizontal component: T sinθ = mrω². The vertical component balances the weight: T cosθ = mg. The radius of the circle is r = L sinθ. Dividing the two equations gives tanθ = (rω²)/g = (L sinθ · ω²)/g, which simplifies to ω² = g/(L cosθ). Thus ω = √(g/(L cosθ)). The period T_p = 2π/ω = 2π √(L cosθ/g). Substituting L = 1.2 m, θ = 30°, cos30° = √3/2 ≈ 0.866, we get T_p = 2π √(1.2 × 0.866 / 9.81) ≈ 2π √(0.106) ≈ 2π × 0.325 = 2.04 s.

沿径向和竖直方向分解力。绳的张力 T 的水平分力提供向心力：T sinθ = mrω²；竖直分力与重力平衡：T cosθ = mg。圆周半径 r = L sinθ。两式相除得 tanθ = (rω²)/g = (L sinθ · ω²)/g，化简便得 ω² = g/(L cosθ)。因此 ω = √(g/(L cosθ))。周期 T = 2π/ω = 2π √(L cosθ/g)。代入 L = 1.2 m，θ = 30°，cos30° = √3/2 ≈ 0.866，得 T = 2π √(1.2 × 0.866 / 9.81) ≈ 2π √(0.106) ≈ 2π × 0.325 = 2.04 s。

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2. Gravitational Field: Satellite Total Energy | 引力场：卫星总能量

A satellite of mass 500 kg orbits Earth at a height where the local gravitational field strength is 8.0 N kg⁻¹. Given Earth’s radius R = 6400 km, determine the satellite’s total mechanical energy.

一颗质量为 500 kg 的卫星在当地引力场强度为 8.0 N·kg⁻¹ 的高度上绕地球运行。已知地球半径 R = 6400 km，求卫星的总机械能。

At the orbital radius r, g_local = GM/r² = 8.0 N kg⁻¹. We find r from g_local/g_surface = (R/r)², with g_surface ≈ 9.81 N kg⁻¹. So r = R √(g_surface/g_local) = 6400 km × √(9.81/8.0) ≈ 6400 × 1.107 = 7085 km = 7.085×10⁶ m.

在轨道半径 r 处，当地的 g = GM/r² = 8.0 N·kg⁻¹。由 g/g₀ = (R/r)²，地面 g₀ ≈ 9.81 N·kg⁻¹，得 r = R √(g₀/g) = 6400 km × √(9.81/8.0) ≈ 6400 × 1.107 = 7085 km = 7.085×10⁶ m。

For a circular orbit, centripetal force is provided by gravity: mv²/r = mg_local, so v² = g_local · r. Kinetic energy KE = ½mv² = ½ m g_local r. Gravitational potential energy U = –GMm/r = –(g_local r²) m / r = –m g_local r. Therefore total mechanical energy E_total = KE + U = –½ m g_local r. Substitute: E_total = –0.5 × 500 × 8.0 × 7.085×10⁶ = –1.417×10¹⁰ J (≈ –1.42×10¹⁰ J). The negative sign indicates a bound orbit.

对于圆轨道，向心力由引力提供：mv²/r = mg，故 v² = g r。动能 KE = ½mv² = ½ m g r。引力势能 U = –GMm/r = –(g r²)m/r = –m g r。因此总机械能 E_total = KE + U = –½ m g r。代入数据：E_total = –0.5 × 500 × 8.0 × 7.085×10⁶ = –1.417×10¹⁰ J（≈ –1.42×10¹⁰ J）。负号表示束缚轨道。

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3. Simple Harmonic Motion: Damped Oscillator | 简谐运动：阻尼振荡

A damped mass–spring system has period 2.0 s and mass 0.20 kg. Its amplitude falls from 10 cm to 5 cm in 20 complete oscillations. Assuming exponential decay A = A₀ e⁻ᵞᵗ, find the damping constant γ and the damping coefficient b.

一个阻尼弹簧振子周期为 2.0 s，质量 0.20 kg。经过 20 次全振动后，振幅由 10 cm 降至 5 cm。假设振幅按 A = A₀ e⁻ᵞᵗ 指数衰减，求阻尼常数 γ 与阻尼系数 b。

The time for 20 oscillations is t = 20 × 2.0 = 40 s. The amplitude ratio gives 0.5 = e⁻ᵞ⁴⁰, so γ = ln 2 / 40 = 0.693 / 40 ≈ 0.0173 s⁻¹. The damping coefficient (force per unit velocity) is b = 2mγ = 2 × 0.20 × 0.0173 = 6.92×10⁻³ kg s⁻¹. The quality factor Q = ω₀/2γ = (π/T)/(2γ) = π/(T·2γ) ≈ π/(2.0 × 0.0346) ≈ 45.4, indicating moderate damping.

20 次振动所用的时间 t = 20 × 2.0 = 40 s。振幅比 0.5 = e⁻ᵞ⁴⁰，故 γ = ln 2 / 40 = 0.693/40 ≈ 0.0173 s⁻¹。阻尼系数（单位速度所受阻力）为 b = 2mγ = 2 × 0.20 × 0.0173 = 6.92×10⁻³ kg·s⁻¹。品质因子 Q = ω₀/2γ = (π/T)/(2γ) ≈ π/(2.0 × 0.0346) ≈ 45.4，属于中等阻尼。

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4. Thermal Physics: RMS Speed of Helium | 热学：氦的方均根速率

Calculate the root-mean-square speed of helium atoms at 300 K. Mass of a helium atom = 6.64×10⁻²⁷ kg, Boltzmann constant k = 1.38×10⁻²³ J K⁻¹.

计算 300 K 下氦原子的方均根速率。氦原子质量 = 6.64×10⁻²⁷ kg，玻尔兹曼常量 k = 1.38×10⁻²³ J·K⁻¹。

The kinetic theory gives average translational kinetic energy per particle: ½ m ⟨c²⟩ = (3/2) kT. Hence the mean square speed ⟨c²⟩ = 3kT / m, and c_rms = √(3kT/m). Substitute: c_rms = √(3 × 1.38×10⁻²³ × 300 / 6.64×10⁻²⁷) = √(1.242×10⁻²⁰ / 6.64×10⁻²⁷) = √(1.87×10⁶) ≈ 1.37×10³ m s⁻¹ (1368 m s⁻¹). This high speed reflects the low atomic mass of helium.

根据分子动理论，每个粒子的平均平移动能为 ½ m ⟨c²⟩ = (3/2) kT。因此均方速率 ⟨c²⟩ = 3kT/m，方均根速率 c_rms = √(3kT/m)。代入数据：c_rms = √(3 × 1.38×10⁻²³ × 300 / 6.64×10⁻²⁷) = √(1.242×10⁻²⁰ / 6.64×10⁻²⁷) = √(1.87×10⁶) ≈ 1.37×10³ m·s⁻¹（1368 m·s⁻¹）。如此高的速率源于氦的原子质量很小。

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5. Electric Fields: Electron Deflection in a Uniform Field | 电场：电子在匀强电场中的偏转

An electron is projected horizontally at 2.0×10⁷ m s⁻¹ into a uniform electric field of strength 5.0 kV m⁻¹ directed vertically downward between two parallel plates. Find its vertical deflection after travelling a horizontal distance of 4.0 cm. (e = 1.60×10⁻¹⁹ C, m_e = 9.11×10⁻³¹ kg)

一电子以 2.0×10⁷ m·s⁻¹ 的水平初速度射入竖直向下的匀强电场，场强为 5.0 kV·m⁻¹，两极板平行。求电子水平前进 4.0 cm 后的竖直偏转量。（e = 1.60×10⁻¹⁹ C，m_e = 9.11×10⁻³¹ kg）

The electron experiences a constant vertical force F = eE, giving acceleration a_y = eE/m. Time to travel 4.0 cm horizontally: t = d / v_x = 0.040 / 2.0×10⁷ = 2.0×10⁻⁹ s. Initial vertical velocity is zero, so vertical displacement y = ½ a_y t² = ½ (eE/m) t². E = 5.0×10³ V m⁻¹. Thus y = 0.5 × (1.60×10⁻¹⁹ × 5.0×10³ / 9.11×10⁻³¹) × (2.0×10⁻⁹)² = 0.5 × (8.0×10⁻¹⁶ / 9.11×10⁻³¹) × 4.0×10⁻¹⁸ = 0.5 × (8.78×10¹⁴) × 4.0×10⁻¹⁸ = 0.5 × 3.512×10⁻³ = 1.76×10⁻³ m ≈ 1.8 mm downwards.

电子受到恒定的竖直力 F = eE，产生竖直加速度 a_y = eE/m。水平前进 4.0 cm 所需时间：t = d / v_x = 0.040 / 2.0×10⁷ = 2.0×10⁻⁹ s。竖直初速度为零，故竖直位移 y = ½ a_y t² = ½ (eE/m) t²。场强 E = 5.0×10³ V·m⁻¹。计算得 y = 0.5 × (1.60×10⁻¹⁹ × 5.0×10³ / 9.11×10⁻³¹) × (2.0×10⁻⁹)² = 0.5 × (8.78×10¹⁴) × 4.0×10⁻¹⁸ = 1.76×10⁻³ m ≈ 1.8 mm，方向向下。

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6. Capacitance: Charge Sharing and Energy Loss | 电容：电荷共享与能量损失

A 10 μF capacitor charged to 200 V is disconnected from the supply and connected in parallel with an uncharged 5 μF capacitor. Find the final common voltage and the total energy stored before and after connection. Explain the energy loss.

一个充电至 200 V 的 10 μF 电容器脱离电源，然后与一个未充电的 5 μF 电容器并联。求并联后的最终电压，以及连接前后储存的总能量，并解释能量损失原因。

Initial charge Q₀ = C₁V₀ = 10×10⁻⁶ × 200 = 2.0×10⁻³ C. After connection, total capacitance C_total = C₁ + C₂ = 15 μF. Charge is conserved, so final voltage V_f = Q₀ / C_total = 2.0×10⁻³ / 15×10⁻⁶ = 133.3 V. Initial energy E_i = ½ C₁ V₀² = 0.5 × 10⁻⁵ × 40000 = 0.20 J. Final energy E_f = ½ C_total V_f² = 0.5 × 15×10⁻⁶ × (400/3)² = 7.5×10⁻⁶ × (160000/9) = 7.5×10⁻⁶ × 17777.8 ≈ 0.1333 J. Energy lost ≈ 0.0667 J. The lost energy is dissipated as heat in the connecting wires and as electromagnetic radiation due to the sudden charge redistribution.

初始电荷 Q₀ = C₁V₀ = 10×10⁻⁶ × 200 = 2.0×10⁻³ C。并联后总电容 C_total = 15 μF。电荷守恒，最终电压 V_f = Q₀ / C_total = 133.3 V。初始能量 E_i = ½ C₁V₀² =

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