📚 Mock Test Walkthrough for AQA Year 13 Physics Unit Test | AQA 十三年级物理单元测试模拟卷解析
Mock tests are an essential part of AQA Year 13 Physics preparation, helping you bridge the gap between knowing the theory and applying it under timed conditions. This walkthrough covers a selection of typical questions from mechanics, fields, thermodynamics, nuclear physics and kinetic theory, highlighting the key reasoning and common pitfalls for each.
模拟测试是 AQA 十三年级物理复习的重要环节，它能帮助你从理解理论跨越到在限时压力下灵活运用。本文解析的选题涵盖力学、场、热力学、核物理与分子运动论，逐一展示关键解题思路和常见易错点。
1. Circular Motion and Orbital Period | 圆周运动与轨道周期
A satellite of mass m orbits a planet of mass M in a circular orbit of radius r. Multiple-choice options give the orbital period as A: 2π√r/GM, B: 2π√r²/GM, C: 2π√r³/GM, D: 2π√r⁴/GM ipetal force: GMm/r² = m v²/r. Substituting v = 2πr/T leads to T² = 4π² r³/(GM), so T = 2π√r³/GM.
一颗质量为 m 的卫星绕质量为 M 的行星做半径为 r 的圆周运动。选择题的周期表达式为 A: 2π√r/GM, B: 2π√r²/GM, C: 2π√r³/GM, D: 2π√r⁴/GM。正确答案是 C。万有引力提供向心力：GMm/r² = m v²/r，代入 v = 2πr/T，得 T² = 4π² r³/(GM)，即 T = 2π√r³/GM。
2. Maximum Kinetic Energy in SHM | 简谐运动中的最大动能
etic energy = ½ m vₘₐₓ² = ½ × 0.50 × (0.10π)² = 0.025π² J ≈ 0.25 J. This also equals the total mechanical energy when the potential energy is zero at equilibrium.
一个0.50 kg的物体以振幅0.10 m、周期2.0 s做简谐运动。角频率 ω = 2π/T = π rad s⁻¹，最大速度 vₘₐₓ = ωA = 0.10π m s⁻¹。最大动能 = ½ m vₘₐₓ² = ½ × 0.50 × (0.10π)² = 0.025π² J ≈ 0.25 J。平衡位置势能为零，此动能即为系统的总机械能。
3. First Law of Thermodynamics Sign Convention | 热力学第一定律中的符号约定
AQA uses ΔU = Q + W, where W is the work done on + 150 = +80 J, so the internal energy increases by 80 J, consistent with a temperature rise.
AQA 采用的公式为 ΔU = Q + W，其中 W 代表对系统做的功。理想气体被压缩，外界对气体做功 150 J，同时气体向环境放热 70 J。因此 W = +150 J，Q = -70 J。ΔU = (-70) + 150 = +80 J，内能增加 80 J，气体温度升高。
4. Orbital Transfers and Gravitational Potential | 轨道转移与引力势能
To raise a satellite from a lower circular orbit r₁ to a higher one r₂, energy must be supplied. Gravitational potential energy is U = -GMm/r. When r 2r) also increases. Although the orbital speed and kinetic energy decrease, the rise in potential energy dominates, so external work is required.
将卫星从低轨道 r₁ 转移到高轨道 r₂ 需要补充能量。引力势能 U = -GMm/r，半径增大时势能负值变小，即势能升高。总机械能 E = -GMm/(2r) 也增大。尽管轨道速度和动能减小，但势能的增加更大，因此必须由外力做功提供能量。
5. Electric Potential at the Midpoint of a Dipole | 偶极子中点电势
ch charge is 0.10 m from the midpoint. V₊ = (8.99×10⁹ × 2.0×10⁻⁶) / 0.10 = 1.80×10⁵ V; V₋ = -1.80×10⁵ V. The net potential is zero. Note that the electric field at the midpoint is not zero, but the potential is.
两个点电荷 +2.0 μC 和 -2.0 μC 相距 0.20 m。中点电势为各电荷电势的标量和。两电荷距中点均为 0.10 m，正电荷贡献 +1.80×10⁵ V，负电荷贡献 -1.80×10⁵ V，总和为零。需注意，中点的电场不为零，但电势为零。
6. Determining Time Constant from a Capacitor Discharge Graph | 由电容器放电图像求时间常数
A capacitor discharges through a resistor, and the current– τ = t₁/₂ / ln 2 ≈ 4.0 / 0.693 = 5.8 s. Alternatively, read the time at which I drops to 37% of I₀ (≈37 μA), which gives τ directly.
电容器通过电阻放电，电流-时间图像呈指数衰减。初始电流 I₀ = 100 μA，经过一个半衰期降至 50 μA，对应时间 t₁/₂ = 4.0 s。时间常数 τ = t₁/₂ / ln 2 ≈ 4.0 / 0.693 = 5.8 s。也可在电流降至初始值 37%（约37 μA）处直接读取 τ。
7. Electron Motion in a Uniform Magnetic Field | 电子在匀强磁场中的运动
An electron (charge e) enters a uniform magnetic field B at speed v perpendicular to the field. The magnetic force F = Bev acts as a centripetal force: Bev = mv²/r, giving the orbital radius r = mv/(eB). The period of circular motion is T = 2πr/v = 2πm/(eB), which is independent of v. This is why all particles with the same charge-to-mass ratio take the same time to complete a loop.
电子以速度 v 垂直进入匀强磁场 B，受到的洛伦兹力 Bev 提供向心力：Bev = mv²/r，得轨道半径 r = mv/(eB)。圆周运动的周期 T = 2πr/v = 2πm/(eB)，与速度无关。所以荷质比相同的粒子旋转一圈所需时间相等。
8. Alpha Decay and Mass–Energy Equivalence | α衰变与质能方程
Uranium-238 undergoes alpha decay as follows: ²³⁸₉₂U →
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