📚 A-Level Physics Unit 2 (Jan 2021): Application Problem Techniques | A-Level物理Unit 2 (2021年1月) 应用题技巧
The January 2021 A-Level Physics Unit 2 paper demanded a strong command of application skills across electricity, waves, materials and quantum phenomena. Many questions moved beyond straightforward recall, requiring students to interpret graphs, manipulate equations in unfamiliar settings and justify practical procedures. This guide breaks down the most effective techniques for tackling the question types that appeared that session.
2021年1月的A-Level物理Unit 2试卷对电学、波、材料及量子现象的应用能力提出了很高要求。大量题目不再停留在简单的公式记忆，而是需要解读图像、在陌生情境中变换方程并论证实验步骤。本文逐一拆解应对该次试卷典型应用题的核心技法。
1. Mastering Internal Resistance and EMF Calculations | 掌握内阻与电动势计算
The relationship ε = I(R + r) is central. In one January 2021 problem, a cell was connected to a variable resistor and terminal p.d. V was recorded against current I. Plotting V on the y‑axis and I on the x‑axis yields a straight line of gradient –r and y‑intercept ε. A frequent trick is to ask for the power delivered to the load: P = I²R, maximised when R = r. Differentiate P = ε²R/(R+r)² or simply state the condition for maximum power transfer. Always convert current to amps and voltage to volts before substituting.
关系式 ε = I(R + r) 是核心。2021年1月有道题把电池与可变电阻连接，记录端电压 V 随电流 I 的变化。以 V 为纵轴、I 为横轴绘图，得到一条斜率为 –r、截距为 ε 的直线。试题常延伸到负载功率 P = I²R，且当 R = r 时输出功率最大。可对 P = ε²R/(R+r)² 求导或直接引用最大功率传输条件。代值前务必将电流统一为安培、电压为伏特。
2. Potential Divider Circuits and Sensor Applications | 电位器电路与传感器应用
A potential divider made from a fixed resistor and a thermistor or LDR is a staple. The output voltage is V_out = V_in × R₂/(R₁+R₂). The January 2021 paper asked candidates to explain how an LDR‑based circuit could act as a light‑activated switch. The technique hinges on identifying whether the sensor sits in the R₂ position: as light intensity falls, the LDR resistance increases, so V_out rises. Use this rise to forward‑bias a transistor or trigger a comparator. Draw clear circuit diagrams and label the variable resistor used to set the switching threshold.
由定值电阻与热敏电阻或光敏电阻组成的分压电路是常见考题，输出电压 V_out = V_in × R₂/(R₁+R₂)。2021年1月卷要求解释基于光敏电阻的电路如何实现光控开关。关键在于识别传感器是否充当 R₂：光强减弱时光敏电阻阻值上升，V_out 随之升高。利用该电压升高使晶体管正偏或触发比较器即可。画图需清晰，并标出用于设定阈值的可变电阻。
3. Stress, Strain and Young Modulus in Real Contexts | 应力、应变与杨氏模量实际情境
Questions often supply a force–extension graph for a wire and ask for the Young modulus E. Use stress = F/A, strain = ΔL/L₀, hence E = (F/A) / (ΔL/L₀). From the graph, pick two points in the straight‑line region to calculate the gradient ΔF/ΔL. The diameter given in millimetres must be converted to metres: A = πd²/4. The January 2021 session included a co‑pper wire where area conversion from mm² to m² caught many out—remember 1 mm² = 10⁻⁶ m². Express E in pascals and never forget to treat the wire as homogeneous.
题目常提供金属丝的力–伸长图并要求求出杨氏模量 E。用应力 = F/A、应变 = ΔL/L₀，则 E = (F/A) / (ΔL/L₀)。在图线直线段取两点计算斜率 ΔF/ΔL。直径须从毫米化为米：A = πd²/4。2021年1月涉及铜丝，众多考生在面积的 mm² 与 m² 换算上出错——牢记 1 mm² = 10⁻⁶ m²。杨氏模量单位为帕斯卡，并须假设丝材均匀。
4. Interpreting Standing Wave Patterns on Strings | 弦上驻波图样解读
For a string fixed at both ends, the nth harmonic frequency is fₙ = (n/2L)√(T/μ), with n = 1,2,3,… The January 2021 paper showed a diagram with several antinodes and asked students to identify the harmonic number and predict the new frequency if the tension were doubled. Count the antinodes: a pattern with three antinodes corresponds to n = 3. Because f ∝ √T, doubling the tension increases frequency by a factor of √2 ≈ 1.41. Also be prepared to calculate μ from mass and length: μ = mass/length, using kilograms per metre.
两端固定的弦上，第 n 次谐频 fₙ = (n/2L)√(T/μ)，n = 1,2,3,…。2021年1月有一题画出含有几个波腹的图样，要求判断谐频阶数并预报张力加倍后的频率。数出波腹数：三个波腹即为 n = 3。利用 f ∝ √T，张力加倍使得频率变至 √2 ≈ 1.41 倍。还要会用 μ = 质量/长度 计算线密度，单位采用千克每米。
5. Using the Diffraction Grating Equation | 使用衍射光栅方程
The grating equation nλ = d sinθ is vital. January 2021 gave the number of lines per millimetre (e.g. 500 lines/mm). Convert to lines per metre: N = 500 × 10³ m⁻¹, then d = 1/N = 2.0 × 10⁻⁶ m. To find the angle for the second‑order blue line (say λ = 450 nm = 4.5 × 10⁻⁷ m), use sinθ = nλ/d. Check whether sinθ ≤ 1; if n is too large, that order is not observable. A common supplementary question asks for the highest observable order nₘₐₓ = floor(d/λ). Always work in consistent metres.
光栅方程 nλ = d sinθ 至为关键。2021年1月给出了每毫米刻线数（如 500 条/毫米）。先化为每米线数 N = 500 × 10³ m⁻¹，得光栅常数 d = 1/N = 2.0 × 10⁻⁶ m。求二级蓝光（设 λ = 450 nm = 4.5 × 10⁻⁷ m）的衍射角时用 sinθ = nλ/d，并检查 sinθ ≤ 1；若 n 过大则该级不可见。常见追问是最大可见级数 nₘₐₓ = floor(d/λ)，全程采用米制单位。
6. Photoelectric Effect: Kinetic Energy and Stopping Potential | 光电效应：动能与遏止电压
Einstein’s photoelectric equation hf = φ + KE_max links frequency and kinetic energy. The January 2021 paper provided a graph of KE_max (or stopping potential V_s) against frequency f. The gradient of KE_max vs f graph is h; the x‑intercept gives the threshold frequency f₀ = φ/h. To convert between KE_max in joules and eV, use 1 eV = 1.6 × 10⁻¹⁹ J. A pitfall is assuming intensity affects KE_max—it does not, as only frequency determines the photon energy. When asked about saturation current, link it to the number of photons, hence intensity.
爱因斯坦光电方程 hf = φ + KE_max 联系频率与动能。2021年1月给出了 KE_max（或遏止电压 V_s）对频率 f 的图线。KE_max–f 图的斜率即为 h，x 轴截距为截止频率 f₀ = φ/h。焦耳与电子伏的换算：1 eV = 1.6 × 10⁻¹⁹ J。常见误区是以为光强会影响最大动能——实际上它只由频率决定。涉及饱和电流时，则将其与光子数（即光强）相关联。
7. Energy Levels and Photon Emission Calculations | 能级与光子发射计算
When an electron drops from level E₂ to E₁, the photon energy is ΔE = |E₂ – E₁| = hf = hc/λ. The January 2021 diagram showed several energy values in eV. To find the wavelength of emitted light: compute ΔE in eV, convert to joules (× 1.6 × 10⁻¹⁹), then λ = hc/ΔE using h = 6.63 × 10⁻³⁴ J·s and c = 3.00 × 10⁸ m·s⁻¹. State the region of the electromagnetic spectrum—for λ ≈ 4.5 × 10⁻⁷ m, it lies in the visible blue region. Pay attention to significant figures and unit prefixes (nm).
电子从能级 E₂ 跃迁至 E₁ 时，光子能量 ΔE = |E₂ – E₁| = hf = hc/λ。2021年1月图中给出若干 eV 值。求波长时，先算出以 eV 为单位的能级差，转为焦耳（× 1.6 × 10⁻¹⁹），再用 λ = hc/ΔE，h = 6.63 × 10⁻³⁴ J·s，c = 3.00 × 10⁸ m·s⁻¹。判断光谱区域——若 λ ≈ 4.5 × 10⁻⁷ m，属于可见光蓝色区。注意有效数字与单位前缀（nm）。
8. Combining Resistors and Equivalent Resistance | 组合电阻与等效电阻
Mixed networks of series and parallel resistors appear regularly. Series: Rₜₒₜ = R₁ + R₂ + … ; parallel: 1/Rₜₒₜ = 1/R₁ + 1/R₂ + … . The January 2021 circuit required three reduction steps. Begin by combining the two parallel resistors to their equivalent Rₚ = (R₁R₂)/(R₁+R₂), then add the series resistor. Redraw the circuit after each step and label the equivalent resistance. Once the total resistance is known, apply V = IR to find the current from the supply, then work backwards to determine branch currents and p.d.s across each component.
串并联混合网络是常规题型。串联：Rₜₒₜ = R₁ + R₂ + …；并联：1/Rₜₒₜ = 1/R₁ + 1/R₂ + …。2021年1月电路需三步化简：先处理两个并联电阻得出等效电阻 Rₚ = (R₁R₂)/(R₁+R₂)，再与串联电阻相加。每化简一步重画电路并标出等效阻值。求得总电阻后，用 V = IR 获得干路电流，反向推算各支路电流与各元件电压。
9. Applying Kirchhoff’s Laws to Multi-loop Circuits
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