Tag: Physics

📚 A-Level Physics Key Derivations from June 2018 Examiner’s Report | A-Level物理核心推导：2018年6月考官报告解读

The June 2018 A-Level Physics Examiner’s Report highlighted that many students lose marks not because they do not know the final formula, but because they cannot show a clear, logical derivation. Mastering derivations is essential for securing top grades in the written papers, especially in questions that explicitly ask “show that” or “derive”. This article revisits several key derivations and consolidates the examiner’s advice on how to present them effectively.

2018年6月的A-Level物理考官报告指出，许多学生丢分并非因为不知道最终公式，而是因为他们无法展示清晰、有逻辑的推导过程。掌握推导过程对于在笔试中取得高分至关重要，尤其是对于那些明确要求“证明”或“推导”的题目。本文将回顾几个重要的推导，并汇总考官对于如何有效展示推导过程的建议。

1. Importance of Derivation in A-Level Physics | A-Level物理中推导的重要性

Derivation questions test your understanding of the underlying principles, not just memory. The examiner expects you to start from fundamental definitions or laws, such as Newton’s second law or definitions of velocity and acceleration, and then mathematically manipulate them to reach the required expression.

推导题考查的是你对底层原理的理解，而不仅仅是记忆。考官期望你从基本定义或定律出发，例如牛顿第二定律或速度和加速度的定义，然后通过数学运算得出所需的表达式。

2. Examiner’s Advice from June 2018 | 2018年6月考官的忠告

The report stressed that many candidates omitted intermediate steps or used “shortcuts” that were not justified. Examiners recommended showing all algebraic steps, stating assumptions (e.g., constant acceleration, no air resistance), and clearly labelling each equation.

报告强调，许多考生省略了中间步骤，或者使用了未说明理由的“捷径”。考官建议展示所有代数步骤，陈述假设（如匀加速度、无空气阻力），并清楚地标记每个方程。

3. Deriving the SUVAT Equations | 推导匀加速运动方程

Start from the definition of acceleration: a = (v − u) / t. Rearranging gives v = u + at. This is the first equation.

从加速度的定义出发：a = (v − u) / t。整理得 v = u + at。这是第一个方程。

To derive s = ut + ½ at², use average velocity: average velocity = (u + v)/2, and displacement s = average velocity × time = ((u + v)/2) t. Substitute v = u + at: s = ((u + u + at)/2) t = (2u + at)/2 × t = ut + ½ at².

要推导 s = ut + ½ at²，使用平均速度：平均速度 = (u + v)/2，位移 s = 平均速度 × 时间 = ((u + v)/2) t。代入 v = u + at：s = ((u + u + at)/2) t = (2u + at)/2 × t = ut + ½ at²。

Then v² = u² + 2as can be derived by eliminating t: from v = u + at, t = (v − u)/a. Substitute into s = ut + ½ at²: s = u((v−u)/a) + ½ a((v−u)/a)² → multiply by 2a: 2as = 2u(v−u) + (v−u)² = 2uv − 2u² + v² − 2uv + u² = v² − u², hence v² = u² + 2as.

然后可以通过消去t推导 v² = u² + 2as：由 v = u + at 得 t = (v − u)/a。代入 s = ut + ½ at²：s = u((v−u)/a) + ½ a((v−u)/a)² → 两边乘以2a：2as = 2u(v−u) + (v−u)² = 2uv − 2u² + v² − 2uv + u² = v² − u²，因此 v² = u² + 2as。

v = u + at, s = (u+v)/2 × t, s = ut + ½ at², v² = u² + 2as

4. Deriving Kinetic Energy (½mv²) | 推导动能公式 (½mv²)

Consider a constant net force F accelerating an object of mass m from rest to speed v over displacement s. Work done = F s. Using Newton’s second law F = ma, and v² = u² + 2as with u=0 gives v² = 2as → a = v²/(2s). Then F = m × v²/(2s). Work done = (m v²/(2s)) × s = ½ m v². This work is stored as kinetic energy.

考虑恒定合外力 F 使质量为 m 的物体从静止加速到速度 v，位移为 s。做功 = F s。利用牛顿第二定律 F = ma，以及 v² = u² + 2as，u=0 得 v² = 2as → a = v²/(2s)。则 F = m × v²/(2s)。功 = (m v²/(2s)) × s = ½ m v²。这个功储存为动能。

Examiners noted that many students incorrectly assume KE = mv² without derivation. Always show the link to work–energy principle.

考官注意到，许多学生错误地假设 KE = mv² 而不加以推导。务必展示功-能原理的关联。

5. Deriving Centripetal Acceleration a = v²/r | 推导向心加速度 a = v²/r

Consider an object moving in a circle of radius r with constant speed v. In a short time Δt, the object moves from point A to B, subtending angle Δθ. The change in velocity vector Δv points toward the centre. Magnitude of Δv ≈ v Δθ (since triangle of velocities is similar to triangle of radii). Δθ = v Δt / r. Thus acceleration a = Δv/Δt = v (v Δt / r)/Δt = v²/r.

考虑一物体在半径为 r 的圆周上以恒定速率 v 运动。在短时间 Δt 内，物体从 A 点运动到 B 点，圆心角为 Δθ。速度矢量的变化 Δv 指向圆心。Δv 的大小 ≈ v Δθ（因为速度矢量三角形与半径三角形相似）。Δθ = v Δt / r。因此加速度 a = Δv/Δt = v (v Δt / r)/Δt = v²/r。

The rigorous derivation uses vector calculus, but the geometric approach is acceptable at A-Level. Always state the assumption of constant speed.

严格推导使用矢量微积分，但在A-Level中几何方法是可以接受的。务必说明速率恒定的假设。

6. Deriving Projectile Range on Horizontal Ground | 推导水平地面上抛体射程

For a projectile launched at speed u at angle θ to horizontal, resolve: u_x = u cos θ, u_y = u sin θ. Time of flight t = 2 u_y / g = 2u sin θ / g. Horizontal range R = u_x × t = (u cos θ) × (2u sin θ / g) = (u² × 2 sin θ cos θ)/g = (u² sin 2θ)/g. Maximum range occurs at θ = 45°.

对于以速度 u、与水平成 θ 角发射的抛体，分解：u_x = u cos θ, u_y = u sin θ。飞行时间 t = 2 u_y / g = 2u sin θ / g。水平射程 R = u_x × t = (u cos θ) × (2u sin θ / g) = (u² × 2 sin θ cos θ)/g = (u² sin 2θ)/g。最大射程出现在 θ = 45°。

The June 2018 report highlighted that many students forgot to double the time to peak, using only u_y/g. Always derive time of flight from vertical motion.

2018年6月的报告指出，许多学生忘记将到达顶点的时间翻倍，只用了 u_y/g。务必从竖直运动推导飞行时间。

7. Deriving Capacitor Discharge Equation Q = Q₀ e^−t/(RC) | 推导电容放电方程 Q = Q₀ e^−t/(RC)

During discharge, current I = −dQ/dt, and I = V/R, also V = Q/C. So −dQ/dt = Q/(RC). Separating variables: dQ/Q = −1/(RC) dt. Integrate both sides: ln Q = −t/(RC) + constant. At t=0, Q=Q₀, so constant = ln Q₀. Thus ln(Q/Q₀) = −t/(RC), or Q = Q₀ e^−t/(RC).

放电过程中，电流 I = −dQ/dt，同时 I = V/R，且 V = Q/C。因此 −dQ/dt = Q/(RC)。分离变量：dQ/Q = −1/(RC) dt。两边积分：ln Q = −t/(RC) + 常数。t=0 时 Q=Q₀，故常数 = ln Q₀。因此 ln(Q/Q₀) = −t/(RC)，即 Q = Q₀ e^−t/(RC)。

Examiners expected clear separation of variables and explicit limits of integration. Missing the negative sign is a common mistake.

考官期望清晰的变量分离和明确的积分限。遗漏负号是一个常见错误。

8. Deriving F = ma from Momentum | 由动量推导 F = ma

Newton’s second law: Force = rate of change of momentum. Momentum p = mv. So F = dp/dt = d(mv)/dt. If mass is constant, F = m dv/dt = m a. This derivation is often required when linking laws.

牛顿第二定律：力 = 动量变化率。动量 p = mv。因此 F = dp/dt = d(mv)/dt。如果质量恒定，F = m dv/dt = m a。这个推导在联系各定律时经常需要。

The report emphasized that candidates should state the assumption of constant mass; otherwise, the product rule must be used.

报告强调，考生应说明质量恒定的假设；否则必须使用乘积法则。

9. Deriving the Period of a Simple Pendulum (T = 2π√(L/g)) | 推导单摆周期 T = 2π√(L/g)

For small angles, restoring force F = −mg sin θ ≈ −mg θ. Using arc displacement x = Lθ, acceleration a = − (g/L) x. This is SHM with ω² = g/L. Period T = 2π/ω = 2π √(L/g).

对于小角度，回复力 F = −mg sin θ ≈ −mg θ。利用弧位移 x = Lθ，加速度 a = − (g/L) x。这是简谐运动，ω² = g/L。周期 T = 2π/ω = 2π √(L/g)。

Examiners noted that credit is lost if students omit the small-angle approximation. Show that sin θ ≈ θ is valid when θ is in radians and small.

考官指出，如果学生遗漏了小角度近似，就会失分。应展示当 θ 以弧度为单位且很小时，sin θ ≈ θ 成立。

10. Deriving the Ideal Gas Equation from Kinetic Theory | 从分子动理论推导理想气体方程

Consider a single molecule in a cubic box of side L. Change in momentum per collision on one wall = 2mv_x. Time between collisions = 2L/v_x. Force on wall = (2mv_x) / (2L/v_x) = mv_x²/L. For N molecules, total force = (m/L) Σ v_x². Pressure p = Force/Area = (m/L³) Σ v_x² = (m/V) Σ v_x². With average velocity squared and using Σ v² = 3 Σ v_x², pV = (1/3) N m ⟨v²⟩. Then with mean kinetic energy = ½ m ⟨v²⟩ = (3/2) kT, we get pV = NkT or pV = nRT.

考虑边长为 L 的立方体中的一个分子。每次与壁碰撞的动量变化 = 2mv_x。碰撞间隔时间 = 2L/v_x。对壁的力 = (2mv_x) / (2L/v_x) = mv_x²/L。对 N 个分子，总力 = (m/L) Σ v_x²。压强 p = 力/面积 = (m/L³) Σ v_x² = (m/V) Σ v_x²。利用速度平方平均值，并由 Σ v² = 3 Σ v_x²，得 pV = (1/3) N m ⟨v²⟩。再由平均动能 = ½ m ⟨v²⟩ = (3/2) kT，可得 pV = NkT 或 pV = nRT。

Although this derivation is not always required in full, the examiner’s report indicated that understanding the steps helps in answering conceptual questions.

尽管这个推导并不总是需要完整给出，但考官报告指出，理解

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📚 A-Level Physics June 2018 Paper 3: Experimental Determination of g | A-Level 物理 2018年6月卷3：重力加速度实验探究

The June 2018 A-Level Physics Paper 3 is a practical skills assessment designed to test candidates’ ability to plan experiments, collect and process data, and evaluate uncertainties. One typical investigation in this paper involves measuring the acceleration of free fall, g, using a steel ball falling through two light gates connected to a timer. The experiment cleverly eliminates the need to know the initial velocity by applying the kinematic equation s = ut + ½gt² and analysing a linear graph of s/t against t.

2018年6月A-Level物理试卷3是一份实验技能试卷，旨在考查考生设计实验、采集处理数据以及评估不确定度的能力。该试卷中一个典型的探究实验是利用钢球下落穿过两个连接计时器的光电门来测量重力加速度g。该实验巧妙地通过运动学方程s = ut + ½gt²，并分析s/t对t的线性图像，消除了必须已知初速度的困难。

1. Overview of the Investigation | 实验探究概述

In this investigation, a small steel ball is released from an electromagnet positioned a few centimetres above a first light gate. The ball interrupts the first light gate, starting a digital timer, and later interrupts a second light gate, stopping the timer. The distance s between the two light gates is varied, and the corresponding fall time t is recorded. Because the ball already has an unknown velocity u when it reaches the first light gate, the data are not directly analysed using s = ½gt². Instead, the equation is rearranged into a linear form: s/t = u + ½ g · t.

在该探究中，一个小钢球从一个位于第一光电门上方几厘米处的电磁铁释放。钢球先后遮断第一和第二个光电门，从而启动和停止数字计时器。改变两光电门之间的距离 s，记录相应的下落时间 t。由于钢球到达第一光电门时已具有未知的速度u，数据不能直接用s = ½gt²处理。因此，将方程变形为线性形式：s/t = u + ½ g · t。

By plotting a graph of s/t (on the y-axis) against t (on the x-axis), the gradient equals ½ g, and the y-intercept equals u. This allows g to be determined without needing to measure u or to ensure the ball is released from exactly zero initial speed at the first light gate. The method also provides a clear route for estimating the uncertainty in g from the best-fit and worst-fit lines.

通过绘制s/t（y轴）对t（x轴）的图像，斜率等于½ g，y轴截距等于u。这样既不需要测量u，也不要求钢球在第一光电门处的初速严格为零，就能够求得g。该方法还提供了通过最佳拟合线和最差拟合线估算g的不确定度的清晰途径。

2. Apparatus and Setup | 实验装置与搭建

The apparatus includes an electromagnet connected to a low-voltage DC supply, a steel ball bearing (approximately 1–2 cm diameter), two light gates interfaced with a digital timer (resolution 0.001 s or better), a metre rule and a vernier calliper for measuring s, a micrometer screw gauge for determining the ball’s diameter, a plumb line to ensure vertical alignment, and a retort stand with clamps to hold the light gates. The second light gate is fixed at a known height, while the first light gate can be moved to vary s.

实验装置包括：连接低压直流电源的电磁铁，一个直径约1–2 cm的钢球，两个与数字计时器（分辨率为0.001 s或更优）连接的光电门，测量s的米尺和游标卡尺，测量钢球直径的螺旋测微器，确保垂直对齐的铅垂线，以及用于固定光电门的铁架台和夹具。第二光电门固定在一个已知高度，第一光电门可以移动以改变s。

The electromagnet must hold the ball securely and release it cleanly without imparting any sideways motion. To minimise the effect of residual magnetism, the current is switched off rapidly using a toggle switch, and the ball is allowed to drop under gravity. The light gates are aligned so that the ball passes through the centre of the infrared beams, and the timer is set to start on the first interruption and stop on the second.

电磁铁必须牢固地吸附钢球，并在切断电流时干净地释放，不能对钢球施加任何侧向运动。为了减小剩磁的影响，使用拨动开关快速断开电流，让钢球在重力作用下下落。两个光电门需要对齐，使钢球通过红外光束的中心位置，计时器设定为第一次遮断时启动、第二次遮断时停止。

3. Experimental Procedure | 实验步骤

1. Use the micrometer to measure the diameter of the steel ball at several different orientations. Record at least five readings, calculate the mean diameter, and determine the absolute uncertainty (e.g. ± ½ range) for the diameter.

1. 使用螺旋测微器在多个不同方向测量钢球直径。至少记录五组读数，计算平均直径，并确定直径的绝对不确定度（例如，± ½ 极差）。

2. Set up the two light gates vertically, one below the other, with the electromagnet positioned just above the top light gate. Use the plumb line to check that the line through the centres of both light gates is vertical.

2. 将两个光电门垂直放置，一个在另一个的下方，电磁铁则紧贴在上方光电门的上方。使用铅垂线检查通过两个光电门中心点的连线是否竖直。

3. Place the first light gate such that the distance s between the two gates is at its maximum (e.g. 0.800 m). Measure s using a metre rule, ensuring your eye is level with the scale to avoid parallax error. Record the value of s with its uncertainty (e.g. ±0.001 m if using a precise scale).

3. 放置第一个光电门，使两光电门之间的距离s为最大预设值（例如0.800 m）。使用米尺测量s，确保视线与刻度齐平以避免视差。记录s值及其不确定度（如使用精密刻度可为±0.001 m）。

4. Switch on the electromagnet, attach the steel ball, and then switch off the electromagnet to release the ball. Record the time t displayed on the timer for the fall between the two light gates. Repeat this measurement three times under the same conditions to obtain t1, t2, t3, and calculate the mean time.

4. 接通电磁铁，吸附钢球，然后切断电磁铁电流释放钢球。记录计时器上显示的两个光电门之间的下落时间t。在相同条件下重复该测量三次，得到t1、t2、t3，并计算平均时间。

5. Decrease the distance s by moving the first light gate downward in about 0.100 m decrements (or as directed in the question). After each adjustment, remeasure s, and repeat the timing experiment as before. Record all readings in a suitable table.

5. 通过向下移动第一个光电门来减小距离s，每次大约减小0.100 m（或按照题目要求）。每次调整后重新测量s，并重复上述计时实验。将所有读数记录在合适的表格中。

6. For each s, calculate the mean time t and the quantity s/t. The raw data and processed columns are now ready for graphing.

6. 对每个s，计算平均时间t和量s/t。原始数据和处理后的列数据现已准备就绪，可以绘图。

4. Data Collection and Table Design | 数据收集与表格设计

A typical results table for this investigation is shown below. The columns must be clearly labelled with the quantity and unit, and uncertainties should be stated in the header. The precision of the recorded values should reflect the resolution of the measuring instruments used.

该探究中典型的实验表格如下所示。各列必须明确标出量和单位，表头中应说明不确定度。记录值的精确度需反映所用测量仪器的分辨率。

For each raw time measurement, the repeat readings typically agree to within ±0.001 s for a digital timer with millisecond resolution. The absolute uncertainty in the mean time is taken as the larger of the half-range or the instrument resolution, while the uncertainty in s/t is propagated from the uncertainties in s and t using the percentage method.

对于每一个时间原始测量值，对于具有毫秒分辨率的数字计时器，重复读数通常在±0.001 s内一致。平均时间的绝对不确定度取半极差与仪器分辨率中的较大者，而s/t的不确定度则由s和t的不确定度通过百分差法传递求得。

5. Graph Plotting and Analysis | 作图与图像分析

A graph of s/t (vertical axis) against t (horizontal axis) should be plotted on standard graph paper. The axes must be labelled with the quantity and unit, and the scales should be chosen so that the data points occupy at least half the grid area in both directions. Each data point is marked with a small cross, and error bars for both s/t and t are added if required by the question.

在标准坐标纸上绘制s/t（纵轴）对t（横轴）的图像。坐标轴必须标出量和单位，所选比例应使数据点在两个方向上都至少占据网格区域的一半。每个数据点用小十字标出，如果题目要求，还需添加s/t和t的误差棒。

As the relationship is linear (s/t = u + ½g · t), a straight best-fit line should be drawn through the points, ensuring an even spread of points above and below the line. Two additional worst-acceptable lines should be drawn: one with the steepest gradient and one with the shallowest gradient that still reasonably fit the data, taking the error bars into account.

由于该关系是线性的（s/t = u + ½g · t），应通过数据点绘制一条最佳拟合直线，并确保点在直线上下均匀分布。另需绘制两条最差可接受线：一条具有最大斜率，另一条具有最小斜率，并且需考虑误差棒，使它们仍能合理地拟合数据。

The gradient of the best-fit line, m = ½g, is calculated by taking a large triangle on the graph. For the worst lines, the gradients m_max and m_min are similarly found. The absolute uncertainty in the gradient, Δm, is given by (m_max – m_min)/2. The value of g is then determined as g = 2m, with an uncertainty Δg = 2Δm.

最佳拟合线的斜率 m = ½g，通过在图线上选取一个大三角形计算得到。对最差线同样求出斜率 m_max 和 m_min。斜率的绝对不确定度 Δm 由 (m_max – m_min)/2 给出。然后由 g = 2m 求出 g 值，不确定度 Δg = 2Δm。

6. Calculating the Acceleration g | 计算重力加速度

From the graph, a typical best-fit gradient might be m = 4.90 m s⁻². This gives an experimental value for g:

由图像得到的典型最佳拟合斜率可能为 m = 4.90 m s⁻²。这给出了实验测得的g值：

g = 2 × 4.90 = 9.80 m s⁻²

If the worst-acceptable slopes are m_max = 5.05 m s⁻² and m_min = 4.75 m s⁻², the uncertainty in the gradient is Δm = (5.05 – 4.75)/2 = 0.15 m s⁻². Hence, Δg = 2 × 0.15 = 0.30 m s⁻². The final result is expressed as g = (9.80 ± 0.30) m s⁻².

如果最差可接受斜率分别为 m_max = 5.05 m s⁻² 和 m_min = 4.75 m s⁻²，则斜率的不确定度为 Δm = (5.05 – 4.75)/2 = 0.15 m s⁻²。因此，Δg = 2 × 0.15 = 0.30 m s⁻²。最终结果表示为 g = (9.80 ± 0.30) m s⁻²。

This value should be compared with the accepted local value of 9.81 m s⁻². The percentage difference is calculated and comments are made on the agreement. The y-intercept (u) from the graph can be checked for consistency with an independent estimate of the velocity at the first light gate, perhaps from v = s/t at the smallest t, or by using the equation v² = u² + 2as.

应将此值与当地公认值9.81 m s⁻²进行比较。计算百分差，并就结果的一致性进行评述。从图像获得的y轴截距u也可以通过与第一光电门处速度的独立估算值（例如用最小t值时的s/t，或利用v² = u² + 2as）进行对比，以检验一致性。

7. Uncertainty Estimation and Propagation | 不确定度估算与传递

Uncertainties arise from several sources in this experiment. The absolute uncertainty in s (Δs) is determined by the metre rule or vernier calliper; a typical value is ±0.001 m. The uncertainty in the mean time t (Δt) combines the reaction time (negligible with light gates) and the spread of repeat readings. For the data above, half-range for t might be 0.0005 s, which is less than the timer resolution, so Δt is taken as ±0.001 s.

该实验的不确定度来自多个来源。s的绝对不确定度（Δs）由米尺或游标卡尺决定，典型值为±0.001 m。平均时间t的不确定度（Δt）则结合了反应时间（使用光电门可忽略）和重复读数的离散度。对于上述数据，t的半极差可能为0.0005 s，小于计时器分辨率，因此Δt取±0.001 s。

The percentage uncertainties in s/t are found by adding the percentage uncertainty in s to the percentage uncertainty in t (%U(s/t) = %U(s) + %U(t)), as division is involved. For example, if s = 0.800 ± 0.001 m, then %U(s) = 0.125%. If t = 0.2477 ± 0.001 s, then %U(t) = 0.404%. Therefore, %U(s/t) ≈ 0.529%. The absolute uncertainty in s/t is then 0.529% of 3.23 m s⁻¹ ≈ 0.02 m s⁻¹. These values are used for error bars on the graph.

s/t的百分不确定度由s的百分不确定度与t的百分不确定度相加得到（%U(s/t) = %U(s) + %U(t)），因为涉及除法运算。例如，若 s = 0.800 ± 0.001 m，则%U(s) = 0.125%。若 t = 0.2477 ± 0.001 s，则%U(t) = 0.404%。因此，%U(s/t) ≈ 0.529%。进而算出s/t的绝对不确定度为3.23 m s⁻¹的0.529%，约等于0.02 m s⁻¹。这些值用于在图像上绘制误差棒。

Propagating these uncertainties through the graph analysis, as described in Section 5, is more reliable than analytical propagation because it includes systematic effects that might shift the best-fit line. The method of extreme lines is the standard approach in A-Level practical papers.

如第5节所述，通过图像分析传递这些不确定度比解析传递更为可靠，因为它涵盖了可能使最佳拟合线偏移的系统效应。极端线法是A-Level实验试卷中的标准方法。

8. Sources of Error and Possible Improvements | 误差来源与改进措施

One significant source of systematic error is the residual magnetism of the electromagnet. After the current is switched off, the ball may still be held momentarily, causing a delay in release and an apparently larger g. Tapping the electromagnet or using a non-magnetic release mechanism can minimise this effect.

一个重要的系统误差来源是电磁铁的剩磁。电流切断后，钢球可能仍被短暂吸住，导致释放延迟，使得测得的g偏大。轻敲电磁铁或使用非磁性释放机构可将此影响降至最低。

Air resistance acts on the ball, especially at higher speeds near the second light gate. This slightly reduces the acceleration, causing a systematic underestimate of g. Using a dense, smooth steel ball and keeping the fall distance moderate helps to reduce the effect.

空气阻力作用于钢球，特别是在接近第二光电门的高速段，这会略微减小加速度，导致g的系统性低估。使用密度大、表面光滑的钢球并保持下落距离适中有助于减小此效应。

Another error arises if the light gates are not perfectly horizontal, so that the ball does not cut the beam at the same point each time, or if the ball’s diameter is comparable to the beam width. Ensuring careful alignment and using a ball diameter large relative to the beam width improves consistency.

如果光电门未能完全保持水平，钢球每次遮断光束的位置将不同，或者钢球直径与光束宽度相当，都会引入误差。仔细对准并使用直径相对于光束宽度较大的钢球可提高重复性。

Parallax error when measuring s can be reduced by using a vernier calliper for smaller separations and by clamping the metre rule firmly in place. Timing errors from the digital timer are usually small, but the timer should be checked for zero error before starting the experiment.

测量s时的视差可通过使用游标卡尺测量较小间距以及将米尺夹紧固定来减小。数字计时器的计时误差通常很小，但在实验开始前应检查计时器的零点误差。

Finally, repeating the experiment for a wider range of s and taking more repeat readings would allow a more reliable estimate of the mean time and a better-defined best-fit line, improving the overall accuracy of g.

最后，在更大的s范围内重复实验，并进行更多次重复读数，可以得到更可靠的平均时间估计和更明确的最佳拟合线，从而提高g的整体准确度。

9. Safety Precautions | 安全注意事项

Although the steel ball is small, it can cause injury if it falls onto a foot or onto sensitive equipment. Ensure that a foam pad or a box of sand is placed beneath the second light gate to catch the ball safely. Keep the area below the electromagnet clear during the release.

尽管钢球很小，但如果掉落到脚上或精密设备上仍可能造成伤害。应在第二光电门下方放置泡沫垫或沙盒，安全接住钢球。释放期间，保持电磁铁下方区域畅通无阻。

When moving the light gates or adjusting the apparatus, take care not to drop the retort stand or to trap fingers between clamps. Electrical connections to the electromagnet should be secure, and the low-voltage DC supply must be treated with respect, avoiding short circuits.

移动光电门或调整装置时，注意不要使铁架台掉落，也不要将手指夹在夹具间。电磁铁的电气连接应牢固，对待低压直流电源也应谨慎，避免短路。

10. Common Mistakes in Paper 3 | 卷3常见错误

Candidates often forget to include units in table headers or to state the absolute uncertainty next to the measured value. In the graph, a frequent error is plotting s against t instead of s/t against t, which would not yield a straight line. Examiners expect the linearised relationship to be clearly identified and used.

考生经常忘记在表格表头中注明单位，或在测量值旁注明绝对不确定度。在图像绘制中，一个常见错误是直接绘制s对t的图像，而非s/t对t，这并不能得到一条直线。考官期望明确识别并使用线性化后的关系式。

When drawing the lines of worst fit, some candidates either ignore the error bars or draw lines that are too steep or too shallow to be plausible. The worst lines must still be reasonably representative of the data spread. Also, the gradient triangle should be large — at least half the length of the drawn line — to minimise read-off error.

在绘制最差拟合线时，一些考生或者忽略误差棒，或者画出斜度过大或过小以至于不合理的最差线。最差线必须仍能合理地代表数据分布。此外，用于计算斜率的三角形应足够大——至少是所画直线长度的一半——以尽可能减小读取误差。

Another mistake is failing to check the y-intercept calculation or misreading the timer display. It is advisable

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📚 Mastering Application Questions in A-Level Physics Unit 4 (Jan 2021) | 精通A-Level物理Unit 4 应用题技巧（2021年1月卷）

The Unit 4 paper of January 2021 is a landmark assessment for A-Level Physics candidates, demanding not just recall of facts but true application of concepts in unfamiliar contexts. From circular motion on banked tracks to charged particles darting through magnetic fields, the paper rewards those who can translate a real‑world scenario into a structured physical model.

2021年1月的Unit 4试卷是A‑Level物理考生的重要里程碑，它不仅要求记忆知识点，更要求在不熟悉的场景中真正应用概念。从倾斜弯道上的圆周运动到带电粒子在磁场中的穿梭，这份试卷青睐那些能够将现实场景转化为结构化物理模型的学生。

1. Understanding the Question and Extracting Information | 理解题意与信息提取

Before reaching for a formula, read the stem twice and highlight every numerical value, unit, and qualitative clue such as “smooth”, “uniform”, “negligible mass” or “from rest”. Jot down the known variables in symbol form, e.g. m = 0.25 kg, v = 12 m s⁻¹, and flag what the question is asking you to find.

在动用任何公式之前，阅读题干两遍，圈出每一个数值、单位以及诸如“光滑”“均匀”“质量可忽略”或“由静止开始”等定性线索。用符号形式记下已知变量，例如 m = 0.25 kg、v = 12 m s⁻¹，并标明题目要求你求解的是什么。

Often a single diagram in the paper contains a wealth of data – lengths, angles, and force arrows. Transfer these onto your own sketch; the act of redrawing helps you spot constraints such as right‑angle triangles or symmetrical paths that aren’t immediately obvious.

试卷上的单一示意图往往蕴藏着海量数据——长度、角度和力箭头。把它们转绘到你自己的草图上；重绘的过程能帮助你发现题干中不太显眼的约束条件，比如直角三角形或对称路径。

A common pitfall is mixing up radius and diameter when the question describes a circular loop or a coil. Always check whether a given dimension refers to the radius or the diameter, and convert accordingly before any calculation.

一个常见陷阱是当题目描述圆形轨道或线圈时把半径和直径搞混。始终检查给定的尺寸指的是半径还是直径，并在计算之前完成相应的换算。

2. Drawing Clear and Labeled Diagrams | 绘制清晰标注的示意图

For motion on a curved path, draw a side‑view or bird’s‑eye diagram with the velocity vector tangent to the curve and the net force pointing toward the centre. Label the weight, normal reaction, friction and tension where relevant – the January 2021 conical pendulum question is a classic example where an incomplete force diagram leads to wrong equations.

对于曲线路径上的运动，画出侧视图或俯视图，速度矢量与曲线相切，合外力指向圆心。标注重力、法向反作用力、摩擦力和拉力——2021年1月卷的圆锥摆题目就是一个典型例子，受力图不全会导致方程出错。

In field questions, sketch the field lines or equipotential surfaces. For electric fields between parallel plates, a quick drawing showing the uniform spacing and the direction from positive to negative immediately reminds you that the field strength E = V/d is constant and the force on a charge is parallel to the field.

在有关场的试题中，画出电场线或等势面。对于平行板之间的电场，简单标出均匀间隔和由正到负的方向，能立刻让你意识到电场强度E = V/d是恒定的，电荷受力平行于电场。

Label all physical quantities directly on the diagram using standard symbols. This prevents you from confusing symbols mid‑calculation and makes it easier to apply Pythagoras’ theorem or trigonometric ratios when resolving vectors.

使用标准符号直接在图上标注所有物理量。这能避免在计算过程中混淆符号，并且在分解矢量时更容易应用勾股定理或三角比。

3. Identifying the Relevant Physical Principles | 识别相关的物理原理

After extracting the data, ask: which area of the syllabus is being tested? Unit 4 of the January 2021 paper weaves together further mechanics, fields and particle physics. A problem about two protons approaching each other likely requires conservation of energy and electric potential energy, not just kinematic equations.

提取数据之后，问自己：这道题在考纲的哪个板块？2021年1月Unit 4试卷把进阶力学、场和粒子物理融合在一起。一个关于两个质子相互靠近的问题很可能需要用到能量守恒和电势能，而不仅仅是运动学公式。

If the scenario mentions “elastic collision” or “bounce”, conservation of kinetic energy and momentum both apply. If the word “sticky” or “embedded” appears, the collision is perfectly inelastic – only momentum is conserved, and you should be ready to calculate the loss of kinetic energy.

如果情景提到“弹性碰撞”或“反弹”，则动能和动量二者均守恒。如果出现“粘在一起”或“嵌入”等字眼，那是完全非弹性碰撞——只有动量守恒，并且你应当准备好计算动能的损失。

When you see a charged particle moving through a magnetic field, immediately recall that the magnetic force acts as a centripetal force, giving BQv = mv²/r. This single relationship is the backbone of many application questions and was examined in the cyclotron context on that paper.

当你看到带电粒子在磁场中运动，立刻想到洛伦兹力提供向心力，即 BQv = mv²/r。这一关系是许多应用题的骨干，在该卷的回旋加速器语境中就考查到了。

4. Choosing and Applying the Right Equations | 选择并应用正确的公式

Write the governing equation in its standard symbolic form before substituting numbers. For instance, for a satellite in circular orbit: gravitational force = centripetal force → GMm/r² = mv²/r. This allows you to cancel m and solve for the required variable elegantly.

代数字之前，先写出标准符号形式的控制方程。例如，对于圆形轨道上的卫星：万有引力 = 向心力 → GMm/r² = mv²/r。这样你可以消掉m，优雅地解出所需变量。

Beware of equations that look similar but apply to different conditions. The formula F = EQ gives the force on a charge in a uniform electric field, while F = kQ₁Q₂/r² deals with point charges. Mixing them is a common mark‑loser in multi‑step problems.

警惕那些看起来相似但适用条件不同的公式。F = EQ 给出的是匀强电场中电荷所受的力，而F = kQ₁Q₂/r² 处理的是点电荷。在多步问题中混淆它们是常见的失分点。

If a problem involves energy transformations, start from a master equation such as W = ΔEk + ΔEp and then expand each term with the appropriate sub‑formula. The January 2021 paper featured a question where a charged particle accelerated through a potential difference gained kinetic energy, directly mapping onto QV = ½mv².

如果问题涉及能量转化，从一个主方程入手，例如 W = ΔEk + ΔEp，然后将每一项用对应的子公式展开。2021年1月卷有一道题，带电粒子经电势差加速获得动能，直接对应QV = ½mv²。

5. Handling Vectors and Directions with Confidence | 自信地处理矢量与方向

Application questions rarely allow you to ignore direction. Define a positive direction at the start – for example “to the right is positive” – and keep that convention throughout the solution. This consistency is crucial when you subtract vectors or calculate impulses.

应用题很少允许你忽略方向。一开始就规定正方向——比如“向右为正”——并在整个解题过程中保持这一约定。当你做矢量减法或计算冲量时，这种一致性至关重要。

When adding non‑collinear forces or momenta, resolve each vector into horizontal and vertical components. Use the relationships Fx = F cos θ and Fy = F sin θ, and then separately apply equilibrium or Newton’s second law in each direction.

当合成非共线的力或动量时，把每个矢量分解为水平和竖直分量。利用关系式Fx = F cos θ 和 Fy = F sin θ，然后分别在每个方向上运用平衡条件或牛顿第二定律。

In magnetic force questions, the direction of the force vector is given by Fleming’s left‑hand rule for conventional current. Take a moment to rotate your hand and confirm the orientation – reversing sign here can cause a cascade of errors in trajectory predictions.

在磁力问题中，力矢量的方向由弗莱明左手定则（用于传统电流）给出。花点时间转一下手确认方向——这里符号搞反会在预测轨迹时引发一连串错误。

6. Mastering Conservation Laws in Depth | 深度掌握守恒定律

Momentum conservation is a vector statement: total initial momentum as a vector equals total final momentum. For a two‑dimensional collision, set up separate equations for the x‑ and y‑components. The 2021 paper tested this with a particle decay scenario where solving simultaneous component equations was the key.

动量守恒是一个矢量陈述：总初动量矢量等于总末动量矢量。对于二维碰撞，要分别对 x 和 y 方向建立方程。2021年试卷通过一个粒子衰变的情景考查了这一点，联立分量方程是解题关键。

Conservation of energy is your bridge between different physical realms. In a nuclear reaction, mass deficit transforms into kinetic energy: Δ m c² = Ek of products. Remember to convert atomic mass units to kilograms properly – missing a factor of 1.66 × 10⁻²⁷ is a frequent slip.

能量守恒是连接不同物理领域的桥梁。在核反应中，质量亏损转化为动能：Δ m c² = 产物的Ek。记得把原子质量单位正确转换为千克——漏掉因子 1.66 × 10⁻²⁷ 是常见失误。

When applying both conservation laws together, first use momentum to find one unknown velocity, then plug into the kinetic energy expression. This stepwise approach prevents you from solving a messy quadratic unnecessarily.

当同时运用两个守恒定律时，可先用动量守恒求出一个未知速度，再代入动能表达式。这种分步策略能避免不必要地求解繁琐的二次方程。

7. Fields and Particle Physics: Linking Theory to Application | 场与粒子物理：将理论联系应用

In a uniform electric field, the path of an electron is parabolic, just like a projectile in a gravitational field. Use the standard SUVAT equations, but with the acceleration replaced by a = Ee/m. The Jan 2021 paper required candidates to calculate the vertical deflection of an electron beam between plates, a direct application of projectile motion analogies.

在匀强电场中，电子的路径是抛物线，就像引力场中的抛体一样。使用标准的SUVAT方程，但加速度替换为a = Ee/m。2021年1月试卷要求考生计算电子束在平板间的竖直偏转，这正是抛体运动类比的直接应用。

Magnetic fields do no work on a charged particle, so the particle’s speed remains constant while its direction changes. This key insight lets you equate centripetal force with magnetic force without worrying about energy changes, and it explains why the cyclotron’s frequency is independent of speed for non‑relativistic particles.

磁场对带电粒子不做功，因此粒子速率保持不变，仅仅是方向改变。这一关键认识让你可以把向心力与磁力直接等同而不必担心能量变化，并且解释了为何非相对论情形下回旋加速器的频率与速率无关。

Particle tracks in bubble chamber photographs offer direct evidence of conservation laws. Curved tracks imply a perpendicular magnetic field, and the radius of curvature gives the momentum: p = BQr. Spirals indicate energy loss, and kinked tracks show decays – these observational skills were implicitly tested in data‑analysis questions.

气泡室照片中的粒子径迹为守恒定律提供了直接证据。弯曲的径迹暗示存在垂直磁场，曲率半径给出动量：p = BQr。螺旋意味着能量损失，折线径迹代表衰变——这些观察技巧在数据分析题中被隐式地考查到。

8. Significant Figures, Unit Conversions and Final Presentation | 有效数字、单位换算与最终呈现

Your final answer must reflect the precision of the data given. If all inputs are given to 3 significant figures, your answer should generally be quoted to 3 s.f. Resist the temptation to copy the full 10‑digit calculator display; it suggests a false precision that examiners penalise.

最终答案必须反映给定数据的精度。如果所有输入量给出的是3位有效数字，答案一般也应当引用3位。克制住照抄计算器10位数显示值的冲动；那暗示着虚假的精度，会被考官扣分。

Units must be consistent throughout. Convert everything to SI base units before plugging into formulas: grams to kilograms, centimetres to metres, microcoulombs to coulombs. A quick unit‑check at the end can catch algebraic blunders – if you expect newtons and get kg m s⁻³, something is off.

单位必须全程一致。在代入公式前把所有量转换为国际基本单位：克换为千克，厘米换为米，微库仑换为库仑。最后进行一次快速单位检验可以捕捉代数错误——如果你预期得到牛顿却得出 kg m s⁻³，那就出问题了。

Express your final answer clearly: box it or underline it, and include the correct unit. In multi‑step “show that” questions, write down the intermediate results so that even if you make a slip, the method marks can be awarded.

清晰呈现最终答案：用方框圈起或下划线，并附上正确单位。在多步的“证明”题中，写下中间结果，这样即使犯了小错，也能获得方法分。

9. Checking Reasonability and Reverse Engineering | 检验合理性并反向推理

Once you have a numerical answer, pause and ask: does this make sense? If you calculate that a proton’s speed is 2 × 10⁸ m s⁻¹, it is close to the speed of light and relativistic corrections might be needed – the problem likely expects you to note that. A tension of 5000 N in a thread meant to hold a 50 g mass is also a red flag.

得出数值答案后，停下来自问：这个结果合理吗？如果你算出质子的速率是 2 × 10⁸ m s⁻¹，这已经接近光速，可能需要相对论修正——题目也许正是期望你指出这一点。一根用来悬挂 50 g 小球的绳子若算出 5000 N 的张力，也是一个危险信号。

You can often check an answer by plugging it back into the original equation or by using an alternative method. For example, in a pair‑production question, calculate the energy of the incoming photon and verify that it matches the sum of the rest mass energies plus kinetic energies.

你通常可以通过把答案代回原方程或使用另一种方法来检验。例如，在电子对生成问题中，计算出给光子的能量，验证它是否等于静止质量能量之和加上动能。

If you find a counter‑intuitive result – say a negative time – re‑examine your sign convention. Negative quantities sometimes have a legitimate meaning (direction), but in time intervals they indicate a setup error.

如果你发现了违背直觉的结果——比如时间为负——重新检查你的正负号规定。负号有时具有合理含义（方向），但在时间间隔中则表明设定有误。

10. Time Management and Exam Strategy for Application Questions | 应用题的时间管理与考试策略

Application questions are often worth the most marks per minute. Scan the paper at the start and note the easier application contexts that you are confident with, such as horizontal circular motion or a single charge in a field, and tackle those first to secure marks and build momentum.

应用题通常每分钟的分值最高。开考时先浏览试卷，标出你有信心的、较简单的应用情景，例如水平圆周运动或场中的单个电荷，优先解决它们以锁定分数并积累做题节奏。

For the final, multi‑page synoptic question, bullet‑point a plan on the side: which principles, which equations, and the order of steps. Spending two minutes on a plan can save ten minutes of fumbling through algebra.

对于最后那道跨越多页的综合性题目，在旁边用要点列出计划：用到哪些原理，哪些方程，以及解题顺序。花两分钟做计划能节省十分钟盲目乱试代数的时间。

Never leave a numerical question blank. Even if you cannot derive the final answer, write down the relevant equations, substitute correct values, and attempt a clear calculation. Structured papers like Unit 4 Jan 2021 reward even partially correct logic generously.

绝不要给数值题留白。即便你无法推得最终答案，也要写出相关的方程，代入正确数值，并尝试明确的计算。像Unit 4 2021年1月这样结构化的试卷，即使部分正确的逻辑也能得到慷慨的分数。

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📚 A-Level Physics Unit 3 Insert (January 2020): Formula Derivations Explained | A-Level物理Unit 3 (2020年1月) 核心公式推导

The Edexcel International A-Level Physics Unit 3 (WPH13/01) examination assesses practical skills, and the January 2020 Insert provided a reference sheet containing key equations for several standard experiments. Understanding the derivations behind these formulas is essential not only for solving problems but also for justifying experimental procedures and analysing uncertainties. This article unpacks the core derivations that appeared in that Insert, offering clear, step‑by‑step explanations for each relationship.

Edexcel International A-Level物理单元3（WPH13/01）考试考查实验技能，2020年1月的附页给出了一张含有多个标准实验关键公式的参考表。理解这些公式背后的推导，不仅对解题至关重要，也有助于合理解释实验步骤和分析不确定度。本文将逐一解读该附页中出现的核心推导，为每一条关系提供清晰、逐步的说明。

1. Free Fall and the Acceleration of Gravity: s = ½ gt² | 自由落体与重力加速度：s = ½ gt²

In a typical free fall experiment, a small dense object (often a steel ball) is dropped from rest through a measured height s. Assuming air resistance is negligible, the acceleration a is constant and equal to g. The relevant SUVAT equation is s = ut + ½ a t². Since the initial velocity u = 0, this simplifies to s = ½ g t².

在典型的自由落体实验中，一个小而密的物体（通常是钢球）从静止释放，通过已知高度 s 下落。假设空气阻力可忽略，加速度 a 恒定且等于 g。适用的运动学公式为 s = ut + ½ a t²。由于初速度 u = 0，简化为 s = ½ g t²。

To extract g, the equation is rearranged to g = 2s / t². A more reliable approach involves taking multiple readings of s and the corresponding fall time t, then plotting s against t². Because s = (½ g) t², the graph is a straight line through the origin with gradient ½ g. Hence g = 2 × gradient. The uncertainty in g follows from the scatter of points and the gradient uncertainty.

为求出 g，将方程改写为 g = 2s / t²。更可靠的方法是采集多组 s 和相应的下落时间 t，然后作 s 对 t² 的图像。由于 s = (½ g) t²，该图为过原点的直线，斜率为 ½ g。因此 g = 2 × 斜率。g 的不确定度可由散点分布以及斜率的不确定度得到。

2. Resistivity of a Wire: ρ = RA / L | 导线电阻率：ρ = RA / L

The resistance R of a uniform metallic wire is directly proportional to its length L and inversely proportional to its cross‑sectional area A. This is summarised as R = ρ L / A, where ρ is the resistivity of the material. For a wire with circular cross‑section of diameter d, the area A = π d² / 4. Substituting gives R = 4 ρ L / (π d²), which can be rearranged to ρ = R π d² / (4 L).

均匀金属导线的电阻 R 与其长度 L 成正比，与横截面积 A 成反比，归纳为 R = ρ L / A，其中 ρ 为材料的电阻率。对直径为 d 的圆形截面导线，面积 A = π d² / 4。代入得 R = 4 ρ L / (π d²)，重排得 ρ = R π d² / (4 L)。

Experimentally, R is found using a voltmeter and ammeter (R = V / I), while d is measured with a micrometer screw gauge. To minimise the effect of random errors, R is measured for different L, and a graph of R against L is plotted. From R = (ρ / A) L, the gradient m = ρ / A, so ρ = m A. If A is known, ρ can be calculated directly from the gradient.

实验中，R 通过伏特计和安培计测得（R = V / I），d 用千分尺测量。为减少随机误差的影响，可测量不同长度 L 对应的 R，并作 R-L 图。由 R = (ρ / A) L，斜率 m = ρ / A，因此 ρ = m A。若面积 A 已知，即可直接从斜率求出 ρ。

3. Young Modulus from Stress and Strain: E = F L / (A ΔL) | 由应力应变求杨氏模量：E = F L / (A ΔL)

Young modulus E is defined as the ratio of tensile stress to tensile strain: E = (F / A) / (ΔL / L) = F L / (A ΔL). Here F is the applied force, A the original cross‑sectional area, L the original length, and ΔL the extension produced.

杨氏模量 E 定义为拉应力与拉应变之比：E = (F / A) / (ΔL / L) = F L / (A ΔL)。式中 F 为外加力，A 为原始横截面积，L 为原始长度，ΔL 为产生的伸长量。

In a typical experiment, a long thin wire is stretched by adding known masses, and the extension is measured with a vernier scale or a traveling microscope. For a wire of diameter d, A = π d² / 4, resulting in E = 4 F L / (π d² ΔL). To obtain a reliable value, a graph of F (or mass m) against ΔL is often drawn. Since F = (E A / L) ΔL, the gradient is E A / L, giving E = gradient × L / A. The uncertainty budget originates from the measurement of d, L, and the precision of the extension readings.

在典型实验中，一根细长的金属丝通过添加已知质量被拉伸，伸长量用游标尺或读数显微镜测量。对于直径为 d 的丝，A = π d² / 4，从而 E = 4 F L / (π d² ΔL)。为获得可靠值，通常作作用力 F（或质量 m）对 ΔL 的图像。因 F = (E A / L) ΔL，斜率为 E A / L，故 E = 斜率 × L / A。不确定度预算来源于直径 d、长度 L 的测量以及伸长读数的精度。

4. Internal Resistance of a Cell: ε = V + Ir | 电池内阻：ε = V + Ir

For a source of emf ε with internal resistance r, the terminal potential difference V across an external load R is given by ε = I (R + r) = I R + I r = V + I r. Rearranging gives V = ε − I r. This is a linear equation of the form y = mx + c, where V is the dependent variable, I is the independent variable, the gradient m = −r, and the y‑intercept c = ε.

对电动势为 ε、内阻为 r 的电源，外接负载 R 时的路端电压 V 满足 ε = I (R + r) = I R + I r = V + I r。重排得 V = ε − I r。这是一个形如 y = mx + c 的线性方程，V 为因变量，I 为自变量，梯度 m = −r，截距 c = ε。

By varying the external resistance and recording corresponding values of V and I, a V‑I graph can be plotted. The gradient is negative and its magnitude gives r, while the vertical intercept gives ε. Alternatively, if a graph of 1/I against R is drawn, from ε = I (R + r) we obtain 1/I = (1/ε) R + r/ε. The gradient is 1/ε and the intercept is r/ε, enabling both ε and r to be determined. Both methods appear in the Insert and are frequently examined.

通过改变外电阻并记录相应的 V 和 I 值，可绘制 V-I 图。梯度为负，其绝对值即为 r，纵轴截距即为 ε。另一种方法是作 1/I 对 R 的图像，由 ε = I (R + r) 可得 1/I = (1/ε) R + r/ε。梯度为 1/ε，截距为 r/ε，由此可同时得到 ε 和 r。这两种方法都出现在附页中，是常考内容。

5. Simple Pendulum and the Determination of g: T = 2π √(L/g) | 单摆与重力加速度测定：T = 2π √(L/g)

For small angular displacements (typically less than about 10°), the period T of a simple pendulum of length L is independent of amplitude and is given by T = 2π √(L / g). Squaring both sides yields T² = 4π² L / g, which can be rearranged to g = 4π² L / T².

在小角度摆动（通常小于约10°）下，摆长为 L 的单摆周期 T 与振幅无关，且由 T = 2π √(L / g) 给出。两边平方得 T² = 4π² L / g，重排可得 g = 4π² L / T²。

Experimentally, the period is measured for several lengths L, and a graph of T² against L is drawn. The equation has the form T² = (4π² / g) L, so the graph is a straight line through the origin with gradient m = 4π² / g. Thus g = 4π² / m. Timings for multiple oscillations (e.g., 20 swings) are used to reduce the random uncertainty in T. The overall percentage uncertainty in g can be estimated from Δg/g = ΔL/L + 2 ΔT/T.

实验中，测量不同摆长 L 对应的周期 T，并作 T²-L 图。该方程具有 T² = (4π² / g) L 的形式，因此图像为过原点的直线，斜率 m = 4π² / g。于是 g = 4π² / m。通常测量多次摆动（例如20次）的时间以减小 T 的随机不确定度。g 的总百分比不确定度可由 Δg/g = ΔL/L + 2 ΔT/T 估算。

6. Speed of Sound in Air Using a Resonance Tube: c = 4 f (l + 0.3d) | 共振管测定空气声速：c = 4 f (l + 0.3d)

A closed‑end resonance tube produces standing waves when the length l of the air column satisfies l + e = n λ / 4, where n is an odd integer (1, 3, 5…) and e is the end correction. For the fundamental frequency f, n = 1, giving l + e = λ / 4. Since c = f λ, it follows that λ = c / f, so c = 4 f (l + e). The end correction e is approximately 0.3 times the diameter d of the tube, leading to the practical formula c = 4 f (l + 0.3d).

闭端共振管中，当空气柱长度 l 满足 l + e = n λ / 4（n 为奇数1,3,5…）时会形成驻波，其中 e 为端部修正。对于基频 f，n = 1，则有 l + e = λ / 4。由 c = f λ 知 λ = c / f，因此 c = 4 f (l + e)。端部修正 e 约等于管直径 d 的0.3倍，于是得到实用公式 c = 4 f (l + 0.3d)。

In the experiment, a tuning fork of known frequency f is held over the open end while the water level is adjusted to vary l. The first position of resonance gives l₁, and further resonances give l₂, l₃… From the differences Δl = l₂ − l₁ = λ / 2, the wavelength can be found without the end correction: λ = 2 Δl, so c = f × 2 Δl. Alternatively, if d is measured, c can be obtained from a single resonance using the end‑correction formula. Both approaches are valid and may be referenced in the Insert.

实验中，将已知频率 f 的音叉置于开口端上方，通过调节水面改变 l。第一次共振位置给出 l₁，随后的共振给出 l₂、l₃……由差值 Δl = l₂ − l₁ = λ / 2 可在无需端部修正的情况下求得波长：λ = 2 Δl，因此 c = f × 2 Δl。另一种方法，若测量了直径 d，则可用含端部修正的公式从单次共振求出 c。两种方法均有效，附页中可能提及。

7. Viscosity by Stokes’ Law: Terminal Velocity Equation | 斯托克斯定律测粘度：终端速度方程

When a small sphere falls through a viscous liquid, it experiences a drag force given by Stokes’ law, Fd = 6π η r v, where η is the dynamic viscosity of the liquid, r is the radius of the sphere, and v is its velocity. At terminal velocity, the net force is zero: weight − buoyancy = drag. Hence (4/3)π r³ ρ₁ g − (4/3)π r³ ρ₂ g = 6π η r v, where ρ₁ is the density of the sphere and ρ₂ is the density of the liquid.

小球在黏性液体中下落时，受到由斯托克斯定律给出的阻力 Fd = 6π η r v，式中 η 为液体的动力黏度，r 为小球半径，v 为下落速度。在终端速度下，合力为零：重力减去浮力等于阻力。因此 (4/3)π r³ ρ₁ g − (4/3)π r³ ρ₂ g = 6π η r v，其中 ρ₁ 为球的密度，ρ₂ 为液体的密度。

Simplifying gives (4/3)π r³ g (ρ₁ − ρ₂) = 6π η r v. Cancelling common factors π and r (r ≠ 0) yields (4/3) r² g (ρ₁ − ρ₂) = 6 η v. Solving for η: η = 2 r² g (ρ₁ − ρ₂) / (9 v). If the terminal velocity v is measured by timing the sphere over a known distance once the motion is uniform, and r and the densities are known, η can be calculated. The Insert may provide this formula directly in the form η = 2r²g(ρs−ρl)/9v, where ρs and ρl denote sphere and liquid densities.

化简得 (4/3)π r³ g (ρ₁ − ρ₂) = 6π η r v。约去公因子 π 和 r（r ≠ 0）得 (4/3) r² g (ρ₁ − ρ₂) = 6 η v。解出 η：η = 2 r² g (ρ₁ − ρ₂) / (9 v)。若通过小球在匀速段通过已知距离的时间来测量终端速度 v，且 r 和密度已知，便可计算 η。附页可能直接以 η = 2r²g(ρs−ρl)/9v 的形式给出该公式，其中 ρs 和 ρl 分别表示球和液体的密度。

8. Propagation of Uncertainties for Derived Quantities | 导出物理量的不确定度传递

When a quantity is calculated from measured variables, its uncertainty is determined by the uncertainties of the inputs. For addition or subtraction (z = x + y or z = x − y), the absolute uncertainties add: Δz = Δx + Δy. For multiplication or division (z = x y or z = x / y), the fractional (or percentage) uncertainties add: Δz / z = Δx / x + Δy / y. For a power relationship (z = xⁿ), the fractional uncertainty becomes Δz / z = |n| Δx / x.

当一个物理量由测量变量计算得出时，其不确定度由各输入量的不确定度决定。对于加减运算（z = x + y 或 z = x − y），绝对不确定度相加：Δz = Δx + Δy。对于乘除运算（z = x y 或 z = x / y），相对（或百分比）不确定度相加：Δz / z = Δx / x + Δy / y。对于幂函数（z = xⁿ），相对不确定度变为 Δz / z = |n| Δx / x。

Consider the pendulum equation g = 4π² L / T². The fractional uncertainty in g is Δg / g = ΔL / L + 2 ΔT / T. If L is measured to ±1 mm and T to ±0.01 s, the percentage uncertainties can be combined to find the overall uncertainty in g. Similarly, for resistivity ρ = RA / L, the uncertainty is built from ΔR, ΔA (which depends on Δd, since A = π d²/4, giving ΔA/A = 2 Δd/d), and ΔL. The Insert often expects students to derive or use such uncertainty relations to comment on the largest source of error.

以单摆方程 g = 4π² L / T² 为例。g 的相对不确定度为 Δg / g = ΔL / L + 2 ΔT / T。若 L 测量至 ±1 mm，T 至 ±0.01 s，可将百分比不确定度组合得出 g 的总不确定度。类似地，对于电阻率 ρ = RA / L，不确定度由 ΔR、ΔA（依赖于 Δd，因 A = π d²/4，故 ΔA/A = 2 Δd/d）和 ΔL 构成。附页常期望学生推导或使用这样的不确定度关系，以判断最大误差来源。

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📚 A-Level Physics Unit 4 Jan 22 Mark Scheme: Experimental Investigation | A-Level 物理 Unit 4 2022年1月评分方案：实验探究精解

This article unpacks the experimental investigation question from the Edexcel International A-Level Physics Unit 4 (WPH14/01) January 2022 examination, using the official mark scheme as a lens. We decode the precise criteria examiners used to award marks for practical planning, measurement, data processing, graphical analysis and evaluation. Whether you are preparing for a future Unit 4 paper or consolidating your practical skills, understanding how marks are allocated is the key to securing top grades in practical-based questions.

本文以 Edexcel 国际 A-Level 物理第四单元（WPH14/01）2022 年 1 月考试的官方评分方案为镜，深度剖析实验探究题。我们将拆解考官在实验计划、测量、数据处理、图形分析和评估环节中使用的精确评分标准。无论你是在准备未来的 Unit 4 考试，还是在巩固实验技能，理解分数是如何分配的都是在实验题中拿到高分的钥匙。

1. Structure of the Experimental Question | 实验题的结构

The Unit 4 experimental question typically carries 12 to 16 marks and follows a predictable structure: recording raw data, tabulating results, processing data (often requiring a linearisation graph), determining a physical quantity, and evaluating the procedure. In the January 2022 mark scheme, marks were split across these sections with a heavy emphasis on valid conclusions and error handling.

第四单元的实验题通常占 12 到 16 分，且遵循一个可预测的结构：记录原始数据、列表整理结果、处理数据（通常需要绘制线性化图像）、确定某个物理量以及评估实验过程。在 2022 年 1 月的评分方案中，分数分散在这些板块，尤其侧重得出有效结论和处理误差的能力。

2. Command Words and What They Demand | 指令词及其要求

Command words in the question paper map directly onto the mark scheme. ‘Measure’ and ‘record’ require accurate values with appropriate units and precision. ‘Plot’ demands correctly labelled axes and sensible scales. ‘Determine’ expects a calculation with the final answer given to the correct number of significant figures. ‘Evaluate’ and ‘suggest’ push you to identify limitations, quantify their effects where possible, and propose realistic improvements. The Jan 22 scheme shows that vague evaluation without specific physics reasoning scores zero.

试卷中的指令词直接对应评分方案中的得分点。’Measure’ 和 ‘record’ 要求给出带有合适单位和精度的准确数值。’Plot’ 要求正确标记坐标轴并使用合理分度。’Determine’ 期望计算出结果，且最终答案保留正确的小数位数。’Evaluate’ 和 ‘suggest’ 则要求你指出局限性，尽可能量化其影响，并提出切合实际的改进措施。2022 年 1 月的评分方案显示，缺乏具体物理推论的模糊评价不会得分。

3. Designing a Valid Investigation: The Simple Pendulum | 设计合理的探究：单摆实验

In the Jan 22 paper, the experimental context involved determining the acceleration of free fall, g, using a simple pendulum. The independent variable was the pendulum length l (measured from the point of suspension to the centre of the bob). The dependent variable was the oscillation period T. Control variables included the amplitude of swing (kept small, below 10°), the mass of the bob, and the room temperature. A well-planned investigation always begins by naming these three variable types explicitly.

在 2022 年 1 月的试卷中，实验背景是用单摆测定自由落体加速度 g。自变量是摆长 l（从悬挂点到摆球中心的距离）。因变量是摆动周期 T。控制变量包括摆幅（始终保持小角度，低于 10°）、摆球质量和室温。一项设计良好的探究总是从明确命名这三类变量开始。

4. Control of Variables and Fair Testing | 变量控制与公平测试

To isolate the relationship between l and T, all other factors must be kept constant. The Jan 22 mark scheme rewarded students who stated that a small amplitude ensures the motion approximates simple harmonic motion, making the period independent of amplitude. Using the same bob eliminated mass as a variable. Conducting all trials in a short time window minimised temperature drift that could alter the length of the string. Marks were deducted if control measures were only listed but not justified in physics terms.

为了孤立 l 与 T 之间的关系，所有其他因素必须保持不变。2022 年 1 月的评分方案奖励那些指出小振幅确保运动近似为简谐运动、从而使周期与振幅无关的学生。使用同一个摆球排除了质量的影响。在短时间内完成所有测量可以最大程度减少因温度变化导致绳长改变的可能。只罗列控制变量而未用物理术语加以论证会被扣分。

5. Measurement Techniques and Instrument Selection | 测量技术与仪器选择

Length l was measured using a metre rule with millimetre markings, and the diameter of the bob was obtained with vernier callipers; the effective length was taken as the distance from the clamp to the bottom of the bob plus its radius. To reduce reaction time uncertainty, the period T was determined by timing 20 complete oscillations with a digital stopwatch and then dividing by 20. The mark scheme accepted repeating each timing three times and averaging, provided the raw data were clearly shown.

摆长 l 使用带有毫米刻度的米尺测量，摆球直径用游标卡尺获取；有效摆长取悬挂点到摆球底端的距离再加上其半径。为了减小反应时间带来的不确定度，周期 T 通过用数字秒表测量 20 次全振动的时间再除以 20 得到。评分方案接受每个时间重复测量三次并取平均值，前提是原始数据必须清晰呈现。

6. Data Presentation and Tabulation | 数据呈现与制表

All recorded data must be entered into a neat table with quantity/unit headers, e.g. ‘l / m’ and ‘T / s’. The Jan 22 mark scheme insisted on consistent decimal places: if l was measured to ±0.001 m, all values in that column had to be shown to three decimal places. Calculated columns such as T² / s² were expected, with values rounded appropriately. No marks were awarded if raw data were missing or if units appeared only next to the numbers rather than in the header.

所有记录的数据都必须整齐地填入表格，并在表头注明量/单位，例如“l / m”和“T / s”。2022 年 1 月的评分方案强调小数点位数一致：若 l 的测量精度为 ±0.001 m，则该列所有数值必须保留三位小数。还应包含计算列，如 T² / s²，数值需适当四舍五入。若缺少原始数据，或单位只写在数字旁边而非表头中，将完全不得分。

7. Graph Plotting and Achieving Linearity | 绘图与实现线性化

The relationship T = 2π√(l/g) is non-linear, so linearisation is essential. Squaring both sides gives the straight-line equation:

T² = (4π² / g) l

Plotting T² on the vertical axis and l on the horizontal axis yields a straight line through the origin. In the Jan 22 scheme, marks were awarded for axes labelled with quantities and units, sensible scales that occupy more than half the grid, accurate plotting of points to within half a small square, and a sharp best-fit line. Failure to linearise correctly was a major mark-losing pitfall.

关系式 T = 2π√(l/g) 是非线性的，因此必须进行线性化。两边平方得到直线方程：

T² = (4π² / g) l

将 T² 作为纵轴、l 作为横轴绘图，会得到一条过原点的直线。2022 年 1 月的方案中，正确标注包含单位的坐标轴、选取超过半幅坐标纸的合理分度、描点精确到半小格以内，并画出清晰的“最佳拟合线”才能拿到绘图分。未能正确线性化是主要的失分陷阱。

8. Determining Gradient and Intercept | 确定斜率与截距

From the line of best fit, the gradient m = Δ(T²) / Δl is calculated using a large triangle that covers at least half of the drawn line. The Jan 22 mark scheme required the triangle vertices to be clearly shown on the graph, with coordinates read to the precision of the scale. The intercept was expected to be zero or very nearly so; a significantly non-zero intercept hinted at a systematic error in length measurement, such as misreading the zero of the metre rule.

在最佳拟合线上，选取一个覆盖所绘直线至少一半长度的大三角形，计算斜率 m = Δ(T²) / Δl。2022 年 1 月的评分方案要求图上清晰标出三角形顶点，坐标读数精确到分度值允许的范围。截距应接近于零；若截距显著不为零，则暗示长度测量中存在系统误差，例如米尺的零点读数错误。

9. Calculating Derived Quantities and Uncertainties | 计算导出量与不确定度

Once the gradient m was found, g was determined from g = 4π² / m. The mark scheme awarded the final mark only if the unit (m s⁻²) was given and the value was sensible (about 9.81 m s⁻²). Percentage uncertainty in g was derived by combining the percentage uncertainty in the gradient, which itself came from the difference between the best-fit and worst-acceptable-fit lines: Δm = |m_best – m_worst|. Many candidates lost marks because they attempted to estimate uncertainty using the scatter of points alone, without drawing the worst-fit line.

得到斜率 m 后，由 g = 4π² / m 计算 g。评分方案只有在给出正确单位（m s⁻²）且数值合理（约 9.81 m s⁻²）时才给最后一步答案分。g 的百分不确定度通过组合斜率的百分不确定度得出，而斜率的不确定度来自最佳拟合线与最差可接受拟合线之差：Δm = |m_best – m_worst|。许多考生因为仅凭点的散布估算不确定度，而没有画出最差拟合线而失分。

10. Evaluating Sources of Error and Limitations | 评估误差来源与局限性

The Jan 22 evaluation section required candidates to distinguish between random and systematic errors. Random errors arose from human reaction time in starting and stopping the stopwatch, which could be reduced by timing more oscillations. Systematic errors included the finite amplitude (even below 10°) causing the period to deviate from the small-angle approximation, and the string possessing some mass and elasticity. High-scoring answers quantified the possible effect, e.g., ‘a 5° amplitude increases T by roughly 0.05%, introducing a systematic underestimate of g by about 0.1%’.

2022 年 1 月的评估部分要求考生区分随机误差和系统误差。随机误差主要来自启动和停止秒表时的反应时间，可通过测量更多个周期的总时间来减小。系统误差包括即使摆角小于 10° 仍会偏离小角度近似，以及细绳本身具有质量和弹性。高分答案能够量化可能的影响，例如“5° 的摆幅会使 T 增加约 0.05%，从而使 g 被系统性低估约 0.1%”。

11. Suggesting Improvements and Further Work | 提出改进措施与后续工作

Suggestions must be practical and linked to the identified limitations. Replacing the stopwatch with a light gate and electronic timer would eliminate human reaction time and improve precision by an order of magnitude. Using a thin, light, inextensible string and a small dense spherical bob would better satisfy the assumptions of the simple pendulum model. The mark scheme rejected vague suggestions like ‘use better equipment’ and required specific physics reasoning for each proposal.

所提建议必须切实可行，并与已指出的局限性相关联。用光门和电子计时器替代秒表可以消除人为反应时间，使精度提高一个数量级。使用细、轻且不可伸长的绳子和一个小而密的球形摆球，能更好地满足单摆模型的假设。评分方案拒绝了“使用更好的设备”这类模糊建议，要求每项提议都要有具体的物理论证。

12. Applying the Mark Scheme: A Worked Example | 应用评分方案：一个实例解析

Imagine a student’s data for l = 0.500, 0.600, 0.700, 0.800, 0.900 m yielded T = 1.42, 1.55, 1.68, 1.79, 1.90 s. The table included T² = 2.02, 2.40, 2.82, 3.20, 3.61 s² with consistent significant figures. The graph of T² against l was a straight line passing close to the origin. The gradient was read as 4.02 s² m⁻¹, giving g = 4π² / 4.02 ≈ 9.82 m s⁻². The worst-fit line gave a gradient of 3.90 s² m⁻¹, leading to Δg/g = (4.02 – 3.90)/4.02 ≈ 3.0%. The evaluation noted that the string was a thick cord, adding significant mass and shifting the centre of mass, explaining why the intercept was not exactly zero. According to the Jan 22 mark scheme, such an answer would secure full marks in every section.

假设一名学生的数据：l = 0.500, 0.600, 0.700, 0.800, 0.900 m，对应 T = 1.42, 1.55, 1.68, 1.79, 1.90 s。表格中附有 T² = 2.02, 2.40, 2.82, 3.20, 3.61 s²，并保持了统一的有效数字。T² 对 l 的图为一条接近过原点的直线。读出的斜率为 4.02 s² m⁻¹，得到 g = 4π² / 4.02 ≈ 9.82 m s⁻²。最差拟合线的斜率为 3.90 s² m⁻¹，由此得出 Δg/g = (4.02 – 3.90)/4.02 ≈ 3.0%。评估部分指出，使用的绳子是一根粗绳，增加了显著质量并移动了质心，这解释了截距为何不恰为零。按照 2022 年 1 月的评分方案，这样一份答案在每个部分都能拿到满分。

Published by TutorHao | Physics Revision Series | aleveler.com

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