📚 Edexcel Physics: Common Pitfall Questions Explained | Edexcel 物理：易错题精讲
In Edexcel A Level Physics, certain concepts consistently trip up students, leading to lost marks even among well-prepared candidates. This article walks you through a selection of commonly misunderstood questions, unpacking the underlying physics, typical errors, and the correct reasoning needed to secure top grades. Each section is designed as a bite-sized revision resource that clarifies why a particular trap exists and how to avoid it.
在 Edexcel A Level 物理中，一些概念反复让学生失分，即使准备充分也难免落入陷阱。本文精选常考易错题，拆解背后的物理原理、典型错误以及正确思路，帮助你锁定高分。每一节都是一个小模块的复习素材，讲清为什么出现这个坑、如何避开。
1. Projectile Confusion: Horizontal and Vertical Independence | 抛体困惑：水平和竖直的独立性
A tennis ball is hit horizontally at 20 m s⁻¹ from a cliff 45 m above the sea. Neglecting air resistance, many students incorrectly assume that the initial vertical velocity is also 20 m s⁻¹, or they use the horizontal speed directly in the vertical motion equations. The correct approach is to treat the two perpendicular components completely separately. The horizontal velocity remains constant at 20 m s⁻¹, and the vertical motion starts from rest (uᵧ = 0) with acceleration g = 9.81 m s⁻² downwards. The time to hit the water is found from s = ½ g t², so 45 = ½ × 9.81 × t², giving t ≈ 3.03 s. The horizontal range is then simply x = uₓ t = 20 × 3.03 ≈ 60.6 m. A common mistake is to use 20 m s⁻¹ in the suvat for the vertical direction, producing a nonsensical time and range.
一个网球以 20 m s⁻¹ 的水平速度从高出海面 45 m 的悬崖上被击出。忽略空气阻力，许多学生会错误地认为初始竖直速度也是 20 m s⁻¹，或直接把水平速度代入竖直运动的方程中。正确做法是严格将两个垂直方向分开：水平速度保持 20 m s⁻¹ 不变，而竖直方向从静止开始 (uᵧ = 0)，加速度 g = 9.81 m s⁻² 向下。落水时间由 s = ½ g t² 求出：45 = ½ × 9.81 × t² → t ≈ 3.03 s。水平位移即为 x = uₓ t = 20 × 3.03 ≈ 60.6 m。典型错误是把 20 m s⁻¹ 放到竖直方向的匀加速公式里，得出荒谬的时间和射程。
2. Resolving Weight on an Inclined Plane | 斜面上重力的分解
Many students draw the weight arrow and then incorrectly resolve it into components parallel and perpendicular to the slope. A typical error is to assign mg cos θ to the parallel component and mg sin θ to the normal reaction. The trick is to align the right-angle triangle correctly. The angle of the incline θ is the angle between the weight vector and the perpendicular to the plane. Therefore, the component of weight down the slope is mg sin θ, and the component perpendicular to the slope is mg cos θ. If friction or tension is involved, always start by drawing a clear free-body diagram with the weight vector pointing straight down, then split it. A numerical illustration: a 5.0 kg mass on a 30° slope experiences a downhill force of 5.0 × 9.81 × sin30° = 24.5 N. If the object is stationary, friction must be 24.5 N up the slope. Checking the normal reaction, it is 5.0 × 9.81 × cos30° = 42.5 N, not simply mg.
很多学生画好重力箭头之后，错误地把重力分解成平行和垂直于斜面的分力。一个典型错误是把 mg cos θ 当作下滑分力，把 mg sin θ 当作支持力。窍门是正确构建直角三角形：斜面倾角 θ 是重力矢量与斜面法线之间的夹角。因此，沿斜面向下的分力为 mg sin θ，垂直于斜面的分力为 mg cos θ。如果涉及摩擦力或张力，务必先画清晰的受力图，重力箭头竖直向下，再分解。数值示例：一个 5.0 kg 的物体置于 30° 斜面上，下滑力为 5.0 × 9.81 × sin30° = 24.5 N。若物体静止，摩擦力应为 24.5 N 沿斜面向上。检查法向反力：5.0 × 9.81 × cos30° = 42.5 N，而不是简单的 mg。
3. Misapplying Newton’s Third Law Pairs | 误用牛顿第三定律的作用力与反作用力
A book rests on a table. Students often state that the weight of the book and the normal contact force from the table form a Newton’s third law pair. This is wrong; those two forces act on the same object (the book) and can be balanced, but a third-law pair must act on different bodies and be of the same type. The correct pair for the book’s weight is the gravitational pull of the book on the Earth. The correct pair for the normal force on the book is the downward normal force that the book exerts on the table. Always check: “If body A exerts a force on body B, then body B exerts an equal and opposite force on body A.” Type and magnitude must match, and they act on different objects.
一本书放在桌面上。学生常说书的重力和桌面对书的支持力是一对牛顿第三定律的作用力与反作用力。这是错误的：这两个力作用在同一个物体（书）上，可以平衡，但第三定律的力对必须作用在不同的物体上并且是同种性质的力。书的重力的正确反作用力是书对地球的引力。支持力的反作用力是书对桌面向下的压力。核查原则：“如果物体 A 对物体 B 施加一个力，那么物体 B 同时对 A 施加等大反向的力。” 类型和大小必须一致，且作用在不同物体上。
4. Confusing Phase and Path Difference in Interference | 干涉中相位差与波程差的混淆
In double-slit interference, a path difference of one full wavelength λ produces constructive interference because the phase difference is 2π. Students sometimes miscount half-wavelength shifts. For a fringe to be dark, the path difference must be an odd multiple of half a wavelength: (m + ½)λ. A common error is to think that any path difference other than mλ will give a dark fringe, ignoring that partial wavelengths also produce intermediate brightness. When converting path difference Δx to phase difference Δφ, use Δφ = (2π/λ) Δx. For sound waves from two loudspeakers, a path difference of 0.85 m for a tone of 680 Hz (λ = speed/frequency = 340/680 = 0.50 m) gives Δx = 1.7λ. The decimal fraction 0.7λ corresponds to a phase difference of 1.4π, resulting in partial destructive interference, not completely silent.
在双缝干涉中，波程差为一个完整波长 λ 时产生相长干涉，因为相位差为 2π。学生有时会数错半波长移动。暗纹条件要求波程差是半波长的奇数倍：(m + ½)λ。常见错误是认为任何不是 mλ 的路径差都会产生暗纹，忽略了部分波长也对应中间亮度。将波程差 Δx 转换为相位差 Δφ 时使用 Δφ = (2π/λ) Δx。对于两个扬声器发出的 680 Hz 声波（λ = 声速/频率 = 340/680 = 0.50 m），若波程差为 0.85 m，则 Δx = 1.7λ。小数部分 0.7λ 对应相位差 1.4π，导致部分削弱干涉，并非完全无声。
5. Potential Dividers and Changing LDR Resistance | 电位器与光敏电阻阻值变化
A circuit contains an LDR in series with a fixed resistor R across a 9.0 V battery. The output voltage is taken across R. As light intensity increases, the LDR resistance decreases. Many students deduce that the output p.d. across R decreases because total resistance decreases. Actually, the decreased LDR resistance makes the fraction across R larger: V_out = (R / (R + R_LDR)) × V_total. When R_LDR drops, the denominator shrinks, so V_out rises. The pitfall is forgetting that it is the ratio that matters, not just the total current. To avoid confusion, treat the potential divider formula directly and sketch the circuit with clearly labeled p.d.s. Always check: if LDR resistance goes to zero, V_out would equal V_total, confirming the logic.
电路由一个 LDR 与一个固定电阻 R 串联，接在 9.0 V 电池两端，输出电压取自 R 两端。随着光强增加，LDR 电阻减小。很多学生推论出 R 两端的电压会减小，因为总电阻减小了。实际上，LDR 电阻降低使得 R 分得的电压比例增大：V_out = (R / (R + R_LDR)) × V_total。当 R_LDR 减小，分母变小，V_out 上升。这个陷阱在于只考虑总电流变化而忽略了比例关系。避免混淆的方法是直接运用分压公式并画出清晰标注电压的电路图。检验：若 LDR 电阻趋近零，V_out 将等于 V_total，印证逻辑。
6. Misreading an I–V Characteristic for a Filament Lamp | 误读灯丝的 I–V 特性曲线
A typical question provides an I–V graph for a filament bulb and asks for the resistance at a specific p.d., say 6.0 V. Many candidates read the current (e.g., 0.40 A) and simply do R = V/I = 6.0/0.40 = 15 Ω. That is correct for that operating point, but the trap lies in then trying to find the resistance at another voltage by assuming R is constant or by drawing a straight line from the origin. The filament’s resistance increases with temperature, so the I–V graph curves downwards (increasing resistance). If the question asks for the p.d. when the resistance is a certain value, you must use the graph to find the corresponding V and I or apply a tangent/ chord correctly. Never take the gradient of the line from origin as 1/R; the correct resistance at a point is the ratio V/I, not the slope.
典型题目给出灯泡的 I–V 曲线并询问 6.0 V 时的电阻。很多考生读出电流（如 0.40 A），简单地用 R = V/I = 6.0/0.40 = 15 Ω 计算。这在那个工作点正确，但陷阱在于之后尝试求另一电压下的电阻时，假设电阻恒定或过原点画直线。灯丝电阻随温度升高而增大，I–V 曲线向下弯曲（电阻递增）。若题目要求电阻为某值时对应的电压，必须从图上找到对应的 V 和 I，或用切线/弦线正确求解。切勿把原点到该点的斜率当作 1/R；某点的电阻是 V/I 比值，而非斜率。
7. Confusing Decay Constant and Half-Life | 衰变常数与半衰期的混淆
The relationship λ = ln 2 / T½ is straightforward, but numerical mistakes thrive. Some students substitute T½ in years into a formula where λ must be in s⁻¹, forgetting unit conversions. For instance, a substance with half-life 5.0 years used in a decay equation A = A₀ e^{−λt} often sees errors when t = 3.0 years. First convert T½ to seconds: 5.0 × 365 × 24 × 3600 = 1.58 × 10⁸ s, then λ = ln2 / 1.58×10⁸ ≈ 4.39×10⁻⁹ s⁻¹. Alternatively, keep time in years: λ = ln2 / 5.0 = 0.1386 year⁻¹, then A = A₀ e^{−0.1386×3.0}. Both are valid provided units are consistent. Another trap: confusing the fraction remaining after n half-lives (1/2ⁿ) with the formula A = A₀ e^{−λt} and mixing them up.
关系式 λ = ln 2 / T½ 很直接，但数值错误频发。有的学生把以年为单位的半衰期直接代入需要 λ 以 s⁻¹ 为单位的公式，遗忘单位换算。例如，半衰期为 5.0 年的物质用于衰变方程 A = A₀ e^{−λt}，当 t = 3.0 年时错误常见。应先将 T½ 转换成秒：5.0 × 365 × 24 × 3600 = 1.58 × 10⁸ s，则 λ = ln2 / 1.58×10⁸ ≈ 4.39×10⁻⁹ s⁻¹。或者保持年为单位：λ = ln2 / 5.0 = 0.1386 year⁻¹，再用 A = A₀ e^{−0.1386×3.0}。只要单位一致，两种都可。另一个陷阱：混淆经过 n 个半衰期后的剩余比例 (1/2ⁿ) 与公式 A = A₀ e^{−λt} 的用法，混用出错。
8. Work Done and Area Under a Force–Extension Graph | 功与力–伸长图下的面积
When a material obeys Hooke’s law, the elastic potential energy stored is ½ F x. Many students forget that this formula only applies when the force is proportional to extension and the graph is a straight line through the origin. If the force–extension graph is curved (e.g., for rubber), the energy stored is the area under the curve, which must be found by counting squares or integration, not by ½ F x. Also, for a loading-unloading cycle, the area between the curves represents the work done against internal friction (hysteresis), often lost as heat. In exam questions, if a sample is stretched up to 0.15 m with final force 30 N but the curve is non-linear, using ½ × 30 × 0.15 = 2.25 J is wrong; the actual area might be 3.0 J. Always count squares carefully.
当材料遵守胡克定律时，储存的弹性势能为 ½ F x。很多学生忘记该公式只适用于力与伸长成正比且图为过原点直线的情况。如果力–伸长图是曲线（如橡胶），储存的能量是曲线下的面积，必须用数格子或积分求解，不能用 ½ F x。此外，在一个加卸载循环中，两曲线之间的面积代表克服内部摩擦（迟滞）所做的功，通常以热能散失。考试题中，若样品拉伸至 0.15 m，末力 30 N 但曲线非线性，用 ½ × 30 × 0.15 = 2.25 J 就是错的；实际面积可能是 3.0 J。务必认真数格子。
9. Sign Errors in Electromagnetic Induction (Lenz’s Law) | 电磁感应中的符号错误（楞次定律）
When explaining the direction of an induced e.m.f., students frequently state the law correctly (“opposes the change”) but then draw current or label poles inconsistently. A bar magnet’s N-pole approaches a coil: the induced current must create a N-pole at the near end to repel the magnet. However, many sketch the induced magnetic field direction as attracting, or they produce the correct pole but then draw the current the wrong way round the coil. Use the right-hand grip rule: for the coil’s near end to be N, current flows anticlockwise when viewed from that end. Another trap is confusing the galvanometer deflection direction with the current direction; always trace the circuit systematically.
在解释感应电动势方向时，学生常正确陈述定律（“反抗变化”），但随后画出电流或标定磁极时却自相矛盾。条形磁铁的 N 极靠近线圈：感应电流必须使线圈近端产生 N 极以排斥磁铁。但许多同学将感应磁场方向画成吸引，或者产生了正确的磁极但电流方向又绕错。使用右手螺旋定则：要使近端为 N 极，从该端观察电流应为逆时针。另一个陷阱是把电流计偏转方向与电流方向混淆；务必系统性地追踪电路。
10. Photoelectric Effect: Frequency Not Intensity | 光电效应：频率而非强度
A classic error is to assert that brighter light will increase the stopping potential. In fact, for a given frequency above the threshold, the maximum kinetic energy of emitted electrons (and thus stopping potential) depends only on frequency according to h f = Φ + K_max. Increasing intensity merely increases the number of photons per second, hence the photocurrent, but does not alter the maximum K.E. of each individual electron. This misunderstanding often appears in graph-plotting questions: the stopping potential V_s versus frequency f yields a straight line of slope h/e, independent of intensity. Students mistakenly draw two lines for two intensities.
一个经典错误是声称更亮的光会增大遏止电势。事实是，对于高于阈值的特定频率，发射电子的最大动能（继而遏止电势）只取决于频率，遵循 h f = Φ + K_max。增大强度仅增加每秒光子数，从而增大光电流，但不改变单个电子的最大动能。这种误解常见于作图题：遏止电势 V_s 对频率 f 图是一条斜率为 h/e 的直线，与强度无关。学生们错误地为两种强度画出两条线。
11. Misunderstanding Momentum in Explosions | 爆炸中的动量误解
Two trolleys initially at rest push apart due to a spring. The total momentum remains zero: m₁ v₁ + m₂ v₂ = 0. Students often forget that the velocities are vectors and must have opposite signs. Taking one direction as positive, the other velocity must be negative. In kinetic energy calculations, they might square the negative velocity and get the right numerical value, but then incorrectly compare magnitudes without considering direction. The kinetic energy ratio is inverse to the mass ratio: KE₁ / KE₂ = m₂ / m₁. Another pitfall is neglecting that the energy released by the spring is the sum of the two kinetic energies, not the kinetic energy of one trolley alone.
两辆原先静止的小车因弹簧而弹开。总动量保持为零：m₁ v₁ + m₂ v₂ = 0。学生常忘记速度是矢量，必须有正负号。取某一方向为正，另一方向的速度应为负。计算动能时，他们可能将负速度平方得到正确数值，但随后比较大小时忽略方向。动能比与质量成反比：KE₁ / KE₂ = m₂ / m₁。另一个易错点是弹簧释放的能量是两辆小车动能之和，而非单单一辆的动能。
12. Standing Waves: Confusing Node and Antinode Definitions | 驻波：波节与波腹定义的混淆
A common exam question asks to measure the wavelength of a stationary wave on a string or in an air column. For a string fixed at both ends, the distance between adjacent nodes is λ/2, not λ. For sound in a tube closed at one end, the distance from the closed end (node) to the first antinode is λ/4. Students frequently multiply by the wrong factor. Also, labeling pressure nodes and displacement nodes can be swapped: in a closed pipe, the closed end is a displacement node but a pressure antinode. Read the question carefully to determine whether the graph represents displacement or pressure variation.
常见考题要求测量弦上或空气柱中驻波的波长。对于两端固定的弦，相邻波节间的距离是 λ/2，不是 λ。对于一端封闭的管内声波，从封闭端（波节）到第一个波腹的距离为 λ/4。学生经常乘错倍数。此外，压强节点与位移节点可能互换：在闭管中，封闭端是位移节点但却是压强波腹。仔细读题，弄清图示是位移变化还是压强变化。
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