📚 A-Level CIE Physics: Mastering Common Pitfall Questions | CIE A-Level 物理易错题精讲
In CIE A-Level Physics, even well-prepared students often stumble on certain questions not because the concepts are unknown, but because of subtle misconceptions, sign errors, or hasty assumptions. This article collects a set of classic mistake-prone problems, each accompanied by a detailed breakdown of the common error and the correct line of reasoning. Working through these examples will help you sharpen your exam technique and avoid costly slip-ups.
在CIE A-Level物理考试中，许多准备充分的学生也常在特定题目上失分，原因往往不是知识盲区，而是细微的误解、符号错误或仓促的假设。本文汇集了一组经典的易错题，每题都配有常见错误的剖析和正确解题思路的详细讲解。认真研读这些实例，将帮助你打磨应试技巧，避免不必要的失分。
1. Kinematics: Sign Conventions in Vertical Throw from a Moving Object | 运动学：从运动物体上抛掷的符号约定
A ball is thrown vertically upwards at 15 m/s from a hot‑air balloon that is ascending at 3 m/s. The balloon is 80 m above the ground at the instant of release. Taking upwards as positive, find the total time until the ball hits the ground. (g = 9.8 m/s²)
一个热气球以 3 m/s 匀速上升，在距地面 80 m 高处向上以 15 m/s 的速度抛出一球。取向上为正方向，求球从抛出到落地所用的总时间。（g = 9.8 m/s²）
Common mistake: Students often set the initial velocity u to +15 m/s or even 0 m/s, forgetting that the ball inherits the balloon’s upward velocity. Another frequent error is taking the displacement s as +80 m because the starting point is above the ground.
常见错误：学生常将初速度 u 设为 +15 m/s 甚至 0，忘记了球具有气球向上的速度。另一个常见错误是将位移 s 设为 +80 m，认为出发点在落地点的上方。
Correct approach: The ball’s initial velocity relative to the ground is u 3 = +18 m/s. Since it ends at the ground 80 m below the launch point, the displacement is s = –80 m. The acceleration is a = –9.8 m/s². Use s = ut + ½ at²:
正确解法：球相对于地面的初速度为 u = +15 + 3 = +18 m/s。由于落地点在抛出点下方 80 m，位移 s = –80 m。加速度 a = –9.8 m/s²。应用 s = ut + ½ at²：
–80 = 18t + ½ (–9.8)t²
Solve the quadratic; discard the negative root to obtain t ≈ 5.8 s. Always assign directions consistently to the vector quantities s, u and a.
解得二次方程，舍去负根，得 t ≈ 5.8 s。务必对位移、初速度和加速度均匀地规定正方向。
2. Newton’s Laws: Tension in Connected Bodies on an Incline | 牛顿定律：斜面连接体中的张力
A block of mass 4.0 kg on a smooth slope inclined at 30° is connected by a light inextensible string over a frictionless pulley to a hanging mass of 3.0 kg. Determine the tension in the string.
倾角 30° 的光滑斜面上放有 4.0 kg 的物块，通过轻绳和光滑滑轮与悬吊的 3.0 kg 重物相连。求绳中张力。
Common mistake: Assuming the tension equals the weight of the hanging mass (29.4 N) or the component of the weight of the block down the slope (19.6 N). Many candidates also forget that both bodies share the same acceleration magnitude.
常见错误：认为绳中张力等于悬挂物的重量（29.4 N）或斜面上物块重力沿斜面分力（19.6 N）。许多考生也忘记了两者的加速度大小相同。
Correct approach: Write two equations of motion. For the hanging mass (downwards positive): 3g – T = 3a (1). For the block on the slope (up the slope positive): T – 4g sin30° = 4a (2). Adding (1) and (2) eliminates T, giving 3g – 4g sin30° = 7a. Solve for a and then substitute back to find T 5 N. The tension is never equal to an isolated weight unless the system is in equilibrium.
正确解法：分别列出运动方程。悬吊物（向下为正）：3g – T = 3a (1)。斜面上物块（沿斜面向上为正）：T – 4g sin30° = 4a (2)。两式相加消去 T，得 3g – 4g sin30° = 7a。解出 a 后代入可得 T ≈ 23.5 N。除非系统平衡，张力绝不等于某一个物体的重量。
3. Work and Energy: Conservation with Non‑Conservative Forces | 功和能：含非保守力的能量守恒
A 2.0 kg block slides down a curved track from rest at a height of 5.0 m. It reaches the bottom with a speed of 8.0 m/s. Calculate the work done against friction during the descent.
一个 2.0 kg 的物块从曲面上 5.0 m 高处由静止滑下，到达底部时速率为 8.0 m/s。求下滑过程中克服摩擦力所做的功。
Common mistake: Ignoring friction and equating the loss in gravitational potential energy to the gain in kinetic energy, then concluding that the speed should be √2gh ≈ 9.9 m/s. The mistake is using mgh = ½ mv² when friction is present.
常见错误：忽略摩擦，将重力势能的减少直接等于动能的增加，然后断言速率应为 √2gh ≈ 9.9 m/s。错误在于有摩擦时仍使用 mgh = ½ mv²。
Correct approach: The work–energy principle for a non‑conservative system: mgh + Wfriction = ½ mv², where Wfriction is negative work done by friction. Rearranging: work done against friction = mgh – ½ mv² = (2.0 × 9.8 × 5.0) – (½ × 2.0 × 8.0²) = 98 – 64 = 34 J. Always account for dissipative forces when they are mentioned or implied.
正确解法：非保守系统的功能原理为：mgh + W摩擦 = ½ mv²，其中 W摩擦 是摩擦所做的负功。移项得克服摩擦做功 = mgh – ½ mv² = (2.0 × 9.8 × 5.0) – (½ × 2.0 × 8.0²) = 98 – 64 = 34 J。只要题目提到或暗示有耗散力，就必须将其计入能量方程。
4. Circular Motion: The Frictionless Banked Curve | 圆周运动：无摩擦斜弯道
A car of mass 1200 kg travels around a frictionless circular bend of radius 80 m that is banked at an angle of 20° to the horizontal. Determine the speed at which the car can negotiate the bend without relying on lateral friction.
质量为 1200 kg 的汽车驶过一半径为 80 m 的无摩擦弯道，弯道向内侧倾斜 20°。求汽车在不依赖侧向摩擦力的情况下安全过弯的速度。
Common mistake: Many learners resolve the normal reaction and then incorrectly set the vertical component equal to the weight, and the horizontal component equal to the centripetal force, but get the angle confused. Others try to use mg sinθ as the centripetal force.
常见错误：许多学生分解支持力后，错误地将竖直分力与重力平衡，水平分力当作向心力，却弄混了角度。也有人尝试用 mg sinθ 作为向心力。
Correct approach: The free‑body diagram shows the normal reaction R inclined at angle θ. Resolving vertically: R cosθ = mg. Horizontally: R sinθ = mv²/r. Dividing the two equations eliminates R and mass: tanθ = v²/(rg). Hence v = √rg tanθ = √80 × 9.8 × tan20° ≈ √784 × 0.364 ≈ 16.9 m/s. The safe speed is mass‑independent.
正确解法：受力分析图中支持力 R 与竖直方向夹角 θ。竖直方向：R cosθ = mg。水平方向：R sinθ = mv²/r。两式相除消去 R 和质量，得 tanθ = v²/(rg)。因此 v = √rg tanθ = √80 × 9.8 × tan20° ≈ √784 × 0.364 ≈ 16.9 m/s。安全速度与质量无关。
5. Electric Fields: Distinguishing Field Strength and Potential | 电场：区分电场强度与电势
Two point charges: Q₁ = +4.0 μC at x = 0 and Q₂ = –2.0 μC at x = 0.60 m. Find the position along the x‑axis where the electric field strength is zero, and another location where the electric potential is zero.
两个点电荷：Q₁ = +4.0 μC 位于 x = 0，Q₂ = –2.0 μC 位于 x = 0.60 m。求 x 轴上电场强度为零的位置，以及电势为零的另一位置。
Common mistake: Believing that zero field and zero potential occur at the same point. Students often add the potentials algebraically to find zero field, or vice versa. They also forget that potential is a scalar while field is a vector.
常见错误：以为零场强和零电势出现在同一点。学生常通过代数相加电势来寻找零场强，或反之。他们也忘记了电势是标量而场强是矢量。
Correct approach: For zero electric field, the vectors due to Q₁ and Q₂ must be equal in magnitude and opposite in direction. This occurs at a point outside the segment joining the charges, closer to the smaller charge: let distance from Q₂ be d, then k × 4.0/(0.60+d)² = k × 2.0/d² → d ≈ 1.46 m, so x = 0.60+1.46 = 2.06 m from origin. For zero potential, the algebraic sum of kQ/r terms is zero: 4.0/x + (–2.0)/(0.60–x) = 0 → x = 0.40 m. The two points are entirely different.
正确解法：电场强度为零时，两电荷产生的场强矢量须大小相等、方向相反。该点位于两电荷连线外，靠近较小的电荷：设距 Q₂ 的距离为 d，则 k × 4.0/(0.60+d)² = k × 2.0/d² → d ≈ 1.46 m，则 x = 0.60+1.46 = 2.06 m。电势为零时，kQ/r 的代数和为零：4.0/x + (–2.0)/(0.60–x) = 0 → x = 0.40 m。两个位置完全不同。
6. DC Circuits: The Loaded Potential Divider | 直流电路：带负载的分压器
A potential divider consists of a 9.0 V battery and two fixed resistors: 600 Ω and 400 Ω in series. A voltmeter of resistance 800 Ω is connected across the 400 Ω resistor. Calculate the voltmeter reading.
一个分压器由 9.0 V 电池和两个固定电阻 600 Ω 与 400 Ω 串联组成。将一个内阻为 800 Ω 的电压表并联在 400 Ω 电阻两端，求电压表的读数。
Common mistake: Using the unloaded potential divider formula Vout = Vtotal ay. This ignores the fact that the voltmeter is in parallel with the 400 Ω resistor, lowering the effective resistance.
常见错误：直接使用空载分压公式 Vout = V总 × (400/(600+400)) = 3.6 V。这忽略了电压表与 400 Ω 电阻并联从而降低了等效电阻。
Correct approach: The parallel combination of 400 Ω and 800 Ω gives an equivalent resistance Rp 400+800) = 266.7 Ω. The total circuit resistance is 600 + 266.7 = 866.7 Ω. Circuit current I = 9.0/866.7 ≈ 0.01038 A. The voltmeter reads the p.d. across the parallel pair: V = I × Rp .77 V. Any measuring instrument with finite resistance affects the circuit being measured.
正确解法：400 Ω 与 800 Ω 的并联等效电阻 Rp = (400×800)/(400+800) = 266.7 Ω。电路总电阻为 600 + 266.7 = 866.7 Ω。回路电流 I = 9.0/866.7 ≈ 0.01038 A。电压表读数为并联部分的电压：V = I × Rp ≈ 0.01038 × 266.7 ≈ 2.77 V。任何有限内阻的测量仪表都会影响被测电路。
7. Waves: Stationary Waves in a Tube Closed at One End | 波：一端封闭管中的驻波
A cylindrical pipe of length 0.850 m is closed at one end. Determine the three lowest resonant frequencies when the speed of sound in air is 340 m/s.
一根长 0.850 m 的圆柱形管子，一端封闭。设声速为 340 m/s，求三个最低的谐振频率。
Common mistake: Applying the formula for a tube open at both ends (fn = nv/(2L)), or incorrectly stating that the fundamental wavelength is 2L (which would be for an open‑open pipe). Remember that a closed end is a displacement node (pressure antinode) and an open end is a displacement antinode.
常见错误：误用两端开口管的公式（fn = nv/(2L)），或错误地认为基频波长为 2L（那是开‑开管的情况）。须牢记封闭端是位移波节（压力波腹），开口端是位移波腹。
Correct approach: For a pipe closed at one end, the standing wave has a node at the closed end and an antinode at the open end. The fundamental (1st harmonic) corresponds to L = λ₁/4 → λ₁ = 4L = 3.40 m, so f₁ = v/λ₁ = 340/3.40 = 100 Hz. Only odd harmonics exist: L = 3λ₃/4 → λ₃ = 4L/3, f₃ = 3v/(4L) = 300 Hz; f₅ = 5v/(4L) = 500 Hz. The frequencies are in the ratio 1 : 3 : 5.
正确解法：一端封闭的管中，封闭端为波节，开口端为波腹。基频（第一谐频）对应 L = λ₁/4 → λ₁ = 4L = 3.40 m，故 f₁ = v/λ₁ = 340/3.40 = 100 Hz。仅存在奇数次谐频：L = 3λ₃/4 → λ₃ = 4L/3，f₃ = 3v/(4L) = 300 Hz；f₅ = 5v/(4L) = 500 Hz。频率比为 1 : 3 : 5。
8. Superposition and Interference: Double‑Slit Fringe Separation | 叠加与干涉：双缝条纹间距
In a Young’s double‑slit experiment using light of wavelength 600 nm, the slit separation is 0.50 mm and the screen is 1.80 m from the slits. Predict how the fringe separation changes if (a) the slit separation is doubled, and (b) the whole apparatus is immersed in water (refractive index 1.33).
在杨氏双缝实验中，使用波长 600 nm 的光，双缝间距 0.50 mm，屏与缝相距 1.80 m。判断以下情况条纹间距如何变化：(a) 双缝间距加倍；(b) 整个装置浸入水中（折射率 1.33）。
Common mistake: Believing that increasing slit separation broadens the fringes. Many also forget that in water the wavelength becomes λ/n, and incorrectly apply the formula with the original wavelength.
常见错误：认为增大缝距会使条纹变宽。还有很多人忘记水中波长变为 λ/n，仍用原波长代入公式计算。
Correct approach: The fringe separation Δy = λD / a, where a is the slit separation. (a) If a is doubled, Δy is halved—the fringes become narrower. (b) In water the wavelength λ' oportional to λ, the fringe separation decreases by a factor of 1.33. Always remember the inverse relationship with slit spacing and direct proportionality with wavelength.
正确解法：条纹间距 Δy = λD / a，其中 a 为缝距。(a) 若 a 加倍，Δy 减半——条纹变窄。(b) 在水中波长 λ' = λ / n = 600/1.33 ≈ 451 nm。因为 Δy 与 λ 成正比，条纹间距减小为原来的 1/1.33。务必牢记与缝距的反比关系和与波长的正比关系。
9. Photoelectric Effect: Intensity, Frequency, and Stopping Potential | 光电效应：光强、频率与遏止电压
face with a work function of 3.0 eV. Calculate the stopping potential. Explain what happens to the stopping potential and the saturation photocurrent if the intensity of the radiation is doubled without changing the frequency.
频率为 1.2 × 10¹⁵ Hz 的紫外光照射在逸出功为 3.0 eV 的金属表面。计算遏止电压。并说明若辐射强度加倍而频率不变，遏止电压和饱和光电流如何变化。
Common mistake: Thinking that increasing intensity increases the kinetic energy of the emitted electrons, and therefore raises the stopping potential. Some also mix units by failing to convert electronvolts to joules or vice versa.
常见错误：认为增大光强会增大逸出电子的动能，从而提高遏止电压。也有人混淆单位，没有将电子伏特与焦耳进行正确换算。
Correct approach: Photon energy E = hf ³⁴ × 1.2×10¹⁵ = 7.96×10⁻¹⁹ J = (7.96×10⁻¹⁹)/(1.60×10⁻¹⁹) ≈ 4.97 eV. Maximum kinetic energy Kmax = E – φ = 4.97 – 3.0 = 1.97 eV. Stopping potential Vs = Kmax ers more photons per second, so the photocurrent doubles, but each photon still has the same energy, so K
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