📚 A-Level WJEC Physics: Common Mistakes and Explanations | A-Level WJEC 物理：易错题精讲
This article tackles some of the most frequent errors students make in WJEC A-Level Physics. By working through these tricky questions, you will deepen your understanding of key concepts and sharpen your exam technique. Each section presents a common mistake, explains why it is wrong, and then shows the correct approach. Using these insights will help you avoid losing marks on topics that appear simple but hide subtle pitfalls.
本文精选了WJEC A-Level物理考试中高频易错题，逐一剖析错误根源并给出正确思路。每个小节针对一个常见误区，先展示典型错误，再解释错误原因，最后演示正确的物理分析和计算方法。掌握这些易错点，能帮助你在考试中有效避坑，提升成绩。
1. Resolving Forces on Inclined Planes | 斜面受力分析误区
A common question asks for the acceleration of a block sliding down a smooth slope inclined at an angle θ to the horizontal. Many candidates write the net force as mg sin θ, but then mistakenly use the mass alone to find acceleration, forgetting that the component of weight along the plane is the only force. However, a more subtle mistake is mixing up sine and cosine when the angle is given differently. The error often occurs when the diagram shows θ between the slope and the vertical. Students blindly apply mg sin θ without checking which angle is given.
一道常见题：光滑斜面倾角θ，求滑块沿斜面下滑的加速度。很多同学会写出合力为 mg sin θ，但有时题目给出的角度是斜面与竖直方向的夹角，若不加辨别直接套用 mg sin θ 就会出错。正确的处理是先明确定义角度，再使用正确的分量。沿斜面方向的重力分量为 mg sin θ，前提是 θ 为斜面与水平面夹角；若给出的是斜面与竖直方向的夹角 φ，则沿斜面方向的分量应写为 mg cos φ。
For example, if a slope makes an angle of 30° with the horizontal, the acceleration is g·sin30° = 4.9 m s⁻². But if the angle given is 30° to the vertical, the correct component is mg cos30°, giving a different value. Always draw a large triangle and label the angle carefully. Then Fr = mg sin(angle between weight and the plane normal) = mg sin(90° – vertical angle) = mg cos(vertical angle).
例如，若斜面与水平面成30°，加速度 a = g·sin30° = 4.9 m s⁻²。若题目却给出斜面与竖直方向成30°，则沿斜面分量为 mg cos30°，数值不同。解决方法：重新画大三角形，标出重力与斜面垂直方向的夹角，再使用三角函数。记住公式 F∥ = mg sin(斜面倾角_与水平面) = mg cos(与竖直方向夹角)。
2. Projectile Motion: Speed at the Top | 抛体运动中最高点的速度
A classic trick question: ‘A ball is thrown at an angle of 40° to the horizontal with initial speed 20 m s⁻¹. What is its speed at the highest point?’ Many students incorrectly answer 0 m s⁻¹, thinking that the ball stops momentarily. In projectile motion, the vertical component of velocity becomes zero at the peak, but the horizontal component remains constant throughout (ignoring air resistance). Therefore, the speed at the highest point is equal to the initial horizontal component: v_top = u cos θ.
一道经典易错题：“一球以初速20m/s、与水平成40°角抛出，最高点速率是多少？”很多学生会答0，认为最高点球静止。但在抛体运动中，最高点只是竖直分速度为零，水平分速度保持不变（忽略空气阻力）。因此最高点速率等于初速的水平分量：v_top = u cos θ。
For u = 20 m s⁻¹ and θ = 40°, v_top = 20 × cos40° ≈ 15.3 m s⁻¹. This error also appears when students are asked to calculate kinetic energy at the top, leading them to put zero. Always remember: horizontal motion is independent of vertical motion. The only influence on the horizontal component is air resistance, which is usually neglected unless stated.
代入数值：u=20 m/s，cos40°≈0.766，因此最高点速率约为15.3 m/s。如果误以为是零，那么计算最高点动能也会出错。务必牢记：水平运动与竖直运动独立，水平速度始终不变（除非计及空气阻力）。
3. Momentum Conservation and Direction | 动量守恒与方向
Examiners test the vector nature of momentum with head‑on collision problems. A typical mistake is ignoring the sign of velocity when a particle rebounds. Consider: ball A (mass 2 kg) moves right at 4 m s⁻¹, collides with stationary ball B (mass 1 kg). After the collision, A rebounds left at 1 m s⁻¹. Find B’s velocity. A student using scalar momentum only might write 2×4 = 2×1 + 1×v, giving v = 6 m s⁻¹. This is wrong because they did not assign a sign for direction.
阅卷者经常考察动量的矢量性。典型错误是忽略反弹速度的符号。例如：A球（2kg）以4m/s向右运动，与静止的B球（1kg）碰撞，碰后A以1m/s向左反弹。求B的速度。有学生会用标量运算：2×4 = 2×1 + 1×v，得出v=6 m/s。这错在没有定义正方向。
Correct method: define ‘right’ as positive. Initial momentum = +2×4 = +8 kg m s⁻¹. After collision, A’s velocity is -1 m s⁻¹ (rebounding). So 8 = 2×(-1) + 1×v ⇒ 8 = -2 + v ⇒ v = 10 m s⁻¹ to the right. Always use a sign convention and represent velocities as vectors. The same rule applies to conservation of momentum in explosions and recoil problems.
正确解法：设向右为正。初始动量 = +2×4 = +8 kg m/s。碰撞后A的速度为-1 m/s。代入动量守恒：8 = 2×(-1) + 1×v 解得 v = 10 m/s，方向向右。务必先规定正方向，用正负号表示方向，然后再代入计算。反冲、爆炸问题也一样。
4. Internal Resistance and Terminal Voltage | 内阻与路端电压
Students often confuse the emf of a cell and its terminal pd. A typical error appears when calculating the power delivered to an external load. For a cell of emf ε and internal resistance r connected to an external resistor R, some candidates incorrectly use ε²/R as the power dissipated in the load. The correct voltage across R is the terminal pd V = ε – Ir, and the current I = ε/(R+r). Thus, the power in the load is I²R = [ε/(R+r)]²R.
很多学生混淆电源电动势与路端电压。错误常出现在计算外电路功率时：对电动势为ε、内阻为r的电源，连接外电阻R，一些人会直接用 ε²/R 计算R上的功率。实际上，R两端的电压是路端电压 V = ε – Ir，而电流 I = ε/(R+r)。因此负载功率 P = I²R = [ε/(R+r)]²R。
Another common slip is failing to recognise that the lost volts (Ir) increase with current, causing the terminal pd to decrease. When the external resistance is half the internal resistance, about one‑third of the emf is dropped inside the source. Always start by drawing the full circuit and labelling total resistance. Use the voltage divider rule only after taking internal resistance into account.
另一个常见疏忽是忘记内电压 Ir 会随电流增大而增大，导致路端电压下降。当外电阻等于内阻一半时，内电压约占总电动势的三分之一。解题时务必先画出全电路，标出总电阻。只有在考虑内阻后，才能正确使用分压关系。
5. Phase Difference and Path Difference in Waves | 波的相位差与路程差
Interference questions frequently cause mistakes when converting path difference to phase difference. A widespread error is to treat one wavelength λ as equivalent to a phase difference of 180° instead of 360°. This results from memorising that ‘half a wavelength leads to destructive interference’ and then confusing the relationship. The correct link is: phase difference Δφ = (2π/λ) × path difference Δx.
干涉问题中，由路程差求相位差时经常出错。普遍错误是把一个波长λ对应于180°相位差而不是360°，这常源于机械记忆“半波长相消干涉”而混淆关系。正确的映射是：相位差 Δφ = (2π/λ) × 路程差 Δx。
For example, if the path difference is 0.75λ, the phase difference is 1.5π rad or 270°, not 135°. This mistake leads students to identify the fringe condition incorrectly. To avoid it, always recall that a full cycle of a wave corresponds to 2π radians, and two waves arriving exactly in phase have a path difference of nλ, not nλ/2.
举例：路程差为0.75λ，相位差应为1.5π rad（270°），而非135°。这个错误会让干涉条纹的判断完全错误。避免方法：始终记住一个完整波周期对应2π弧度，同相到达的路程差必须是 nλ，不是 nλ/2。
6. Photoelectric Effect: Misunderstanding Intensity | 光电效应：光强的误解
Many students state that increasing the intensity of incident light will increase the kinetic energy of the emitted photoelectrons. This is a fundamental misconception. The maximum kinetic energy K_max depends only on the frequency of the light and the work function φ of the metal: K_max = hf – φ. Intensity influences the number of photons per second, thus affecting the photocurrent, not the maximum energy per electron.
很多学生认为增大入射光强会提高光电子动能，这是根本性的误解。最大动能 K_max 仅取决于光的频率和金属的逸出功：K_max = hf – φ。光强只影响单位时间的光子数目，从而改变光电流的大小，并不改变单个电子的最大动能。
A typical exam question gives the work function of sodium as 2.3 eV and asks what happens when the light intensity is doubled while keeping the wavelength above the threshold. Students erroneously expect the electrons to be emitted with greater speed. Instead, if the frequency is below the threshold no electrons are emitted regardless of intensity. Above threshold, doubling intensity simply doubles the rate of electron emission. Write this as a firm rule.
典型考题：钠的逸出功为2.3eV，如果用波长大于截止波长的光照射，加倍光强会怎样？学生常错答为电子发射速度增大。正确是：频率低于截止频率时，无论多强光都不会有电子逸出；频率足够高时，加倍光强只加倍电子发射率。牢记公式 K_max = hf – φ。
7. Simple Harmonic Motion: Where Acceleration is Greatest | 简谐运动：加速度最大点
A surprisingly common error is confusing the positions of maximum velocity and maximum acceleration. Students often believe the oscillator moves fastest at the extreme displacement, because they think the restoring force is ‘pushing’ hardest there. In SHM, acceleration is proportional to displacement from equilibrium and directed towards the centre: a = –(2πf)²x. Thus magnitude of acceleration is maximum at maximum displacement, and zero at equilibrium. Velocity, on the other hand, is maximum at equilibrium and zero at the extremes.
一个惊人普遍的错误是混淆速度最大点和加速度最大点。很多学生以为振子在最大位移处运动最快，因为他们觉得此处回复力最强、“推”得最猛。简谐运动中，加速度与位移成正比且指向平衡位置：a = –(2πf)²x。因此加速度大小在最大位移处最大，在平衡位置为零。而速度则在平衡位置最大，在极端点为零。
Consider a mass–spring system oscillating horizontally. At x = A, acceleration a_max = (2πf)²A. The speed is zero because the mass momentarily stops to change direction. To avoid this mistake, sketch the energy changes: at extremes, all energy is elastic potential, zero kinetic; at equilibrium, all energy is kinetic. Always relate the gradient of the velocity–time graph to acceleration.
考虑水平弹簧振子。在 x=A 处，加速度最大 a_max = (2πf)²A，但速度为零，因为振子在此处瞬间停下掉头。为避开错误，可画能量转化图：极端点弹性势能最大、动能为零；平衡位置动能最大。也可从v-t图斜率理解加速度。
8. Energy Levels and Spectral Line Calculations | 能级与谱线计算
When calculating the wavelength of a photon emitted during an electron transition, a common error is using the wrong energy difference. Some students subtract the upper level energy from the lower level energy (E_lower – E_upper), obtaining a negative value and then dropping the sign. This often leads to an inverted frequency. The photon energy must equal E_upper – E_lower (positive). Moreover, they frequently forget to convert electronvolts to joules before using E = hc/λ.
计算电子跃迁发射光子的波长时，常犯错误是用 E_lower – E_upper 作差，出现负值后丢掉符号。这会导致频率计算错误。光子能量应等于高能级减去低能级：ΔE = E_upper – E_lower（正值）。另一个常见疏忽是忘记将电子伏特转换成焦耳再代入 E = hc/λ。
For hydrogen, the transition from n=3 to n=2 releases 1.89 eV. Students sometimes directly use 1.89 in λ = hc/E without conversion. Correct: E = 1.89 × 1.60×10⁻¹⁹ J, then λ = (6.63×10⁻³⁴ × 3.00×10⁸) / (3.02×10⁻¹⁹) ≈ 6.58×10⁻⁷ m. Also note that the ground state is the most negative; an excited electron has less negative energy. Always check the magnitude order.
例如氢原子n=3跃迁到n=2释放能量1.89eV。学生有时直接把1.89代入λ=hc/E而不单位换算。正确：E = 1.89×1.60×10⁻¹⁹ J，然后算得波长约6.58×10⁻⁷ m。还要注意基态能量最负，激发态能量负得更少，ΔE必为正。检查数量级。
9. Particle Physics: Conservation Laws in Decay | 粒子物理：衰变中的守恒律
WJEC specifications require applying conservation of lepton number, baryon number and charge to particle decays. A typical mistake is forgetting that a neutrino or antineutrino must be emitted to conserve lepton number. For instance, a student may write the beta-minus decay of a neutron as n → p + e⁻, missing the antineutrino. This violates lepton number conservation because the left side has lepton number 0, while the right side has +1 (electron only).
WJEC大纲要求利用轻子数、重子数与电荷守恒分析粒子衰变。典型错误是遗漏中微子或反中微子，导致轻子数不守恒。例如写中子β⁻衰变为 n → p + e⁻，漏写反中微子，这样左边轻子数为0，右边电子轻子数为+1，不守恒。
Correct equation: n → p + e⁻ + ν̅_e. (ν̅_e is an electron antineutrino with lepton number –1.) Total lepton number of products = +1 –1 = 0, matching the neutron. Likewise, in kaon and pion decays, always check strangeness is conserved in strong interactions but can change by 1 in weak decays. Using a table of properties helps prevent these errors.
正确方程：n → p + e⁻ + ν̅_e （反电子中微子轻子数为-1），产物总轻子数=+1-1=0，守恒。同样，在K介子和π介子衰变中，需注意奇异数在强相互作用中守恒，在弱作用中可改变±1。列表格辅助检查能有效避免遗漏。
10. Connected Bodies and Tension Errors | 连接体与张力错误
Problems involving two masses connected by a light string over a pulley, with one mass hanging and the other on a smooth table, often lead to mistakes in calculating the tension. A frequent error is to write T = m_hanging g as the tension, completely ignoring the acceleration of the system. Tension is not equal to the hanging weight unless the system is in equilibrium or the mass of the string is significant.
两个物体通过轻绳跨过滑轮连接，一个悬挂，一个在光滑桌面上，求张力时的错误率很高。常见错误是直接令绳张力 T = m_hanging g，完全忽略系统加速运动。实际上，只有当系统平衡或绳有质量时张力才等于悬挂物重力。
Proper method: treat the whole system to find acceleration a = (m_hang × g) / (m_hang + m_table), then isolate one mass to find T. For the mass on the table, T = m_table × a. Substituting gives T = (m_table × m_hang × g) / (m_table + m_hang). The tension is always less than m_hang g when the system accelerates. Doing this step blindly leads to inconsistent values in such two‑step problems.
正确方法：先整体法求加速度 a = (m_hang × g) / (m_hang + m_table)，再隔离物体求T。对桌面上物体有 T = m_table × a，代入得 T = (m_table × m_hang × g)/(m_table + m_hang)。加速时张力一定小于悬挂物重力。忽视加速度直接写T=mg是典型失分点。
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