📚 AS Physics Common Mistake Questions Explained | AS 物理：易错题精讲
In AS Physics, marks are often lost not through a lack of understanding, but by falling into predictable traps set by examiners. This article highlights the most common mistakes students make across mechanics, waves, electricity, and materials, and shows you how to think clearly and avoid them.
在AS物理考试中，学生失分往往不是因为概念不懂，而是掉进了出题人设计的常见陷阱。本文汇集了力学、波、电学和材料等领域中最典型的错误，并教你如何理清思路、避开这些失分点。
1. Confusing Newton’s Third Law Pairs with Equilibrium Forces | 混淆牛顿第三定律作用力与平衡力
A typical question: A book rests on a table. Many students state that the weight of the book and the upward normal force from the table form an action–reaction pair according to Newton’s third law. This is wrong. These two forces act on the same object (the book), so they are equilibrium forces, not a third-law pair.
一个典型问题：一本书静止在桌面上。许多学生会说书的重力和桌面向上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，因此它们是一对平衡力，而不是第三定律的力对。
The correct third-law pair for the book’s weight is: the Earth pulls the book downwards, and the book pulls the Earth upwards with an equal and opposite gravitational force. The pair for the normal force is: the table pushes the book upwards, and the book pushes the table downwards.
正确的第三定律力对是：书的重力源于地球对书的引力，同时书对地球施加等大反向的引力；支持力的力对是：桌子向上推书，书向下压桌子。
Always ask: ‘Do the two forces act on different objects?’ If yes, they can be a third-law pair. If not, they may be equilibrium forces.
始终问自己：“这两个力是否作用在不同的物体上？”若是，它们可能是一对作用力与反作用力；若不是，则可能是平衡力。
2. Misapplying SUVAT Equations with Incorrect Direction | 运动学公式中忘记设定正方向
Consider a ball thrown vertically upwards with an initial speed of 10 m s⁻¹. Students are asked to find the time to reach the highest point. They often write v = u + at with v = 0, u = 10, a = 9.8 and obtain t ≈ 1.02 s. This gives the correct magnitude but loses the physics if they treat acceleration due to gravity as positive without defining a direction.
设想一个被竖直上抛的小球，初速度10 m s⁻¹。要求计算到达最高点所需的时间。很多学生直接套用v = u + at，取v = 0，u = 10，a = 9.8，得出t ≈ 1.02 s。数值正确，但如果他们没有明确正方向，直接将重力加速度当作正值代入，物理过程就会混乱。
The safe approach: choose upwards as positive. Then u = +10 m s⁻¹, a = −9.8 m s⁻², v = 0. Using v = u + at gives 0 = 10 − 9.8t, so t = 1.02 s. The sign convention ensures that displacement, velocity and acceleration are all consistent. Never ignore the vector nature of these quantities.
安全的做法是：先设定向上为正方向。那么u = +10 m s⁻¹，a = −9.8 m s⁻²，v = 0。代入v = u + at得0 = 10 − 9.8t，因此t = 1.02 s。这样的符号规范保证了位移、速度和加速度的一致性。永远不要忽略运动量的矢量本质。
Another related error is using s = ut + ½at² with a positive a for the upward journey, which would give an increasing distance instead of a decreasing one. Always check whether your sign choice makes physical sense.
另一个常见错误是在上抛过程中仍使用正的加速度a代入s = ut + ½at²，这会导致位移随时间不减反增。一定要检查你的符号设定在物理上是否合理。
3. Assuming Frictional Force Always Equals μN | 认为摩擦力总等于 μN
A block rests on a rough slope inclined at an angle θ to the horizontal. The question asks for the frictional force acting on the block. Many students immediately write f = μR or f = μmg cosθ. This is only valid if the block is about to slip or is sliding. For a stationary block, the frictional force is simply whatever is required to maintain equilibrium – it could be less than the maximum static friction μₛR.
一个木块静止在倾角为θ的粗糙斜面上。题目要求计算木块受到的摩擦力。许多学生立刻写下f = μR 或 f = μmg cosθ。这只有在木块即将滑动或正在滑动时才成立。对于静止的木块，摩擦力只需要满足平衡条件即可，其大小可能远小于最大静摩擦力μₛR。
Correct reasoning: resolve forces along the slope. The component of weight down the slope is mg sinθ. Since the block is stationary, the frictional force must exactly balance this, so f = mg sinθ, directed up the slope. Only calculate the maximum possible friction if you need to check whether the block will slide.
正确的思路：沿斜面分解力。重力沿斜面的分量为mg sinθ。由于木块静止，摩擦力必须恰好与之平衡，因此f = mg sinθ，方向沿斜面向上。只有当需要判断物块是否会滑动时，才去计算最大静摩擦力。
4. Forgetting Internal Resistance in Circuit Calculations | 电路计算中忽略电源内阻
A cell of e.m.f. 12 V and internal resistance 2 Ω is connected to an external 4 Ω resistor. Students often calculate the current as I = V / R = 12 / 4 = 3 A. This ignores the internal resistance, which is a very common mistake.
一个电动势为12 V、内阻为2 Ω的电池连接着一个4 Ω的外电阻。很多学生直接计算电流为I = V / R = 12 / 4 = 3 A。这就忽略了电源内阻，是极为常见的错误。
The correct total resistance in the circuit is R_total = r + R = 2 + 4 = 6 Ω. Then current I = ε / R_total = 12 / 6 = 2 A. The terminal p.d. across the cell is then V = ε − Ir = 12 − 2×2 = 8 V. Questions frequently ask for terminal p.d., so always check whether you need to subtract the ‘lost volts’ across the internal resistance.
正确的电路总电阻为R_total = r + R = 2 + 4 = 6 Ω，因此电流I = ε / R_total = 12 / 6 = 2 A。电池的端电压则为V = ε − Ir = 12 − 2×2 = 8 V。考题经常要求计算端电压，所以永远要记得从电动势中减去内阻上的“损耗电压”。
5. Misunderstanding Path Difference and Phase Difference in Waves | 波的路程差与相位差对应关系出错
When two coherent sources produce an interference pattern, students often mix up the conditions for constructive and destructive interference. A common mistake is to think that a path difference of λ gives destructive interference. In fact, constructive interference occurs when the path difference is a whole number of wavelengths: nλ (n = 0, 1, 2…). Destructive interference occurs when the path difference is an odd multiple of half-wavelengths: (n + ½)λ.
两个相干波源产生干涉图样时，学生经常混淆加强和减弱的条件。常见错误是认为路程差为一个波长λ时得到减弱。实际上，当路程差为波长的整数倍nλ（n = 0, 1, 2…）时，发生加强干涉；当路程差为半波长的奇数倍(n + ½)λ时，发生减弱干涉。
The phase difference Δφ is related to path difference Δx by Δφ = (2π/λ) × Δx. So a path difference of λ gives a phase difference of exactly 2π, meaning the waves arrive in phase (constructive). Many students incorrectly associate a full-wavelength path difference with a phase difference of π.
相位差Δφ与路程差Δx的关系为Δφ = (2π/λ) × Δx。因此，路程差为一个λ时，相位差正好是2π，表示波到达时同相（加强）。许多学生错误地认为一整波长的路程差对应相位差π。
6. Incorrect Use of Young’s Double-Slit Formula | 错误使用杨氏双缝干涉公式
The fringe spacing Δx in a double-slit experiment is given by Δx = λD / a, where λ is the wavelength, D is the distance from the slits to the screen, and a is the separation between the two slits. A frequent error is confusing the slit separation a with the slit width (which affects diffraction, not the fringe spacing).
杨氏双缝实验中的条纹间距公式为Δx = λD / a，其中λ是波长，D是双缝到屏幕的距离，a是双缝之间的间距。一个常见的混淆是把双缝间距a错当成单缝的缝宽（缝宽影响衍射效果，并不决定条纹间距）。
Another pitfall is using the formula when the small-angle approximation no longer holds, e.g. when the fringes are observed at large angles. However, at AS level the approximation is usually valid, but students must still ensure that they are using the correct slit-to-screen distance and consistent units (e.g. everything in metres).
另一个易错点是当小角度近似不再成立时仍套用该公式；不过AS阶段通常满足近似条件。但学生仍需确保代入的是正确的双缝到屏幕的距离，并且单位统一（例如全部使用米）。
When a question asks how the fringe spacing changes when the slit separation is halved, the answer is that Δx doubles (since Δx ∝ 1/a). Students sometimes mistakenly think that doubling a will double Δx.
当题目问双缝间距减半时条纹间距如何变化，答案是Δx加倍（因为Δx与1/a成正比）。学生有时会错误地认为增大a会使Δx变大。
7. Confusing Electrical Power Formulas | 混淆电功率公式 P = I²R 与 P = V²/R
The two expressions for electrical power — P = I²R and P = V²/R — are both correct for a resistor, but they are useful in different contexts. A common mistake is to use P = V²/R when the resistor is in series with another, and the voltage across it is not the full supply voltage.
电阻的电功率有两个常用表达式——P = I²R 和 P = V²/R，二者都正确，但分别适用于不同情境。一个常见错误是：当电阻与另一电阻串联时，仍使用P = V²/R，而它两端的电压并不是电源的总电压。
For example, two identical lamps rated ‘6 V, 12 W’ are connected in series to a 6 V supply. Many students calculate the total resistance of one lamp from its rating (R = V²/P = 6²/12 = 3 Ω), then say each lamp will dissipate 12 W. Actually, in series each lamp gets only 3 V, so the power dissipated in each is (3 V)² / 3 Ω = 3 W, not 12 W. The correct way is to recognise that the current through the series combination is I = 6 V / 6 Ω = 1 A, and then power per lamp = I²R = 1² × 3 = 3 W.
比如，两个标有“6 V, 12 W”的相同灯泡串联后接在6 V电源上。很多学生根据额定值求出一个灯泡的电阻R = V²/P = 6²/12 = 3 Ω，然后就认为每个灯泡仍消耗12 W。实际上，串联后每个灯泡只分得3 V电压，因此每个灯泡的实际功率为(3 V)² / 3 Ω = 3 W，而不是12 W。正确的做法是求出串联总电流 I = 6 V / 6 Ω = 1 A，再通过P = I²R = 1² × 3 = 3 W计算单个灯泡的功率。
Always check which quantity — current or voltage — is the same for the components you are analysing, and choose the appropriate power formula.
务必先判断你正在分析的元件上，是电流相同还是电压相同，再选用合适的功率公式。
8. Misinterpreting Momentum Conservation in Collisions | 碰撞问题中动量守恒的矢量处理错误
The principle of conservation of momentum is straightforward — total momentum before collision equals total momentum after, provided no external resultant force acts. However, students frequently treat momentum as a scalar. In two-dimensional collisions or when objects move in opposite directions, you must assign a positive direction.
动量守恒定律本身很简单——只要系统合外力为零，碰撞前的总动量等于碰撞后的总动量。但学生常常将动量当作标量处理。在二维碰撞中，或当物体反向运动时，必须先规定正方向。
Example: a 2 kg trolley moving at 3 m s⁻¹ to the right collides and sticks to a 1 kg trolley moving at 2 m s⁻¹ to the left. Find the common velocity after the collision. A wrong calculation: momentum before = 2×3 + 1×2 = 6 + 2 = 8 kg m s⁻¹, so velocity = 8 / 3 = 2.67 m s⁻¹. The mistake is ignoring the direction of the second trolley. Taking right as positive, its velocity is −2 m s⁻¹, giving total momentum = 2×3 + 1×(−2) = 6 − 2 = 4 kg m s⁻¹, so final velocity = 4 / 3 = 1.33 m s⁻¹ to the right.
例题：一辆质量2 kg的小车以3 m s⁻¹的速度向右运动，与一辆质量1 kg、以2 m s⁻¹向左运动的小车相撞并粘在一起。求碰撞后的共同速度。错误计算：碰前总动量 = 2×3 + 1×2 = 6 + 2 = 8 kg m s⁻¹，因此速度 = 8 / 3 = 2.67 m s⁻¹。错误在于忽略了第二辆小车的运动方向。若取向右为正，则第二辆车速度为−2 m s⁻¹，总动量 = 2×3 + 1×(−2) = 6 − 2 = 4 kg m s⁻¹，最后速度 = 4 / 3 = 1.33 m s⁻¹，方向向右。
Always state the positive direction clearly at the start of the problem to avoid sign errors.
解题一开始就要明确正方向，这样才能避免符号错误。
9. Forgetting That Centripetal Force is a Net Force | 忘记向心力是合力而非独立力
A classic mistake appears in questions about a mass swinging in a vertical circle. At the lowest point, students often write the tension in the string as T = mv²/r, completely forgetting the weight of the mass. The correct statement is that the centripetal force is the net force towards the centre, which at the bottom is T − mg = mv²/r, so T = mg + mv²/r.
在竖直圆周运动的问题中常见一个典型错误：在最低点时，学生常常写出绳的拉力 T = mv²/r，完全忽略了重力。正确的理解是，向心力是指向圆心的合力，在最低点应有 T − mg = mv²/r，因此 T = mg + mv²/r。
At the highest point, the net force is T + mg = mv²/r. If the mass is moving just fast enough to keep the string taut, then T = 0 and mg = mv²/r, giving the critical speed. The mistake is to think that there is an extra ‘centripetal force’ provided by something else; in reality, it comes from real forces like tension, gravity, friction or the normal force.
在最高点，合力为 T + mg = mv²/r。若物体刚好能维持圆周运动，绳的拉力T = 0，此时 mg = mv²/r 可求出临界速度。错误在于认为存在一个额外的“向心力”；实际上向心力是由真实的力（拉力、重力、摩擦力或支持力）充当的。
10. Incorrectly Splitting Initial Velocity in Projectile Motion | 抛体运动初速度分解错误
In projectile problems, the time of flight depends solely on the vertical component of the initial velocity, provided the launch and landing are at the same height. Students often mistakenly use the total initial speed to find the time. For a projectile launched at speed u at an angle θ to the horizontal, the vertical component is u sinθ. Use this with v = u sinθ + a t (with a = −g) to find the time to maximum height, and then double it for the total flight time when the landing level is the same.
在抛体运动中，若发射点和落地点高度相同，飞行时间仅取决于初速度的竖直分量。学生常错误地用初速度的合速度来计算时间。一个以速度u、与水平方向成θ角射出的抛体，其竖直分量为u sinθ。用v = u sinθ + a t（a = −g）求得到达最高点的时间，再乘以2即得总飞行时间。
Another frequent error is horizontal displacement: it is simply horizontal velocity × total time, i.e. (u cosθ) × t_total. Students sometimes treat the horizontal motion as accelerated, but in the absence of air resistance, the horizontal component remains constant.
另一个常见错误是计算水平位移：水平位移即为水平速度 × 总时间，即(u cosθ) × t_total。有学生误以为水平方向也有加速度，但在忽略空气阻力的前提下，水平速度始终不变。
11. Misusing Stress, Strain and the Young Modulus | 应力、应变与杨氏模量概念混淆
A wire is stretched by a force. Questions often ask what happens to the Young modulus when the wire is made longer or thinner. The Young modulus is a material property and does not depend on the dimensions of the sample. Many students believe a thinner wire has a larger Young modulus because it stretches more easily – but it has a smaller cross-sectional area, so for the same force it experiences a larger stress and therefore a larger strain, but the ratio stress/strain (the Young modulus) remains unchanged for the same material.
一根金属丝被拉伸。题目常问若将丝做长或做细，杨氏模量如何变化。杨氏模量是材料的固有属性，不依赖于样品的尺寸。很多学生认为细的金属丝更容易拉伸，因此杨氏模量更大——实际上，细丝的截面积更小，在相同拉力下应力更大，从而应变更大，但应力与应变的比值（即杨氏模量）对于同一种材料保持不变。
To find the Young modulus, use E = (F L) / (A ΔL). Experimental errors often arise from measuring the extension ΔL too early, or not allowing the wire to return to its original length between loads (checking for elastic limit). Always plot stress vs strain, and the gradient gives E.
计算杨氏模量的公式为E = (F L) / (A ΔL)。实验误差常来自过早记录伸长量ΔL，或在每次加力之间未让金属丝回复原长（以检查是否超过弹性极限）。一定要画应力–应变图，其斜率即为E。
12. Adding Resistors in Series and Parallel Incorrectly | 串并联等效电阻计算粗心
The formulas are well known: R_total = R₁ + R₂ + … for series, and 1/R_total = 1/R₁ + 1/R₂ + … for parallel. Still, calculation slips are common. In particular, for two resistors in parallel, the product-over-sum shortcut R_total = (R₁ × R₂) / (R₁ + R₂) is valid, but students often use it when more than two resistors are in parallel, or they forget to invert the result after summing reciprocals.
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