Example Responses for AQA International AS Physics | AQA国际AS物理示例答题

📚 Example Responses for AQA International AS Physics | AQA国际AS物理示例答题

This article shows you how to turn an ordinary answer into a full-mark answer for PH01 Unit 1. We will examine the command words, the structure of calculations, and full worked examples that follow exactly what the mark scheme rewards.

这篇文章将向你展示如何把普通作答变成PH01 Unit 1的满分作答。我们会分析指令词、计算题结构，并给出完全符合评分标准得分点的完整例题。

1. Understanding Command Words | 理解指令词

Every question begins with a command word that tells you exactly how much detail the examiner expects. For example, ‘state’ may need only a value, while ‘explain’ requires a reason with a cause-and-effect link.

每一道题都以指令词开头，它告诉考官期待多少细节。例如“写出”可能只需要一个数值，而“解释”则需要包含因果关系的理由。

Define – write one exact sentence. Use the scientific term inside the definition, for example: ‘The Young modulus is the ratio of tensile stress to tensile strain.’

定义 —— 写一个准确的句子，并在定义中使用科学术语，例如：“杨氏模量是拉伸应力与拉伸应变之比。”
State – no explanation is needed. A short phrase, equation or numerical answer is enough.

写出/给出 —— 不需要解释，一个短语、公式或数值答案即可。
Calculate – show the equation, substitute the numbers, and give the answer with an appropriate unit.

计算 —— 写出公式，代入数值，并给出带有正确单位的答案。
Show that – start from a given or known equation and reach the stated result. Keep at least one extra significant figure until the end so that your rounding does not create a misleading answer.

证明/验证 —— 从给定公式或已知公式出发，推得题目给出的结果。过程中保留至少一位额外有效数字，避免因四舍五入造成误导性结果。
Explain – use ‘because’ to link a physics principle to the situation in the question.

解释 —— 使用“因为”将物理原理与题目情境联系起来。
Suggest – apply known ideas to an unfamiliar situation. You may use approximate values or reasonable assumptions.

提出/建议 —— 将已有知识运用到陌生情境中，可以使用近似值或合理假设。

2. Structure of a Short Answer | 简答题的作答结构

A two-mark ‘explain’ question is not answered well by a single long sentence. The examiner is looking for two separate points: the physics idea and the link to the evidence in the question.

一道2分的“解释”题并不适合用一句冗长的话来作答。考官期待两个独立得分点：物理原理，以及与题干证据的联系。

For example: ‘A stream of electrons passes through a narrow gap. Explain, in terms of its wavelength, why the electron beam diffracts.’ A strong answer is: ‘Moving electrons have a de Broglie wavelength comparable to the width of the gap. When the gap is about the same size as the wavelength, the electron wave spreads out into the shadow region, so diffraction is observed.’

例如：“一束电子穿过一个狭缝。试从波长的角度解释电子束发生衍射的原因。”一个高分答案是：“运动的电子具有德布罗意波长，其大小与狭缝宽度可比。当缝宽接近波长时，电子波会扩展进入阴影区域，因此观察到衍射现象。”

Write your answers in the same order as the mark scheme. If a question says ‘state and explain’, first state the fact, then give the reason. Do not bury the answer inside a long narrative.

作答顺序应与评分标准一致。如果题目要求“写出并解释”，先陈述事实，再给理由，不要把答案埋没在长篇叙述中。

3. Structure of a Calculation Answer | 计算题的解答结构

For calculation questions in PH01, follow the sequence: equation, substitution, answer, unit. Even if your final number is wrong, you can still earn method marks from the equation and substitution.

对于PH01的计算题，顺序应为：公式、代入、答案、单位。即使最终数值错误，只要写出公式并正确代入，仍能获得方法分。

v = u + at

Suppose u = 2.0 m s⁻¹, a = 1.5 m s⁻² and t = 4.0 s. The correct presentation is: v = 2.0 + 1.5 × 4.0 = 8.0 m s⁻¹. Never write the answer without showing where it came from, because the examiner needs to see the substitution.

设u = 2.0 m s⁻¹，a = 1.5 m s⁻²，t = 4.0 s。正确书写是：v = 2.0 + 1.5 × 4.0 = 8.0 m s⁻¹。切勿只写答案而不展示来源，考官需要看到代入过程。

Check that the unit is consistent with the equation. If mass is given in grams, convert to kilograms. If length is given in cm, convert to metres before substituting into a formula that produces metres.

检查单位是否与公式一致。若质量以克给出，先换算成千克；若长度以厘米给出，在代入以米为单位的公式前先换算为米。

4. Worked Example 1: Measurements and Uncertainty | 实例1：测量与不确定度

Question: A student measures the diameter d of a wire with a micrometer and records d = 0.48 mm ± 0.01 mm. Calculate the cross-sectional area A of the wire and state the absolute uncertainty in A.

题目：学生用千分尺测量金属丝直径d，记录d = 0.48 mm ± 0.01 mm。计算金属丝的横截面积A，并给出A的绝对不确定度。

Full answer: Convert the diameter to metres: d = 0.48 × 10⁻³ m. The cross-sectional area is

满分答案：先将直径换算为米：d = 0.48 × 10⁻³ m。横截面积为

A = πd² / 4 = π × (0.48 × 10⁻³)² / 4 = 1.81 × 10⁻⁷ m²

The percentage uncertainty in d is (0.01 / 0.48) × 100% = 2.08%. Because A ∝ d², the percentage uncertainty in A is twice this value, 4.17%. The absolute uncertainty in A is 4.17% × 1.81 × 10⁻⁷ = 7.5 × 10⁻⁹ m², so the final answer is

直径的百分不确定度为(0.01 / 0.48) × 100% = 2.08%。由于A ∝ d²，A的百分不确定度是其两倍，即4.17%。A的绝对不确定度为4.17% × 1.81 × 10⁻⁷ = 7.5 × 10⁻⁹ m²，因此最终结果为

A = (1.81 ± 0.08) × 10⁻⁷ m²

Notice how the answer quotes the uncertainty to one significant figure and matches the decimal place of the measurement. Do not quote the central value to more than three significant figures unless instructed.

注意答案将不确定度保留一位有效数字，并与主值的小数位对齐。除非题目要求，否则主值不要超过三位有效数字。

5. Worked Example 2: Motion Graphs | 实例2：运动学图像

Question: An object moves in a straight line. For the first 4 s it accelerates uniformly from rest to 2 m s⁻¹. It then travels at constant velocity for 4 s, and decelerates uniformly to rest over the next 2 s. Calculate the total distance travelled.

题目：一物体沿直线运动。前4 s内从静止匀加速到2 m s⁻¹，随后以恒速运动4 s，最后2 s内匀减速至静止。求物体通过的总距离。

Full answer: Draw a velocity–time graph. The distance is the area under the graph:

满分答案：画出速度–时间图像。距离等于图像下的面积：

Distance = ½ × 4 × 2 + 4 × 2 + ½ × 2 × 2 = 4 + 8 + 2 = 14 m

The first term is the triangular area while the object accelerates, the second term is the rectangle at constant speed, and the third term is the triangle during deceleration. Give the unit ‘m’ clearly; a number without a unit loses the mark.

第一项是加速阶段的三角形面积，第二项是匀速阶段的矩形面积，第三项是减速阶段的三角形面积。务必写明单位“m”，没有单位的数字会失分。

For graph questions, you can also gain marks by labelling both axes correctly with quantity and unit, for example ‘velocity / m s⁻¹’ and ‘time / s’.

在图像题中，正确标注两个轴的物理量及单位也能得分，例如“velocity / m s⁻¹”和“time / s”。

6. Worked Example 3: Energy and Work | 实例3：能量与做功

Question: A car of mass 1200 kg accelerates from 10 m s⁻¹ to 20 m s⁻¹ along a horizontal road. Friction and air resistance are negligible. Calculate the work done by the engine and the average driving force if the acceleration takes place over 200 m.

题目：一辆质量1200 kg的汽车在水平路面上从10 m s⁻¹加速到20 m s⁻¹。忽略摩擦和空气阻力。求发动机所做的功，以及若加速距离为200 m时的平均牵引力。

Full answer: The work done equals the change in kinetic energy:

满分答案：做功等于动能变化量：

W = ΔE_k = ½mv² − ½mu² = ½ × 1200 × 20² − ½ × 1200 × 10² = 240 000 − 60 000 = 180 000 J

The average driving force is obtained from W = Fs:

平均牵引力由W = Fs得到：

F = W / s = 180 000 / 200 = 900 N

Use the equation W = Fs only when F is constant along the direction of motion. If the force is at an angle θ to the displacement, use W = Fs cos θ. Mentioning this in an answer can secure an extra explanation mark.

只有当力沿运动方向且恒定时才能使用W = Fs。如果力与位移方向夹角为θ，应使用W = Fs cos θ。在答案中注明这一点，可能额外获得解释分。

7. Worked Example 4: Waves | 实例4：波动

Question: A progressive wave travels along a rope with a wavelength of 0.80 m and a frequency of 250 Hz. Calculate the speed of the wave.

题目：一列行波沿绳子传播，波长为0.80 m，频率为250 Hz。计算波的传播速度。

Full answer: Use the wave equation v = fλ, so v = 250 × 0.80 = 200 m s⁻¹.

满分答案：使用波速公式v = fλ，因此v = 250 × 0.80 = 200 m s⁻¹。

For a ‘describe how you would measure’ part, state the method: hold a metre rule parallel to the rope and take a photograph of the wave; measure the distance between adjacent crests. Repeat for several separations and divide by the number of wavelengths to reduce uncertainty.

对于“描述你如何测量”的部分，应写明方法：将米尺平行于绳子放置并拍摄波的照片；测量相邻波峰之间的距离。多次测量多个间隔后除以波长个数，以减小不确定度。

Do not confuse frequency with wave speed. Frequency is determined by the source, not the medium; wave speed is determined by the medium. The equation v = fλ connects them through the wavelength.

不要混淆频率与波速。频率由波源决定，而非介质；波速由介质决定。公式v = fλ通过波长将两者联系起来。

8. Units, Prefixes and Significant Figures | 单位、前缀与有效数字

Unit 1 frequently uses prefixes such as nano (n = 10⁻⁹), micro (μ = 10⁻⁶), milli (m = 10⁻³), centi (c = 10⁻²), kilo (k = 10³) and mega (M = 10⁶). You must convert before doing calculations.

Unit 1经常使用前缀：纳n = 10⁻⁹、微μ = 10⁻⁶、毫m = 10⁻³、厘c = 10⁻²、千k = 10³、兆M = 10⁶。计算前必须完成换算。

For example, a microwave wavelength might be 12 cm. Convert to metres: 12 cm = 12 × 10⁻² m = 0.12 m. If you leave it as 12 cm inside the wave equation, the speed will come out as cm Hz instead of m s⁻¹, and the unit will reveal the error.

例如，微波波长可能是12 cm。换算为米：12 cm = 12 × 10⁻² m = 0.12 m。如果直接在波速公式中使用12 cm，波速就会变成cm·Hz而不是m s⁻¹，而单位本身就会暴露出错误。

Significant figures also matter. If the question gives data to two significant figures, the final answer should normally also be to two significant figures. The exception is when the first digit is 1, where keeping three is often acceptable.

有效数字同样重要。若题干数据为两位有效数字，最终答案通常也应为两位有效数字。例外情况是当首位为1时，保留三位通常可以接受。

9. Graph Work: Labelling, Lines and Gradients | 图表：标注、连线和斜率

In a plotting question, six marks usually come from: both axes labelled correctly, suitable scales, all points plotted accurately, a best-fit straight line, a large gradient triangle, and a correct calculation of the gradient with units.

绘图题通常有6分：两轴正确标注、刻度合适、所有点准确描出、拟合直线、大三角形取斜率、斜率单位正确。

Label axes with ‘quantity/unit’, for example ‘extension / mm’. Choose scales so that the plotted points occupy more than half of the grid. Do not force the line through the origin unless the data and physics require it.

坐标轴应写成“物理量/单位”，例如“extension / mm”。刻度选择应使点占据网格一半以上。除非数据和物理原理要求，不要强行让直线经过原点。

When finding the gradient, choose two points on the drawn line, not two plotted data points. The two points should be far apart so the percentage reading uncertainty is small. Show the calculation:

计算斜率时，应在所画直线上取两点，而不是直接用数据点。两点相距越远，读数百分不确定度越小。写出计算过程：

gradient = Δy / Δx = (y₂ − y₁) / (x₂ − x₁)

If the graph is a straight line through the origin, the gradient often represents a physical quantity such as force per extension, which is the spring constant k.

如果图像是过原点的直线，其斜率常代表某个物理量，例如力除以伸长量为劲度系数k。

10. Describing a Practical Procedure | 实验过程描述题

A common PH01 extended response asks you to describe how to determine a physical quantity. For example: ‘Describe how to determine the Young modulus of a metal wire, including any precautions.’

PH01常见的论述题要求你描述测得某个物理量的实验步骤。例如：“描述测量金属丝杨氏模量的方法，包括注意事项。”

A full-mark answer would include: measure the original length L of the wire with a metre rule, measure the diameter d at several points along the wire using a micrometer and average, load the wire with known masses, measure the extension ΔL using a marker and a travelling microscope or vernier scale, calculate stress = F/A and strain = ΔL/L, and plot a stress–strain graph whose gradient is the Young modulus.

满分答案应包括：用米尺测量金属丝原长L；用千分尺沿金属丝多处测量直径d并取平均；通过已知质量加载；用标记和读数显微镜或游标尺测量伸长量ΔL；计算应力 = F/A和应变 = ΔL/L；绘制应力–应变图，其斜率即为杨氏模量。

Add precautions such as: keep the load below the elastic limit, use a long wire so that the extension is measurable, avoid parallax when reading scales, and check that the wire is straight before each reading.

同时补充注意事项：载荷保持在弹性限度内；使用长金属丝以保证伸长量可测；读数时避免视差；每次读数前确保金属丝绷直。

Each precaution must be linked to why it improves accuracy. A vague ‘be careful’ earns no mark; ‘measure diameter several times to reduce random errors’ is a credit-worthy point.

每一项注意事项都必须说明为何能提高准确性。笼统的“小心

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Example Responses for AQA International AS Physics | AQA国际AS物理示例答题