📚 Cracking the AS Physics Unit 1 Jan 19 Experimental Investigation | 攻克 AS 物理单元 1 2019 年 1 月实验探究题
The January 2019 Edexcel AS Physics Unit 1 paper presented a compelling practical investigation: using two light gates and a timer to measure the acceleration of a freely falling ball. This type of question tests not only your grasp of kinematic equations but also your ability to handle raw data, evaluate uncertainties, and suggest meaningful improvements. Whether you are revising for your mocks or preparing for the final exam, a thorough walkthrough of this past-paper investigation will sharpen your analytical skills and boost your confidence.
2019 年 1 月 Edexcel AS 物理单元 1 试卷中，展示了一个引人入胜的实验探究：利用两个光门和一个计时器来测量自由下落小球的加速度。这类题目既考查你对运动学方程的掌握，也考验你处理原始数据、评估不确定度和提出有意义的改进措施的能力。无论你是在为模拟考做准备，还是为最终考试冲刺，仔细拆解这道真题探究都将提升你的分析能力并增强应试信心。
1. Overview of the Investigation | 实验探究概述
The goal was to determine the acceleration due to gravity, g, by releasing a metal ball from rest and recording its motion through two light gates placed one above the other. A data logger or timer measured the short time intervals during which the ball interrupted each beam, as well as the time taken to travel from the first gate to the second.
实验的目标是通过将金属球从静止释放并记录其经过两个一上一下放置的光门时的运动，来测定重力加速度 g。数据采集器或计时器测量了小球遮挡每个光束的短暂时间间隔，以及小球从第一个光门运动到第二个光门所经历的时间。
Although a single set of readings was provided in the printed question, the underlying method is identical to a full-scale multiple-measurement experiment. You are expected to use the fundamental definitions of average velocity for a very short time interval, treat it as instantaneous velocity, and then apply the equations of uniformly accelerated motion.
虽然试卷中仅提供了一组读数，但其基础方法与完整的多组测量实验完全相同。你应当运用极短时间间隔内平均速度的定义，将其视为瞬时速度，然后再应用匀加速直线运动的方程。
2. Apparatus and Setup | 实验装置与设置
A typical setup includes a rigid stand with two light gates, a steel or brass ball, a millisecond timer connected to the gates, a metre rule, and a micrometer or vernier calipers for measuring the ball’s diameter. The upper gate is placed just below the release point so that the ball is already moving when it enters the first beam, but its initial velocity is not required to be zero – we only need the velocities at each gate and the transit time between them.
典型的装置包括一个带两个光门的稳固支架、一个钢球或铜球、连接到光门的毫秒计时器、一把米尺以及用于测量小球直径的千分尺或游标卡尺。上方光门放置在释放点下方不远处，使小球在进入第一束光线时已经开始运动，但我们并不要求其初速度为零——只需知道小球经过每个光门时的速度和两光门之间的传输时间。
The diameter of the ball is critical because the light-gate timing, tgate, is the interval during which the ball’s leading edge enters the beam until its trailing edge leaves. Dividing the diameter by this ‘interruption time’ gives the average velocity. Provided the beam is very narrow and the interruption time is sufficiently short, this average velocity closely approximates the instantaneous velocity at the midpoint of the gate.
小球的直径至关重要，因为光门计时 tgate 是从小球前沿进入光束到后沿离开光束的时间间隔。用直径除以这个“遮光时间”便可得到平均速度。只要光束足够窄且遮光时间足够短，该平均速度就非常接近光门中点处的瞬时速度。
3. Measuring the Diameter of the Ball | 测量球的直径
In the January 2019 paper, the diameter d was given as 2.00 cm, recorded to three significant figures. Most students would have measured this with a micrometer screw gauge, which can read to ±0.01 mm. If you are repeating such an experiment, take several readings in different orientations to check for spherical symmetry and reduce random error.
在 2019 年 1 月的试卷中，直径 d 给出为 2.00 cm，保留三位有效数字。大多数学生会使用可读至 ±0.01 mm 的千分尺来测量该数据。如果你亲自重复这样的实验，应从不同方向多次测量，以检查球体的对称性并减小随机误差。
Always record the diameter in metres before substituting into velocity equations. Here, d = 0.0200 m. A common slip is to forget converting centimetres to metres, which would throw the velocity values off by a factor of 100.
在代入速度方程前，务必将直径换算为米。此处 d = 0.0200 m。一个常见的疏忽是忘记将厘米转换为米，这会使速度值偏离 100 倍。
4. Instantaneous Velocity at Each Gate | 每个光门处的瞬时速度
The velocity at Gate 1 is found using v1 = d / t1, where t1 is the time the ball takes to pass through the first beam. Similarly, v2 = d / t2 for the second gate. The exam question supplied t1 = 0.0150 s and t2 = 0.0090 s.
光门 1 处的速度用 v1 = d / t1 计算，其中 t1 是小球穿过第一道光束的时间。同理，光门 2 处的速度 v2 = d / t2。试题给出的数据为 t1 = 0.0150 s 和 t2 = 0.0090 s。
Performing the arithmetic:
进行算术运算：
v1 = 0.0200 m ÷ 0.0150 s = 1.333 m s⁻¹
v2 = 0.0200 m ÷ 0.0090 s = 2.222 m s⁻¹
Notice that the ball’s speed increases substantially between the two gates, consistent with acceleration due to gravity.
请注意，小球在两光门之间的速度显著增加，这与重力加速度的作用一致。
5. Time Between the Gates | 光门之间的时间
The timer also recorded the interval t between the instant the ball entered the first gate and the instant it entered the second gate. In the paper, t = 0.400 s. This is not the time the ball spends between the gates; it is the time from the start of the first interruption to the start of the second interruption.
计时器还记录了小球进入第一个光门的瞬间到进入第二个光门的瞬间之间的时间间隔 t。在试卷中，t = 0.400 s。这并不是小球在两光门之间停留的总时间，而是从第一次遮光开始到第二次遮光开始的时间。
Understanding this distinction is crucial. If the question asked for the time the ball is actually between the gates, you would need to subtract half of t1 and half of t2, but since the acceleration calculation relies on the change in velocity over the interval between the velocity measurement points, using the given t is correct for the equation v2 = v1 + g t.
理解这一区别至关重要。如果题目要求的是小球实际处于两光门之间的时间，你需要减去 t1 和 t2 各自的一半，但因为加速度计算依赖于两个速度测量点之间的速度变化时间，使用所给的 t 对于方程 v2 = v1 + g t 来说是正确的。
6. Deriving the Equation for g | 推导重力加速度 g 的方程
Since the ball moves with constant acceleration (we assume air resistance is negligible), we can use the first SUVAT equation in the direction of motion:
由于小球做匀加速运动（我们假设空气阻力可忽略），我们可以沿运动方向使用第一个 SUVAT 方程：
v = u + g t
Taking the velocity at Gate 1 as the initial velocity u and the velocity at Gate 2 as the final velocity v, we can rearrange to isolate g:
将光门 1 处的速度视为初速度 u，光门 2 处的速度视为末速度 v，我们可以重新排列以分离出 g：
g = (v2 − v1) / t
This is the key working equation for the investigation. It does not require the distance between the gates, which simplifies the experimental procedure. The student only needs to measure three time intervals and the ball’s diameter.
这就是本探究的核心工作方程。它并不需要两光门之间的距离，从而简化了实验步骤。学生只需测量三个时间间隔和小球的直径即可。
7. Data, Substitution and Calculated Value of g | 数据、代入与 g 的计算值
Let us collect the given data in a table for clarity.
让我们将给定数据整理成表格以便清晰查看。
| Quantity | Symbol | Value | Unit |
|---|---|---|---|
| Diameter of ball | d | 0.0200 | m |
| Time through Gate 1 | t1 | 0.0150 | s |
| Time through Gate 2 | t2 | 0.0090 | s |
| Time between gates | t | 0.400 | s |
| Velocity at Gate 1 | v1 = d / t1 | 1.333 | m s⁻¹ |
| Velocity at Gate 2 | v2 = d / t2 | 2.222 | m s⁻¹ |
| Calc. acceleration | g = (v2 − v1) / t | 2.22 | m s⁻² |
Substituting the values yields g ≈ 2.22 m s⁻², which is far smaller than the accepted 9.81 m s⁻². This discrepancy is deliberate – the exam expects you to recognise that the experiment, as conducted, suffers from significant systematic errors.
代入数值计算得到 g ≈ 2.22 m s⁻²，远小于公认值 9.81 m s⁻²。这一偏差是有意设置的——试题期望你认识到该实验在实施过程中存在显著的系统误差。
8. Plotting a Graph for More Reliable Results | 绘制图表以获得更可靠的结果
While the paper presented a single data set, a better experimental strategy involves repeating the measurement with the second gate at different positions below the first gate. If the distance s between the gates is measured, you can record several pairs of (s, v2² − v1²).
尽管试卷中只呈现了一组数据，但更优化的实验策略是在第一个光门下方不同位置重复测量第二个光门。若测量了两光门之间的距离 s，你可以记录多组 (s, v2² − v1²) 数据对。
From v2² = v1² + 2 g s, a plot of (v2² − v1²) on the y-axis against s on the x-axis yields a straight line through the origin with slope equal to 2 g. This method reduces the impact of random timing errors and allows you to judge reproducibility from scatter.
由 v2² = v1² + 2 g s 可知，以 (v2² − v1²) 为 y 轴、以 s 为 x 轴作图，会得到一条过原点的直线，其斜率等于 2 g。该方法减弱了随机计时误差的影响，并让你能够根据离散程度判断数据的可重复性。
Even if the graph shows a systematic offset, the slope can still provide an accurate g if the offset is constant. This is a key advantage of graphical analysis over single-point calculations.
即使图中显示出系统偏移，只要偏移恒定，斜率仍可给出准确的 g 值。这是图形分析法相对于单点计算的一个关键优势。
9. Sources of Uncertainty and Error | 不确定度与误差来源
The vast difference between the calculated 2.22 m s⁻² and 9.81 m s⁻² demands a discussion of error sources.
计算值 2.22 m s⁻² 与 9.81 m s⁻² 的巨大差异要求我们对误差来源进行讨论。
1. Air resistance and drag. A small but significant upward force opposes the motion, reducing the resultant acceleration. The effect is pronounced for a small, light ball, especially at higher speeds.
1. 空气阻力与曳力。 一个虽然不大但不可忽略的向上的力阻碍了运动，减小了合加速度。对于体积小、质量轻的球，尤其是速度较高时，这种效应尤为明显。
2. Measurement of t1 and t2. The times are very short (of the order of milliseconds). Any uncertainty in the timer, or a slight misalignment of the beam so that it is not perpendicular to the motion, can change the effective interruption path length. The finite beam width also means the average velocity is not exactly the instantaneous velocity at the gate centre.
2. t1 和 t2 的测量。 这两段时间极短
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