Modelling the Bounce Height of a Ball: A Case Study in Action | 弹跳球高度建模案例分析实战

📚 Modelling the Bounce Height of a Ball: A Case Study in Action | 弹跳球高度建模案例分析实战

In this case study, we bring together key concepts from the AQA Year 12 Mathematics syllabus to analyse a real-world scenario: modelling how the bounce height of a ball decays after repeated drops. You will follow the entire modelling cycle, from collecting experimental data to fitting a non-linear function, linearising it using logarithms, applying least-squares regression, and finally using calculus to interpret the model in a continuous context. This hands-on approach mirrors the style of applied questions found in your AS examinations, where reasoning, calculation, and interpretation are equally valued.

在这个案例研究中，我们将综合运用 AQA 12 年级数学课程中的核心知识，分析一个真实的场景：建立球多次反弹后高度衰减的数学模型。你将亲历完整的建模循环——从实验数据收集到拟合非线性函数，利用对数进行线性化，应用最小二乘回归，最后借助微积分在连续情形下解释模型。这种动手实操的方式模拟了 AS 考试中的应用题风格，题目同时重视推理、计算和解释。

1. Setting Up the Experiment and Collecting Data | 实验设置与数据收集

Imagine a physics laboratory where a small rubber ball is dropped from a fixed height of 150 cm onto a hard, flat surface. A high-speed camera records the maximum height reached after each successive bounce. The experiment is repeated several times to minimise random error, and the average heights are recorded for bounce numbers 0, 1, 2, 3, 4 and 5. Here, bounce number 0 refers to the initial drop height. The recorded data are shown in the table below.

设想一个物理实验室，一只小橡胶球从 150 cm 的固定高度自由下落到坚硬平整的地面上。高速摄影机记录下每次弹跳后达到的最大高度。实验重复多次以减小随机误差，并记录弹跳次数 0, 1, 2, 3, 4, 5 对应的平均高度。其中弹跳次数 0 表示初始下落高度。记录数据见下表。

Bounce number, n	0	1	2	3	4	5
Height h (cm)	150	120	96	77	62	50

The independent variable is the bounce number n, and the dependent variable is the height h. At this stage, we simply observe that height decreases with each bounce, but not at a constant rate. The drop appears to follow some curved pattern, which suggests an exponential decay rather than a linear relationship.

自变量是弹跳次数 n，因变量是高度 h。在这个阶段，我们仅观察到高度随弹跳次数的增加而降低，但并非匀速变化。下降似乎呈某种曲线模式，这暗示着指数衰减而非线性关系。

In AQA applied questions, you are often given a table of values and asked to propose a suitable model. Before plunging into calculations, it is always wise to sketch a scatter diagram. This visual check helps you decide whether to try a linear, quadratic, exponential or trigonometric model.

在 AQA 应用题中，你通常会得到一份数值表并被要求提出合适的模型。在进行计算之前，绘制散点图永远是明智之举。这种可视化的检验有助于你决定是尝试线性、二次、指数还是三角函数模型。

---

2. Drawing a Scatter Diagram and Observing Trends | 绘制散点图并观察趋势

Plot the points (n, h) on graph paper or using a graphical calculator. The first point is (0, 150), followed by (1, 120), (2, 96), (3, 77), (4, 62) and (5, 50). The curve is clearly downward-sloping and convex, meaning that the rate of decrease becomes smaller as n increases. This shape is characteristic of exponential decay, where the height is multiplied by a constant factor between successive bounces.

在坐标纸或图形计算器上标出点 (n, h)。第一个点是 (0, 150)，接着是 (1, 120), (2, 96), (3, 77), (4, 62) 和 (5, 50)。曲线明显向右下倾斜且呈凸状，意味着随着 n 增大，递减的速度越来越慢。这种形状是指数衰减的典型特征，即相邻弹跳间高度乘以一个恒定因子。

: 120/150 = 0.8, 96/120 = 0.8, 77/96 ≈ 0.802, 62/77 ≈ 0.805, 50/62 ≈ 0.806. Apart from small experimental fluctuations, the ratio is approximately constant at 0.8. This strongly supports an exponential model of the form h = H₀ rⁿ, or equivalently h = H₀ e⁻ᵏⁿ with k = -ln r.

如果数据遵循幂函数或二次函数，曲率会有所不同。检查相邻高度之比可以得到进一步证据：120/150 = 0.8, 96/120 = 0.8, 77/96 ≈ 0.802, 62/77 ≈ 0.805, 50/62 ≈ 0.806。除了较小的实验波动，比率大致恒定在 0.8。这强有力地支持形如 h = H₀ rⁿ 的指数模型，或等价地 h = H₀ e⁻ᵏⁿ，其中 k = -ln r。

At AS level, you are not expected to derive the model purely from physical laws, but you should recognise that repeated multiplication by a factor leads to exponential behaviour. The physical interpretation is that the ball loses a fixed percentage of its energy on each impact, which translates into a constant rebound ratio.

在 AS 阶段，并不要求你完全从物理定律推导模型，但你应当认识到，反复乘以一个因子会导致指数行为。物理解释是球每次撞击地面时损失固定百分比的能量，这便转化为恒定的反弹比率。

---

3. Choosing an Exponential Decay Model | 选择指数衰减模型

Based on the pattern observed, we propose the model

根据观察到的模式，我们提出如下模型

h = H₀ e⁻ᵏⁿ

arameter H₀ is simply the height at bounce zero. The decay constant k determines how rapidly the height falls; a larger k means a steeper drop.

其中 H₀ 表示理论初始高度（应接近 150 cm），k 为正的衰减常数。当 n = 0 时，h = H₀ e⁰ = H₀，因此参数 H₀ 就是零次弹跳时的高度。衰减常数 k 决定了高度下降的速率；k 越大，下降越急剧。

, we would expect k = -ln(0.8) ≈ 0.223. This gives a ballpark figure to compare with the more precise estimate we will obtain through regression.

另一种写法为 h = H₀ rⁿ，其中 r 是公比。比较两种形式可知 r = e⁻ᵏ。鉴于我们大致恒定的比率为 0.8，预期 k = -ln(0.8) ≈ 0.223。这提供了一个大致的数值，可与稍后通过回归得到的更精确估计相比较。

In an examination, you might be told to assume an exponential model and then asked to estimate its parameters. The standard technique is to linearise the relationship by taking natural logarithms, which transforms the curve into a straight line and allows you to use least-squares regression.

在考试中，你可能会被要求假定为指数模型，然后估计其参数。标准方法是通过取自然对数将关系线性化，这样曲线就转化为一条直线，进而可以使用最小二乘回归。

---

广告 / Sponsored

4. Linearising the Model Using Logarithms | 利用对数线性化模型

Take the natural logarithm of both sides of the model equation:

对模型方程两边取自然对数：

ln h = ln(H₀ e⁻ᵏⁿ) = ln H₀ + ln(e⁻ᵏⁿ) = ln H₀ - k n

f we plot Y against X, we should obtain approximately a straight line with negative slope.

令 Y = ln h，X = n。关系式变为 Y = ln H₀ - k X，这是一条形如 Y = a + b X 的直线方程，其中 a = ln H₀，b = -k。因此，若以 Y 对 X 作图，我们应当得到一条近似直线，其斜率为负。

Calculate the natural logarithms of the h-values from our table:

计算表格中 h 值的自然对数：

n	0	1	2	3	4	5
h	150	120	96	77	62	50
Y = ln h	5.011	4.788	4.564	4.344	4.127	3.912

These Y-values are given to three decimal places. Notice how the logarithms decrease roughly by the same amount each time: the differences are about 0.223, 0.224, 0.220, 0.217, 0.215. This confirms the linear trend and validates the choice of an exponential model.

这些 Y 值精确到三位小数。注意到对数值每次减少的量大致相同：差值分别约为 0.223, 0.224, 0.220, 0.217, 0.215。这证实了线性趋势，并验证了指数模型的选择是合理的。

---

5. Calculating the Least-Squares Regression Line | 计算最小二乘回归直线

t minimises the sum of squared vertical deviations. The standard formulae from the AQA formula booklet are:

现在我们视 n 为自变量 X，ln h 为因变量 Y。共有六对数据。目标是找到最佳拟合直线 Y = a + b X，使得竖直偏差的平方和最小。AQA 公式手册中的标准公式为：

b = (n ΣXY - ΣX ΣY) / (n ΣX² - (ΣX)²)

a = Ȳ - b X̄

4+4.127+3.912 = 26.746. ΣXY = (0×5.011)+(1×4.788)+(2×4.564)+(3×4.344)+(4×4.127)+(5×3.912) = 0 + 4.788 + 9.128 + 13.032 + 16.508 + 19.560 = 63.016

首先计算必要的和。此处 n = 6（数据点数目），ΣX = 0+1+2+3+4+5 = 15。ΣX² = 0²+1²+2²+3²+4²+5² = 55。ΣY = 5.011+4.788+4.564+4.344+4.127+3.912 = 26.746。 564)+(3×4.344)+(4×4.127)+(5×3.912) = 0 + 4.788 + 9.128 + 13.032 + 16.508 + 19.560 = 63.016

Substitute into the formula for b:

代入公式求 b：

b = (6 × 63.016 - 15 × 26.746) / (6 × 55 - 15²)

= (378.096 - 401.190) / (330 - 225)

= (-23.094) / 105

≈ -0.2199

Now the mean of X, X̄ = 15/6 = 2.5, and the mean of Y, Ȳ = 26.746/6 ≈ 4.4577. Then the intercept a = Ȳ - b X̄ = 4.4577 - (-0.2199)×2.5 = 4.4577 + 0.54975 = 5.0075.

X 的均值 X̄ = 15/6 = 2.5，Y 的均值 Ȳ = 26.746/6 ≈ 4.4577。因此截距 a = Ȳ - b X̄ = 4.4577 - (-0.2199)×2.5 = 4.4577 + 0.54975 = 5.0075。

Thus the least-squares regression line for the transformed data is:

于是转换后数据的最小二乘回归直线为：

ln h = 5.0075 - 0.2199 n

This line closely fits our logarithmic points. The negative slope indicates the decay constant k is approximately 0.22, consistent with our earlier rough estimate.

该直线与我们的对数点高度吻合。负斜率表明衰减常数 k 约为 0.22，与我们早先的粗略估计一致。

---

6. Determining the Model Parameters H₀ and k | 确定模型参数 H₀ 和 k

n equation ln h = a + b n, we identify ln H₀ = a = 5.0075 and -k = b ≈ -0.2199. Therefore,

由回归方程 ln h = a + b n，可得 ln H₀ = a = 5.0075，且 -k = b ≈ -0.2199。因此，

H₀ = e⁵.⁰⁰⁷⁵ ≈ 149.5 cm

k = 0.2199 ≈ 0.220 (to 3 s.f.)

The estimated initial height of 149.5 cm is pleasingly close to the known drop height of 150 cm. The small discrepancy arises from experimental error and the fact that we are fitting a smooth curve to slightly scattered data. The model that best describes our observations is therefore:

估计的初始高度 149.5 cm 令人满意地接近已知的 150 cm 下落高度。微小的差异源于实验误差以及我们用平滑曲线拟合略有波动的数据这一事实。因此，最能刻画观测结果的模型为：

h = 149.5 e⁻⁰.²²⁰ⁿ

, because e⁻⁰.²²⁰ ≈ 0.802. This tells us that, on average, the ball rebounds to about 80.2% of its previous height after each bounce.

如果偏好比率形式，可写作 h = 149.5 × (0.802)ⁿ，因为 e⁻⁰.²²⁰ ≈ 0.802。这告诉我们，平均而言，每次弹跳后球反弹至前次高度的约 80.2%。

In AQA exams, you may be asked to interpret the constants in context. Here, 149.5 is the model's predicted initial height, and the constant 0.220 determines the rate of energy loss per bounce. Always relate parameters back to the real-world situation to score interpretation marks.

在 AQA 考试中，你可能需要解释常数的实际含义。此处，149.5 是模型预测的初始高度，常数 0.220 决定了每次弹跳的能量损失率。务必把参数与实际情境联系起来，以拿到解释分。

---

7. Validating the Model and Making Predictions | 模型验证与预测

eights for each n and comparing with the observed values. For n = 3, the model gives h = 149.5 e⁻⁰.²²⁰׳ = 149.5 e⁻⁰.⁶⁶⁰ ≈ 149.5 × 0.5169 = 77.3 cm, which matches the observed 77 cm very well. For n = 5, the prediction is 149.5 e⁻¹.¹⁰⁰ ≈ 149.5 × 0.3329 ≈ 49.8 cm, close to the measured 50 cm.

好的数学模型不仅要拟合已有数据，还应能做出可靠的预测。我们可以计算每个 n 对应的预测高度并与观测值比较，以此来检验模型。对于 n = 3，模型给出 h = 149.5 e⁻⁰.²²⁰׳ = 149.5 e⁻⁰.⁶⁶⁰ ≈ 149.5 × 0.5169 = 77.3 cm，与观测值 77 cm 非常吻合。对于 n = 5，预测值为 149.5 e⁻¹.¹⁰⁰ ≈ 149.5 × 0.3329 ≈ 49.8 cm，接近于实测的 50 cm。

We can also extrapolate to future bounces. What height would we expect at the 8th bounce? Substituting n = 8 gives h = 149.5 e⁻⁰.²²⁰×⁸ = 149.5 e⁻¹

Published by TutorHao | Year 12 Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）