Case Study Practical Exercise: Bungee Jump Analysis | 案例分析实战演练：蹦极跳跃分析

📚 Case Study Practical Exercise: Bungee Jump Analysis | 案例分析实战演练：蹦极跳跃分析

In OCR A Level Physics, applying theoretical knowledge to real-world scenarios is a crucial skill. This case study walks you through a structured analysis of a bungee jump, integrating mechanics, materials, energy conservation and experimental thinking. By solving the problem step by step, you will see how multiple topics connect and learn to present answers that meet examination standards.

在 OCR A Level 物理中，将理论知识应用到真实场景是一项关键技能。本案例以一次蹦极跳跃为主线，带你进行系统化的分析，综合运用力学、材料、能量守恒和实验思维。通过逐步解决问题，你会看到不同主题如何相互关联，并学会写出符合考试要求的解答。

1. Scenario and Given Data | 场景与已知数据

A bungee jumper of mass 70 kg leaps from a bridge 65 m above water. The rope has an unstretched length L₀ = 25 m and a spring constant k = 180 N m⁻¹, obeying Hooke’s law until failure. Assume air resistance is negligible for the main calculation. Determine key physical quantities and assess safety.

一位质量为 70 kg 的蹦极者从距水面 65 m 的桥上跳下。蹦极绳原长 L₀ = 25 m，劲度系数 k = 180 N m⁻¹，在断裂前始终遵循胡克定律。主要计算中忽略空气阻力。试确定关键物理量并评估安全性。

Mass m = 70 kg
Gravitational field strength g = 9.81 m s⁻²
Unstretched rope length L₀ = 25 m
Spring constant k = 180 N m⁻¹
Bridge height above water = 65 m

质量 m = 70 kg；重力场强度 g = 9.81 m s⁻²；绳原长 L₀ = 25 m；劲度系数 k = 180 N m⁻¹；桥面距水面 65 m。

2. Physical Model and Assumptions | 物理模型与假设

We treat the jumper as a point mass. The rope is massless and obeys Hooke’s law: F = kΔx when stretched. The motion is split into two phases: free fall for the first 25 m, followed by deceleration when the rope extends. Energy losses due to air drag, internal rope heating and attachment point movement are initially ignored. The bridge and rope attachment are rigid.

我们把蹦极者视为质点。蹦极绳质量不计，伸长时遵循胡克定律 F = kΔx。整个过程分为两个阶段：前 25 m 的自由下落，随后绳子伸长做减速运动。初步忽略空气阻力、绳内耗和挂点移动造成的能量损失，桥面与绳固定点视为刚性连接。

3. Free-Fall Phase Analysis | 自由下落阶段分析

Until the rope becomes taut, the jumper is in free fall under gravity. The vertical distance travelled is L₀ = 25 m. Using v² = u² + 2as with u = 0, a = g, we find the speed just as the rope starts to stretch:

在绳子拉紧之前，蹦极者在重力作用下自由下落。竖直下落距离为 L₀ = 25 m。由 v² = u² + 2as，初速度 u = 0，加速度 a = g，可以算得绳子刚刚开始伸长时的速度：

v₁ = √(2 g L₀) = √(2 × 9.81 × 25) ≈ 22.15 m s⁻¹

This speed is about 80 km h⁻¹ and occurs after a fall time of t = √(2L₀/g) ≈ 2.26 s.

这个速度大约为 80 km h⁻¹，出现的时间约为 t = √(2L₀/g) ≈ 2.26 s。

4. Rope Extension and Elastic Potential Energy | 绳子的伸展与弹性势能

Once the rope length exceeds L₀, it behaves as a spring. If the extension is x, the elastic potential energy stored is Eₚ,elastic = ½ k x². Meanwhile, the jumper continues to lose gravitational potential energy as they descend further. The total mechanical energy is conserved (ignoring losses).

一旦绳长超过 L₀，它就表现出弹簧特性。若伸长量为 x，储存的弹性势能为 Eₚ,elastic = ½ k x²。同时，蹦极者继续下落，重力势能进一步减少。不计损耗时，总机械能守恒。

5. Energy Conservation to Find Maximum Extension | 利用能量守恒求最大伸长量

At the lowest point, the jumper’s velocity is momentarily zero. Taking the bridge level as the reference for gravitational potential energy, the total drop distance is h = L₀ + x_max. Energy conservation gives:

在最低点，蹦极者的速度瞬时为零。以桥面为重力势能零点，总下落距离为 h = L₀ + x_max。能量守恒给出：

m g (L₀ + x_max) = ½ k x_max²

Substituting values: 70 × 9.81 × (25 + x) = 0.5 × 180 × x² → 686.7(25 + x) = 90 x². Rearranging: 90 x² − 686.7 x − 17167.5 = 0. Solving the quadratic yields x_max ≈ 18.14 m. The total drop is about 43.14 m, well above the water (clearance ~21.9 m).

代入数值：70 × 9.81 × (25 + x) = 0.5 × 180 × x² → 686.7(25 + x) = 90 x²。整理得 90 x² − 686.7 x − 17167.5 = 0。解二次方程得 x_max ≈ 18.14 m，总下落距离约 43.14 m，远高于水面（净空约 21.9 m）。

6. Calculating Maximum Speed | 计算最大速度

Maximum speed occurs when the net force on the jumper is zero, i.e. when the upward elastic force equals the weight. This happens at an extension x_eq where k x_eq = m g, so x_eq = m g / k = 686.7 / 180 ≈ 3.82 m. Using energy conservation up to that point:

当蹦极者所受合力为零时，速度达到最大值，即向上的弹力等于重力时。此时伸长量 x_eq 满足 k x_eq = m g，故 x_eq = m g / k = 686.7 / 180 ≈ 3.82 m。对该点使用能量守恒：

m g (L₀ + x_eq) = ½ m v_max² + ½ k x_eq²

Computing: Gravitational potential energy lost = 686.7 × 28.82 ≈ 19788 J. Elastic energy stored = ½ × 180 × (3.82)² ≈ 1310 J. Hence kinetic energy = 19788 − 1310 = 18478 J, giving v_max ≈ 22.98 m s⁻¹. Interestingly, this is slightly higher than the free-fall speed at L₀ because gravity still does net positive work.

计算：重力势能减少量 = 686.7 × 28.82 ≈ 19788 J，弹性势能 = ½ × 180 × (3.82)² ≈ 1310 J，因此动能 = 19788 − 1310 = 18478 J，v_max ≈ 22.98 m s⁻¹。有趣的是，这略高于在 L₀ 处的自由落体速度，因为重力仍在做净正功。

7. Maximum Acceleration and Force Analysis | 最大加速度与受力分析

At the lowest point, the rope force reaches its maximum: F_max = k x_max = 180 × 18.14 ≈ 3265 N. The net upward force on the jumper is F_net = F_max − m g = 3265 − 686.7 ≈ 2578 N, producing an upward acceleration:

在最低点，绳子的拉力达到最大值：F_max = k x_max = 180 × 18.14 ≈ 3265 N。蹦极者所受的净向上力为 F_net = F_max − m g = 3265 − 686.7 ≈ 2578 N，产生的向上加速度为：

a_max = F_net / m = 2578 / 70 ≈ 36.8 m s⁻²

Expressed in terms of g, this is about 3.75g. The typical safe limit for bungee jumps is below 5g, so this design is acceptable from a dynamic loading perspective.

以 g 为单位，约为 3.75g。蹦极运动中通常的安全上限低于 5g，所以从动态载荷角度看该设计是可接受的。

8. Safety Assessment and Material Limits | 安全评估与材料极限

Apart from acceleration, we must ensure the rope does not exceed its elastic limit. If the rope’s cross-sectional area is known, we can compute stress σ = F_max / A. Suppose the rope has a diameter of 12 mm (A ≈ 1.13×10⁻⁴ m²), then σ ≈ 3265 / 1.13×10⁻⁴ ≈ 28.9 MPa. Typical nylon climbing ropes have yield strengths around 50–80 MPa, so the rope remains safe. An appropriate safety factor (yield stress / working stress) of at least 2 is desirable. The clearance to the water also provides a safety margin.

除加速度外，还必须确保绳子不超出弹性极限。若已知绳横截面积，可计算应力 σ = F_max / A。假设绳直径 12 mm（A ≈ 1.13×10⁻⁴ m²），则 σ ≈ 3265 / 1.13×10⁻⁴ ≈ 28.9 MPa。尼龙攀登绳的屈服强度通常在 50–80 MPa 左右，因此该绳是安全的。要求安全系数（屈服应力/工作应力）至少达到 2 为宜。距水面的净空同样提供了安全余量。

9. Experimental Verification: Determining the Rope’s Spring Constant | 实验验证：测定绳子的劲度系数

In practice, k must be measured. A simple method is to suspend known masses from a sample of the rope and measure the extension Δx. Plotting force (mg) against extension gives a straight line through the origin for a Hookean material; the gradient equals k. The experiment must control temperature and avoid over-stretching. Repeated measurements reduce random uncertainties.

实际中需要测量 k。简单方法是在一段绳样上悬挂已知质量，测量伸长量 Δx。对遵循胡克定律的材料，拉力 (mg) 对伸长的图像为一条过原点的直线，其斜率即为 k。实验需控制温度并避免过度拉伸。多次重复测量可减小随机不确定度。

10. Error and Uncertainty Analysis | 误差与不确定性分析

Uncertainties in length measurement (ruler ±1 mm), mass (balance ±0.1 kg) and the assumption g = 9.81 ± 0.01 m s⁻² propagate into k. For the bungee jump, uncertainties in k, m and L₀ affect the calculated x_max and a_max. An exam-style question might ask you to estimate percentage uncertainties and comment on the reliability of the safety conclusions. You can use the propagation formula: if y = a × b, then %Uy ≈ %Ua + %Ub.

长度测量（直尺 ±1 mm）、质量（天平 ±0.1 kg）以及 g = 9.81 ± 0.01 m s⁻² 的假设都会引入不确定度，并传递到 k。在蹦极分析中，k、m 和 L₀ 的不确定度会影响计算出的 x_max 和 a_max。考试题可能要求你估算百分不确定度，并评论安全结论的可靠性。可使用传递公式：若 y = a × b，则 %Uy ≈ %Ua + %Ub。

11. Extension: Effect of Air Resistance | 拓展讨论：空气阻力的影响

In reality, air resistance acts opposite to velocity and dissipates mechanical energy. During free fall, the jumper would reach a lower speed than 22.15 m s⁻¹, and the maximum extension would be reduced because less kinetic energy is available to be converted into elastic strain energy. The maximum acceleration would also be slightly lower. However, the rope still undergoes many cycles; damping gradually reduces the amplitude. Qualitatively, the true x_max < 18.14 m, making the jump even safer, but the time to settle might differ.

实际上，空气阻力与速度反向，会耗散机械能。自由下落过程中，蹦极者达到的速度会低于 22.15 m s⁻¹，最大伸长量也会因可用于转化为弹性势能的动能减少而降低，最大加速度也会略低。尽管如此，绳子仍会经历多次振荡，阻尼逐渐减小振幅。定性而言，真实的 x_max < 18.14 m，使得蹦极更安全，但稳定下来所需时间可能不同。

12. Conclusion and Exam Tips | 总结与考试技巧

This case study demonstrates how to link kinematics, Newton’s laws, energy conservation, materials and experimental physics in one coherent problem. For OCR exams, always define your system, state assumptions clearly, show energy or force equations and solve algebraically before substituting numbers. Comment on the physical meaning of results (e.g. whether 3.75g is safe). Practice drawing free-body diagrams and energy flow diagrams. Finally, be ready to evaluate the impact of ignoring air resistance and to discuss experimental uncertainties.

本案例展示了如何在一个连贯的问题中联系运动学、牛顿定律、能量守恒、材料和实验物理。在 OCR 考试中，务必定义系统、清晰说明假设、先列出能量或力的方程并在代入数字前完成代数推导。对结果的物理意义（如 3.75g 是否安全）进行评述。多练习画受力图和能量流动图。最后，要准备好评估忽略空气阻力的影响，并讨论实验不确定度。

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Case Study Practical Exercise: Bungee Jump Analysis | 案例分析实战演练：蹦极跳跃分析