📚 Interdisciplinary Problems in SQA Physics | SQA 物理跨学科综合题型训练
Interdisciplinary problem-solving is now a core part of the SQA National 5 and Higher Physics curriculum. Real‑world challenges rarely fit neatly into one subject, so exam questions increasingly combine physics with mathematics, chemistry, biology, engineering, and earth science. This article provides structured practice across key cross‑topic themes, helping you build the flexible thinking and analytical skills needed for top marks. Each section unpacks a typical problem type, outlines the underlying physics, and shows how to integrate knowledge from other disciplines.
跨学科问题解决如今是 SQA National 5 和 Higher 物理课程的核心部分。现实世界的挑战很少能整齐地归入单一学科，因此考试题目日益将物理与数学、化学、生物、工程学以及地球科学相结合。本文围绕关键的跨主题领域提供结构化训练，帮助你培养获得高分所需的灵活思维和分析能力。每一节会剖析一类典型问题，勾勒背后的物理原理，并展示如何融合其他学科的知识。
1. Vectors in Navigation and Engineering | 导航与工程中的向量
An aircraft flies at 240 m s⁻¹ on a bearing of 070° while a steady wind blows from 310° at 50 m s⁻¹. Resolve the velocity vectors into perpendicular components and find the resultant ground velocity.
一架飞机以 240 m s⁻¹ 沿方位角 070° 飞行，同时有稳定的风从 310° 以 50 m s⁻¹ 吹来。将速度矢量分解为垂直分量，并求出合速度。
Begin by drawing a vector diagram. The plane’s airspeed has a direction of 070° (clockwise from north). In standard Cartesian coordinates, north is 0° measured anticlockwise, so 070° corresponds to an angle of 20° anticlockwise from the east. Careful conversion between bearing and Cartesian systems is essential here – this mixes physics with mathematical conventions from navigation.
首先画出矢量图。飞机的空速方向为 070°（自北顺时针），在标准笛卡尔坐标系中北为 0°（逆时针），因此 070° 相当于自东逆时针 20°。此处需要在方位制与笛卡尔坐标系之间仔细转换——这融合了物理与导航中的数学惯例。
Resolve the plane’s velocity: vₓ = 240 cos 20° ≈ 225.5 m s⁻¹ east, vy = 240 sin 20° ≈ 82.1 m 310°, meaning it blows toward 130° (bearing). Convert: 130° bearing → Cartesian angle 90° – 130°? Think: bearing 130° is 40° south of east, so wind vector: wₓ = 50 cos 40° ≈ 38.3 m s⁻¹ east, wy = –50 sin 40° ≈ –32.1 m s⁻¹ (south). Add components: resultant vₓ = 225.5 + 38.3 = 263.8 m s⁻¹, vy = 82.1 – 32.1 = 50.0 m s⁻¹. Magnitude = √263.8² + 50.0² ≈ 268 m s⁻¹. Direction: θ = tan⁻¹(50.0 / 263.8) ≈ 10.7° north of east, bearing = 90° – 10.7° = 079.3°.
分解飞机速度：vₓ = 240 cos 20° ≈ 225.5 m s⁻¹ 向东，vy = 240 sin 20° ≈ 82.1 m s⁻¹ 向北。风来自 310°，即吹向 130°（方位角）。转换：方位 130° 为东偏南 40°，故风矢量：wₓ = 50 cos 40° ≈ 38.3 m s⁻¹ 向东，wy = –50 sin 40° ≈ –32.1 m s⁻¹（向南）。分量相加：合速度 vₓ = 263.8 m s⁻¹，vy = 50.0 m s⁻¹。速率 = √263.8² + 50.0² ≈ 268 m s⁻¹，方向为东偏北 10.7°，方位角 = 079.3°。
This type of problem appears in engineering contexts (crane loads, river crossings) and requires confident handling of trigonometry and angle measures.
这类问题出现在工程情境中（起重机负载、渡河问题），需要熟练运用三角学与角度度量。
2. Kinetic Energy and Stopping Distances with Data Analysis | 动能与刹车距离及数据分析
A car of mass 1200 kg is travelling at 13.4 m s⁻¹ (30 mph). The driver brakes and the car skids to rest. The measured skid marks are 14.2 m long. Use the work–energy theorem to calculate the average braking force. How would the stopping distance change if the initial speed doubled? Link to mathematics: plot a graph of thinking distance, braking distance, and total stopping distance versus speed for a range of speeds.
一辆质量为 1200 kg 的汽车以 13.4 m s⁻¹（30 mph）行驶。驾驶员刹车后汽车滑行至停止，测得滑痕长 14.2 m。利用功能定理计算平均制动力。如果初速度加倍，刹车距离会如何变化？联系数学：绘制反应距离、刹车距离和总停车距离随速度变化的图像。
Physics: initial kinetic energy Eₖ = ½ m v² = ½ × 1200 × (13.4)² = 107 736 rk done by friction W = F × d. Assuming all kinetic energy is dissipated, F = Eₖ / d = 107 736 / 14.2 ≈ 7580 N. This is the average friction force. If speed doubles to 26.8 m s⁻¹, Eₖ = ½ × 1200 × (26.8)² = 430 944 y four times the original. With the same braking force, stopping distance d = Eₖ / F = 430 944 / 7580 ≈ 56.8 m, which is four times the original distance (14.2 m × 4 ≈ 56.8 m). This quadratic relationship (d ∝ v²) is a core outcome of combining kinematics and energy – a clear crossover with algebra and graph plotting.
物理：初始动能 Eₖ = ½ m v² = ½ × 1200 × (13.4)² = 107 736 J。摩擦力做功 W = F × d。假设动能全部耗尽，F = Eₖ / d = 107 736 / 14.2 ≈ 7580 N。此为平均摩擦力。若速度加倍至 26.8 m s⁻¹，动能变为原来的 4 倍，刹车距离也变为原来的 4 倍，即 56.8 m。这种二次方关系（d ∝ v²）是运动学与能量结合的核心成果——明显涉及代数与图像绘制。
From a mathematics perspective, construct a table: speed v (m s⁻¹), thinking distance dt = v × reaction time (say 0.7 s), braking distance db = v² / (2 × deceleration), total dtotal = dt + db. Plotting these three curves shows the non‑linear growth of stopping distance, linking directly to road safety campaigns and data interpretation skills.
从数学角度出发，构建表格：速度 v，反应距离 dt = v × 反应时间（假设 0.7 s），刹车距离 db = v² / (2 × 减速度)，总距离 dtotal = dt + db。绘制这三条曲线可展示停车距离的非线性增长，直接联系道路安全宣传与数据解读能力。
3. Radioactive Decay and Biological Half‑Life | 放射性衰变与生物半衰期
A patient is injected with a technetium‑99m tracer for a bone scan. The physical half‑life of Tc‑99m is 6.0 hours, and its biological half‑life in the body is 12 hours. Determine the effective half‑life and the fraction of the initial activity remaining after 18 hours. Explain why effective half‑life is always shorter than both physical and biological half‑lives.
一位患者被注射了用于骨扫描的锝-99m 示踪剂。Tc‑99m 的物理半衰期为 6.0 小时，其在体内的生物半衰期为 12 小时。求有效半衰期以及 18 小时后初始活度的残留分数。解释为什么有效半衰期总是短于物理半衰期和生物半衰期。
This problem sits at the physics–biology interface. The effective decay constant λ_eff = λ_phys + λ_biol. Using half‑lives: T½_phys = 6.0 h → λ_phys = ln 2 / 6.0 ≈ 0.1155 h⁻¹. T½_biol = 12 h → λ_biol = ln 2 / 12 ≈ 0.05776 h⁻¹. So λ_eff = 0.1733 h⁻¹. Then T½_eff = ln 2 / λ_eff ≈ 0.693 / 0.1733 ≈ 4.0 h. The effective half‑life (4.0 h) is indeed shorter than both contributions. After 18 h, number of effective half‑lives = 18 / 4.0 = 4.5. Remaining fraction = (½)4.5 = 1 / (24.5) = 1 / (22.6) ≈ 0.044 or 4.4%.
这个问题处于物理与生物学的交界。有效衰变常量 λ_eff = λ_phys + λ_biol。计算半衰期：T½_phys = 6.0 h，λ_phys = ln 2 / 6.0 ≈ 0.1155 h⁻¹；T½_biol = 12 h，λ_biol = ln 2 / 12 ≈ 0.05776 h⁻¹。λ_eff = 0.1733 h⁻¹，有效半衰期 T½_eff ≈ 4.0 小时。确实比两个贡献都短。18 小时后，有效半衰期个数为 4.5，残留分数约为 4.4%。
Biologically, the body excretes the tracer at a certain rate, while the nuclei themselves decay. The total removal rate is the sum of the two independent processes. This is analogous to combining probabilities or rates in chemistry kinetics, showing how interdisciplinary reasoning directly yields the answer.
生物学上，身体以一定速率排出示踪剂，同时原子核自身也在衰变。总移除速率是两个独立过程之和。这类似于化学动力学中概率或速率的合成，显示了跨学科推理如何直接得出答案。
4. Electrical Power and Heat Transfer in a Domestic Kettle | 家用电水壶的电功率与热传递
An electric kettle rated at 2200 W contains 1.5 kg of water at 18 °C. Assuming 85% efficiency, calculate the time needed to bring the water to boiling point (100 °C). Specific heat capacity of water c = 4180 J kg⁻¹ tle were used at a high altitude where atmospheric pressure lowers the boiling point to 92 °C? Link to geography: discuss the impact of altitude on cooking times.
一个额定功率为 2200 W 的电水壶装有 1.5 kg 温度为 18 °C 的水。假设效率为 85%，计算将水加热至沸点（100 °C）所需的时间。水的比热容 c = 4180 J kg⁻¹ K⁻¹。如果在高海拔地区使用，大气压降低导致沸点降至 92 °C，时间会如何变化？联系地理：讨论海拔对烹饪时间的影响。
Physics: energy required to heat water = m c Δθ = 1.5 × 4180 × (100 – 18) = 1.5 × 4180 × 82 = 514 140 J. The kettle delivers useful power Puseful = 0.85 × 2200 = 1870 W. Time t = Energy / Power = 514 140 / 1870 ≈ 275 s (about 4 min 35 s). At altitude with boiling point 92 °C, Δθ = 92 – 18 = 74 °C, energy = 1.5 × 4180 × 74 = 463 980 J. Time = 463 980 / 1870 ≈ 248 f food often takes longer at altitude because the lower temperature reduces reaction rates – a chemistry and biology consideration beyond the pure physics of boiling water.
物理：加热水所需能量 = m c Δθ = 1.5 × 4180 × 82 = 514 140 J。水壶有用功率 Puseful = 0.85 × 2200 = 1870 W。时间 t = 能量 / 功率 ≈ 275 s（约 4 分 35 秒）。在高海拔，沸点 92 °C，Δθ = 74 °C，能量 = 463 980 J，时间约为 248 s，节省约 27 秒。看似更短，但烹饪食物通常更费时，因为较低的温度降低了反应速率——这涉及化学和生物学考量，超越了单纯的烧水物理。
Geographically, the boiling point drops by about 1 °C for every 285 m of altitude gain. This affects energy use in mountainous regions and is a classic example of physics–geography integration in SQA questions.
地理上，海拔每升高约 285 m，沸点下降约 1 °C。这会影响山区的能源使用，是 SQA 试题中物理与地理结合的经典例子。
5. Moments and Biomechanics: The Human Forearm | 力矩与生物力学：人体前臂
A person holds a 5.0 kg dumbbell in the hand with the forearm horizontal. The biceps muscle attaches 4.0 cm from the elbow joint, and the centre of mass of the forearm plus hand (mass 2.0 kg) is 15 cm from the elbow. The dumbbell is held 35 cm from the elbow. Calculate the force exerted by the biceps muscle to maintain equilibrium. Explain why this force is much larger than the total weight supported.
一个人手持 5.0 kg 哑铃，前臂保持水平。肱二头肌附着在距肘关节 4.0 cm 处，前臂和手的质心（质量 2.0 kg）距肘 15 cm。哑铃握于距肘 35 cm 处。计算为保持平衡肱二头肌需施加的力，并解释为何该力远大于所支撑的总重量。
biology. Taking moments about the elbow joint: clockwise moments = (forearm weight × 0.15 m) + (dumbbell weight × 0.35 m) = (2.0 × 9.8 × 0.15) + (5.0 × 9.8 × 0.35) = 2.94 + 17.15 = 20.09 N m. The biceps provides an anticlockwise moment: Fbiceps × 0.04 m. For equilibrium, Fbiceps × 0.04 = 20.09 → Fbiceps ≈ 502 N. This is about 502 N compared to the total weight of (2.0+5.0)×9.8 = 68.6 N – o close to the pivot. This biological design trades force for speed and range of movement.
这是一个生物学中的杠杆系统经典案例。对肘关节取矩：顺时针力矩 =（前臂重 × 0.15 m）+（哑铃重 × 0.35 m）= 20.09 N m。肱二头肌提供逆时针力矩：F_二头肌 × 0.04 m。平衡时 F_二头肌 ≈ 502 N。这与总重量 68.6 N 相比超过 7 倍。原因是肌肉的力臂很短：由于作用点靠近支点，需要施加很大的力。这种生物设计以力换取了速度和运动范围。
Such calculations frequently appear in SQA questions linking physics to sports science or physiotherapy, and they require careful identification of forces and perpendicular distances.
此类计算经常出现在 SQA 将物理与运动科学或物理治疗结合的试题中，需要谨慎识别力与垂直距离。
6. Refractive Index and Optical Fibres: Communication Technology | 折射率与光纤：通信技术
An optical fibre has a core of refractive index 1.48 and a cladding of refractive index 1.44. Calculate the critical angle for total internal reflection at the core–cladding boundary. If the fibre is bent, discuss how the radius of curvature affects signal loss, linking to the mathematics of geometry and engineering tolerances.
一根光纤的纤芯折射率为 1.48，包层折射率为 1.44。计算芯-包层界面的全内反射临界角。若光纤被弯曲，讨论曲率半径如何影响信号损耗，并联系几何数学与工程容差。
Using Snell’s law: critical angle θ_c = sin⁻¹(n_cladding / ncore ry at an angle greater than 76.7° to the normal will be totally internally reflected. When the fibre is bent, incident angles can fall below the critical angle, causing light to escape. The minimum bend radius Rmin can be approximated using the formula Rmin = dcore / (2 × (ncore/n_cladding – 1)), where dcore is the core diameter. This formula is derived from geometric optics and circular geometry – a direct link to mathematics.
使用斯涅尔定律：临界角 θ_c = sin⁻¹(1.44 / 1.48) ≈ 76.7°。任何以大于 76.7° 入射到边界的全内反射。当光纤弯曲时，入射角可能低于临界角，导致光逸出。最小弯曲半径 Rmin 可由 Rmin = dcore / (2 × (ncore/n_cladding – 1)) 近似给出，该式源自几何光学与圆的几何——与数学直接关联。
For a typical single‑mode fibre with core diameter 9 µm, Rmin = 9 µm / (2 × (1.48/1.44 – 1)) = 9 µm / (2 × 0.0278) ≈ 162 µ his interplay of physics and engineering mathematics is a fertile area for exam questions.
对于典型单模光纤，芯径 9 µm，Rmin ≈ 162 µm。实际上安装限制要大得多（数厘米）以留出安全余量。这种物理与工程数学的相互作用是考试出题的沃土。
7. Gas Laws and Chemical Reactions: The Airbag System | 气体定律与化学反应：安全气囊系统
s rapidly, producing nitrogen gas: 2NaN₃ → 2Na + 3N₂. A typical driver airbag contains 130 g of NaN₃. Calculate the volume of nitrogen produced at 1.01 × 10⁵ Pa and 300 K, assuming ideal gas behaviour. Use stoichiometry and the ideal gas equation.
安全气囊通过叠氮化钠（NaN₃）迅速分解产生氮气而充气：2NaN₃ → 2Na + 3N₂。一个典型的驾驶员气囊含 130 g NaN₃。计算在 1.01 × 10⁵ Pa 和 300 K 下产生的氮气体积，假设为理想气体。需运用化学计量学与理想气体方程。
Stoichiometry (chemistry): molar mass of NaN₃ = 65.0 g mol⁻¹. Moles of NaN₃ = 130 / 65.0 = 2.00 mol. From the reaction, 2 mol NaN₃ produce 3 mol N₂, so moles of N₂ = 3.00 mol. Physics: using pV = nRT, V = nRT / p = 3.00 × 8.31 × 300 / 101 000 ≈ 0.0740 m³ = 74.0 L. In practice, the gas is hot initially (≈600 y. This problem links the mole concept, chemical equation balancing, and the ideal gas law – a classic cross‑over tested in SQA.
化学计量学：NaN₃ 摩尔质量为 65.0 g mol⁻¹，其物质的量 = 130 / 65.0 = 2.00 mol。按反应式，产生 3.00 mol N₂。物理：使用 pV = nRT，V = 3.00 × 8.31 × 300 / 101 000 ≈ 0.0740 m³ = 74.0 L。实际上气体初始是热的（约 600 K），体积会更大，但随后迅速冷却。此题将摩尔概念、化学方程式配平与理想气体定律结合在一起——SQA 常考的跨学科题型。
8. Seismic Waves and Earth Structure: Physics Meets Geology | 地震波与地球结构：物理遇见地质
P‑waves travel at about 8.0 km s⁻¹ in the Earth’s mantle, while S‑waves travel at 4.5 km s⁻¹. If a seismograph records P‑waves 120 s before S‑waves from an earthquake, how far away is the epicentre? Discuss the evidence for a liquid outer core based on S‑wave shadow zones.
P 波在地幔中的传播速度约为 8.0 km s⁻¹，S 波为 4.5 km s⁻¹。如果一个地震仪记录到某次地震的 P 波比 S 波早 120 秒，震中距离有多远？试基于 S 波阴影区讨论液态外地核的证据。
Let the distance be d. Time for P‑wave: tP = d / 8.0; for S‑wave: tS = d / 4.5. The difference Δt = tS – tP = d(1/4.5 – 1/8.0) = 120 s. Compute 1/4.5 = 0.2222, 1/8.0 = 0.125, difference = 0.0972 s km⁻¹. So d = 120 / 0.0972 ≈ 1235 a-based approach is common in geophysics and uses algebraic manipulation from mathematics.
设距离为 d。P 波时间 tP = d / 8.0，S 波时间 tS = d / 4.5。时间差 Δt = d(1/4.5 – 1/8.0) = 120 s。计算 1/4.5 = 0.2222，1/8.0 = 0.125，差值为 0.0972 s km⁻¹。于是 d = 120 / 0.0972 ≈ 1235 km。这种基于公式的方法常见于地球物理，并运用数学中的代数处理。
Geology–physics link: S‑waves are transverse and cannot travel through liquids. Their absence beyond an angular distance of 103° to 180° from an earthquake indicates a liquid outer core. P‑wave refraction at the core–mantle boundary produces a P‑wave shadow zone between about 103° and 142°. By combining wave theory and properties of materials, geophysicists mapped the Earth’s interior – a perfect fusion of physics and geology.
地质与物理的联系：S 波是横波，不能在液体中传播。它们在震中角距 103° 至 180° 之后的缺失表明存在液态外地核。P 波在核‑幔边界折射产生约 103° 至 142° 的 P 波阴影区。通过结合波动理论与材料性质，地球物理学家绘制出地球内部结构——这是物理与地质的完美融合。
9. Ohm’s Law, Resistivity, and Material Science | 欧姆定律、电阻率与材料科学
A wire of length 2.5 m and cross‑sectional area 0.15 mm² has a resistance of 0.85 esistivity of the material. Use the data to identify the material from a table (copper 1.7 × 10⁻⁸ Ω m, constantan 49 × 10⁻⁸ Ω m, nichrome 100 × 10⁻⁸ Ω resistors, linking their low temperature coefficient of resistivity to thermal stability.
一根长 2.5 m、横截面积 0.15 mm² 的导线，电阻为 0.85 Ω。求材料的电阻率，并利用数据从表格中识别该材料（铜 1.7 × 10⁻⁸ Ω m，康铜 49 × 10⁻⁸ Ω m，镍铬合金 100 × 10⁻⁸ Ω m）。解释为何康铜等合金用于标准电阻，并将其低电阻温度系数与热稳定性相联系。
Using R = ρ L / A, rearrange: ρ = R A / L. Area A = 0.15 mm² = 0.15 × 10⁻⁶ m². ρ = 0.85 × 0.15×10⁻⁶ / 2.5 = (0.1275×10⁻⁶) / 2.5 = 5.1×10⁻⁸ Ω (1.7×10⁻⁸), closer to constantan (49×10⁻⁸) but actually lower. It might be a special alloy – much lower temperature coefficient of resistance (α). Constantan’s α is about ±0.002×10⁻³ K⁻¹, compared to copper’s 3.9×10⁻³ stance in precision electronics, a crucial topic in material science for physics students.
由 R = ρ L / A 重新排列：ρ = R A / L。面积 A = 0.15 mm² = 0.15 × 10⁻⁶ m²。ρ = 0.85 × 0.15×10⁻⁶ / 2.5 ≈ 5.1×10⁻⁸ Ω m。此值远离纯铜，更接近康铜，但略低，可能为某种特殊合金。无论如何，合金通常具有更高电阻率和低得多的电阻温度系数。康铜的 α 约为 ±0.002×10⁻³ K⁻¹，而铜的为 3.9×10⁻³ K⁻¹。这种对温度的极低依赖性保证了精密电子设备中的稳定阻值，是物理学生需要掌握的材料科学关键内容。
10. Projectile Motion and Sports Science | 抛体运动与运动科学
A basketball player shoots with an initial velocity of 8.2 m s⁻¹ at an angle of 55° to the horizontal. The basket is at a horizontal distance of 6.0 m and height 3.05 m. The ball is released at a height of 2.0 m. Determine whether the shot scores, and find the angle needed to maximise the probability of scoring from that position. Discuss the role of spin (backspin) and its effect on range – a link to fluid dynamics (Magnus effect).
一名篮球运动员以 8.2 m s⁻¹ 的初速、与水平成 55° 角投篮。篮筐水平距离 6.0 m，高度 3.05 m。出手高度为 2.0 m。判断该球能否命中，并求出从该位置提高命中概率所需的最佳角度。讨论旋转（后旋）的作用及其对射程的影响——与流体动力学（马格纳斯效应）的联系。
Resolve initial velocity: vx = 8.2 cos 55° ≈ 4.70 m s⁻¹, vy = 8.2 sin 55° ≈ 6.72 m s⁻¹. Time to reach x = 6.0 m: t = 6.0 / 4.70 ≈ 1.277 s. Vertical position at that time: y = y₀ + vy t – ½ g t² = 2.0 + (6.72×1.277) – (4.9×1.277²) = 2.0 + 8.58 – 7.99 = 2.59 m. The basket is at 3.05 too low). To hit, the angle must be adjusted. The complete trajectory equation y = x tanθ – (g x²) / (2 v² cos²θ) + y₀ can be solved for θ given y = 3.05 m – ossible angle (a steeper arc). This combines trigonometric equations and quadratic formula – pure applied mathematics.
分解初速：vx ≈ 4.70 m s⁻¹，vy ≈ 6.72 m s⁻¹。到达 x = 6.0 m 所用时间 t ≈ 1.277 s。此时垂直位置 y = 2.0 + (6.72×1.277) – (4.9×1.277²) = 2.0 + 8.58 – 7.99 = 2.59 m。篮筐高 3.05 m，故该球偏低未中。调整角度，使用轨迹方程 y = x tanθ – (g x²) / (2 v² cos²θ) + y₀，令 y = 3.05 m 得到关于 tanθ 的二次方程，可解得更佳角度（更陡的弧线）。这融合了三角方程与二次公式——纯粹的应用数学。
Sports science adds another layer: backspin creates a pressure difference according to Bernoulli’s principle (Magnus effect), adding lift and flattening the trajectory, which increases the margin for error. Exam questions might ask you to explain the effect in terms of air flow and pressure – a crossover to fluid dynamics.
运动科学增添了另一层次：后旋根据伯努利原理产生压力差（马格纳斯效应），提供额外升力并使轨迹更为平直，增大了容错空间。试题可能要求用气流和压力解释该效应——与流体动力学的交叉。
11. Energy Transfers in a Roller Coaster: Engineering Design | 过山车中的能量转换：工程设计
A roller coaster car of mass 500 kg starts from rest at a height of 40 m. It descends into a circular loop of radius 12 m. Neglecting friction, find the speed at the top of the loop and calculate the normal reaction force at that point. Explain what happens if the velocity is too low, and link the design constraint to the concept of centripetal force and g‑force experienced by riders. How would an engineer use data logging (light gates, accelerometers) to verify the design?
一辆质量为 500 kg 的过山车从 40 m 高处由静止出发，驶入一个半径为 12 m 的圆形回环。忽略摩擦，求回环顶点的速率并计算该点的法向反作用力。解释若速度过低会发生什么，并将设计约束与向心力和乘客承受的 g 力概念相联系。工程师会如何借助数据记录（光门、加速度计）验证设计？
Using energy conservation: initial GPE = mgh = 500 × 9.8 × 40 = 196 000 J. At loop top, height = 2r = 24 m, so GPE =
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