📚 Year 13 CCEA Physics: Case Study Practical Exercises | CCEA Year 13 物理：案例分析实战演练
In CCEA Year 13 Physics, case study questions go beyond simple recall – they test your ability to apply principles to unfamiliar, real-world scenarios. This article presents a series of practical exercises designed to sharpen your analytical skills. Each case study is broken down into clear steps: identifying given data, selecting relevant equations, solving systematically, and reflecting on the physical meaning of the result. Work through these examples to build confidence for your AS examinations.
在 CCEA Year 13 物理中，案例分析题不仅考查记忆，更考查你将物理原理应用于陌生的实际情境的能力。本文提供了一系列实战演练，旨在强化你的分析技能。每个案例都按照清晰的步骤展开：识别已知数据、选取相关方程、系统求解并反思结果的物理意义。通过这些练习，你将建立应对 AS 考试的信心。
1. Case Study 1: Stopping Distance of a Car | 案例 1：汽车的刹车距离
A car is travelling at 20 m/s on a straight road. The driver sees an obstacle and applies the brakes, causing a uniform deceleration of 6.0 m/s². We examine the thinking distance (reaction time 0.70 s) and the braking distance to determine whether the car stops in time.
一辆汽车以 20 m/s 的速度在笔直道路上行驶。司机发现障碍物并踩下刹车，产生 6.0 m/s² 的匀减速度。我们分析反应距离（反应时间 0.70 s）和制动距离，以判断汽车是否能及时停下。
Step 1: Calculate thinking distance. During the reaction time, the car continues at constant speed: s₁ = u × t = 20 × 0.70 = 14 m.
步骤 1：计算反应距离。在反应时间内，汽车保持匀速：s₁ = u × t = 20 × 0.70 = 14 m。
Step 2: Determine braking distance using v² = u² + 2as, with v = 0, u = 20 m/s, a = -6.0 m/s². Rearranging: 0 = (20)² + 2×(-6.0)×s₂ → s₂ = (400)/(12) = 33.3 m.
步骤 2：利用 v² = u² + 2as 求制动距离，其中 v = 0，u = 20 m/s，a = -6.0 m/s²。整理得：0 = (20)² + 2×(-6.0)×s₂ → s₂ = (400)/(12) = 33.3 m。
Step 3: Total stopping distance = s₁ + s₂ = 14 + 33.3 = 47.3 m. If the obstacle is 45 m away, the collision is unavoidable under these conditions. This highlights the importance of speed limits and road conditions.
步骤 3：总停车距离 = s₁ + s₂ = 14 + 33.3 = 47.3 m。若障碍物在 45 m 外，在此条件下碰撞无法避免。这凸显了限速和路面状况的重要性。
2. Case Study 2: Block Sliding Down an Incline | 案例 2：滑块沿斜面下滑
A 4.0 kg block slides from rest down a rough plane inclined at 25° to the horizontal. The coefficient of kinetic friction is 0.30. We need to find the acceleration and the speed after 2.5 m of travel.
一个 4.0 kg 的滑块从静止开始沿粗糙斜面下滑，斜面与水平面夹角 25°。动摩擦因数为 0.30。求加速度和滑行 2.5 m 后的速度。
Step 1: Resolve weight: component down the plane = mg sinθ = 4.0×9.8×sin25° ≈ 16.6 N. Normal reaction N = mg cosθ = 4.0×9.8×cos25° ≈ 35.5 N.
步骤 1：分解重力：沿斜面的分力 = mg sinθ = 4.0×9.8×sin25° ≈ 16.6 N。法向反力 N = mg cosθ = 4.0×9.8×cos25° ≈ 35.5 N。
Step 2: Frictional force f = μₖN = 0.30×35.5 = 10.65 N opposing motion. Net force down the plane Fₙₑₜ = 16.6 – 10.65 = 5.95 N.
步骤 2：摩擦力 f = μₖN = 0.30×35.5 = 10.65 N，方向与运动相反。沿斜面的净力 Fₙₑₜ = 16.6 – 10.65 = 5.95 N。
Step 3: Acceleration a = Fₙₑₜ/m = 5.95/4.0 = 1.49 m/s². Then using v² = u² + 2as with u=0: v² = 0 + 2×1.49×2.5 = 7.45, so v ≈ 2.73 m/s. Note that the acceleration is independent of mass only if friction is neglected.
步骤 3：加速度 a = Fₙₑₜ/m = 5.95/4.0 = 1.49 m/s²。再利用 v² = u² + 2as，u=0：v² = 0 + 2×1.49×2.5 = 7.45，因此 v ≈ 2.73 m/s。注意，仅当忽略摩擦时加速度才与质量无关。
3. Case Study 3: Roller Coaster Loop-the-Loop | 案例 3：过山车圆环通道
A roller coaster car is released from rest at a height h above the ground and enters a vertical circular loop of radius 8.0 m. To stay on the track, the car must maintain a minimum speed at the top of the loop. We determine the minimum release height.
一辆过山车从地面以上高度 h 处由静止释放，进入半径为 8.0 m 的竖直圆环。为保持在轨道上，小车必须在圆环顶部保持最小速度。求最小释放高度。
Step 1: At the top of the loop, the critical condition is that the normal reaction is just zero, so weight provides centripetal force: mg = mv²/r → vₘᵢₙ = √(gr) = √(9.8×8.0) = √78.4 ≈ 8.86 m/s.
步骤 1：在圆环顶部，临界条件为法向反力恰好为零，重力提供向心力：mg = mv²/r → vₘᵢₙ = √(gr) = √(9.8×8.0) = √78.4 ≈ 8.86 m/s。
Step 2: Apply conservation of mechanical energy between release point and top of loop. Taking the ground as zero potential: mgh = ½mv² + mg(2r). Cancel m: gh = ½v² + 2gr.
步骤 2：在释放点和圆环顶部之间应用机械能守恒。以地面为零势能面：mgh = ½mv² + mg(2r)。消去 m：gh = ½v² + 2gr。
Step 3: Substitute v = √(gr): gh = ½(gr) + 2gr = 2.5gr → h = 2.5r = 2.5×8.0 = 20.0 m. The car must start at least 20 m above the ground. This result is independent of mass and assumes no energy losses.
步骤 3：代入 v = √(gr)：gh = ½(gr) + 2gr = 2.5gr → h = 2.5r = 2.5×8.0 = 20.0 m。小车必须从至少 20 m 高处释放。该结果与质量无关，且假设无能量损失。
4. Case Study 4: Bullet and Block Inelastic Collision | 案例 4：子弹与物块的非弹性碰撞
A 15 g bullet is fired horizontally into a stationary wooden block of mass 2.5 kg suspended by light strings. The bullet embeds itself, and the block swings upward, rising through a vertical height of 12 cm. We find the initial speed of the bullet.
一颗 15 g 的子弹水平射入一个用轻绳悬挂、质量为 2.5 kg 的静止木块。子弹嵌入其中，木块向上摆动，垂直升高 12 cm。求子弹的初始速度。
Step 1: Treat the process in two stages: perfectly inelastic collision, then pendulum swing. First, for the swing use conservation of energy to find speed of block+bullet just after collision: ½(m+M)v² = (m+M)gh → v = √(2gh) = √(2×9.8×0.12) = √2.352 = 1.53 m/s.
步骤 1：将过程分为两个阶段：完全非弹性碰撞，随后摆锤摆动。首先，对于摆动利用能量守恒求碰撞后瞬间木块+子弹的速度：½(m+M)v² = (m+M)gh → v = √(2gh) = √(2×9.8×0.12) = √2.352 = 1.53 m/s。
Step 2: For the collision, momentum is conserved in the horizontal direction: m_bullet × u = (m_bullet + M) × v. Here m_bullet = 0.015 kg, M = 2.5 kg, v = 1.53 m/s.
步骤 2：对于碰撞，水平方向动量守恒：m_bullet × u = (m_bullet + M) × v。此处 m_bullet = 0.015 kg，M = 2.5 kg，v = 1.53 m/s。
Step 3: Solve for u: u = [(0.015+2.5)×1.53] / 0.015 = (2.515×1.53)/0.015 = 3.847/0.015 ≈ 256.5 m/s. The bullet’s initial speed is about 260 m/s, a typical value. Kinetic energy is not conserved; most is dissipated as heat and deformation.
步骤 3：求解 u：u = [(0.015+2.5)×1.53] / 0.015 = (2.515×1.53)/0.015 = 3.847/0.015 ≈ 256.5 m/s。子弹的初始速度约为 260 m/s，属于典型值。动能并不守恒，大部分以热和形变形式耗散。
5. Case Study 5: Complex Resistor Network | 案例 5：复杂电阻网络
Consider a circuit with a 12 V battery connected to three resistors: R1 = 10 Ω in series with a parallel combination of R2 = 20 Ω and R3 = 30 Ω. Determine the total current drawn from the battery and the power dissipated in R2.
考虑一个电路，12 V 电池连接到三个电阻：R1 = 10 Ω 串联于 R2 = 20 Ω 和 R3 = 30 Ω 的并联组合。求电池提供的总电流以及 R2 消耗的功率。
Step 1: Calculate equivalent resistance of the parallel branch: 1/R_para = 1/20 + 1/30 = (3+2)/60 = 5/60 → R_para = 12 Ω.
步骤 1：计算并联支路的等效电阻：1/R_para = 1/20 + 1/30 = (3+2)/60 = 5/60 → R_para = 12 Ω。
Step 2: Total circuit resistance R_total = R1 + R_para = 10 + 12 = 22 Ω. Total current I = V/R_total = 12/22 = 0.545 A (approx).
步骤 2：总电路电阻 R_total = R1 + R_para = 10 + 12 = 22 Ω。总电流 I = V/R_total = 12/22 = 0.545 A（约）。
Step 3: The voltage across the parallel branch is V_para = I × R_para = 0.545×12 = 6.54 V. The current through R2 is I₂ = V_para/R₂ = 6.54/20 = 0.327 A. Power in R2: P = I₂²R₂ = (0.327)²×20 ≈ 2.14 W. Alternatively, P = V_para²/R₂ = (6.54)²/20 ≈ 2.14 W. This step-by-step breakdown helps avoid errors in more intricate networks.
步骤 3：并联支路的电压 V_para = I × R_para = 0.545×12 = 6.54 V。流过 R2 的电流 I₂ = V_para/R₂ = 6.54/20 = 0.327 A。R2 的功率：P = I₂²R₂ = (0.327)²×20 ≈ 2.14 W。或 P = V_para²/R₂ = (6.54)²/20 ≈ 2.14 W。这种逐步分解有助于在更复杂的网络中避免错误。
6. Case Study 6: Young’s Double-Slit Wavelength Determination | 案例 6：杨氏双缝干涉测定波长
In a Young’s double-slit experiment, monochromatic light is incident on two slits separated by 0.40 mm. The interference pattern is observed on a screen 2.5 m away, and the distance between the central bright fringe and the fourth bright fringe is 14.0 mm. Determine the wavelength of the light.
在杨氏双缝实验中，单色光照射在间距为 0.40 mm 的双缝上。在 2.5 m 远的屏幕上观察干涉图样，中央亮纹与第四级亮纹之间的距离为 14.0 mm。求光的波长。
Step 1: Recall the double-slit equation λ = ay/D, where a is slit separation, y is fringe separation for adjacent bright fringes, D is screen distance. Here y must be the fringe spacing, not the total distance to the fourth fringe.
步骤 1：回顾双缝方程 λ = ay/D，其中 a 为缝距，y 为相邻亮纹间距，D 为屏距。此处 y 必须是条纹间距，而非到第四级条纹的总距离。
Step 2: The fourth bright fringe corresponds to n=4 (with central n=0). Thus total distance from centre = 4 fringe spacings: 4y = 14.0 mm → y = 3.50 mm = 3.50×10⁻³ m. Slit separation a = 0.40 mm = 4.0×10⁻⁴ m, D = 2.5 m.
步骤 2：第四级亮纹对应 n=4（中央为 n=0）。因此从中央算起的总距离为 4 个条纹间距：4y = 14.0 mm → y = 3.50 mm = 3.50×10⁻³ m。缝距 a = 0.40 mm = 4.0×10⁻⁴ m，D = 2.5 m。
Step 3: Substituting: λ = (4.0×10⁻⁴ × 3.50×10⁻³) / 2.5 = (1.4×10⁻⁶)/2.5 = 5.6×10⁻⁷ m = 560 nm. This is in the visible green-yellow region. Accurate measurement of fringe spacing is crucial for reliable results; using multiple fringes reduces the percentage uncertainty.
步骤 3：代入得：λ = (4.0×10⁻⁴ × 3.50×10⁻³) / 2.5 = (1.4×10⁻⁶)/2.5 = 5.6×10⁻⁷ m = 560 nm。该波长位于可见光的绿-黄区域。精确测量条纹间距对可靠结果至关重要；利用多个条纹可减小百分误差。
7. Case Study 7: Photoelectric Effect Data Analysis | 案例 7：光电效应数据分析
In a photoelectric experiment, a student records the stopping voltage Vₛ for various frequencies f of incident light on a metal surface. The data are: (6.0×10¹⁴ Hz, 0.48 V), (7.0×10¹⁴ Hz, 0.90 V), (8.0×10¹⁴ Hz, 1.32 V), (9.0×10¹⁴ Hz, 1.74 V). Use this to find Planck’s constant and the work function.
在光电效应实验中，学生记录下不同频率 f 的入射光照射某金属表面时的遏止电压 Vₛ。数据为：(6.0×10¹⁴ Hz, 0.48 V)、(7.0×10¹⁴ Hz, 0.90 V)、(8.0×10¹⁴ Hz, 1.32 V)、(9.0×10¹⁴ Hz, 1.74 V)。利用这些数据求出普朗克常数和逸出功。
Step 1: Apply Einstein’s photoelectric equation: eVₛ = hf – φ, where e = 1.60×10⁻¹⁹ C. Thus Vₛ = (h/e)f – φ/e. This is a linear relationship of the form y = mx + c, with slope = h/e.
步骤 1：应用爱因斯坦光电方程：eVₛ = hf – φ，其中 e = 1.60×10⁻¹⁹ C。因此 Vₛ = (h/e)f – φ/e。这是形如 y = mx + c 的线性关系，斜率 = h/e。
Step 2: Calculate the slope from any two data points, e.g. (6.0, 0.48) and (9.0, 1.74): slope = (1.74 – 0.48) / (9.0×10¹⁴ – 6.0×10¹⁴) = 1.26 / (3.0×10¹⁴) = 4.20×10⁻¹⁵ V·s.
步骤 2：从任意两个数据点计算斜率，例如 (6.0, 0.48) 和 (9.0, 1.74)：斜率 = (1.74 – 0.48) / (9.0×10¹⁴ – 6.0×10¹⁴) = 1.26 / (3.0×10¹⁴) = 4.20×10⁻¹⁵ V·s。
Step 3: Then h = slope × e = 4.20×10⁻¹⁵ × 1.60×10⁻¹⁹ = 6.72×10⁻³⁴ J·s, close to the accepted 6.63×10⁻³⁴ J·s. The intercept c = Vₛ – (slope)f; using the first point: 0.48 = (4.20×10⁻¹⁵)(6.0×10¹⁴) – φ/e → 0.48 = 2.52 – φ/e → φ/e = 2.04 → φ = 2.04×1.60×10⁻¹⁹ = 3.26×10⁻¹⁹ J (≈2.04 eV). The negative intercept confirms the threshold frequency f₀ = φ/h.
步骤 3：则 h = 斜率 × e = 4.20×10⁻¹⁵ × 1.60×10⁻¹⁹ = 6.72×10⁻³⁴ J·s，接近公认值 6.63×10⁻³⁴ J·s。截距：利用第一点数据，0.48 = (4.20×10⁻¹⁵)(6.0×10¹⁴) – φ/e → 0.48 = 2.52 – φ/e → φ/e = 2.04 → φ = 2.04×1.60×10⁻¹⁹ = 3.26×10⁻¹⁹ J（约 2.04 eV）。负截距证实了截止频率 f₀ = φ/h。
8. Case Study 8: Isothermal Compression of an Ideal Gas | 案例 8：理想气体的等温压缩
A cylinder fitted with a frictionless piston contains 0.020 mol of an ideal gas at 300 K. The initial volume is 500 cm³. The gas is compressed slowly so that its temperature remains constant until the pressure reaches 2.5 × 10⁵ Pa. Find the final volume and the work done on the gas.
一个装有光滑活塞的气缸内含有 0.020 mol 理想气体，温度为 300 K。初始体积为 500 cm³。缓慢压缩气体，使其温度保持不变，直到压强达到 2.5 × 10⁵ Pa。求最终体积和对气体做的功。
Step 1: Use the ideal gas law pV = nRT. Initial pressure p₁ = nRT/V₁ = (0.020×8.31×300) / (500×10⁻⁶) = (49.86) / (5.0×10⁻⁴) = 9.97×10⁴ Pa.
步骤 1：用理想气体状态方程 pV = nRT。初始压强 p₁ = nRT/V₁ = (0.020×8.31×300) / (500×10⁻⁶) = (49.86) / (5.0×10⁻⁴) = 9.97×10⁴ Pa。
Step 2: Since temperature is constant, Boyle’s law applies: p₁V₁ = p₂V₂. Thus V₂ = p₁V₁/p₂ = (9.97×10⁴ × 5.0×10⁻⁴) / (2.5×10⁵) = 49.86 / 2.5×10⁵ = 1.994×10⁻⁴ m³ = 199 cm³.
步骤 2：由于温度恒定，玻意耳定律适用：p₁V₁ = p₂V₂。因此 V₂ = p₁V₁/p₂ = (9.97×10⁴ × 5.0×10⁻⁴) / (2.5×10⁵) = 49.86 / 2.5×10⁵ = 1.994×10⁻⁴ m³ = 199 cm³。
Step 3: Work done on the gas during isothermal compression is W = -nRT ln(V₂/V₁). Here V₂/V₁ = 199/500 = 0.398. So W = -0.020×8.31×300 × ln(0.398) = -49.86 × (-0.921) = +45.9 J. The positive sign indicates work is done on the gas, increasing its internal energy? No – in an isothermal process, the internal energy of an ideal gas stays constant; the work done on the gas is converted to heat expelled to the surroundings. This exercise reinforces the connection between gas laws and thermodynamics.
步骤 3：等温压缩过程中对气体做的功为 W = -nRT ln(V₂/V₁)。此处 V₂/V₁ = 199/500 = 0.398。因此 W = -0.020×8.31×300 × ln(0.398) = -49.86 × (-0.921) = +45.9 J。正号表示对气体做功。这是否增加了内能？不——理想气体在等温过程中内能保持不变；对气体做的功转化为热量向环境释放。此练习强化了气体定律与热力学之间的联系。
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