📚 Year 12 WJEC Physics: Case Study Practice | 12年级WJEC物理：案例分析实战演练
Welcome to this revision article designed to sharpen your problem-solving skills for WJEC Year 12 Physics. Through a series of carefully chosen case studies, you will learn how to apply physical principles to unfamiliar contexts, break down complex scenarios into manageable steps, and present your reasoning with clarity. Each case mirrors the type of extended question you may encounter in an exam, blending conceptual understanding with quantitative analysis.
欢迎阅读这篇为WJEC 12年级物理设计的复习文章，旨在提升你的解题能力。通过一系列精选的案例分析，你将学会如何将物理原理应用于陌生情境，将复杂场景分解为可管理的步骤，并清晰地展示推理过程。每个案例都模拟了考试中可能遇到的扩展题型，融合了概念理解与定量分析。
1. Kinematics in Free Fall with Drag | 自由落体与空气阻力运动学
A skydiver of mass 78 kg jumps from an aircraft. The vertical velocity data recorded during the first 20 seconds show that the acceleration decreases from 9.7 m s⁻² to 0.2 m s⁻². Analyse the motion using Newton’s second law and the concept of resultant force.
一名质量为78 kg的跳伞者从飞机上跳下。前20秒记录的竖直速度数据显示加速度从9.7 m s⁻²降至0.2 m s⁻²。利用牛顿第二定律和合力的概念分析该运动。
Initially, the only significant vertical force is weight W = mg, giving a resultant force F = 78 × 9.81 ≈ 765 N downward, so acceleration a ≈ 9.8 m s⁻². As speed builds up, air resistance D increases until D approaches W. The net force F_net = W − D, hence a = (W − D)/m. When D = W, a = 0 and terminal velocity is reached. The measured a = 0.2 m s⁻² indicates D is nearly equal to W. The slight remaining acceleration suggests a small imbalance, probably because the skydiver has not yet adopted a perfectly stable spread-eagle position.
最初，显著的竖直力只有重力 W = mg，产生向下的合力 F = 78 × 9.81 ≈ 765 N，因此加速度 a ≈ 9.8 m s⁻²。随着速度增加，空气阻力 D 增大，直到 D 接近 W。净力 F_net = W − D，故 a = (W − D)/m。当 D = W 时，a = 0，达到终极速度。测得的 a = 0.2 m s⁻² 表明 D 几乎等于 W。剩余的微小加速度表明存在小幅不平衡，可能因为跳伞者尚未采取完全稳定的展开姿势。
If the terminal velocity is 55 m s⁻¹, the resistive force at that speed equals weight, 765 N. For a quadratic drag model, D = ½ C ρ A v², we can estimate the effective area A if the drag coefficient C is known. In exam questions, you may be asked to calculate drag force or sketch the acceleration-time graph. Always label forces clearly.
若终极速度为55 m s⁻¹，该速度下的阻力等于重力765 N。对于二次阻力模型 D = ½ C ρ A v²，若已知阻力系数 C，可估算有效面积 A。在考题中，可能要求计算阻力或绘制加速度-时间图像。务必清晰标注受力。
2. Projectile Motion: Kicked Football | 抛体运动：踢足球
A footballer kicks a ball from ground level with an initial speed of 22 m s⁻¹ at an angle of 35° to the horizontal. Calculate the time of flight, maximum height, and horizontal range. Ignore air resistance.
一名足球运动员从地面以22 m s⁻¹的初速度、与水平方向成35°角踢出足球。计算飞行时间、最大高度和水平射程。忽略空气阻力。
Resolve initial velocity: u_x = 22 cos 35° ≈ 18.0 m s⁻¹, u_y = 22 sin 35° ≈ 12.6 m s⁻¹. Time to peak: using v_y = u_y − gt, with v_y = 0 at peak, t_peak = u_y / g = 12.6 / 9.81 ≈ 1.28 s. Total flight time T = 2 × t_peak ≈ 2.57 s. Maximum height H = u_y² / (2g) = (12.6²) / (2 × 9.81) ≈ 8.1 m. Range R = u_x × T = 18.0 × 2.57 ≈ 46.3 m.
分解初速度：u_x = 22 cos 35° ≈ 18.0 m s⁻¹，u_y = 22 sin 35° ≈ 12.6 m s⁻¹。上升至最高点时间：利用 v_y = u_y − gt，最高点 v_y = 0，t_peak = u_y / g = 12.6 / 9.81 ≈ 1.28 s。总飞行时间 T = 2 × 1.28 ≈ 2.57 s。最大高度 H = u_y² / (2g) = (12.6²) / (2 × 9.81) ≈ 8.1 m。射程 R = u_x × T = 18.0 × 2.57 ≈ 46.3 m。
Note that the horizontal component of velocity remains constant. The path is symmetric only if launch and landing are at the same height. In WJEC exams, you may be required to derive an equation for the trajectory y(x) or to calculate the velocity vector at impact.
注意速度的水平分量保持不变。只有当发射点和落地点高度相同时，路径才是对称的。在WJEC考试中，可能要求推导轨迹方程 y(x) 或计算落地时的速度矢量。
3. Connected Particles on a Slope | 斜面上的连接体
Two blocks are connected by a light inextensible string passing over a smooth pulley. Block A (3.0 kg) rests on a rough 25° incline (coefficient of kinetic friction μₖ = 0.20), while block B (1.5 kg) hangs freely. The system is released from rest. Determine the acceleration and the tension in the string.
两个物块由一根轻质不可伸长的绳子连接，跨过一个光滑滑轮。物块A（3.0 kg）置于粗糙的25°斜面上（动摩擦系数 μₖ = 0.20），物块B（1.5 kg）自由悬挂。系统从静止释放。求加速度和绳中张力。
Draw free-body diagrams. For A on the incline: weight component down the plane = m_A g sin θ, friction f = μₖ R, normal reaction R = m_A g cos θ. Net force on A: T − m_A g sin θ − f = m_A a. For B: m_B g − T = m_B a. Add equations: m_B g − m_A g sin θ − μₖ m_A g cos θ = (m_A + m_B) a. Substitute numbers: m_A g sin 25° = 3.0 × 9.81 × 0.423 ≈ 12.44 N; m_A g cos 25° = 3.0 × 9.81 × 0.906 ≈ 26.67 N; friction = 0.20 × 26.67 ≈ 5.33 N. Then 1.5 × 9.81 − 12.44 − 5.33 = (4.5) a → 14.715 − 17.77 = −3.055 N, a = −0.679 m s⁻². Negative sign means block A slides down the plane, opposite to assumed direction. Magnitude of acceleration = 0.68 m s⁻². Tension T from block B: T = m_B (g − a) = 1.5 × (9.81 + 0.68) ≈ 15.7 N (taking downward as positive for B, so a is negative in original assumption; adjust accordingly).
绘制受力图。对于斜面上的A：沿斜面向下的重力分力 = m_A g sin θ，摩擦力 f = μₖ R，法向反力 R = m_A g cos θ。A的合力：T − m_A g sin θ − f = m_A a。对于B：m_B g − T = m_B a。两式相加得：m_B g − m_A g sin θ − μₖ m_A g cos θ = (m_A + m_B) a。代入数值：m_A g sin 25° ≈ 12.44 N；m_A g cos 25° ≈ 26.67 N；摩擦力 ≈ 5.33 N。则 14.715 − 12.44 − 5.33 = 4.5 a → a ≈ −0.68 m s⁻²。负号表示A沿斜面向下滑动，与原假设方向相反。加速度大小为0.68 m s⁻²。由B求张力：T = m_B (g − a) = 1.5 × (9.81 + 0.68) ≈ 15.7 N（需根据正方向调整）。
This case highlights the importance of sign conventions and checking whether the assumption of direction is correct. Always state the direction clearly in the final answer.
此案例强调了符号约定的重要性，并需要检查假设的方向是否正确。在最终答案中务必明确说明方向。
4. Energy Analysis of a Roller Coaster | 过山车的能量分析
A roller coaster car of mass 420 kg is pulled to a height of 45 m above ground and then released. At the lowest point its speed is measured as 27 m s⁻¹. Calculate the energy lost to friction and the average resistive force if the track length from top to bottom is 85 m.
一辆质量为420 kg的过山车被拉到离地45 m高处后释放。在最低点测得速度为27 m s⁻¹。计算摩擦损耗的能量，以及若从顶到底轨道长度为85 m时的平均阻力。
Initial gravitational potential energy E_p = mgh = 420 × 9.81 × 45 ≈ 185,409 J. Kinetic energy at bottom E_k = ½ m v² = 0.5 × 420 × 27² = 210 × 729 = 153,090 J. Energy lost ΔE = 185,409 − 153,090 = 32,319 J. Work done against friction = F_friction × s, so average resistive force F = ΔE / s = 32,319 / 85 ≈ 380 N. This assumes constant friction along the track.
初始重力势能 E_p = mgh = 420 × 9.81 × 45 ≈ 185,409 J。底部动能 E_k = ½ m v² = 0.5 × 420 × 27² = 153,090 J。损失的能量 ΔE = 185,409 − 153,090 = 32,319 J。克服摩擦力做的功 = F_摩擦 × s，因此平均阻力 F = ΔE / s = 32,319 / 85 ≈ 380 N。此计算假设全程摩擦力恒定。
You may also be asked to find the efficiency = (useful output / input) × 100% = (153,090 / 185,409) × 100% ≈ 82.6%. Such problems blend energy conservation with real-world losses, a common exam theme.
你可能还被要求计算效率 = (有用输出 / 输入) × 100% = (153,090 / 185,409) × 100% ≈ 82.6%。这类问题将能量守恒与现实损耗相结合，是常考的题型。
5. Momentum and Inelastic Collision: Car Crash | 动量与非弹性碰撞：车祸
A car of mass 950 kg travelling at 18 m s⁻¹ collides with a stationary car of mass 1200 kg. The cars lock together and slide with locked wheels across a tarmac surface (μₖ = 0.65). Determine the speed just after collision and the distance slid before stopping.
一辆质量为950 kg、速度为18 m s⁻¹的汽车与一辆静止的1200 kg汽车相撞。两车锁在一起并车轮抱死滑过柏油路面（μₖ = 0.65）。求碰撞后的瞬间速度以及停止前滑行的距离。
Using conservation of momentum: m₁ u₁ + m₂ × 0 = (m₁ + m₂) v → 950 × 18 = 2150 × v → v = 17,100 / 2150 ≈ 7.95 m s⁻¹. The friction force on combined wreck: f = μₖ R = μₖ (m₁ + m₂) g = 0.65 × 2150 × 9.81 ≈ 13,702 N. Deceleration a = f / combined mass = μₖ g = 0.65 × 9.81 ≈ 6.38 m s⁻². Using v² = u² + 2as with v = 0, s = (v² − u²) / (2 × (−a)) = u² / (2a) = (7.95²) / (2 × 6.38) ≈ 63.2 / 12.76 ≈ 4.95 m.
应用动量守恒：m₁ u₁ + m₂ × 0 = (m₁ + m₂) v → 950 × 18 = 2150 × v → v ≈ 7.95 m s⁻¹。结合体的摩擦力：f = μₖ R = μₖ (m₁ + m₂) g = 0.65 × 2150 × 9.81 ≈ 13,702 N。减速度 a = f / 总质量 = μₖ g = 6.38 m s⁻²。利用 v² = u² + 2as，v = 0，s = u² / (2a) ≈ (7.95²) / (2 × 6.38) ≈ 4.95 m。
Note that the deceleration is independent of mass and equals μₖ g. In a collision analysis, you must clearly state whether the collision is elastic or inelastic. Here kinetic energy is not conserved: initial KE = ½ × 950 × 18² ≈ 154 kJ, final KE = ½ × 2150 × 7.95² ≈ 68 kJ, loss ≈ 86 kJ converted to deformation and heat.
注意减速度与质量无关，等于 μₖ g。在碰撞分析中，必须明确说明碰撞是弹性的还是非弹性的。此处动能不守恒：初始动能约154 kJ，末动能约68 kJ，损失约86 kJ转化为形变和热能。
6. Interpreting a Stress-Strain Curve | 应力-应变曲线解读
A sample of copper wire of original length 2.00 m and diameter 0.50 mm is stretched until fracture. The force-extension data are converted into stress and strain. Key points on the graph: elastic limit at stress 180 MPa, strain 0.0012; yield point at stress 200 MPa; ultimate tensile strength 250 MPa at strain 0.20; fracture at strain 0.45. Calculate the Young modulus for the elastic region and the work done per unit volume up to the elastic limit.
一根原长2.00 m、直径0.50 mm的铜丝被拉伸至断裂。力-伸长量数据已转换为应力和应变。曲线上的关键点：弹性极限应力180 MPa、应变0.0012；屈服点应力200 MPa；抗拉强度250 MPa、应变0.20；断裂时应变0.45。计算弹性区的杨氏模量以及弹性极限前单位体积做的功。
Young modulus E = stress / strain in the linear region. Using elastic limit point: E = 180 × 10⁶ Pa / 0.0012 = 1.50 × 10¹¹ Pa = 150 GPa. Work done per unit volume = area under stress-strain curve up to elastic limit, which is a triangle for linear elastic: ½ × stress × strain = 0.5 × 180 × 10⁶ × 0.0012 = 108,000 J m⁻³. This is the elastic strain energy stored per unit volume.
杨氏模量 E = 线性区的应力 / 应变。使用弹性极限点：E = 180×10⁶ Pa / 0.0012 = 1.50×10¹¹ Pa = 150 GPa。单位体积做功 = 弹性极限前应力-应变曲线下的面积，线性弹性区为三角形：½ × 应力 × 应变 = 0.5 × 180×10⁶ × 0.0012 = 108,000 J m⁻³。这是单位体积储存的弹性应变能。
Knowing the cross-sectional area A = π (0.25×10⁻³)² ≈ 1.96×10⁻⁷ m², you could find the force at elastic limit F = stress × A ≈ 35.3 N, and total elastic energy = work per volume × original volume = 108,000 × (A × 2.00) ≈ 0.042 J. WJEC particularly values the ability to extract information from graphs.
已知横截面积 A ≈ 1.96×10⁻⁷ m²，可求出弹性极限时的力 F = 35.3 N，总弹性能 = 单位体积功 × 原始体积 ≈ 0.042 J。WJEC特别重视从图中提取信息的能力。
7. Potential Divider with Thermistor | 含热敏电阻的分压电路
A potential divider consists of a 12 V battery, a fixed resistor R₁ = 4.7 kΩ, and an NTC thermistor R₂ in series. At room temperature (20°C), the thermistor resistance is 10 kΩ. Calculate the output voltage V_out across R₂ and explain how it changes if the temperature rises to 50°C, where R₂ drops to 3.0 kΩ.
一个分压电路由12 V电池、固定电阻 R₁ = 4.7 kΩ 和负温度系数热敏电阻 R₂ 串联组成。在室温（20°C）下，热敏电阻阻值为10 kΩ。计算 R₂ 两端的输出电压 V_out，并解释当温度升至50°C（R₂ 降至3.0 kΩ）时它会如何变化。
Using the potential divider formula: V_out = V_in × [R₂ / (R₁ + R₂)]. At 20°C: V_out = 12 × [10,000 / (4700 + 10000)] = 12 × (10/14.7) ≈ 8.16 V. At 50°C: V_out = 12 × [3000 / (4700 + 3000)] = 12 × (3/7.7) ≈ 4.68 V. The output voltage decreases as temperature rises because thermistor resistance falls, reducing the share of the total voltage.
使用分压公式：V_out = V_in × [R₂ / (R₁ + R₂)]。20°C时：V_out = 12 × [10000 / (4700+10000)] ≈ 8.16 V。50°C时：V_out = 12 × [3000 / (4700+3000)] ≈ 4.68 V。输出电压随温度升高而降低，因为热敏电阻阻值下降，分得的电压比例减小。
This circuit can be used as a temperature sensor. If R₁ and R₂ positions are swapped, the output would increase with temperature. Draw a circuit diagram carefully, and always consider tolerance and meter loading in practical questions.
该电路可用作温度传感器。若交换 R₁ 和 R₂ 的位置，输出电压将随温度升高而增加。仔细绘制电路图，并在实验题中考虑容差和电表负载效应。
8. Resistivity of a Metallic Conductor | 金属导体的电阻率
An experiment is performed to determine the resistivity of nichrome wire. A 1.20 m length of wire with diameter 0.315 mm is connected in a circuit. When the current is 0.52 A, the potential difference across the wire is 3.85 V. Calculate the resistivity and compare the value to the accepted figure of 1.10 × 10⁻⁶ Ω m.
进行一项实验以测定镍铬合金丝的电阻率。一段长1.20 m、直径0.315 mm的导线接入电路。当电流为0.52 A时，导线两端电势差为3.85 V。计算电阻率，并与公认值1.10 × 10⁻⁶ Ω m 比较。
Resistance R = V / I = 3.85 / 0.52 ≈ 7.40 Ω. Cross-sectional area A = π (d/2)² = π × (0.315 × 10⁻³ / 2)² ≈ π × (0.1575 × 10⁻³)² = 7.79 × 10⁻⁸ m². Using ρ = RA / L = 7.40 × 7.79 × 10⁻⁸ / 1.20 ≈ 4.80 × 10⁻⁷ Ω m. The measured value is about 0.48 × 10⁻⁶ Ω m, which is lower than the accepted value. Possible reasons: wire diameter larger than nominal, temperature rise, or measurement errors. The accepted value corresponds to a wire of length slightly shorter or area smaller?
电阻 R = V / I = 3.85 / 0.52 ≈ 7.40 Ω。横截面积 A = π (d/2)² ≈ 7.79 × 10⁻⁸ m²。根据 ρ = RA / L = 7.40 × 7.79 × 10⁻⁸ / 1.20 ≈ 4.80 × 10⁻⁷ Ω m。测量值约为0.48 × 10⁻⁶ Ω m，低于公认值。可能原因：导线实际直径大于标称值、温度升高或测量误差。公认值对应的是稍短或更细的导线。
Always express resistivity in Ω m, and be able to design an experiment to minimise uncertainty. Use a micrometer for diameter and measure potential difference with a high-resistance voltmeter. Plot R vs L to find resistivity from gradient if multiple lengths are used.
电阻率单位始终用Ω m，并能够设计实验以减小不确定度。用千分尺测直径，用高阻电压表测电势差。若使用多个长度，可绘制 R-L 图，通过斜率求电阻率。
9. Young’s Double-Slit Interference | 杨氏双缝干涉
In a Young’s double-slit experiment using a laser of wavelength 635 nm, the slits separation is 0.25 mm and the screen is 2.0 m away. Calculate the fringe spacing and predict how the pattern changes if the experiment is submerged in water (refractive index n = 1.33).
在杨氏双缝实验中，使用波长635 nm的激光，双缝间距0.25 mm，屏幕距离2.0 m。计算条纹间距，并预测若将实验装置浸入水中（折射率 n = 1.33）条纹会如何变化。
Fringe spacing Δy = λ D / s, where λ = 635 × 10⁻⁹ m, D = 2.0 m, s = 0.25 × 10⁻³ m. Δy = (635 × 10⁻⁹ × 2.0) / (0.25 × 10⁻³) = 5.08 × 10⁻³ m = 5.08 mm. In water, the wavelength of light becomes λ’ = λ / n = 635 / 1.33 ≈ 477 nm. The fringe spacing will reduce in proportion: Δy’ = λ’ D / s = (477 × 10⁻⁹ × 2.0) / (0.25 × 10⁻³) ≈ 3.82 mm. The pattern retains the same overall intensity distribution but becomes compressed.
条纹间距 Δy = λ D / s = (635×10⁻⁹ × 2.0) / (0.25×10⁻³) ≈ 5.08 mm。在水中，光波长变为 λ’ = λ / n = 635 / 1.33 ≈ 477 nm。条纹间距将按比例减小：Δy’ ≈ 3.82 mm。图案保持相同强度分布但变得密集。
This case links wave optics with refraction. You may also need to describe the effect of using white light: central white fringe and coloured fringes on either side, with blue closer to center. For WJEC, ensure you can derive the path difference condition for constructive interference: d sin θ = nλ.
此案例将波动光学与折射联系起来。你可能还需描述使用白光的效果：中央白色条纹，两侧彩色条纹，蓝色更靠近中心。对WJEC来说，确保能推导出相长干涉的条件：d sin θ = nλ。
10. Photoelectric Effect and Threshold Frequency | 光电效应与截止频率
Ultraviolet light of frequency 1.0 × 10¹⁵ Hz strikes a clean zinc plate. The work function of zinc is 4.3 eV. Determine whether electrons are emitted, and if so, their maximum kinetic energy in joules. Also calculate the de Broglie wavelength of these fastest photoelectrons.
频率为1.0 × 10¹⁵ Hz的紫外光照射到洁净的锌板上。锌的功函数为4.3 eV。判断是否有电子发射，如有，计算其最大动能（以焦耳为单位）。同时计算这些最快光电子的德布罗意波长。
Photon energy E_photon = h f = 6.63 × 10⁻³⁴ × 1.0 × 10¹⁵ = 6.63 × 10⁻¹⁹ J. Convert work function to joules: φ = 4.3 eV × 1.60 × 10⁻¹⁹ J/eV = 6.88 × 10⁻¹⁹ J. Since E_photon < φ, no electrons are emitted. The threshold frequency f₀ = φ / h = 6.88 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ ≈ 1.04 × 10¹⁵ Hz, which is higher than the given frequency. Therefore, photoelectric emission does not occur.
光子能量 E_光子 = h f = 6.63×10⁻³⁴ × 1.0×10¹⁵ = 6.63×10⁻¹⁹ J。功函数转换为焦耳：φ = 4.3 eV × 1.60×10⁻¹⁹ J/eV = 6.88×10⁻¹⁹ J。由于 E_光子 < φ，无电子发射。截止频率 f₀ = φ / h ≈ 1.04×10¹⁵ Hz，高于给定频率。因此不发生光电发射。
If a light source with frequency 1.2 × 10¹⁵ Hz were used instead, then E_k max = h f − φ = (7.96 × 10⁻¹⁹ J) − (6.88 × 10⁻¹⁹ J) = 1.08 × 10⁻¹⁹ J. de Broglie wavelength λ = h / p = h / √(2 m E_k) ≈ 6.63×10⁻³⁴ / √(2 × 9.11×10⁻³¹ × 1.08×10⁻¹⁹) ≈ 1.5 × 10⁻⁹ m. This demonstrates the wave-particle duality concept.
若换用频率为1.2×10¹⁵ Hz的光源，则 E_k max = h f − φ = 1.08×10⁻¹⁹ J。德布罗意波长 λ = h / p ≈ 1.5×10⁻⁹ m。这体现了波粒二象性概念。
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