题型概述：从条件建立力学模型Question type: building a mechanics model
Round 2 力学题通常把同一情境分成相互衔接的长题：先解释或证明，再读图、算数，最后讨论模型边界。本页选取 2024 Q2、2025 Q3 与 Q4。两份原卷均要求 3 小时完成四题，建议每题约 45 分钟，只说明各题分值相近，未印具体分值；本页不补造官方分数，顶部统计 23 项数值核对。常见失分点是忽略平衡条件、将平均速度当瞬时速度，以及在大振幅问题中直接使用小角度近似。Round 2 mechanics uses linked parts: explanation or proof, graph reading and calculation, then discussion of model limits. This module selects 2024 Q2 and 2025 Q3–Q4. Both papers allow three hours for four questions, about 45 minutes each, and say the questions carry similar marks without printing numerical allocations. No official marks are invented here; the header tracks 23 numerical checks. Typical errors are missing equilibrium conditions, confusing average with instantaneous speed, and applying a small-angle approximation at large amplitude.
1先判定约束，再选择方程Identify constraints before choosing equations
牛顿第二定律始终成立，但 L = mg 只适用于无竖直加速度的情况。平飞时迎角改变会迫使速度改变，不能只改变一个系数而把其他条件都固定。Newton’s second law always applies, but L = mg requires zero vertical acceleration. In level flight, changing angle forces a speed change; one cannot change just a coefficient while fixing every other variable.
L − mg = maᵧ; L = ½C_LρAv²; D/L = C_D/C_L先写“什么保持不变”，再代数值。Write down what remains fixed before substituting numbers.
2量纲与幂律Dimensions and power laws
量纲分析将复杂过程约束为幂律，但不能求出无量纲常数。若位移随时间为幂函数，微分后才能得到瞬时速度；指数也直接告诉我们变量加倍时的变化倍数。Dimensional analysis constrains a process to a power law but cannot determine a dimensionless coefficient. Differentiate a displacement power law to obtain instantaneous speed; its exponent also tells you the effect of doubling a variable.
y = Cxⁿ ⇒ y(2x)/y(x) = 2ⁿ; d(Ctⁿ)/dt = nCtⁿ⁻¹冲击波半径的时间指数是 2/5；其速度指数则是 −3/5。The shock radius has time exponent 2/5, while its speed has exponent −3/5.
3能量、转动惯量与近似Energy, inertia and approximations
能量守恒可以给出大振幅单摆在每个位置的角速度，周期再由时间积分求出。复摆还要考虑转动惯量；重心位置相同，并不意味着周期相同。Energy conservation gives angular speed at every position even at large amplitude; integrate time to find the period. For a compound pendulum, inertia matters too: the same centre of mass does not imply the same period.
½mℓ²θ̇² = mgℓ(cosθ − cosθ₀); Iθ̈ = Γ使用 sinθ ≈ θ 时，应明确角度以弧度表示并且足够小。State that θ is small and in radians when using sinθ ≈ θ.
子题型导航Sub-type navigation
| 子题型Sub-type | 识别特征Recognise it | 核心步骤Core steps | 真题Question |
|---|---|---|---|
| 升力与阻力Lift and drag | 系数图、迎角、平飞Coefficient graphs, angle, level flight | 受力平衡 → 速度 → 功率Force balance → speed → power | 2024 Q2 |
| 量纲与变化率Dimensions and rates | 未知指数、照片标尺Unknown powers, photograph scale | 匹配量纲 → 求导 → 检查边界Match dimensions → differentiate → test limits | 2025 Q3 |
| 非线性振动Nonlinear oscillations | 大角度、周期积分、复摆Large angle, period integral, compound pendulum | 能量 → 换元 → AGM / 转动方程Energy → substitution → AGM / rotation | 2025 Q4 |