📚 BPhO Physics Competition: Key Formulas & Problem-Solving Strategies | BPhO物理竞赛：常用公式与解题思路
BPhO Round 1 and Round 2 are among the most demanding pre-university physics examinations. Success depends on applying a small set of core formulas with clear physical reasoning and careful mathematics.
BPhO 第一轮和第二轮是极具挑战性的大学预科物理竞赛。取得高分的关键，是在扎实的数学基础上，熟练运用少数核心公式并保持清晰的物理直觉。
1. Kinematics and Newton's Laws | 运动学与牛顿定律
Constant-acceleration problems appear in nearly every BPhO paper. The first step is always to list the known variables and choose the equation that links the target quantity.
匀加速运动问题几乎每年都会出现。第一步永远是把已知量列出来，再选择联系目标量的那条运动学公式。
v = u + at, s = ut + ½at², v² = u² + 2as
eleration depends on time or position, return to the definitions: a = dv/dt and v = dx/dt.
这三个方程只适用于匀加速运动。当加速度随时间或位置变化时，应回到定义式：a = dv/dt，v = dx/dt。
Newton' rces to kinematics. On BPhO diagrams, draw forces, resolve them into components, and choose positive directions explicitly.
牛顿第二定律 F = ma 是联系力与运动的桥梁。在 BPhO 的受力图中，先画出所有力、分解到坐标轴，并明确正方向。
Weight: F = mg, where g is the gravitational field strength.
重力：F = mg，其中 g 为重力场强度。
Friction: f ≤ μN; static friction adjusts up to its maximum value.
摩擦力：f ≤ μN；静摩擦力会在零到最大值之间自动调整。
Spring: F = −kx for small displacements from equilibrium.
弹簧：F = −kx，适用于偏离平衡位置的小位移。
2. Work, Energy and Power | 功、能量和功率
Whenever forces do work, energy conservation often gives the quickest route to velocity or height.
只要有力做功，能量守恒常常是求速度或高度的最快路径。
W = Fs cosθ, KE = ½mv², GPE = mgh
Work done by a constant force is the product of force, displacement and the cosine of the angle between them. The net work equals the change in kinetic energy: Wnet = ΔKE.
恒力做功等于力、位移以及二者夹角余弦的乘积。合外力做的总功等于动能变化：Wnet = ΔKE。
In one dimension, this relation lets you find force from a potential-energy graph, a favourite BPhO task.
对于保守力，F = −dU/dx。一维情况下，可以由势能曲线求力，这是 BPhO 的常考题型。
verage power is total work divided by total time. Remember that Fv is only instantaneous if both F and v are instantaneous values.
瞬时功率为 P = Fv，平均功率是总功除以总时间。注意 Fv 里的 F 和 v 都必须是瞬时值。
3. Rotational Dynamics | 转动动力学与角动量
Rotational mechanics generalises linear mechanics to spinning bodies. Always choose a reference axis and keep it fixed throughout the problem.
转动动力学是直线力学的推广。解题时先确定参考轴，然后全程保持一致。
τ = rF sinθ, I = Σmr², τ = Iα
Torque τ equals moment of inertia I multiplied by angular acceleration α. The moment of inertia depends on the chosen axis and the distribution of mass.
力矩 τ 等于转动惯量 I 乘以角加速度 α。转动惯量取决于所选转轴和质量分布。
The parallel-axis theorem is essential for composite bodies: I = Icm + Md², where d is the distance from the centre-of-mass axis.
平行轴定理对组合物体非常重要：I = Icm + Md²，式中 d 是质心轴与新轴之间的距离。
ernal torque is zero, angular momentum is conserved, even during collisions and explosions.
角动量 L = Iω。若合外力矩为零，则角动量守恒，即使在碰撞或爆炸过程中也成立。
For pure rolling without slipping, vcm = rω, and total kinetic energy is KE = ½mv² + ½Iω².
无滑动纯滚动时，vcm = rω，总动能为平动动能加转动动能：KE = ½mv² + ½Iω²。
4. Gravitation and Orbits | 万有引力与轨道运动
Gravitation is one of the most common contexts for circular-motion and energy questions in BPhO.
万有引力是 BPhO 中圆周运动与能量问题最常见的背景之一。
F = GMm/r², g = GM/r²
The gravitational force acts along the line joining the centres. For spherical objects, treat the whole mass as being concentrated at the centre.
引力沿两球心的连线方向。处理球体时，可把质量视为集中在球心。
ntial is V = −GM/r. The negative sign matters: energy increases when separation increases.
引力势能 U = −GMm/r，引力势 V = −GM/r。负号很重要：间距增大时势能增大。
For a circular orbit, speed is v = √GM/r. Substituting v = 2πr/T gives Kepler's third law: T² = (4π²/GM)r³.
圆形轨道速率 v = √GM/r。代入 v = 2πr/T 可得开普勒第三定律：T² = (4π²/GM)r³。
Escape speed from radius R is v = √2GM/R, obtained by setting the total energy to zero.
从半径 R 处逃离的逃逸速度为 v = √2GM/R，令总能量为零即可导出。
5. Oscillations and Simple Harmonic Motion | 简谐振动
n acceleration is proportional to displacement and directed towards equilibrium: a = −ω²x.
简谐运动出现在加速度与位移成正比且始终指向平衡位置时：a = −ω²x。
x = A cos(ωt + φ), v = −Aω sin(ωt + φ), a = −ω²x
All three equations describe the same motion; choose the one that matches the given initial conditions.
三个方程描述同一种运动；根据初始条件选择使用哪一个。
For a mass on a spring, T = 2π√m/k. For a simple pendulum with small amplitude, T = 2π√L/g.
弹簧振子周期 T = 2π√m/k。小角度单摆周期
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