📚 Derivation of Key Formulae in AS Mechanics | AS 力学关键公式推导
Understanding how the key equations of mechanics are derived strengthens your grasp of physical principles and boosts confidence in applying them to exam problems. This revision guide walks you through the step-by-step derivations of the most important formulae covered in the AS Physics (IAL Unit 1) syllabus, from kinematic relationships to work-energy theorems and momentum conservation. Each derivation is presented with clear reasoning and paired bilingual explanations to support both English and Chinese learners.
理解力学核心公式是如何推导出来的，可以加深你对物理原理的掌握，并增强你解题时的信心。这份复习指南将带你逐步推导 AS 物理（IAL Unit 1）大纲中最重要的公式，涵盖运动学关系、功能定理和动量守恒等内容。每条推导都附有清晰的推理和配对的双语解释，以帮助中英文学习者。
1. Deriving v = u + at | 推导 v = u + at
For an object moving with constant acceleration, the average acceleration a is defined as the rate of change of velocity. If the initial velocity is u and the final velocity after time t is v, then a = (v – u) / t.
对于一个匀加速运动的物体，平均加速度 a 定义为速度的变化率。若初速度为 u，经过时间 t 后的末速度为 v，则 a = (v – u) / t。
Rearranging this definition by multiplying both sides by t gives v – u = a t. Adding u to both sides yields the first kinematic equation.
将该定义式两边同乘以 t，得到 v – u = a t。两边再加 u，就得到第一个运动学方程。
v = u + at
This equation applies whenever the acceleration is uniform and motion is in a straight line.
该方程适用于加速度恒定且直线运动的情况。
2. Displacement from a Velocity-Time Graph: s = ½(u+v)t | 由速度-时间图推导 s = ½(u+v)t
The area under a velocity-time graph represents the displacement s of the object. For constant acceleration, the graph is a straight line from u to v over time t, forming a trapezium.
速度-时间图下的面积表示物体的位移 s。对于匀加速运动，该图是从 u 到 v 的一条直线，形成一个梯形。
The area of a trapezium is the average of the parallel sides multiplied by the distance between them. Hence, s = ((u + v) / 2) × t.
梯形的面积等于平行边的平均值乘以它们之间的距离。因此，s = ((u + v) / 2) × t。
s = ½ (u + v) t
This displacement formula is valid only when acceleration is constant, and it is often used alongside the first kinematic equation.
这个位移公式仅在加速度恒定时有效，通常与第一个运动学方程配合使用。
3. Deriving s = ut + ½at² | 推导 s = ut + ½at²
We can eliminate the final velocity v from the displacement equation by substituting v = u + a t into s = ½ (u + v) t.
我们可以将 v = u + a t 代入 s = ½ (u + v) t，从而消去末速度 v。
Substituting gives s = ½ (u + u + a t) t = ½ (2u + a t) t. Expanding the brackets leads directly to the required form.
代入后得到 s = ½ (u + u + a t) t = ½ (2u + a t) t。展开括号后直接得到所需形式。
s = ut + ½ a t²
This equation gives the displacement of a uniformly accelerated object starting with initial velocity u, without needing the final velocity.
该方程给出了以初速度 u 开始的匀加速物体在时间 t 内的位移，无需知道末速度。
4. Deriving v² = u² + 2as | 推导 v² = u² + 2as
Sometimes it is useful to relate velocity and displacement without involving time. We can derive this relation by eliminating t from the earlier equations.
有时候我们需要在不涉及时间的情况下将速度与位移联系起来。我们可以通过消去 t 来推导这一关系。
From v = u + a t, we have t = (v – u) / a. Substitute this into s = ½ (u + v) t.
由 v = u + a t，得 t = (v – u) / a。将其代入 s = ½ (u + v) t。
This gives s = ½ (u + v) × (v – u) / a = (v² – u²) / (2a). Multiplying both sides by 2a yields the final expression.
得到 s = ½ (u + v) × (v – u) / a = (v² – u²) / (2a)。两边同乘 2a 即得最终表达式。
v² = u² + 2 a s
This powerful equation connects the squares of the initial and final velocities with the displacement and constant acceleration.
这个重要的方程将初末速度的平方与位移和恒定加速度联系起来。
5. Newton’s Second Law F = ma from Momentum | 由动量推导牛顿第二定律 F = ma
Newton formulated his second law in terms of momentum. The linear momentum p of a body is defined as p = m v.
牛顿用动量来表述第二定律。物体的线动量 p 定义为 p = m v。
The resultant force acting on a body equals the rate of change of its momentum: F = dp/dt = d(m v)/dt.
作用在物体上的合力等于其动量的变化率：F = dp/dt = d(m v)/dt。
If the mass m remains constant, it can be taken outside the derivative: F = m (dv/dt). Since dv/dt = a, we obtain the familiar form.
如果质量 m 保持不变，可将其提出微商符号外：F = m (dv/dt)。因为 dv/dt = a，我们便得到常见形式。
F = m a
This equation is valid only for systems with constant mass, but it is the cornerstone of classical mechanics.
该方程仅适用于质量恒定的系统，但它是经典力学的基石。
6. Impulse-Momentum Theorem | 冲量-动量定理
From Newton’s second law in the form F = dp/dt, we can write F dt = dp. Integrating both sides over a time interval Δt gives the impulse-momentum relationship.
由牛顿第二定律 F = dp/dt 的形式，可写为 F dt = dp。在时间间隔 Δt 内积分，便得到冲量-动量关系。
If the net force is constant, the integral simplifies: F Δt = Δp. The left side, F Δt, is defined as the impulse J.
如果合力恒定，积分简化为：F Δt = Δp。左边 F Δt 定义为冲量 J。
J = F Δt = m v – m u
This theorem states that the impulse delivered to an object equals the change in its momentum. It is extremely useful for analysing collisions and impacts.
该定理表明，物体受到的冲量等于其动量的变化。它在分析碰撞和冲击时极为有用。
7. Derivation of Kinetic Energy: KE = ½mv² | 动能公式推导：KE = ½mv²
Consider a constant net force F acting on a body of mass m over a displacement s in the direction of the force. The work done W is F s.
考虑一个质量为 m 的物体，在恒定的合力 F 作用下沿力方向发生位移 s。所做的功 W = F s。
Using F = m a and the kinematic relation v² = u² + 2 a s, we can rewrite a s = (v² – u²) / 2. Therefore, W = m a s = m × (v² – u²) / 2.
利用 F = m a 和运动学关系 v² = u² + 2 a s，可将 a s 改写为 (v² – u²) / 2。因此，W = m a s = m × (v² – u²) / 2。
This simplifies to W = ½ m v² – ½ m u². The quantity ½ m v² is defined as the kinetic energy KE.
化简为 W = ½ m v² – ½ m u²。量 ½ m v² 被定义为动能 KE。
KE = ½ m v²
The work-energy theorem emerges directly: the net work done on an object equals its change in kinetic energy.
直接得到了功能定理：对物体所做的净功等于其动能的变化量。
8. Conservation of Momentum from Newton’s Third Law | 由牛顿第三定律推导动量守恒
Consider two bodies A and B isolated from external forces, colliding for a short time Δt. Let F_AB be the force exerted by A on B, and F_BA the force exerted by B on A.
考虑两个不受外力的物体 A 和 B，在短时间 Δt 内发生碰撞。设 F_AB 为 A 施于 B 的力，F_BA 为 B 施于 A 的力。
By Newton’s third law, these forces are equal in magnitude and opposite in direction: F_AB = -F_BA. Multiplying by Δt gives the impulses: F_AB Δt = – F_BA Δt.
根据牛顿第三定律，这两个力大小相等、方向相反：F_AB = -F_BA。同乘 Δt 得到冲量：F_AB Δt = – F_BA Δt。
Using the impulse-momentum theorem, the impulse on each body equals its change in momentum. Hence Δp_B = – Δp_A, which implies Δp_A + Δp_B = 0.
运用冲量-动量定理，每个物体受到冲量等于其动量变化。因此 Δp_B = – Δp_A，意味着 Δp_A + Δp_B = 0。
Therefore, the total momentum before collision equals the total momentum after collision: m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂.
所以，碰撞前的总动量等于碰撞后的总动量：m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂。
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
This principle is fundamental to all collision and explosion analyses in mechanics.
这一原理是力学中所有碰撞与爆炸分析的基础。
9. Relation between Momentum and Kinetic Energy | 动量与动能的关系
Momentum p and kinetic energy KE can be related directly, which is helpful when analysing motion without knowing mass separately.
动量 p 和动能 KE 可以直接关联，这在无需单独知道质量的情况下分析运动时很有用。
Start with p = m v and KE = ½ m v². Solve for v from the momentum: v = p / m. Substitute into the kinetic energy expression.
由 p = m v 和 KE = ½ m v² 出发。由动量解出 v：v = p / m。代入动能表达式。
KE = ½ m (p / m)² = p² / (2m). Rearranging gives the useful form p² = 2 m KE.
KE = ½ m (p / m)² = p² / (2m)。整理后得到有用形式 p² = 2 m KE。
p² = 2 m KE
This equation shows that for a given mass, a larger kinetic energy means a larger momentum, and it is frequently used in particle physics and collision problems.
该方程表明，在给定质量下，动能越大动量也越大，常用于粒子物理和碰撞问题中。
10. Deriving Centripetal Acceleration a = v²/r | 向心加速度 a = v²/r 的推导
An object moving in a circle of radius r at constant speed v experiences a change in direction. Consider two velocity vectors at points A and B separated by a small angle Δθ and a short arc length v Δt.
一个物体以恒定速率 v 在半径为 r 的圆上运动，方向在变化。考虑在 A 点和 B 点处的两个速度矢量，它们所夹的小角度为 Δθ，弧长为 v Δt。
The change in velocity Δv points approximately towards the centre. For small Δθ, the magnitude of Δv can be found from similar triangles: Δv / v = chord AB / r ≈ v Δt / r.
速度的变化量 Δv 近似指向圆心。对于小角度 Δθ，Δv 的大小可由相似三角形求得：Δv / v = 弦长 AB / r ≈ v Δt / r。
Dividing through by Δt gives Δv / Δt = v² / r. Taking the limit as Δt → 0 yields the instantaneous acceleration, directed towards the centre.
两边除以 Δt 得 Δv / Δt = v² / r。取 Δt → 0 的极限，即得指向圆心的瞬时加速度。
a = v² / r
This centripetal acceleration is essential for understanding circular motion, orbiting satellites, and banked tracks.
这个向心加速度对于理解圆周运动、轨道卫星和倾斜弯道至关重要。
11. Projectile Motion: Trajectory Equation | 抛体运动：轨迹方程
Consider a projectile launched with speed u at an angle θ to the horizontal. The horizontal component of velocity is u cos θ, and the vertical component is u sin θ.
考虑一个以速率 u、与水平方向成 θ 角发射的抛体。水平分速度为 u cos θ，竖直分速度为 u sin θ。
With negligible air resistance, the horizontal motion is uniform: x = (u cos θ) t. The vertical motion is uniformly accelerated by gravity: y = (u sin θ) t – ½ g t².
忽略空气阻力时，水平方向为匀速运动：x = (u cos θ) t。竖直方向受重力匀加速：y = (u sin θ) t – ½ g t²。
Eliminate time t using t = x / (u cos θ) from the horizontal equation. Substitute this into the y-equation.
利用水平方程 t = x / (u cos θ) 消去时间 t，并代入 y 方程。
This gives y = (u sin θ) (x / (u cos θ)) – ½ g (x² / (u² cos² θ)) = x tan θ – (g x²) / (2 u² cos² θ).
得到 y = (u sin θ) (x / (u cos θ)) – ½ g (x² / (u² cos² θ)) = x tan θ – (g x²) / (2 u² cos² θ)。
y = x tan θ – (g x²) / (2 u² cos² θ)
This parabolic equation describes the path of a projectile and is very useful for determining range, maximum height, and trajectory shape.
这个抛物线方程描绘了抛体的轨迹，对于确定射程、最大高度和轨迹形状非常有用。
12. Elastic Potential Energy Stored in a Spring | 弹簧储存的弹性势能
For a spring obeying Hooke’s law within its elastic limit, the force F required to extend or compress it by a distance x is F = k x, where k is the spring constant.
在弹性限度内遵守胡克定律的弹簧，将其拉伸或压缩距离 x 所需的力为 F = k x，其中 k 是劲度系数。
The work done in stretching the spring is the area under the force-extension graph. Since the force increases linearly from 0 to F, the graph forms a triangle of base x and height F.
拉伸弹簧所做的功是力-伸长图下的面积。由于力从 0 线性增加至 F，该图形是个底为 x、高为 F 的三角形。
Thus, the work done W = ½ F x. Substituting F = k x gives
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