📚 AS Physics Unit 1 June 2019: Key Formula Derivations | AS物理单元1 2019年6月：关键公式推导
The June 2019 AS Physics Unit 1 paper challenged students to show a deep understanding of mechanics and materials by deriving core equations. This article unpacks the step-by-step logic behind the most important derivations, from the equations of motion to the expression for elastic potential energy. Mastering these not only prepares you for similar questions but also builds a rock-solid foundation for the rest of the A-level course.
2019年6月的AS物理单元1试卷通过推导核心方程来考查学生对力学和材料的深刻理解。本文逐步拆解了从运动方程到弹性势能表达式等最重要推导背后的逻辑。掌握这些内容不仅能让你从容应对类似题目，还能为后续A-level课程打下坚实的基础。
1. Deriving the First Equation of Motion: v = u + at | 推导第一个运动方程：v = u + at
We start from the definition of acceleration: acceleration is the rate of change of velocity.
我们从加速度的定义出发：加速度是速度的变化率。
If an object accelerates uniformly from initial velocity u to final velocity v in time t, the acceleration a is given by a = (v – u)/t.
若一个物体在时间 t 内从初速度 u 匀加速到末速度 v，则加速度 a 表示为 a = (v – u)/t。
Rearranging this expression immediately yields the first equation of motion.
将此式重新整理就直接得到第一个运动方程。
v = u + a t
This linear relationship is the cornerstone of all uniformly accelerated motion calculations.
这个线性关系是所有匀加速运动计算的基石。
2. Deriving the Second Equation: s = ut + ½at² | 推导第二个方程：s = ut + ½at²
For an object moving with uniform acceleration, the displacement s can be found using the average velocity.
对于匀加速运动的物体，位移 s 可用平均速度求出。
The average velocity is (u + v)/2, and displacement is s = average velocity × time.
平均速度是 (u + v)/2，而位移 s = 平均速度 × 时间。
We already know that v = u + a t from the first equation. Substituting this into the average velocity expression gives s = [u + (u + a t)]/2 × t.
由第一方程已知 v = u + a t，将其代入平均速度表达式可得 s = [u + (u + a t)]/2 × t。
Simplifying the right-hand side leads directly to the second equation.
化简右端直接导出第二个运动方程。
s = u t + ½ a t²
3. Deriving the Third Equation: v² = u² + 2as | 推导第三个方程：v² = u² + 2as
We can eliminate time t from the first two equations to obtain a relationship between velocity and displacement.
我们可以从前面两个方程中消去时间 t，得到速度与位移的关系。
From the first equation, t = (v – u)/a. Substitute this into the second equation s = u t + ½ a t².
由第一方程得 t = (v – u)/a，将其代入第二方程 s = u t + ½ a t²。
After algebraic manipulation: s = u(v – u)/a + ½ a [(v – u)/a]², which simplifies to s = (v² – u²)/(2a).
经过代数运算：s = u(v – u)/a + ½ a [(v – u)/a]²，化简后得到 s = (v² – u²)/(2a)。
Multiplying both sides by 2a gives the compact and powerful third equation.
两边乘以 2a 即得到这个简洁而强大的第三方程。
v² = u² + 2 a s
4. Deriving Kinetic Energy: KE = ½mv² | 推导动能：KE = ½mv²
Kinetic energy is the energy of motion. To derive its standard formula, consider the work done to accelerate a mass from rest to speed v.
动能是运动的能量。为推导其标准公式，考虑把一个质量从静止加速到速度 v 所做的功。
Work done W = F s, and by Newton’s second law F = m a.
做的功 W = F s，根据牛顿第二定律 F = m a。
For an object starting from rest, u = 0, the third equation gives v² = 2 a s, so a s = v²/2.
物体从静止开始，u = 0，由第三方程得 v² = 2 a s，即 a s = v²/2。
Substituting F and s into the work expression: W = m a × s = m × (a s) = m × (v²/2) = ½ m v².
将 F 和 s 代入功的表达式：W = m a × s = m × (a s) = m × (v²/2) = ½ m v²。
This net work done is stored as kinetic energy, hence:
这个净功储存为动能，因此：
KE = ½ m v²
5. Deriving Gravitational Potential Energy: GPE = mgh | 推导重力势能：GPE = mgh
Gravitational potential energy is the work done against gravity to lift an object to a height h.
重力势能是将物体提升至高度 h 过程中克服重力所做的功。
To raise a mass m at constant speed, you must apply a force equal to its weight, F = m g, over the vertical displacement h.
要匀速提升质量 m，你必须施加一个与重力相等的力 F = m g，并经历竖直位移 h。
Since work W = force × distance moved in the direction of the force, W = m g × h.
由于功 W = 力 × 沿力方向的位移，有 W = m g × h。
Assuming no energy losses, this work is stored as gravitational potential energy.
假设没有能量损失，这个功便储存为重力势能。
GPE = m g h
Note that g is the gravitational field strength (9.81 N/kg on Earth’s surface), and h is measured vertically.
注意 g 是重力场强度（地球表面为 9.81 N/kg），h 沿竖直方向测量。
6. Deriving the Impulse–Momentum Theorem: FΔt = Δp | 推导冲量–动量定理：FΔt = Δp
Newton’s second law expresses the relationship between force and change in momentum.
牛顿第二定律表达了力与动量变化的关系。
Starting from the definition of momentum p = m v, the rate of change of momentum for constant mass is m Δv/Δt.
从动量定义 p = m v 出发，质量恒定时动量变化率是 m Δv/Δt。
Since a = Δv/Δt, we have Δp/Δt = m a = F (net force).
因为 a = Δv/Δt，所以有 Δp/Δt = m a = F（合外力）。
Multiplying both sides by the time interval Δt gives the impulse–momentum relationship.
两边乘以时间间隔 Δt 便得到冲量–动量关系。
F Δt = m v − m u = Δ p
This shows that the impulse of a force equals the change in momentum it produces, a key principle in collision and safety analysis.
这表明力的冲量等于它产生的动量变化，这是碰撞和安全分析中的关键原理。
7. Deriving the Work–Energy Principle | 推导功能原理
The work–energy principle states that the net work done on an object equals its change in kinetic energy.
功能原理指出，对物体所做的净功等于其动能的变化量。
Consider a mass m initially at speed u, experiencing a constant net force F over displacement s, reaching final speed v.
考虑一个初始速度为 u 的质量 m，在位移 s 上受到恒定的合外力 F，最终速度变为 v。
From F = m a and a = (v² − u²)/(2s) (from the third equation), the net work is W = F s = m × (v² − u²)/(2s) × s = ½ m v² − ½ m u².
由 F = m a 及 a = (v² − u²)/(2s)（来自第三方程），净功为 W = F s = m × (v² − u²)/(2s) × s = ½ m v² − ½ m u²。
Thus:
因此：
Wnet = Δ KE = ½ m v² − ½ m u²
This principle is widely used to solve problems where forces do work and speed changes, linking mechanics and energy conservation.
该原理广泛用于求解力做功并导致速度变化的问题，将力学与能量守恒联系起来。
8. Deriving Hooke’s Law and Elastic Potential Energy | 推导胡克定律和弹性势能
For a spring or any elastic material within its limit of proportionality, the extension x is directly proportional to the applied force.
对于弹簧或任何处于比例极限内的弹性材料，伸长量 x 与施加的力成正比。
This is expressed as Hooke’s law: F = k x, where k is the spring constant.
这表示为胡克定律：F = k x，其中 k 是劲度系数。
To find the elastic potential energy stored, we note that the force is not constant but increases linearly with extension.
为求所储存的弹性势能，需注意力并非恒定，而是随伸长量线性增加。
The work done stretching the spring is the area under the force–extension graph, a triangle of height F and base x.
拉长弹簧所做的功是力–伸长量图像下的面积，即一个高为 F、底为 x 的三角形。
Hence W = average force × extension = ½ F × x = ½ (k x) × x = ½ k x².
因此 W = 平均力 × 伸长量 = ½ F × x = ½ (k x) × x = ½ k x²。
Eelastic = ½ k (Δx)²
This energy is stored as elastic potential energy and is recoverable when the spring returns to its original length.
该能量以弹性势能形式储存，弹簧恢复原长时可释放。
9. Deriving Power as Force × Velocity: P = Fv | 推导功率公式：P = Fv
Power is defined as the rate of doing work or transferring energy.
功率定义为做功或能量传递的速率。
If a constant force F moves an object through a small displacement Δs in time Δt, the work done is ΔW = F Δs.
若恒力 F 在时间 Δt 内使物体产生小位移 Δs，则做功为 ΔW = F Δs。
The instantaneous power is then P = ΔW/Δt = F (Δs/Δt) = F v, where v is the instantaneous velocity in the direction of the force.
瞬时功率即为 P = ΔW/Δt = F (Δs/Δt) = F v，其中 v 是沿力方向的瞬时速度。
P = F v
This relationship explains why a car engine must deliver more power at higher speeds to overcome resistive forces, even if the driving force remains constant.
这个关系解释了为什么即使驱动力恒定，汽车发动机也必须在高速时提供更大功率以克服阻力。
10. Deriving Young Modulus from Stress and Strain | 从应力和应变推导杨氏模量
Young modulus E quantifies the stiffness of a material and is defined as the ratio of tensile stress to tensile strain.
杨氏模量 E 量化材料的刚度，定义为拉伸应力与拉伸应变的比值。
Tensile stress σ is the force applied per unit cross-sectional area: σ = F / A.
拉伸应力 σ 是单位横截面积上施加的力：σ = F / A。
Tensile strain ε is the extension per unit original length: ε = ΔL / L₀.
拉伸应变 ε 是单位原长的伸长量：ε = ΔL / L₀。
Therefore Young modulus is written as:
因此杨氏模量写作：
E = σ / ε = (F / A) / (ΔL / L₀) = F L₀ / (A ΔL)
This expression allows the determination of the stiffness of a wire from simple measurements of force, length, and extension, a classic experimental derivation in the AS syllabus.
该表达式使得仅通过测量力、长度和伸长量即可测定金属丝的刚度，这是AS大纲中经典的实验推导。
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