📚 AS Physics Unit 1 June 2022: Key Formula Derivations | AS物理单元1 2022年6月关键公式推导
This article revisits the essential formula derivations that appeared in or are fundamental to the AS Physics Unit 1 June 2022 exam paper. Mastery of these derivations not only secures marks in structured questions but also deepens your understanding of mechanics and materials. Each section pairs a step-by-step derivation with practical insights, directly reflecting the style of reasoning expected in the examination.
本文重温在2022年6月AS物理单元1试卷中出现或作为基础的关键公式推导。掌握这些推导不仅能在结构化问题中拿分，还能加深你对力学与材料的理解。每一节将逐步推导与实用见解配对，直接反映考试中期望的推理风格。
1. Derivation of the SUVAT Equations | SUVAT方程的推导
The SUVAT equations for uniform acceleration are derived from the definitions of velocity and acceleration. Consider an object moving with initial velocity u, constant acceleration a, and final velocity v over time t. The average velocity is (u+v)/2, so displacement s = average velocity × t gives s = ((u+v)/2)t. Combine with v = u + at to eliminate v, yielding s = ut + ½at².
匀加速运动的SUVAT方程源于速度和加速度的定义。考虑一个物体以初速度u、恒定加速度a运动，t时间后达到末速度v。平均速度为(u+v)/2，因此位移 s = 平均速度 × t，即 s = ((u+v)/2)t。结合 v = u + at 消去v，得到 s = ut + ½at²。
Squaring both sides of v = u + at and substituting from the displacement equation gives v² = u² + 2as. This equation is useful when time is unknown. The equation s = vt – ½at² is derived similarly by expressing u in terms of v. These five SUVAT equations are the core of kinematics in Unit 1.
将 v = u + at 两边平方，并代入位移方程可得出 v² = u² + 2as。该方程在时间未知时很有用。通过用v表示u，类似地可推导出 s = vt – ½at²。这五个SUVAT方程是单元1运动学的核心。
2. Deriving Acceleration from Free-Fall Data | 从自由落体数据推导加速度
In the June 2022 paper, an experiment using a free-fall apparatus was analyzed. A ball is dropped from rest, and timing gates measure the time to fall a known distance. Assuming g is constant, we can use s = ½gt² for u=0. Plotting s against t² gives a straight line through the origin whose gradient is ½g, allowing g to be determined without needing v.
在2022年6月试卷中，分析了使用自由落体仪器的实验。小球从静止下落，计时门测量下落已知距离所用的时间。假设g恒定且u=0，可使用 s = ½gt²。绘制s对t²的图会得到一条过原点的直线，其斜率为½g，从而无需末速度v即可求出g。
Another approach uses v² = 2gs, where v is measured via light gates. Plotting v² against s yields gradient 2g. The derivation also requires an understanding that the distance s is the height fallen, and that air resistance may cause a slight reduction in the measured acceleration. Students must justify why the line should pass through the origin and how systematic errors are minimised.
另一种方法使用 v² = 2gs，其中v通过光门测量。绘制v²对s的图，斜率为2g。这一推导还需要理解距离s是下落高度，以及空气阻力可能导致实测加速度略微减小。学生必须说明为什么图线应过原点以及如何减小系统误差。
3. Newton’s Second Law and Experimental Verification | 牛顿第二定律与实验验证
Newton’s second law states that the resultant force F on a body of constant mass m is directly proportional to its acceleration a, and F = ma. The derivation of this proportionality often appears in questions involving trolleys on a track. A trolley is pulled by a mass hanging over a pulley, causing a tension T in the string. Applying Newton’s second law to the trolley and the hanging mass gives T = Ma and mg – T = ma, which combine to a = mg/(M+m).
牛顿第二定律指出，作用在质量为m的物体上的合力F与其加速度a成正比，即F = ma。在涉及轨道小车的问题中常出现这种比例关系的推导。小车由悬挂在滑轮上的重物拉动，绳中张力为T。分别对小车和悬挂质量应用牛顿第二定律：T = Ma，mg – T = ma，联立得到 a = mg/(M+m)。
To verify F ∝ a, the total mass of the system is kept constant while the accelerating force is changed by transferring masses from the trolley to the hanger. The acceleration is measured, and a graph of force versus acceleration is plotted, which should be a straight line through the origin. The derivation shows that the gradient of this line is the total mass (M + m).
为了验证 F ∝ a，保持系统总质量不变，通过将质量从小车转移到挂钩上来改变加速力。测量加速度，绘制力与加速度的关系图，应为一条过原点的直线。推导表明该直线的斜率即为总质量 (M + m)。
4. Derivation of Kinetic Energy Formula | 动能公式的推导
The formula for kinetic energy, Eₖ = ½mv², is derived from the work-energy principle. A constant net force F acts on a body of mass m over a displacement s. The work done W = Fs. Using Newton’s second law F = ma and the SUVAT equation v² = u² + 2as, we can substitute a from F/m to get v² – u² = 2(F/m)s, so Fs = ½m(v² – u²). If the body starts from rest, u=0, so work done = ½mv², which is defined as the kinetic energy.
动能公式 Eₖ = ½mv² 源于功-能原理。一个恒定的合外力F作用在质量为m的物体上，位移为s。做功 W = Fs。利用牛顿第二定律 F = ma 和SUVAT方程 v² = u² + 2as，代入 a = F/m 可得 v² – u² = 2(F/m)s，因此 Fs = ½m(v² – u²)。若物体从静止开始，u=0，则做功为 ½mv²，这就是动能的定义。
In exam questions, students are often asked to derive the kinetic energy as the area under a force–velocity graph or to apply the work–energy theorem to find the speed of a falling object. Understanding this derivation helps in problems involving energy transfers, such as collisions or the conversion of gravitational potential energy to kinetic energy.
在考试问题中，常要求学生将动能推导为力–速度图下的面积，或应用功–能定理求下落物体的速率。理解这一推导有助于解决涉及能量转移的问题，例如碰撞或重力势能向动能的转换。
5. Conservation of Momentum from Newton’s Third Law | 牛顿第三定律动量守恒推导
The principle of conservation of momentum is derived from Newton’s third law and second law. Consider two objects A and B that exert forces on each other during a collision. By Newton’s third law, F_AB = –F_BA. The force acting on A is F_AB = Δp_A/Δt, and on B is F_BA = Δp_B/Δt. Equating and cancelling Δt gives Δp_A = –Δp_B, so total change in momentum is zero. Hence, in the absence of external forces, total momentum before = total momentum after.
动量守恒定律源自牛顿第三定律和第二定律。考虑两个物体A和B在碰撞中互相施加力。根据牛顿第三定律，F_AB = –F_BA。作用于A的力为 F_AB = Δp_A/Δt，作用于B的力为 F_BA = Δp_B/Δt。二者相等并消去Δt，得到 Δp_A = –Δp_B，因此总动量变化为零。所以在无外力时，作用前总动量等于作用后总动量。
This derivation is frequently examined when analysing collisions and explosions in Unit 1. In the June 2022 paper, a question required students to use the conservation of momentum to calculate the final velocity of a combined trolley system. The derivation also underlies impulse calculations, where impulse = change in momentum = FΔt.
在单元1分析碰撞和爆炸时，常考察这一推导。在2022年6月试卷中，有一道题要求运用动量守恒计算组合小车系统的最终速度。该推导也是冲量计算的基础，冲量 = 动量变化 = FΔt。
6. Elastic Potential Energy in Hooke’s Law | 胡克定律中的弹性势能
When a spring obeys Hooke’s law, the force F is directly proportional to the extension x: F = kx. The work done in stretching the spring is stored as elastic potential energy. Because the force is not constant but increases linearly from zero to F, the average force is F/2. Thus work done = average force × extension = (½F)x. Substituting F = kx gives elastic potential energy E = ½kx². This derivation is often shown as the area under the force–extension graph, which is a triangle of base x and height kx.
当弹簧遵守胡克定律时，力F与伸长量x成正比：F = kx。拉伸弹簧所做的功以弹性势能的形式储存。由于力不是恒定的，而是从零线性增加到F，平均力为F/2。因此做功 = 平均力 × 伸长量 = (½F)x。代入 F = kx 得到弹性势能 E = ½kx²。这一推导常表示为力–伸长图下的面积，即以x为底、kx为高的三角形面积。
In the context of material testing, the work done per unit volume is related to the area under the stress–strain curve. For a linear-elastic material, the elastic strain energy per unit volume = ½ × stress × strain, which leads directly to the derivation of the Young modulus from tensile tests.
在材料测试的语境下，单位体积的做功与应力–应变曲线下的面积有关。对于线弹性材料，单位体积的弹性应变能 = ½ × 应力 × 应变，这直接引导出从拉伸试验推导杨氏模量。
7. Stress, Strain, and Young Modulus in Materials | 材料中的应力、应变与杨氏模量
Stress (σ) is defined as the force applied per unit cross-sectional area: σ = F/A. Strain (ε) is the extension per unit original length: ε = ΔL/L₀. The Young modulus (E) is the ratio of stress to strain in the linear region: E = σ/ε = (F/A) / (ΔL/L₀) = FL₀/(AΔL). The derivation of this formula requires careful measurement of the original length, diameter (to find area), and the extension under known loads.
应力（σ）定义为单位横截面积上的力：σ = F/A。应变（ε）是单位原长的伸长量：ε = ΔL/L₀。杨氏模量（E）是线弹性区域内应力与应变之比：E = σ/ε = (F/A) / (ΔL/L₀) = FL₀/(AΔL)。该公式的推导需要精确测量原长、直径（以计算面积）以及在已知载荷下的伸长量。
In a typical AS practical, a wire is loaded with incremental weights and the extension is recorded. The gradient of the force–extension graph is k, the spring constant. The Young modulus can then be expressed as E = kL₀/A. This derivation connects the macroscopic behaviour of a spring to a material property independent of the wire’s dimensions, which was heavily tested in the June 2022 Unit 1 materials section.
在典型的AS实验中，对细丝逐级加载并记录伸长量。力–伸长图的斜率是弹簧常数k。杨氏模量可表示为 E = kL₀/A。这一推导将弹簧的宏观行为与独立于细丝尺寸的材料属性联系起来，在2022年6月单元1的材料部分有大量考查。
8. Resolving Vectors and Equilibrium Conditions | 向量分解与平衡条件
In many mechanics problems, forces must be resolved into perpendicular components. For a force F at an angle θ to the horizontal, the horizontal component is F cos θ and the vertical component is F sin θ. The derivation follows from the definitions of sine and cosine in a right-angled triangle. When an object is in equilibrium, the vector sum of forces is zero, meaning the sum of horizontal components = 0 and the sum of vertical components = 0.
在许多力学问题中，必须将力分解为垂直分量。对于与水平方向成θ角的力F，水平分量为F cos θ，垂直分量为F sin θ。这一推导遵循直角三角形中正弦和余弦的定义。当物体处于平衡状态时，力的矢量和为零，即水平分量之和=0，垂直分量之和=0。
For three forces in equilibrium, they can be represented as the sides of a triangle, leading to the Lami’s theorem derivation. This approach was essential to solve inclined plane problems on the June 2022 paper, where resolving weight into components parallel and perpendicular to the slope (mg sin θ and mg cos θ) allowed calculation of normal reaction and friction.
对于三个力平衡的情形，它们可以表示为三角形的三条边，从而引出拉密定理的推导。在2022年6月试卷中，这一方法对解决斜面问题至关重要，将重力分解为平行和垂直于斜面的分量（mg sin θ 和 mg cos θ），可计算法向反力和摩擦力。
Deriving expressions for acceleration down a frictionless incline yields a = g sin θ, which comes from the component of weight along the slope. This shows how formula derivation from basic principles is a recurring theme in the AS examination.
对于无摩擦斜面，推导出沿斜面加速度为 a = g sin θ，这源于重力沿斜面的分量。这表明从基本原理出发进行公式推导是AS考试中反复出现的主题。
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