📚 AS Physics Unit 2 Formula Derivations | AS物理第二单元公式推导
Welcome to this revision guide focused on key formula derivations for AS Physics Unit 2. Mastering these derivations not only helps you answer ‘show that’ and structured questions in the January 2020-style paper but also deepens your understanding of fundamental physics principles. We will walk through essential derivations from mechanics, materials, and waves, presenting each step clearly and linking the physics to the mathematics.
欢迎来到这份专注于AS物理第二单元关键公式推导的复习指南。掌握这些推导不仅有助于你在类似2020年1月试卷的‘证明题’和结构化问题中作答，还能加深你对基本物理原理的理解。我们将从力学、材料和波中精选核心推导，清晰展示每一步，并将物理与数学联系起来。
1. The SUVAT Equation s = ut + ½ at² | 匀加速运动方程 s = ut + ½ at² 推导
The derivation of s = ut + ½ at² begins with the definition of average velocity. For an object moving with uniform acceleration a, initial velocity u, and final velocity v, the displacement s is average velocity multiplied by time. Average velocity is (u + v)/2. We also know from the definition of acceleration that v = u + at. Substituting v into the average velocity expression gives (u + u + at)/2 = u + (at)/2. Multiplying by time t yields s = (u + (at)/2) × t = ut + ½ at². This equation relates displacement directly to initial velocity, acceleration, and time without needing final velocity.
s = ut + ½ at² 的推导始于平均速度的定义。对于以匀加速度 a 运动、初速度为 u、末速度为 v 的物体，位移 s 是平均速度乘以时间。平均速度为 (u + v)/2。由加速度定义可知 v = u + at。将 v 代入平均速度表达式得 (u + u + at)/2 = u + (at)/2。再乘以时间 t 得到 s = (u + (at)/2) × t = ut + ½ at²。该方程将位移直接与初速度、加速度和时间联系起来，无需末速度。
2. The SUVAT Equation v² = u² + 2as | 速度-位移公式 v² = u² + 2as 推导
To derive v² = u² + 2as, we start from two basic equations: v = u + at and s = ut + ½ at². First, rearrange v = u + at to express time as t = (v – u)/a. Substitute this expression for t into the displacement equation: s = u[(v – u)/a] + ½ a[(v – u)/a]². Simplifying gives s = (uv – u²)/a + (v² – 2uv + u²)/(2a). Taking common denominator 2a, we get s = (2uv – 2u² + v² – 2uv + u²)/(2a) = (v² – u²)/(2a). Multiplying both sides by 2a gives 2as = v² – u², and rearranging leads to v² = u² + 2as. This formula is particularly useful when time is not given or required.
为推导 v² = u² + 2as，我们从两个基本方程出发：v = u + at 和 s = ut + ½ at²。首先，将 v = u + at 变形为 t = (v – u)/a。将这个 t 的表达式代入位移方程：s = u[(v – u)/a] + ½ a[(v – u)/a]²。化简得 s = (uv – u²)/a + (v² – 2uv + u²)/(2a)。取公分母 2a，得到 s = (2uv – 2u² + v² – 2uv + u²)/(2a) = (v² – u²)/(2a)。两边同乘 2a 得 2as = v² – u²，移项后即为 v² = u² + 2as。当题目未给出或不需求时间时，该公式尤为有用。
3. Newton’s Second Law and Momentum: F = Δp/Δt | 牛顿第二定律与动量：F = Δp/Δt 推导
Newton’s second law is often expressed as F = ma, but its more fundamental form relates force to the rate of change of momentum. Momentum p is defined as mass m × velocity v, so Δp = m(v – u) if mass is constant. Acceleration a = (v – u)/Δt, thus m a = m (v – u)/Δt = Δp/Δt. Therefore, resultant force F = Δp/Δt. In situations where mass changes, such as a rocket ejecting fuel, we must use this general form. For constant mass problems, F = ma is perfectly valid and can be derived directly from F = Δp/Δt assuming m is constant.
牛顿第二定律常表示为 F = ma，但其更基本的形式将力与动量变化率联系起来。动量 p 定义为质量 m × 速度 v，因此如果质量恒定，Δp = m(v – u)。加速度 a = (v – u)/Δt，于是 m a = m (v – u)/Δt = Δp/Δt。因此，合力 F = Δp/Δt。在质量变化的情况下，如火箭喷射燃料，我们必须使用这一普遍形式。对于质量恒定的问题，F = ma 完全有效，并可直接从 F = Δp/Δt 中假设 m 恒定推导得出。
4. Principle of Conservation of Momentum | 动量守恒定律推导
Consider two objects A and B colliding in an isolated system (no external forces). During collision, according to Newton’s third law, the force exerted by A on B (F_AB) is equal and opposite to the force exerted by B on A (F_BA). From Newton’s second law in momentum form, F_AB = Δp_B/Δt and F_BA = Δp_A/Δt. Since F_AB = -F_BA, we have Δp_B/Δt = -Δp_A/Δt, which implies Δp_A + Δp_B = 0. Therefore, the total change in momentum is zero, meaning total momentum before collision equals total momentum after collision: m_A u_A + m_B u_B = m_A v_A + m_B v_B. This conservation law is a cornerstone for solving collision and explosion problems.
考虑两个物体 A 和 B 在孤立系统（无外力）中碰撞。碰撞过程中，根据牛顿第三定律，A 对 B 的力 F_AB 与 B 对 A 的力 F_BA 大小相等、方向相反。由动量形式的牛顿第二定律，F_AB = Δp_B/Δt，F_BA = Δp_A/Δt。因为 F_AB = -F_BA，所以 Δp_B/Δt = -Δp_A/Δt，即 Δp_A + Δp_B = 0。因此动量总变化量为零，意味着碰撞前总动量等于碰撞后总动量：m_A u_A + m_B u_B = m_A v_A + m_B v_B。该守恒定律是解决碰撞和爆炸问题的基石。
5. Kinetic Energy Formula: Ek = ½ m v² | 动能公式 Ek = ½ m v² 推导
Kinetic energy is the energy possessed by a body due to its motion. To derive Ek = ½ m v², consider work done by a constant resultant force F acting on an object initially at rest (u=0) over a displacement s. Work done W = F s. Using F = m a and v² = u² + 2as with u=0 gives v² = 2as, so a = v²/(2s). Substituting into the work expression: W = m × (v²/(2s)) × s = ½ m v². This work done on the object transfers energy to it in the form of kinetic energy. Thus kinetic energy Ek = ½ m v². For an object starting with initial velocity u, the change in kinetic energy is ΔEk = ½ m v² – ½ m u².
动能是物体因运动而具有的能量。为推导 Ek = ½ m v²，考虑一个恒定的合力 F 作用在原本静止的物体（u=0）上，使其发生位移 s。做功 W = F s。利用 F = m a 以及 v² = u² + 2as（其中 u=0）得 v² = 2as，故 a = v²/(2s)。代入功的表达式：W = m × (v²/(2s)) × s = ½ m v²。这个对物体做的功将能量以动能形式传递给物体。因此动能 Ek = ½ m v²。对于有初速度 u 的物体，动能变化量为 ΔEk = ½ m v² – ½ m u²。
6. Gravitational Potential Energy: Ep = mgh | 重力势能 Ep = mgh 推导
Gravitational potential energy near the Earth’s surface is derived from work done against gravity. To lift an object of mass m through a height h at constant velocity, the upward force must exactly balance the weight mg. The work done by this lifting force is W = force × distance = mg × h. This work is stored as gravitational potential energy, giving Ep = mgh. The derivation assumes the gravitational field is uniform and g is constant. In examination questions, students may be asked to show that the change in potential energy equals mgh, with proper attention to the direction of force and displacement.
地表附近的重力势能通过克服重力做功推导。要以恒定速度将质量为 m 的物体提升高度 h，向上的力必须恰好等于重力 mg。提升力所做的功为 W = 力 × 距离 = mg × h。这个功存储为重力势能，即 Ep = mgh。该推导假设重力场均匀且 g 恒定。在考试问题中，学生可能被要求证明势能变化等于 mgh，需注意力和位移的方向。
7. Hooke’s Law and Elastic Potential Energy: F = kx and E = ½ k x² | 胡克定律和弹性势能：F = kx 和 E = ½ k x² 推导
Hooke’s law states that the extension x of a spring is directly proportional to the applied force F, as long as the elastic limit is not exceeded, giving F = kx where k is the spring constant. To find the elastic potential energy stored, consider the work done in stretching the spring. Since the force varies linearly from 0 to F = kx, the average force is ½ kx. Work done = average force × extension = (½ kx) × x = ½ k x². More rigorously, using integration: work = ∫₀ˣ F dx = ∫₀ˣ kx dx = ½ k x². This energy is recoverable as the spring returns to its original length.
胡克定律指出，只要不超过弹性限度，弹簧的伸长量 x 与所施加的力 F 成正比，即 F = kx，其中 k 为弹性系数。为求储存的弹性势能，考虑拉伸弹簧所做的功。由于力从 0 线性增加至 F = kx，平均力为 ½ kx。做的功 = 平均力 × 伸长量 = (½ kx) × x = ½ k x²。更严格地，使用积分：功 = ∫₀ˣ F dx = ∫₀ˣ kx dx = ½ k x²。当弹簧恢复原长时，这部分能量可以被释放。
8. Young’s Modulus from Stress and Strain | 通过应力和应变推导杨氏模量
Young’s modulus E quantifies the stiffness of a material. It is defined as tensile stress divided by tensile strain: E = (F/A) / (ΔL/L) where F is force applied, A is cross-sectional area, ΔL is extension, and L is original length. To derive the formula used in a typical experiment, we combine this with Hooke’s law for a wire: F = (EA/L) × ΔL. The gradient of a force-extension graph is k = EA/L. From this, E = kL/A. Students may need to derive expressions for gradient and uncertainties in E. For a wire under test, measurements of diameter (hence A), length L, and accurate extension allow calculation of the Young modulus.
杨氏模量 E 量化了材料的刚性。它定义为拉伸应力除以拉伸应变：E = (F/A) / (ΔL/L)，其中 F 是施加的力，A 是横截面积，ΔL 是伸长量，L 是原始长度。为推导典型实验所用公式，将其与金属丝的胡克定律结合：F = (EA/L) × ΔL。力-伸长量图的斜率为 k = EA/L。由此，E = kL/A。学生可能需要推导斜率表达式和 E 的不确定度。对于测试中的金属丝，通过测量直径（得到 A）、长度 L 和精确伸长量即可计算出杨氏模量。
9. Wave Speed Equation: v = fλ | 波速公式 v = fλ 推导
The wave equation v = fλ relates wave speed v, frequency f, and wavelength λ. The derivation is based on the definition of these quantities. Frequency f is the number of complete oscillations per second, so the time for one complete wave cycle (period T) is 1/f. In one period, a wave crest travels exactly one wavelength λ. Therefore, speed v = distance / time = λ / T. Substituting T = 1/f gives v = λ × f. This relationship holds for all types of waves, including electromagnetic, sound, and water waves, provided the medium does not cause dispersion that alters speed with frequency.
波动方程 v = fλ 联系了波速 v、频率 f 和波长 λ。推导基于这些量的定义。频率 f 是每秒完整振动的次数，因此一次完整波动周期所需时间（周期 T）为 1/f。在一个周期内，一个波峰恰好传播一个波长 λ 的距离。因此，波速 v = 距离 / 时间 = λ / T。代入 T = 1/f 得到 v = λ × f。该关系适用于所有类型的波，包括电磁波、声波和水波，前提是介质不引起使波速随频率变化的色散。
10. Double-Slit Fringe Spacing: Δx = λD/s | 双缝干涉条纹间距公式 Δx = λD/s 推导
Young’s double-slit experiment produces an interference pattern of bright and dark fringes. The spacing between adjacent bright fringes Δx (fringe width) can be derived using path difference and small-angle approximation. For two slits separated by distance s, and a screen at distance D (much larger than s), the path difference between the two waves arriving at a point on the screen at an angle θ is s sin θ. For the first bright fringe (constructive interference), path difference = λ. Using the small-angle approximation sin θ ≈ tan θ ≈ x/D, where x is distance from central maximum, we have s × (x/D) = λ, so x = λD/s. The fringe spacing Δx between the central and first bright fringe is also λD/s, and this is the same for all adjacent fringes. This formula allows measurement of wavelength of light.
杨氏双缝实验产生明暗相间的干涉条纹。相邻亮纹的间距 Δx（条纹宽度）可利用光程差和小角度近似推导。两缝间距为 s，屏到缝的距离为 D（远大于 s），到达屏上某点的两列波的光程差在与中心方向夹角为 θ 时为 s sin θ。对于第一级亮纹（相长干涉），光程差 = λ。利用小角近似 sin θ ≈ tan θ ≈ x/D，其中 x 是偏离中央极大的距离，得到 s × (x/D) = λ，即 x = λD/s。中央亮纹与第一级亮纹之间的间距 Δx 也为 λD/s，且所有相邻条纹间距均相同。该公式可用于测量光的波长。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导