📚 PH01 International AS Physics Paper (9 May 2023): Key Concepts Explained | 2023年5月PH01国际AS物理试卷核心概念解析
The PH01 International AS Physics paper from 9 May 2023 is a crucial assessment covering mechanics, materials and fluid dynamics. This article breaks down the key concepts that appeared in the paper, providing clear explanations and examples to support your revision. Use these notes to deepen your understanding and tackle similar problems with confidence.
2023年5月9日的PH01国际AS物理试卷是一场涵盖力学、材料和流体动力学的重要测评。本文深入解析试卷中出现的核心概念，提供清晰的解释与实例，助你巩固复习。利用这些笔记加深理解，自信地应对同类习题。
1. Scalar and Vector Quantities | 标量与矢量
Many exam questions require you to distinguish between scalar quantities (magnitude only) and vector quantities (magnitude and direction). In mechanics, common scalars are distance, speed, mass, energy and time; vectors include displacement, velocity, acceleration, force, momentum and weight.
许多考题要求你区分标量（仅有大小）和矢量（大小和方向）。力学中常见的标量有路程、速率、质量、能量和时间；矢量包括位移、速度、加速度、力、动量和重量。
When adding vectors acting along the same line, use a sign convention. For forces at right angles, Pythagoras and trigonometry give the resultant: R = √(F₁² + F₂²), with direction θ = tan⁻¹(F₂/F₁) from the horizontal. The examination may ask for resolution of a single force into perpendicular components, such as weight down a slope: component parallel to slope = mg sin θ, perpendicular = mg cos θ.
共线矢量相加时需规定正方向。互成直角的力则用勾股定理和三角比求合力：R = √(F₁² + F₂²)，方向与水平夹角 θ = tan⁻¹(F₂/F₁)。试卷可能要求分解单个力，例如斜面上重力：沿斜面分力 = mg sin θ，垂直于斜面分力 = mg cos θ。
An aeroplane flying in a crosswind experiences a resultant velocity found by vector addition; this same skill is tested in relative velocity and navigation problems.
一架飞机在有侧风的情况下飞行，其合速度由矢量加法求得；相对速度和导航类问题考查的正是同一技能。
2. Motion Graphs and the SUVAT Equations | 运动图像与SUVAT方程
Interpreting displacement–time (s–t), velocity–time (v–t) and acceleration–time (a–t) graphs is a fundamental skill. On a v–t graph the gradient gives acceleration, and the area between the graph and the time axis gives the change in displacement. Curved sections indicate non‑constant acceleration.
解读位移–时间（s–t）、速度–时间（v–t）和加速度–时间（a–t）图像是一项基本功。在v–t图上，斜率表示加速度，图线与时间轴围成的面积表示位移变化量。曲线段表示加速度不恒定。
When acceleration is constant, the four SUVAT equations apply. Always choose a positive direction and use s for displacement, u for initial velocity, v for final velocity, a for acceleration and t for time. The equations are:
加速度恒定时可使用四个SUVAT方程。务必规定正方向，用s表示位移、u表示初速度、v表示末速度、a表示加速度、t表示时间。方程如下：
v = u + at
v = u + at
s = ut + ½at²
s = ut + ½at²
v² = u² + 2as
v² = u² + 2as
s = ½(u + v)t
s = ½(u + v)t
A typical PH01 question provides a v–t graph and asks for total distance travelled and maximum acceleration. Remember to split the graph into straight‑line segments, calculate individual areas and gradients, and sum the areas for total distance.
一道典型的PH01题会给出v–t图，要求总路程和最大加速度。记住将图分成直线段，分别计算面积与斜率，并累加面积求总路程。
3. Forces, Free‑body Diagrams and Equilibrium | 力、受力图与平衡
Newton’s laws are at the heart of PH01 mechanics. Always draw a free‑body diagram showing all forces acting on the body: weight (mg), normal reaction, tension, friction, applied forces and upthrust where relevant. Use F = ma in the direction of the net force.
牛顿定律是PH01力学的核心。务必画出受力图，标出作用在物体上的所有力：重力（mg）、支持力、张力、摩擦力、外加力以及必要时考虑浮力。沿合力方向应用F = ma。
Equilibrium occurs when the resultant force is zero and the resultant moment about any pivot is zero. For a beam or a ladder problem, take moments about a point where an unknown force acts to simplify the equation. Use ∑F↑ = ∑F↓ and ∑ clockwise moments = ∑ anticlockwise moments.
当合力为零且对于任何支点的合力矩为零时物体处于平衡。对于横梁或梯子问题，选择未知力作用点作为支点求矩可简化方程。使用∑向上的力 = ∑向下的力以及∑顺时针力矩 = ∑逆时针力矩。
On a slope, the normal reaction is R = mg cos θ, and the limiting friction before sliding is F_max = μR, where μ is the coefficient of static friction. If the object is sliding, use the kinetic coefficient.
在斜面上，支持力为R = mg cos θ，滑动前的最大静摩擦力为F_max = μR，μ为静摩擦系数。若物体正在滑动，则用动摩擦系数。
4. Energy, Work and Power | 能量、功与功率
Work done by a constant force is W = F s cos θ, where θ is the angle between the force and the displacement. The area under a force–displacement graph gives the work done, even if the force varies.
恒力做功为W = F s cos θ，θ是力与位移的夹角。力–位移图下方面积即为所做的功，即使力是变化的也适用。
The principle of conservation of energy states that energy cannot be created or destroyed, only transferred. In mechanical systems, gravitational potential energy GPE = mgh and kinetic energy KE = ½mv² are frequently interconverted. Accounting for energy dissipated as heat enables you to write energy balance equations.
能量守恒定律指出能量不能被创造或消灭，只能转移。机械系统中，重力势能GPE = mgh和动能KE = ½mv²经常相互转化。计入以热量形式耗散的能量便可写出能量平衡方程。
Power is the rate of doing work: P = W / t or, for constant force and velocity, P = Fv. A car moving at constant maximum speed has engine power P = F_drag × v, useful for terminal‑velocity calculations.
功率是做功的速率：P = W / t，若力与速度恒定，则P = Fv。以恒定最大速度行驶的汽车，其发动机功率P = F_drag × v，有助于计算终极速度。
5. Momentum and Impulse | 动量与冲量
Linear momentum is p = mv and is a vector. The principle of conservation of momentum states that, in the absence of external forces, the total momentum of a system remains constant. Collisions and explosions are solved by equating total momentum before and after the event.
线动量为p = mv，是矢量。动量守恒定律指出，在无外力作用的系统中，总动量保持不变。碰撞与爆炸问题通过令事件前后总动量相等来求解。
Impulse J = FΔt = Δp. The area under a force–time graph equals the impulse. In questions involving rebounding balls, the change in momentum is Δp = m(v₁ − (−v₂)) = m(v₁ + v₂), and the average force is F_avg = Δp / Δt.
冲量J = FΔt = Δp。力–时间图下方的面积等于冲量。在涉及反弹球的题目中，动量变化量为Δp = m(v₁ − (−v₂)) = m(v₁ + v₂)，平均力为F_avg = Δp / Δt。
Elastic and inelastic collisions are distinguished by whether kinetic energy is conserved. In a perfectly elastic collision both momentum and kinetic energy are conserved; in an inelastic collision some kinetic energy is transformed into other forms.
弹性碰撞和非弹性碰撞的区别在于动能是否守恒。完全弹性碰撞中动量和动能均守恒；非弹性碰撞中部分动能转化为其他形式。
6. Stress, Strain and Young’s Modulus | 应力、应变与杨氏模量
Stress is defined as the force per unit cross‑sectional area: σ = F / A (unit Pa). Strain is the extension per unit original length: ε = ΔL / L₀ (dimensionless). Young’s modulus E = stress / strain = (F L₀) / (A ΔL), valid only within the linear elastic region where Hooke’s law applies.
应力定义为单位横截面积上的力：σ = F / A（单位Pa）。应变是单位原长的伸长量：ε = ΔL / L₀（无量纲）。杨氏模量E = 应力 / 应变 = (F L₀) / (A ΔL)，仅在符合胡克定律的线弹性区域内有效。
A typical question provides a force–extension graph for a wire and asks for Young’s modulus. First convert force to stress and extension to strain, or directly use E = (F L₀) / (A ΔL). Ensure units are in metres and pascals.
典型考题给出金属丝的力–伸长量图像，要求求杨氏模量。可先将力转换为应力，伸长量转换为应变，或直接用E = (F L₀) / (A ΔL)。务必注意单位用米和帕斯卡。
The gradient of the linear part of a stress–strain graph equals Young’s modulus. Identifying the elastic limit, yield point and ultimate tensile stress is essential for materials characterisation.
应力–应变图直线部分的斜率等于杨氏模量。识别弹性极限、屈服点和极限抗拉强度对于材料表征至关重要。
7. Elastic and Plastic Behaviour | 弹性与塑性行为
Elastic deformation is reversible: the material returns to its original shape when the load is removed. Plastic deformation is permanent. The area under a force–extension graph represents work done; the area between loading and unloading curves equals the energy dissipated as heat.
弹性形变是可逆的：撤去负荷后材料恢复原状。塑性形变是永久的。力–伸长图下方的面积代表所做的功；加载与卸载曲线之间的面积等于以热量形式耗散的能量。
The spring constant k for a wire of original length L₀, cross‑sectional area A and Young’s modulus E is k = EA / L₀. For two springs in series, 1/k_total = 1/k₁ + 1/k₂; in parallel, k_total = k₁ + k₂.
对于原长L₀、截面积A和杨氏模量E的金属丝，其劲度系数k = EA / L₀。两弹簧串联时1/k_total = 1/k₁ + 1/k₂；并联时k_total = k₁ + k₂。
Brittle materials break soon after the elastic limit without noticeable plastic deformation, whereas ductile materials show a large plastic region. You should be able to sketch and interpret stress–strain curves for both categories.
脆性材料在弹性极限后很快断裂，无明显塑性形变；而延性材料展现出大范围的塑性区域。应能绘出并解读这两类材料的应力–应变曲线。
8. Fluid Flow and Viscosity | 流体流动与黏性
Viscosity is the internal friction of a fluid that resists flow. The viscous drag force on a small sphere moving at low Reynolds numbers is given by Stokes’ law: F_drag = 6π η r v, where η is the dynamic viscosity, r the radius and v the terminal velocity.
黏性是流体阻碍流动的内摩擦力。在低雷诺数下，小球所受的黏性阻力由斯托克斯定律给出：F_drag = 6π η r v，η为动力黏度，r为半径，v为匀速速度。
When a sphere falls through a fluid, it reaches terminal velocity when weight = upthrust + viscous drag. This leads to (4/3)π r³ ρ_s g = (4/3)π r³ ρ_f g + 6π η r v, which can be rearranged to find η or v.
小球在流体中下落时，当重力 = 浮力 + 黏性阻力，便达到终极速度。由此得到(4/3)π r³ ρ_s g = (4/3)π r³ ρ_f g + 6π η r v，整理后可求η或v。
For streamlined objects at high speed, turbulent flow dominates and the drag is proportional to v². Laminar flow, in contrast, gives drag proportional to v. Students should recognise the difference and know that a low viscosity fluid provides less resistance.
对于高速流线型物体，湍流占主导地位，阻力与v²成正比。层流中的阻力则与v成正比。考生应能分辨这两种情况，并知道低黏度流体阻力较小。
9. Terminal Velocity and Falling Objects | 终极速度与落体
An object falling through a fluid initially accelerates under gravity. As the speed rises, the resisting force (drag) increases until the net force becomes zero. At this point the velocity is constant – the terminal velocity.
物体在流体中下落时，起初在重力作用下加速。随着速度增大，阻力也增大，直到合力为零。此时速度恒定，即终极速度。
Sketching the velocity–time graph for a falling sphere shows a curve that starts at zero gradient and tends asymptotically to the terminal velocity. Exam questions often ask to explain the shape using Newton’s second law: mg − F_drag − upthrust = ma.
绘制下落小球的速度–时间图像，曲线从零斜率开始，渐近趋向终极速度。考题常要求用牛顿第二定律解释图形形状：mg − F_drag − upthrust = ma。
Comparing terminal velocities in different fluids, or for spheres of different radii, tests understanding of the balance of forces. For a given sphere, a more viscous fluid yields a lower terminal speed, as drag builds more quickly.
比较不同流体中或不同半径小球的终极速度，考查的是对力平衡的理解。对于同一小球，较黏的流体因阻力更快增大而导致更低的终极速度。
10. Experimental Determination of Young’s Modulus | 杨氏模量的实验测定
The standard experiment uses a long wire (e.g. 2 m of copper or steel) clamped at one end, with a known load applied via a pulley. The extension is measured using a vernier scale or a micrometer screw gauge on a marker. Stress is calculated from F/A and strain from ΔL/L₀.
标准实验使用一根长钢丝（例如2 m铜丝或钢丝），一端固定，通过滑轮施加已知负荷。使用游标尺或千分尺测量标记处的伸长量。应力由F/A计算，应变由ΔL/L₀计算。
Plotting stress against strain gives a straight line through the origin for the elastic region; its gradient is Young’s modulus. Ensure you record the wire’s diameter in several places to find the mean cross‑sectional area, reducing uncertainty.
画出应力–应变图，弹性区域为过原点的直线；其斜率为杨氏模量。务必在多处测量金属丝直径，求得平均截面积，以减小不确定度。
Safety and procedural accuracy are common parts of the question: use a control wire to compensate for thermal expansion, add weights gently to avoid impulse, and measure the original length with a metre rule while the wire is taut but unloaded.
实验安全与操作精度是常见考点：使用补偿线消除热膨胀影响，轻轻增加砝码以避免冲量，在金属丝拉直但未加载时用米尺测量原长。
11. Energy Efficiency and Power Output | 能量效率与输出功率
Efficiency is the ratio of useful energy output to total energy input, often expressed as a percentage: Efficiency = (useful power out / total power in) × 100%. In mechanical systems, losses occur through friction, air resistance and heat.
效率是有用输出能量与总输入能量之比，常以百分比表示：效率 = (有用输出功率 / 总输入功率) × 100%。机械系统中，摩擦、空气阻力及发热引起能量损耗。
Questions linking an electric motor lifting a mass require you to equate electrical power P = IV to the rate of gain of gravitational potential energy mg Δh / t, accounting for efficiency. This type of question often appears in the PH01 paper as a synoptic exercise.
结合电动机提升质量的问题，需将电功率P = IV等同于重力势能增加率mg Δh / t，并考虑效率因子。此类题目常作为综合性练习出现在PH01试卷中。
When a battery drives a motor that winds a load, the output mechanical power is F v where v is the load’s speed. Comparing F v with IV gives the system’s overall efficiency.
当电池驱动电动机卷起重物时，输出机械功率为F v，v为重物速度。比较F v与IV即可得出系统总效率。
12. Momentum in Two Dimensions and Explosions | 二维动量与爆炸
If two particles collide and fly apart at an angle, conservation of momentum applies separately in two perpendicular directions. Resolve the momenta before and after into x‑ and y‑components, then form two equations. This skill is directly tested in vector‑based collision problems.
若两个粒子碰撞后以某一夹角飞散，动量守恒在两个垂直方向上分别成立。将碰撞前后的动量沿x和y方向分解，列出两个方程。此技能在矢量碰撞问题中直接考查。
For an explosion (e.g. a stationary firework splitting into two or three fragments), the total initial momentum is zero. The vector sum of the fragments’ momenta must be zero, allowing you to find an unknown mass or velocity.
对于爆炸（例如静止的烟花炸成两片或三片），初始总动量为零。碎片动量的矢量总和必为零，借此可求未知的质量或速度。
Constructing a closed triangle of momentum vectors is often the fastest solution. Remember that momentum is a vector and that directions are vital – a common error is to forget the sign when a fragment reverses direction.
构建动量矢量封闭三角形通常是最快的解法。牢记动量是矢量，方向至关重要——常见错误是碎片反向运动时忘记符号。
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