📚 A-Level Physics Unit 1 Jan 2021: Core Concepts Breakdown | A-Level 物理单元1 2021年1月核心概念解析
The January 2021 A-Level Physics Unit 1 paper tests a wide range of foundational topics in mechanics and materials. Understanding these concepts deeply is essential not only for answering examination questions with confidence but also for building a solid groundwork for more advanced topics in Units 2, 4 and 5. This article breaks down the key concepts that appeared in that sitting, explaining the underlying physics with clear examples and providing the relevant equations using standard notation. Whether you are reviewing your performance or preparing for a resit, these concise explanations will help you consolidate your knowledge and avoid common pitfalls.
2021年1月的A-Level物理单元1试卷涵盖了力学与材料学中广泛的基础知识点。深刻理解这些概念，不仅能让你在考试中自信作答，更是为进阶单元2、4和5的学习打下坚实基础。本文对此次真题中出现的核心概念进行逐一解析，用清晰实例阐述底层物理原理，并用标准符号列出关键公式。无论你是在回顾作答情况还是准备补考，这些精炼的讲解都能帮助你巩固知识，避开常见失分点。
1. Vectors and Scalars | 矢量与标量
A scalar quantity has magnitude only, while a vector possesses both magnitude and direction. In Unit 1, common scalars include distance, speed, mass, energy and time; typical vectors are displacement, velocity, acceleration, force and momentum. The ability to resolve a vector into perpendicular components – usually horizontal and vertical – is fundamental for tackling equilibrium problems, inclined planes and projectile motion.
标量只有大小，矢量则同时具有大小和方向。在单元1中，常见的标量有路程、速率、质量、能量和时间；典型的矢量包括位移、速度、加速度、力和动量。将矢量分解为互相垂直的分量（通常为水平和竖直方向）是解决平衡问题、斜面运动和抛体运动的基础技能。
vectors, you must account for their directions. For perpendicular vectors, use Pythagoras' e/adjacent) to find the direction. For non-perpendicular vectors, resolve each into components, sum the components separately, and then recombine.
矢量相加时必须考虑方向。对于互相垂直的矢量，用勾股定理求合矢量的大小，用三角函数（tan θ = 对边/邻边）求方向。对于不垂直的矢量，先将每个矢量分解为分量，再分别对分量求和，最后合成。
2. Uniformly Accelerated Motion | 匀加速运动
The four suvat equations link displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t) for motion in a straight line with constant acceleration. They appear in nearly every mechanics problem in the paper.
四个匀加速运动方程（suvat）将位移(s)、初速度(u)、末速度(v)、加速度(a)和时间(t)联系起来，适用于加速度恒定的直线运动。这份试卷中几乎每一道力学题都会用到它们。
v = u + a t
s = u t + ½ a t²
s = ½ (u + v) t
v² = u² + 2 a s
is thrown upwards and you choose upward as positive, then acceleration due to gravity a = – , especially when calculating displacement and velocity at different points of the motion.
代入数值前一定要先规定正方向。如果以向上为正，那么重力加速度 a = –9.81 m s⁻²。很多学生因为符号混乱而丢分，尤其是在计算运动轨迹上不同点的位移和速度时。
3. Projectile Motion | 抛体运动
Projectile motion is treated as two independent perpendicular motions: constant horizontal velocity and uniformly accelerated vertical motion under gravity. The horizontal and vertical components are computed separately and time is the common link.
抛体运动可视为两个相互独立的垂直运动：水平方向匀速直线运动，竖直方向受重力作用做匀加速运动。水平与竖直分量分别计算，时间则是关联两者的桥梁。
ht is found from vertical motion using s = u t + ½ a t² with vertical displacement often zero for symmetric landings. Maximum height occurs when vᵧ = 0.
对于以初速度 v₀、仰角 θ 水平抛出的物体：水平分量 vₓ = v₀ cos θ；竖直分量 vᵧ = v₀ sin θ。飞行时间由竖直运动方程 s = u t + ½ a t² 求得，对称落点情况下竖直位移常为零。物体到达最高点时 vᵧ = 0。
4. Newton's Laws and Free-Body Diagrams | 牛顿定律与受力图
Newton' t force acts. The second law, F = m a, links resultant force, mass and acceleration. The third law highlights that forces come in pairs: if body A exerts a force on body B, then B exerts an equal and opposite force on A.
牛顿三定律是单元1力学的核心。第一定律：物体在没有合外力作用时，保持静止或匀速直线运动状态。第二定律 F = m a 将合外力、质量和加速度联系起来。第三定律指出力成对出现：若A对B施加了力，则B同时对A施加大小相等、方向相反的力。
all forces acting on a single body: weight, normal reaction, tension, friction, driving forces. Resolve forces along the direction of acceleration and perpendicular to it. The January 2021 paper featured problems with connected particles and tension in strings, where isolating each mass and writing the equations of motion is the key.
在应用 F = m a 之前，画受力分析图至关重要。标出作用在单个物体上的所有力：重力、支持力、张力、摩擦力、牵引力。沿加速度方向和垂直于加速度的方向分解力。2021年1月的试卷中含有连接体与绳中张力的问题，隔离每一个物体并列出运动方程是解题关键。
5. Momentum and Impulse | 动量与冲量
ple of conservation of momentum states that in a closed system with no external forces, the total momentum before a collision or explosion equals the total momentum after. This is routinely tested in questions involving snooker balls, trolleys, or ballistic pendulums.
动量 p = m v 是矢量。动量守恒定律指出，在无外力的封闭系统中，碰撞或爆炸前的总动量等于碰撞后的总动量。这一考点常出现在涉及台球、小车或冲击摆的问题中。
Impulse is the change in momentum, given by F Δt = Δp. The area under a force– m a graph and relate it to velocity change. Remember that impulse and momentum are vectors – direction matters.
冲量等于动量的变化量，公式为 F Δt = Δp。力-时间图像下方的面积代表冲量。在1月的试卷中，学生需从图像计算冲量并关联速度变化。牢记冲量和动量都是矢量——方向不可忽略。
6. Work, Energy and Power | 功、能量与功率
d as gravitational potential energy (Eₚ = m g h) or kinetic energy (Eₖ = ½ m v²). The work–energy principle is an alternative to Newton's laws: the net work done on an object equals its change in kinetic energy.
恒力做功 W = F s cos θ，其中 s 为位移，θ 为力与位移间的夹角。能量可以储存为重力势能 (Eₚ = m g h) 或动能 (Eₖ = ½ m v²)。功能原理是牛顿定律的另一种表达：物体所受合外力做的功等于动能的变化量。
e ratio of useful output power to total input power. In the exam, energy methods often simplify problems with varying forces or curved paths, so consider them as a first approach.
功率是做功的快慢，P = W / t。若物体在恒力作用下做匀速直线运动，功率也可写作 P = F v。效率等于有用输出功率与总输入功率之比。考试中，对于变力或曲线路径问题，能量方法常能简化计算，不妨优先考虑。
7. Stress, Strain and the Young Modulus | 应力、应变与杨氏模量
rise how a material deforms under load. The Young modulus E is the ratio of stress to strain within the linear elastic region: E = σ / ε.
应力 σ = F / A（单位横截面积上的力），应变 ε = ΔL / L₀（伸长量与原长之比）。这两个量描述了材料在载荷下的变形响应。杨氏模量 E 是线弹性区域内应力与应变的比值：E = σ / ε。
A steep stress–strain graph indicates a stiff material (high E). The gradient of the initial straight line gives the Young modulus. In the January 2021 question, students were asked to determine E from a force–extension graph, requiring conversion to stress and strain. Always check units: Pa or N m⁻².
应力-应变图初始直线陡峭说明材料刚性大（E值高）。初始直线的斜率即为杨氏模量。2021年1月试题要求学生从力-伸长图求出E值，需要先转换为应力和应变。务必注意单位：帕斯卡 (Pa) 或 N m⁻²。
8. Elastic and Plastic Deformation | 弹性与塑性变形
Elastic deformation is reversible: the material returns to its original length when the load is removed. Plastic deformation is permanent. The limit of proportionality marks the end of the linear relationship between stress and strain; beyond it, Hooke's law no longer applies. The elastic limit is slightly beyond this point – up to it, the material still behaves elastically, even if the graph curves.
弹性变形是可逆的：撤去载荷后材料恢复原长。塑性变形是永久变形。比例极限是应力-应变线性关系的终点，越过此点胡克定律不再适用。弹性极限略高于比例极限——在此点之下即使曲线弯曲，材料仍表现为弹性。
Students often confuse the terms 'limit of proportionality', 'elastic limit' and 'yield point'. The yield point is where the material suddenly extends with little or no additional load. The January paper included a graph with a distinct yield point and asked for the identification of different regions, so be precise with your terminology.
学生经常混淆“比例极限”“弹性极限”和“屈服点”。屈服点是指材料在几乎不增加载荷的情况下突然伸长的位置。1月试卷中出现了一张带有明显屈服点的图像，要求标出不同区域，因此术语表述必须准确。
9. Fluid Statics: Pressure and Upthrust | 流体静力学：压强与浮力
, where ρ is fluid density. This arises because of the weight of the fluid above. Upthrust (buoyancy) on an object immersed in a fluid equals the weight of fluid displaced – Archimedes' principle.
流体中深度 h 处的压强 p = ρ g h，ρ 为流体密度。这是由上方流体的重量引起的。浸在流体中的物体所受的浮力等于它排开流体的重量——这就是阿基米德原理。
ght. For a submerged object, the net upward force is upthrust minus weight. The January 2021 paper featured a question where a cylinder was partially submerged, requiring students to equate upthrust and weight to find the density of the cylinder. Drawing a free-body diagram with weight and upthrust clarifies the equilibrium condition.
对于漂浮物体，浮力等于重力。对于完全浸没的物体，净向上力为浮力减重力。2021年1月试卷中有一道圆柱体部分浸没的题目，学生需令浮力与重力平衡来求圆柱密度。画出重力与浮力的受力示意图可清晰展示平衡条件。
10. Viscous Drag and Terminal Velocity | 粘性阻力与终端速度
it experiences a drag force that opposes motion. For a sphere at low speeds (laminar flow), Stokes' is viscosity, r radius and v speed. As speed increases, drag increases until it balances the driving force, resulting in terminal velocity.
物体在流体中运动时会受到与运动方向相反的阻力。对于低速运动的小球（层流），斯托克斯定律给出阻力 F = 6π η r v，其中 η 为粘滞系数，r 为半径，v 为速度。随着速度增大，阻力增加直至与驱动力平衡，此时物体达到终极速度。
For a sphere falling through a viscous liquid, terminal velocity is reached when weight – paper required students to apply this balance condition to find viscosity. Important: upthrust must be subtracted from weight because the fluid provides a buoyant force independent of velocity.
球体在粘滞液体中下落时，当重力减浮力等于粘性阻力时即达到终极速度。1月试卷要求运用该平衡条件求粘滞系数。注意：计算合力时一定要先减去浮力，因为浮力是流体施加的、与速度无关的向上力。
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