📚 A-Level AQA Physics: Dynamics Key Points Review | A-Level AQA 物理：动力学 考点精讲
Dynamics is the study of the forces that cause motion and the resulting changes in motion. For AQA A-Level Physics, a solid grasp of Newton’s laws, momentum, impulse, and collisions is essential. This revision guide walks you through every key topic, with clear explanations and exam-oriented insights.
动力学研究引起运动的力以及运动的变化。对于 AQA A-Level 物理，牢固掌握牛顿定律、动量、冲量和碰撞至关重要。这篇复习指南将带你梳理每一个核心考点，提供清晰的解释和应考心得。
1. Newton’s First Law of Motion | 牛顿第一定律
Newton’s first law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by a resultant external force. In other words, if the net force on a body is zero, its velocity does not change. This is often called the law of inertia.
牛顿第一定律指出，除非受到合外力的作用，物体将保持静止或匀速直线运动状态。换句话说，如果物体所受合外力为零，其速度就不会改变。这通常被称为惯性定律。
Inertia is the tendency of an object to resist changes in its motion. The mass of an object is a measure of its inertia.
惯性是物体抵抗运动状态改变的倾向。物体的质量是惯性的量度。
Consider a book on a table: gravity pulls down and the normal reaction pushes up. The two forces cancel, so the book stays at rest. If the table is pulled smoothly, the book may slide due to friction but without a net force it would keep moving.
想象桌面上的一本书：重力向下拉，法向反作用力向上推。两力抵消，书保持静止。如果桌面被平稳地拉动，书可能因摩擦力而滑动，但若没有净力，它将一直运动下去。
2. Newton’s Second Law and Momentum | 牛顿第二定律与动量
Newton’s second law, in its most general form, states that the net force on a body equals the rate of change of its momentum: F = dp/dt. For an object of constant mass m, this simplifies to F = m a, where a is the acceleration.
牛顿第二定律的最普遍形式是：物体所受的合外力等于其动量的变化率：F = dp/dt。对于质量 m 恒定的物体，这简化为 F = m a，其中 a 是加速度。
Momentum p is defined as the product of mass and velocity: p = m v. It is a vector quantity, so direction must be taken into account. Momentum has units of kg m s⁻¹ or N s.
动量 p 定义为质量与速度的乘积：p = m v。它是矢量，因此必须考虑方向。动量的单位是 kg·m·s⁻¹ 或 N·s。
F = d(mv)/dt = m dv/dt = m a (constant m)
When using F = m a, remember that F represents the resultant force in a particular direction. Always resolve forces first, then apply the equation to the appropriate direction.
使用 F = m a 时，切记 F 代表某一方向上的合外力。务必先分解力，再将方程应用于相应方向。
3. Newton’s Third Law | 牛顿第三定律
Newton’s third law tells us that if body A exerts a force on body B, body B exerts an equal and opposite force on body A. These two forces are of the same type, act on different bodies, and are always paired.
牛顿第三定律告诉我们，如果物体A对物体B施加一个力，那么物体B同时会对物体A施加一个大小相等、方向相反的力。这两个力属于同一类型，作用在不同的物体上，并且总是成对出现。
For instance, when you push against a wall, the wall pushes back on you with the same magnitude. The Earth pulls the Moon gravitationally; the Moon pulls the Earth back. In a rocket launch, the exhaust gases are pushed backward, and the gases push the rocket forward.
例如，当你推墙时，墙以同样大小的力反推你。地球以引力吸引月球，月球也以引力吸引地球。在火箭发射中，废气被向后推出，气体同时将火箭向前推。
An important exam point: action–reaction pairs do not cancel each other because they act on different objects. Only forces on the same object can cancel.
一个重要考点：作用力与反作用力不会相互抵消，因为它们作用在不同物体上。只有作用在同一物体上的力才能抵消。
4. Impulse and the Impulse-Momentum Theorem | 冲量与动量定理
Impulse J is the product of a force F and the time interval Δt during which it acts: J = F Δt. When the force varies, the impulse equals the area under a force–time graph.
冲量 J 是力 F 与其作用时间 Δt 的乘积：J = F Δt。当力变化时，冲量等于力–时间图像下的面积。
The impulse–momentum theorem states that the impulse delivered to an object equals its change in momentum: J = Δp = m v – m u. This is a direct consequence of Newton’s second law integrated over time.
动量定理指出，物体所受的冲量等于其动量的变化量：J = Δp = m v – m u。这是牛顿第二定律对时间积分的直接结果。
F Δt = m v – m u
This relationship is extremely useful for solving problems where a large force acts over a short time, such as hitting a ball, a crash test, or kicking a football. Always set a positive direction and treat velocities as vectors.
这一关系在解决大力短时作用的问题时极其有用，比如击球、碰撞测试或踢足球。务必设定正方向，并将速度作为矢量处理。
5. Conservation of Momentum | 动量守恒
The principle of conservation of momentum states that for a system with no external resultant force, the total momentum before an event equals the total momentum after the event. Momentum is conserved in all collisions and explosions, provided external influences are negligible.
动量守恒定律指出，对于一个无合外力的系统，事件发生前的总动量等于事件发生后的总动量。在一切碰撞和爆炸中，只要外部影响可忽略，动量总是守恒的。
For a two-body collision, we write: m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂, where u are initial velocities and v are final velocities. Again, direction is crucial; use positive and negative signs consistently.
对于两体碰撞，可写为：m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂，其中 u 为初速度，v 为末速度。同样，方向至关重要；要始终如一地使用正负号。
This conservation law is particularly powerful because it allows you to find one unknown velocity without knowledge of the detailed forces during the interaction. Always start by diagramming the system and labelling velocities.
动量守恒定律特别强大，因为它让你无需知道相互作用期间力的细节，即可求出一个未知速度。解题时务必从系统图和标注速度开始。
6. Elastic and Inelastic Collisions | 弹性碰撞与非弹性碰撞
Collisions are classified by what happens to kinetic energy. In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, momentum is conserved but total kinetic energy decreases.
碰撞根据动能的变化分类。在弹性碰撞中，动量和动能都守恒。在非弹性碰撞中，动量守恒但总动能减少。
A perfectly inelastic collision is one where the bodies stick together and move with a common final velocity. Here, kinetic energy is not conserved; some is transformed into heat, sound, or deformation.
完全非弹性碰撞是指物体碰撞后粘在一起，以共同速度运动。此时动能不守恒，一部分能量转化为热、声或形变能。
For AQA, you are expected to apply conservation of momentum to both types and to use energy principles to solve elastic collisions. In an elastic head-on collision between equal masses, the bodies simply exchange velocities.
AQA 要求你能够在两种碰撞类型中应用动量守恒，并用能量原理求解弹性碰撞问题。对于质量相等的正碰弹性碰撞，两物体不过是交换速度。
Elastic: ½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²
Often you will solve a problem by first writing the momentum conservation equation and then, if elastic, adding the kinetic energy equation. Check each case carefully to avoid mixing inelastic and elastic assumptions.
解题时通常先列出动量守恒方程，若为弹性碰撞则再加上动能方程。务必仔细审题，避免将非弹性假设与弹性假设相混淆。
7. Explosions and Recoil | 爆炸与反冲
An explosion can be thought of as an inelastic collision in reverse. Initially, the total momentum is zero. After the explosion, the fragments fly apart, and the vector sum of their momenta remains zero.
爆炸可以看作是一种反向的非弹性碰撞。起初总动量为零。爆炸后，碎片向四面八方飞去，它们的动量矢量和仍为零。
For a stationary rocket, when fuel is ejected downward at high speed, the rocket recoils upward. The forward momentum gained by the rocket equals the backward momentum of the ejected gases. Neglecting external forces, momentum is conserved.
对于静止的火箭，当燃料向下高速喷出时，火箭向上反冲。火箭获得的向前动量等于喷出气体的向后动量。忽略外力时，动量守恒。
0 = m₁ v₁ + m₂ v₂ → m₁ v₁ = – m₂ v₂
Recoil problems often involve guns and bullets. A bullet gains high velocity, while the heavy gun recoils much slower. The negative sign indicates opposite directions.
反冲问题常涉及枪和子弹。子弹获得高速度，而沉重的枪向后反冲速度很慢。负号表示方向相反。
In exam questions, identify that initial total momentum is zero. Use symbols carefully and be ready to give speed, velocity, or magnitude as required.
在考题中，要识别出初始总动量为零。谨慎使用符号，并根据要求给出速率、速度或大小。
8. Force–Time Graphs | 力-时间图像
A force–time graph plots the net force acting on an object against time. The area under the graph represents the impulse. Whether force is constant or varying, area = ∫ F dt = Δp.
力-时间图像描绘了作用在物体上的合外力随时间的变化。图像下的面积代表冲量。无论力是恒定的还是变化的，面积 = ∫ F dt = Δp。
For a constant force, the graph is a horizontal line and impulse = F Δt. For a triangle or trapezoid, use area formulas. In many impact situations, the force rises rapidly and falls, approximated by shapes.
对于恒力，图像是一条水平线，冲量 = F Δt。对于三角形或梯形，使用面积公式。在许多碰撞情形中，力快速地上升又下降，可近似为简单形状。
Typical AQA question: find the change in momentum from a given force–time graph, or calculate average force during an impact using impulse = area.
典型的 AQA 考题：根据给出的力-时间图像求动量变化，或利用冲量 = 面积计算碰撞过程中的平均作用力。
When the force is not constant, you can often be asked to estimate the area by counting squares, then link to Δp and final velocity.
当力不恒定时，常要求你通过数格子的方式估算面积，再联系 Δp 和末速度。
9. Friction and Inclined Planes | 摩擦力与斜面
Friction is a force that opposes relative motion between surfaces. The maximum static friction and kinetic friction are each proportional to the normal reaction R: f ≤ μ R (static) or f = μ R (kinetic), where μ is the coefficient of friction.
摩擦力是阻碍表面间相对运动的力。最大静摩擦力和动摩擦力都与法向反作用力 R 成正比：f ≤ μ R（静摩擦）或 f = μ R（动摩擦），μ 为摩擦系数。
On an inclined plane, the weight mg must be resolved into components parallel and perpendicular to the slope: mg sin θ down the slope, mg cos θ perpendicular. The normal reaction R = mg cos θ when on a smooth slope, but can change with other forces.
在斜面上，重力 mg 必须分解为平行和垂直于斜面的分量：沿斜面向下的 mg sin θ，垂直于斜面的 mg cos θ。对于光滑斜面，法向反作用力 R = mg cos θ，但若有其他力时会改变。
Σ F parallel = m a → mg sin θ – f = m a
If the object moves at constant velocity, acceleration a = 0, so friction exactly balances the parallel component of weight. When the object is on the point of sliding, use the maximum static friction.
若物体匀速运动，加速度 a = 0，则摩擦力恰好与重力平行分量平衡。物体即将滑动时，应用最大静摩擦力。
Always draw a clear free-body diagram, label all forces, and then write two perpendicular equations: one along the slope, one normal to it. This systematic approach prevents sign errors.
始终绘制清晰的受力图，标注所有力，然后列出两个垂直方向上的方程：一个沿斜面方向，一个垂直于斜面。这种系统方法可以避免符号错误。
10. Connected Particles (Tension and Pulleys) | 连接体问题（张力与滑轮）
When two or more bodies are connected by a light, inextensible string, they accelerate with the same magnitude. The tension in the string is constant throughout, provided the string is light and the pulley is smooth and massless.
当两个或多个物体由轻质不可伸长的绳子连接时，它们加速度大小相同。若绳子轻质，滑轮光滑且无质量，绳内张力处处相等。
In a typical problem, treat each particle separately, apply F = m a in the direction of motion, and solve the simultaneous equations. Tension pulls each body towards the string; weight acts downward.
在典型问题中，需分别对每个物体处理，在运动方向上应用 F = m a，然后解联立方程。张力将物体拉向绳子方向；重力竖直向下。
For a mass m₁ on a smooth table connected over a pulley to a hanging mass m₂: for m₁ horizontally: T = m₁ a; for m₂ vertically: m₂ g – T = m₂ a. Solve to find a and T.
对于光滑桌面上质量为 m₁、绕过滑轮连接有悬挂质量 m₂ 的系统：对 m₁ 水平方向：T = m₁ a；对 m₂ 竖直方向：m₂ g – T = m₂ a。解出 a 和 T。
a = (m₂ g) / (m₁ + m₂) T = (m₁ m₂ g) / (m₁ + m₂)
If the table has friction, introduce f = μ R opposing motion. Similarly, for inclined planes, resolve weight components. Always align your sign convention with the direction of motion.
如果桌面有摩擦，引入与运动方向相反的 f = μ R。同理，对于斜面，要分解重力分量。务必使符号约定与运动方向一致。
11. Resolving Forces and Free-Body Diagrams | 力的分解与受力分析图
A free-body diagram shows all the external forces acting on a single body, depicted as arrows originating from the object. It is the starting point for almost every dynamics problem.
受力分析图展示作用在单个物体上的所有外力，表示为从物体出发的箭头。它是几乎每道动力学问题的起点。
Resolving a force means breaking it into perpendicular components, usually horizontal and vertical, or parallel and perpendicular to a slope. Use trigonometric functions: FX = F cos θ, FY = F sin θ.
分解力是指将其拆分为垂直的分量，通常沿水平/竖直方向，或沿斜面/垂直于斜面。使用三角函数：FX = F cos θ，FY = F sin θ。
When the body is in equilibrium, the net force in any direction is zero. Use Σ Fₓ = 0 and Σ F_y = 0 to find unknown forces. For an accelerating body, Σ F = m a in the direction of motion.
当物体平衡时，任一方向上的合外力为零。利用 Σ Fₓ = 0 和 Σ F_y = 0 求未知力。对于加速物体，在运动方向上 Σ F = m a。
Common errors: forgetting to include weight, confusing mass and weight, incorrectly identifying the normal reaction, or double-counting forces. Practise with a variety of setups, including strings at angles and friction forces.
常见错误：遗漏重力，混淆质量与重量，错误判断法向反作用力，或重复计算力。要通过多种情境练习，包括绳子成角度以及有摩擦力的情况。
12. Exam Tips and Common Mistakes | 考试技巧与常见错误
AQA dynamics questions often combine momentum, energy, and force concepts in single multi-step problems. Always read carefully: is the collision elastic? Are we asked for speed or velocity? Is energy conserved?
AQA 动力学问题常将动量、能量和力的概念合并为多步求解题。务必仔细审题：是弹性碰撞吗？要求的是速率还是速度？能量守恒吗？
Never forget to define a positive direction at the start, especially when dealing with vectors. Momentum and velocity signs are a primary source of mark loss. Use arrows in diagrams to remind yourself.
永远不要忘记在开始时定义正方向，尤其在处理矢量时。动量和速度的符号是失分的主要原因。用图中的箭头提醒自己。
Units: mass in kg, acceleration in m s⁻², force in N, momentum in kg m s⁻¹. Convert grams to kilograms where needed. In impulse problems, time is in seconds.
单位：质量用 kg，加速度用 m·s⁻²，力用 N，动量用 kg·m·s⁻¹。必要时要将克转换为千克。冲量问题中时间单位是秒。
Avoid confusing ‘impulse’ with ‘momentum’. Impulse is F Δt; momentum is m v. The theorem links them: Δp = impulse. Make sure you can explain this link in words, as 6‑mark explanation questions are common.
避免混淆“冲量”和“动量”。冲量是 F Δt；动量是 m v。定理将它们联系起来：Δp = 冲量。确保能用语言解释这一联系，因为 6 分解释题很常见。
When solving connected-particle questions, state that tension is constant because the string is light and the pulley is smooth. This small justification can earn marks.
在解连接体问题时，要说明张力恒定，因为绳子轻质且滑轮光滑。这一简短的论证可以赢得分数。
Finally, practise with past papers. Dynamics topics appear recurrently, and familiarity with the style of command words like ‘state’, ‘explain’, ‘calculate’ and ‘determine’ will build confidence.
最后，多练历年真题。动力学内容反复出现，熟悉“state”、“explain”、“calculate”、“determine”等指令词的形式会树立信心。
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