AS AQA Further Mathematics 9665 Mechanics | AS AQA 进阶数学 9665 力学精讲

📚 AS AQA Further Mathematics 9665 Mechanics | AS AQA 进阶数学 9665 力学精讲

This topic test guide is designed for candidates sitting the OxfordAQA International AS Level Further Mathematics (9665) Mechanics paper. Mechanics in Further Mathematics extends the core A-level ideas of kinematics, dynamics, energy, and momentum into more rigorous, multi-stage problems. Success depends not only on knowing formulas, but on recognising which physical principle applies, drawing clear diagrams, and maintaining consistent sign conventions.

本篇专题测试指南专为参加 OxfordAQA 国际 AS 进阶数学（9665）力学试卷的考生编写。进阶数学中的力学部分将 A-level 核心课程中的运动学、动力学、能量与动量概念，延伸为更严谨、多步骤的综合问题。取得高分的关键不仅在于熟记公式，更在于识别应运用哪条物理原理、绘制清晰的受力图，并始终保持统一的符号约定。

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1. Kinematics with Variable Acceleration | 变加速度运动学

In AS Further Mechanics, displacement \(x\), velocity \(v\), and acceleration \(a\) are often given as functions of time \(t\). You must be fluent in differentiating and integrating with respect to time. The key relationships are: \(v = \frac{dx}{dt}\), \(a = \frac{dv}{dt} = \frac{d^2x}{dt^2}\), and \(x = \int v \, dt\), \(v = \int a \, dt\).

在 AS 进阶力学中，位移 x、速度 v 和加速度 a 通常表示为时间 t 的函数。你必须熟练地对时间求导和积分。核心关系为：v = dx/dt，a = dv/dt = d²x/dt²；反之，x = ∫v dt，v = ∫a dt。

When integrating, always include the constant of integration and use the initial conditions to find it. A common error is forgetting that, for example, the area under a velocity-time graph gives displacement, while the area under a speed-time graph gives distance. These are different when the velocity changes sign.

积分时务必加上积分常数，并利用初始条件确定其值。一个常见错误是忽略：速度-时间图像下的面积表示位移，而速率-时间图像下的面积表示路程。当速度变号时，两者并不相同。

A particle moves such that \(a = 6t – 4\) m s⁻². Given that when \(t=1\), \(v=2\) m s⁻¹ and \(x=0\), find the displacement at \(t=2\). Integrating: \(v = 3t² – 4t + C\); using \(v(1)=2\) gives \(C=3\), so \(v = 3t² – 4t + 3\). Integrating again: \(x = t³ – 2t² + 3t + D\); using \(x(1)=0\) gives \(D=-2\). Thus \(x(2) = 8 – 8 + 6 – 2 = 4\) m.

质点以 a = 6t − 4 m s⁻² 运动。已知 t=1 时，v=2 m s⁻¹ 且 x=0，求 t=2 时的位移。先积分得 v = 3t² − 4t + C；由 v(1)=2 得 C=3，故 v = 3t² − 4t + 3。再积分得 x = t³ − 2t² + 3t + D；由 x(1)=0 得 D=−2。因此 x(2) = 8 − 8 + 6 − 2 = 4 m。

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2. Newton’s Second Law with Variable Forces | 变力作用下的牛顿第二定律

Newton’s second law is \(F = ma\). In Further Mechanics, the resultant force \(F\) may itself be a function of \(t\), \(x\), or \(v\). When \(F\) is a function of \(t\), integrate directly: \(m \frac{dv}{dt} = F(t)\). When \(F\) is a function of \(x\), use the identity \(a = v \frac{dv}{dx}\). When \(F\) is a function of \(v\), separate variables or use \(F = mv \frac{dv}{dx}\).

牛顿第二定律为 F = ma。在进阶力学中，合外力 F 本身可能是 t、x 或 v 的函数。当 F 是 t 的函数时，直接积分：m dv/dt = F(t)。当 F 是 x 的函数时，利用恒等式 a = v dv/dx。当 F 是 v 的函数时，分离变量或使用 F = mv dv/dx。

Consider a particle of mass 2 kg moving under a force \(F = 4t + 6\) N. At \(t=0\), the particle is at rest. Find the velocity at \(t=3\). Using \(2 \frac{dv}{dt} = 4t + 6\), so \(\frac{dv}{dt} = 2t + 3\). Integrating: \(v = t² + 3t + C\); \(v(0)=0\) gives \(C=0\). Thus \(v(3) = 9 + 9 = 18\) m s⁻¹.

质量为 2 kg 的质点在力 F = 4t + 6 N 作用下运动。t=0 时质点为静止。求 t=3 时的速度。由 2 dv/dt = 4t + 6，得 dv/dt = 2t + 3。积分得 v = t² + 3t + C；v(0)=0 得 C=0。因此 v(3) = 9 + 9 = 18 m s⁻¹。

When the force depends on displacement, integrate \(m v \frac{dv}{dx} = F(x)\). This often leads to a velocity-squared expression. Always check the direction of motion before assigning signs.

当力与位移有关时，对 m v dv/dx = F(x) 积分。这通常会得到含 v² 的表达式。在确定正负号之前，务必先判断运动方向。

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3. Momentum and Impulse in Two Dimensions | 二维动量与冲量

Momentum is a vector quantity: \(\mathbf{p} = m\mathbf{v}\). Impulse is defined as the change in momentum: \(\mathbf{I} = m\mathbf{v} – m\mathbf{u}\), and is also equal to the integral of force with respect to time: \(\mathbf{I} = \int \mathbf{F} \, dt\). In two dimensions, you must treat the \(x\) and \(y\) components separately.

动量是矢量：p = m v。冲量定义为动量的变化量：I = m v − m u，同时也等于力对时间的积分：I = ∫F dt。在二维问题中，必须将 x 分量与 y 分量分别处理。

For a particle of mass 3 kg moving with velocity \((4\mathbf{i} – 2\mathbf{j})\) m s⁻¹, a force of \((6\mathbf{i} + 3\mathbf{j})\) N acts for 2 seconds. The impulse is \(\mathbf{I} = \mathbf{F} t = (12\mathbf{i} + 6\mathbf{j})\) N s. The new momentum is \(3(4\mathbf{i} – 2\mathbf{j}) + (12\mathbf{i} + 6\mathbf{j}) = (24\mathbf{i} + 0\mathbf{j})\), giving a final velocity of \((8\mathbf{i})\) m s⁻¹.

质量为 3 kg 的质点以速度 (4i − 2j) m s⁻¹ 运动，力 (6i + 3j) N 作用了 2 秒。冲量 I = F t = (12i + 6j) N s。新的动量为 3(4i − 2j) + (12i + 6j) = (24i + 0j)，故末速度为 (8i) m s⁻¹。

• Draw a diagram showing initial and final momentum vectors. 绘制初末动量矢量图。
• Resolve into perpendicular components. 分解为互相垂直的分量。
• Apply conservation of momentum separately in each direction for collisions. 碰撞时在每个方向上分别应用动量守恒。

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4. Work, Energy, and Power | 功、能与功率

Work done by a constant force is \(W = Fs\), where \(s\) is the distance moved in the direction of the force. For a variable force, \(W = \int F \, dx\). Kinetic energy is \(\frac{1}{2}mv²\). Gravitational potential energy is \(mgh\). The work-energy principle states that the net work done equals the change in kinetic energy.

恒力做功为 W = Fs，其中 s 为沿力方向移动的距离。变力做功为 W = ∫F dx。动能为 ½mv²。重力势能为 mgh。功能原理指出：合外力所做的总功等于动能的变化量。

Power is the rate of doing work: \(P = \frac{dW}{dt} = Fv\). For a vehicle moving at constant speed up a slope, the engine power must balance both the work against gravity and any resistive forces. If a car of mass 1200 kg moves up a slope inclined at \(5°\) at a constant speed of 20 m s⁻¹ with resistance 400 N, the tractive force satisfies \(F = 400 + 1200 \times 9.8 \times \sin 5°\).

功率是做功的速率：P = dW/dt = Fv。当车辆以恒定速度上坡时，发动机功率必须同时克服重力做功和阻力。若质量为 1200 kg 的汽车以 20 m s⁻¹ 的恒定速度驶上倾角 5° 的斜坡，阻力为 400 N，则牵引力满足 F = 400 + 1200 × 9.8 × sin 5°。

P = Fv

Remember that when using the work-energy principle for a system of connected particles, internal forces do equal and opposite work and cancel out. This simplifies many pulley problems.

请记住，在连接体系统中使用功能原理时，内力做等大反向的功而相互抵消。这大大简化了许多滑轮问题。

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5. Coefficient of Restitution | 恢复系数

For two spheres colliding directly along their line of centres, the coefficient of restitution \(e\) is defined as \(e = \frac{v_2 – v_1}{u_1 – u_2}\), where \(u_1, u_2\) are the initial velocities and \(v_1, v_2\) are the final velocities in the positive direction. The value of \(e\) satisfies \(0 \le e \le 1\). For perfectly elastic collisions, \(e=1\); for perfectly inelastic collisions, \(e=0\).

对于沿连心线发生正碰的两个球，恢复系数 e 定义为 e = (v₂ − v₁)/(u₁ − u₂)，其中 u₁、u₂ 为初速度，v₁、v₂ 为沿正方向的末速度。e 的取值范围为 0 ≤ e ≤ 1。完全弹性碰撞时 e = 1；完全非弹性碰撞时 e = 0。

For example, a sphere A of mass 2 kg moving at 5 m s⁻¹ collides directly with sphere B of mass 3 kg moving at 2 m s⁻¹ in the same direction. Given \(e = 0.5\), find the velocities after collision. Conservation of momentum: \(2(5) + 3(2) = 2v_A + 3v_B\), so \(2v_A + 3v_B = 16\). Restitution: \((v_B – v_A)/(5 – 2) = 0.5\), so \(v_B – v_A = 1.5\). Solving simultaneously gives \(v_A = 2.3\) m s⁻¹ and \(v_B = 3.8\) m s⁻¹.

例如：球 A 质量 2 kg，以 5 m s⁻¹ 运动；球 B 质量 3 kg，以 2 m s⁻¹ 同向运动。给定 e = 0.5，求碰撞后速度。动量守恒：2(5) + 3(2) = 2v_A + 3v_B，即 2v_A + 3v_B = 16。恢复系数：(v_B − v_A)/(5 − 2) = 0.5，即 v_B − v_A = 1.5。联立解得 v_A = 2.3 m s⁻¹，v_B = 3.8 m s⁻¹。

Relative speed of separation = e × Relative speed of approach

分离相对速度 = e × 接近相对速度

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6. Oblique Collisions with a Fixed Wall | 斜碰固定平面

When a sphere collides obliquely with a smooth fixed wall, the component of velocity parallel to the wall remains unchanged, while the perpendicular component reverses in direction and is reduced by the factor \(e\). This gives two equations:

当球与光滑固定平面发生斜碰时，平行于平面的速度分量保持不变，而垂直分量方向反转且大小变为原来的 e 倍。由此得到两个方程：

v sin β = u sin α

v cos β = e u cos α

where \(\alpha\) is the angle of approach to the normal, \(\beta\) is the angle of departure from the normal, \(u\) is the approach speed, and \(v\) is the departure speed. These equations are essential for solving trajectory-rebound problems.

其中 α 为入射方向与法线的夹角，β 为反弹方向与法线的夹角，u 为入射速率，v 为反弹速率。这两个方程是解决反弹轨迹问题的关键。

A useful derived result is \(\tan \beta = \frac{\tan \alpha}{e}\). If \(e < 1\), the rebound angle is steeper relative to the normal than the approach angle.

一个有用的派生结论是 tan β = tan α / e。若 e < 1，则反弹方向相对法线比入射方向更陡。

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7. Connected Particles on Inclined Planes | 斜面上的连接体

Problems with two particles connected by a light inextensible string over a smooth pulley require you to treat the system as a whole while also considering individual equations of motion. For a particle on a rough inclined plane connected to a hanging particle, the equations are:

由轻绳绕过光滑滑轮连接两个质点的题目，需要你既把系统视为整体，又分别对每个质点列运动方程。对于粗糙斜面上的质点与悬挂质点相连的情形，方程为：

For the hanging mass \(m_1\): \(m_1 g – T = m_1 a\).

对悬挂质量 m₁：m₁g − T = m₁a。

For the mass on the plane \(m_2\): \(T – m_2 g \sin \theta – \mu R = m_2 a\), where \(R = m_2 g \cos \theta\).

对斜面上的质量 m₂：T − m₂g sin θ − μR = m₂a，其中 R = m₂g cos θ。

Add the two equations to eliminate \(T\) and solve for \(a\). Once \(a\) is known, substitute back to find \(T\). Always determine the direction of friction by considering which way the system is about to move.

将两式相加消去 T，即可解出 a。求得 a 后再代回求绳中张力 T。务必先判断系统将向哪个方向运动，再确定摩擦力的方向。

a = [m₁g − m₂g sin θ − μm₂g cos θ] / (m₁ + m₂)

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8. Elastic Strings and Hooke’s Law | 弹性绳与胡克定律

Hooke’s law states that the tension in an elastic string is proportional to its extension: \(T = \frac{\lambda x}{l}\), where \(l\) is the natural length, \(x\) is the extension, and \(\lambda\) is the modulus of elasticity. The elastic potential energy stored in a stretched string is \(E = \frac{\lambda x²}{2l}\).

胡克定律指出：弹性绳中的张力与伸长量成正比，即 T = λx/l，其中 l 为自然长度，x 为伸长量，λ 为弹性模量。弹性绳拉伸时储存的弹性势能为 E = λx²/(2l)。

For an elastic string of natural length 1.5 m and modulus 60 N, a 3 kg particle is attached and the string is vertical. At equilibrium, the tension equals the weight: \(T = 3 \times 9.8 = 29.4\) N. Using \(T = \lambda x / l\), we get \(x = 29.4 \times 1.5 / 60 = 0.735\) m. The equilibrium length is therefore \(1.5 + 0.735 = 2.235\) m.

弹性绳自然长度 1.5 m，弹性模量 60 N，悬挂 3 kg 质点。平衡时张力等于重力：T = 3 × 9.8 = 29.4 N。由 T = λx/l 得 x = 29.4 × 1.5 / 60 = 0.735 m。因此平衡长度为 1.5 + 0.735 = 2.235 m。

For energy calculations, the total energy of a particle on an elastic string is the sum of kinetic energy, gravitational potential energy, and elastic potential energy. This total remains constant if no external work is done.

在能量计算中，弹性绳上质点的总能量等于动能、重力势能与弹性势能之和。若无外力做功，该总量保持不变。

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9. Simple Harmonic Motion | 简谐运动

Simple harmonic motion (SHM) occurs when acceleration is proportional to and directed towards a fixed point: \(a = -\omega² x\), where \(\omega\) is the angular frequency and \(x\) is the displacement from equilibrium. The general solution is \(x = A \cos(\omega t) + B \sin(\omega t)\), or equivalently \(x = R \cos(\omega t – \phi)\).

简谐运动（SHM）发生在加速度与位移成正比且方向指向固定中心点时：a = −ω²x，其中 ω 为角频率，x 为相对平衡位置的位移。通解为 x = A cos(ωt) + B sin(ωt)，等价于 x = R cos(ωt − φ)。

Key results include: maximum speed at equilibrium \(v_{\max} = \omega A\), and \(v² = \omega²(A² – x²)\). The time period of a particle on an elastic string in vertical SHM is \(T = 2\pi \sqrt{\frac{m}{\lambda/l}} = 2\pi \sqrt{\frac{m l}{\lambda}}\).

重要结论包括：平衡位置处速率最大，v_max = ωA；以及 v² = ω²(A² − x²)。竖直弹性绳上质点做 SHM 的周期为 T = 2π√(m l / λ)。

When a particle oscillates on a smooth horizontal table attached to one end of an elastic string, the period is independent of amplitude. This is a hallmark of SHM. Recognising SHM instantly saves time: look for \(a \propto -x\).

当质点在光滑水平桌面上连接弹性绳一端振动时，周期与振幅无关。这是 SHM 的标志性特征。快速识别 SHM 能节省时间：看到 a ∝ −x 即可判断。

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10. Motion in a Vertical Circle | 竖直圆周运动

For a particle moving in a vertical circle of radius \(r\), the radial acceleration is \(\frac{v²}{r}\) directed towards the centre. At any point where the string or rod makes an angle \(\theta\) with the downward vertical, the radial equation is \(T – mg \cos \theta = \frac{mv²}{r}\).

质点在竖直平面内沿半径为 r 的圆周运动时，径向加速度为 v²/r，方向指向圆心。在细绳或杆与竖直向下方向成角 θ 的位置，径向方程为 T − mg cos θ = mv²/r。

Use conservation of energy to relate the speed at different heights. For a particle released from rest at the horizontal position, at a general angle \(\theta\), the loss in gravitational potential energy is \(mgr \cos \theta\), so \(\frac{1}{2}mv² = mgr \cos \theta\).

利用能量守恒将不同高度处的速率联系起来。对于从水平位置静止释放的质点，在任意角 θ 处，重力势能的减少为 mgr cos θ，故 ½mv² = mgr cos θ。

For the particle to just complete a vertical circle on a string, the condition at the top is \(T \ge 0\), which means \(\frac{mv²}{r} \ge mg\). Combining with energy conservation gives the minimum speed at the bottom: \(v_{\text{bottom}} = \sqrt{5gr}\).

要使绳端质点恰好完成完整竖直圆周运动，最高点处需满足 T ≥ 0，即 mv²/r ≥ mg。结合能量守恒可得最低点的最小速率：v_最低 = √(5gr)。

v_min = √(5gr)

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11. Exam Strategy and Common Pitfalls | 应试策略与常见误区

Many marks are lost to sign errors and unresolved vectors. Always define a positive direction at the start of each question, and stick to it throughout. For problems involving friction, first check whether the system is moving or on the point of moving: use \(F \le \mu R\) for equilibrium and \(F = \mu R\) for limiting friction.

许多分数因符号错误和矢量分解不当而丢失。务必在每道题开头定义正方向，并全程保持一致。涉及摩擦力的题目，先判断系统是静止还是处于临界运动状态：平衡时用 F ≤ μR，临界摩擦力用 F = μR。

• Read the question twice. Identify whether the string is elastic or inelastic; whether the surface is smooth or rough. 题目读两遍。判断绳是弹性还是非弹性；表面是光滑还是粗糙。
• Sketch a labelled diagram. Include all forces, dimensions and velocities. 画带标注的示意图。标明所有力、尺寸和速度。
• Check units. Mixed units are a frequent source of errors; convert to SI before calculating. 检查单位。单位混用是常见错误来源；计算前先转换为国际单位制。
• Interpret the answer. If a velocity comes out negative, it simply means the direction is opposite to your chosen positive direction. 解读答案。若速度为负，仅表示方向与所选正方向相反。

For the topic test, allocate about 1.5 minutes per mark. If a question involves many parts, the earlier parts are usually designed to guide the later calculations. Use results from part (a) in part (b); do not restart from first principles unless necessary.

在专题测试中，建议每题约用时 1.5 分钟/分。若题目分多问，前几问通常为后续计算作铺垫。应使用第 (a) 问的结果来解决第 (b) 问；除非必要，不要从头重新计算。

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12. Practice Questions with Worked Solutions | 练习与详解

Question 1. A particle of mass 0.5 kg moves along a straight line under the action of a force \(F = 10 – 2t\) N. At \(t = 0\), the particle is at rest. Find the maximum velocity attained.

题目 1。质量为 0.5 kg 的质点沿直线运动，受力 F = 10 − 2t N。t = 0 时静止。求达到的最大速度。

Solution 1. Using \(m \frac{dv}{dt} = F\): \(0.5 \frac{dv}{dt} = 10 – 2t\), so \(\frac{dv}{dt} = 20 – 4t\). Integrating: \(v = 20t – 2t² + C\). At \(t=0\), \(v=0\), so \(C=0\). Maximum velocity occurs when acceleration is zero: \(20 – 4t = 0\), giving \(t = 5\) s. Therefore \(v_{\max} = 20(5) – 2(25) = 100 – 50 = 50\) m s⁻¹.

详解 1。由 m dv/dt = F 得 0.5 dv/dt = 10 − 2t，所以 dv/dt = 20 − 4t。积分得 v = 20t − 2t² + C。t = 0 时 v = 0，故 C = 0。最大速度出现在加速度为零时：20 − 4t = 0，得 t = 5 s。因此 v_max = 20(5) − 2(25) = 100 − 50 = 50 m s⁻¹。

Question 2. A sphere moving at 6 m s⁻¹ collides obliquely with a smooth wall. The direction of motion makes 30° with the wall. Given \(e = 0.6\), find the speed and direction after impact.

题目 2。小球以 6 m s⁻¹ 的速度与光滑墙面斜碰。运动方向与墙面成 30°角。已知 e = 0.6，求碰撞后的速度与方向。

Solution 2. The angle with the normal is \(90° – 30° = 60°\). Parallel component: \(v \sin \beta = 6 \sin 60° = 5.196\) m s⁻¹. Normal component: \(v \cos \beta = e (6 \cos 60°) = 0.6 \times 3 = 1.8\) m s⁻¹. Thus \(v = \sqrt{5.196² + 1.8²} = \sqrt{27 + 3.24} = \sqrt{30.24} \approx 5.50\) m s⁻¹. The angle to the wall satisfies \(\tan \theta_{\text{wall}} = 5.196 / 1.8 \approx 2.887\), giving about 70.9° with the wall after impact.

详解 2。与法线的夹角为 90° − 30° = 60°。平行分量：v sin β = 6 sin 60° = 5.196 m s⁻¹。法线分量：v cos β = e(6 cos 60°) = 0.6 × 3 = 1.8 m s⁻¹。因此 v = √(5.196² + 1.8²) = √(27 + 3.24) = √30.24 ≈ 5.50 m s⁻¹。与墙面的夹角满足 tan θ_墙 = 5.196 / 1.8 ≈ 2.887，故碰撞后与墙面夹角约为 70.9°。

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