📚 AP Physics Mechanics: Exam Problem Analysis | AP 物理力学：真题解析
Mastering AP Physics Mechanics requires not only understanding fundamental principles but also practicing with real exam-style problems. In this guide, we analyze typical multiple-choice and free-response questions covering kinematics, Newton's laws, energy, momentum, rotation, and more. Each problem is broken down step by step, with clear explanations in both English and Chinese to help bilingual learners solidify their problem-solving skills and ace the AP exam.
掌握AP物理力学不仅需要理解基本原理，还需要通过真题进行练习。本文分析了涵盖运动学、牛顿定律、能量、动量、转动等内容的典型选择题和解答题。每个问题都进行逐步解析，并用中英双语清晰解释，帮助双语学习者巩固解题技巧，在AP考试中夺得高分。
1. Kinematics: Constant Acceleration Motion | 运动学：匀加速运动
A common AP problem involves a car starting from rest, accelerating uniformly for a certain time, then cruising at constant speed, and finally decelerating to a stop. For instance: A car accelerates from rest at 2 m/s² for 5 s, maintains constant velocity for 10 s, then decelerates at 2 m/s² until rest. Find the total displacement.
一个常见的AP考题涉及一辆车从静止开始，匀加速一段时间，然后匀速行驶，最后匀减速至停止。例如：汽车从静止以2 m/s²的加速度加速5秒，保持匀速10秒，再以2 m/s²的减速度减速至停止。求总位移。
ly from 0 to v = a×t = 2×5 = 10 m/s. The area of this triangle is ½ × base × height = ½ × 5 × 10 = 25 m. During constant velocity, the area is a rectangle: 10 m/s × 10 s = 100 m. During deceleration, the time to stop is Δv / a = 10 / 2 = 5 s, and the area under the v-t graph is another triangle: ½ × 5 × 10 = 25 m. Total displacement = 25 + 100 + 25 = 150 m.
解法：首先，绘制速度-时间图。v-t图下的面积即为位移。加速阶段，速度从0线性增加到 v = a×t = 2×5 = 10 m/s，三角形面积为 ½ × 底 × 高 = ½ × 5 × 10 = 25 m。匀速阶段，面积为矩形：10 m/s × 10 s = 100 m。减速阶段，停止所需时间 t = Δv / a = 10 / 2 = 5 s，v-t图下的三角形面积：½ × 5 × 10 = 25 m。总位移 = 25 + 100 + 25 = 150 m。
2. Newton's Laws: Inclined Plane with Friction | 牛顿定律：斜面与摩擦
A block of mass m slides down a rough incline of angle θ with coefficient of kinetic friction μ. Find its acceleration. This problem tests free-body diagrams and component resolution.
一个质量为m的物块沿倾角θ的粗糙斜面下滑，动摩擦系数为μ。求加速度。此题考查受力图和力的分解。
cos θ perpendicular. Normal force N = mg cos θ. Kinetic friction f = μN = μ mg cos θ, acting up the plane. Net force down the plane: Fnet = mg sin θ - μ mg cos θ. Using Newton's second law, a = Fnet / m = g(sin θ - μ cos θ). If sin θ > μ cos θ, the block accelerates down.
解法：画受力图。重力mg向下。分解为沿斜面和垂直斜面的分量：沿斜面向下为 mg sin θ，垂直斜面为 mg cos θ。支持力 N = mg cos θ。动摩擦力 f = μN = μ mg cos θ，方向沿斜面向上。沿斜面向下的合力：Fnet = mg sin θ - μ mg cos θ。由牛顿第二定律，a = Fnet / m = g(sin θ - μ cos θ)。若 sin θ > μ cos θ，滑块加速下滑。
3. Work, Energy, Power: Spring Compression | 功、能、功率：弹簧压缩
onless surface hits a spring with k = 200 N/m. Find the maximum compression of the spring.
一个常见的能量问题：质量为0.5 kg的物块以2 m/s的速度在光滑水平面上滑动，撞上一个劲度系数k = 200 N/m的弹簧。求弹簧的最大压缩量。
elastic potential energy stored in spring: U = ½ k x². Set ½ k x² = 1 J → ½ × 200 × x² = 1 → 100 x² = 1 → x = 0.1 m = 10 cm. So maximum compression is 0.1 m.
解法：利用机械能守恒。初始动能 KE = ½ m v² = ½ × 0.5 × (2)² = 1 J。在最大压缩时，所有动能转化为弹簧的弹性势能：U = ½ k x²。因此 ½ k x² = 1 J → ½ × 200 × x² = 1 → 100 x² = 1 → x = 0.1 m = 10 cm。最大压缩量为0.1 m。
4. Momentum and Impulse: Collision Analysis | 动量与冲量：碰撞分析
AP exams often ask: Cart A (mass m, velocity v) collides elastically with stationary Cart B (mass 2m). Find velocities after collision. For elastic collisions, both momentum and kinetic energy are conserved.
AP考试常考：小车A（质量m，速度v）与静止的小车B（质量2m）发生弹性碰撞。求碰后速度。弹性碰撞中，动量和动能均守恒。
Solution: Let vA' and vB' be final velocities. Momentum conservation: m v + 0 = m vA' + 2m vB' → v = vA' + 2 vB'. (1) Kinetic energy conservation: ½ m v² = ½ m vA'² + ½ (2m) vB'² → v² = vA'² + 2 vB'². (2) Solve simultaneously. From (1), vA' = v - 2 vB'. Substitute into (2): v² = (v - 2v_B')² + 2 vB'² = v² - 4 v vB' + 4 vB'² + 2 vB'² → 0 = -4 v vB' + 6 vB'² → vB'(6 vB' - 4 v) = 0. Non-zero solution: vB' = (2/3) v. Then vA' /3) v. So Cart A rebounds with 1/3 v in opposite direction, Cart B moves forward at 2/3 v.
解法：设碰后速度分别为 vA' 和 vB'。动量守恒：m v = m vA' + 2m vB' → v = vA' + 2 vB'。 (1) 动能守恒：½ m v² = ½ m vA'² + ½ (2m) vB'² → v² = vA'² + 2 vB'²。 (2) 联立求解。由(1)得 vA' = v - 2 vB'。代入(2)：v² = (v - 2v_B')² + 2 vB'² = v² - 4 v vB' + 4 vB'² + 2 vB'² → 0 = -4 v vB' + 6 vB'² → vB'(6 vB' - 4 v) = 0。非零解：vB' = (2/3) v。则 vA' = v - 2×(2/3 v) = - (1/3) v。因此小车A以1/3 v反向弹回，小车B以2/3 v向前运动。
5. Circular Motion & Gravitation: Satellite Orbit | 圆周运动与引力：卫星轨道
Calculate the orbital speed of a satellite orbiting Earth at altitude h. Given: Earth radius R, mass M, gravitational constant G. Neglect air resistance.
计算一颗在高度 h 处绕地球运行的卫星的轨道速度。已知地球半径R、质量M、引力常数G，忽略空气阻力。
= G M / (R+h). Thus orbital speed v = √(G M / (R+h)). For a low Earth orbit (h << R), v ≈ √G M / R. On Earth's surface, g = G M / R², so v ≈ √g R ≈ 7.9 km/s (first cosmic velocity).
解法：向心力由万有引力提供：G M m / (R+h)² = m v² / (R+h)。消去m，得 v² = G M / (R+h)。因此轨道速度 v = √(G M / (R+h))。对于近地轨道 (h << R)，v ≈ √G M / R。在地表，g = G M / R²，故 v ≈ √g R ≈ 7.9 km/s（第一宇宙速度）。
6. Rotational Motion: Rolling without Slipping | 转动：无滑滚动
an incline of angle θ. Find the acceleration of its center of mass. (Moment of inertia I = ⅖ mR²)
一个质量为m、半径为R的实心球沿倾角θ的斜面滚下且不发生滑动。求其质心加速度。（转动惯量 I = ⅖ mR²）
ling without slipping, a = α R. Thus α = a / R. Substitute: f R = I (a/R) → f = I a / R² = (⅖ mR²) a / R² = ⅖ m a. Plug into translation: m a = m g sin θ - ⅖ m a → (1 + ⅖ g sin θ. Note this is less than g sin θ (sliding without friction would give a = g sin θ).
解法：受力：重力分量 mg sin θ 沿斜面向下，静摩擦力 f 沿斜面向上（提供力矩）。平动：m a = mg sin θ - f。转动：力矩 τ = f R = I α。无滑滚动条件 a = α R，故 α = a / R。代入：f R = I (a/R) → f = I a / R² = (⅖ mR²) a / R² = ⅖ m a。代入平动方程：m a = m g sin θ - ⅖ m a → (1 + ⅖) m a = m g sin θ → (7/5) a = g sin θ → a = (5/7) g sin θ。注意这
Published by TutorHao | AP Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图