Show all supporting work on scratch paper. Each item is labeled with its Mechanics topic and whether a calculator is permitted on that AP Exam section. Take $g = 9.8~\mathrm{m/s^2}$ unless a problem says otherwise.请在草稿纸上写出所有辅助计算过程。每道题均标注了对应的力学主题，以及该 AP 考试部分是否允许使用计算器。除题目另有说明外，取 $g = 9.8~\mathrm{m/s^2}$。
Q1EASY2.1 Newton's First Law2.1 牛顿第一定律No Calculator
A puck slides on a frictionless horizontal surface with constant velocity. Which of the following must be true?一个冰球在无摩擦水平面上以匀速滑行。下列说法中哪一项必然正确？
(A)A net force acts on the puck in the direction of motion.沿运动方向有合力作用于冰球。
(B)The puck experiences no forces at all.冰球不受任何力的作用。
(C)The vector sum of all forces on the puck is zero.冰球所受所有力的矢量和为零。
(D)The puck must be accelerating.冰球一定在加速。
Q2EASY2.2 Newton's Second Law2.2 牛顿第二定律No Calculator
A net force of $12~\mathrm{N}$ acts on a $3~\mathrm{kg}$ block. The block's acceleration is$12~\mathrm{N}$ 的合力作用于一个 $3~\mathrm{kg}$ 的物块。该物块的加速度为
(A) $0.25~\mathrm{m/s^2}$
(B) $4~\mathrm{m/s^2}$
(C) $9~\mathrm{m/s^2}$
(D) $36~\mathrm{m/s^2}$
Q3EASY2.3 Newton's Third Law2.3 牛顿第三定律No Calculator
A car pushes a heavier truck along a road, accelerating both. By Newton's third law, the truck pushes back on the car with a force that is一辆汽车在路上推动一辆更重的卡车，使两者均加速。根据牛顿第三定律，卡车对汽车的反作用力
(A)larger than the car's force on the truck (because the truck is heavier).大于汽车对卡车的力（因为卡车更重）。
(B)smaller than the car's force on the truck (because the car is moving the system).小于汽车对卡车的力（因为汽车在推动整个系统）。
(C)equal in magnitude and opposite in direction to the car's force on the truck.与汽车对卡车的力大小相等、方向相反。
(D)equal only when the system has zero acceleration.仅当系统加速度为零时才与汽车对卡车的力相等。
Q4EASY2.4 Static Friction2.4 静摩擦力No Calculator
A $5.0~\mathrm{kg}$ block sits at rest on a horizontal surface with $\mu_s = 0.40$. A horizontal force of $12~\mathrm{N}$ is applied. Take $g = 10~\mathrm{m/s^2}$. Which statement best describes the situation?一个 $5.0~\mathrm{kg}$ 的物块静止在水平面上，静摩擦系数 $\mu_s = 0.40$。施加 $12~\mathrm{N}$ 的水平力，取 $g = 10~\mathrm{m/s^2}$。下列说法中哪一项最准确描述了该情况？
(A)The block accelerates at $2.4~\mathrm{m/s^2}$.物块以 $2.4~\mathrm{m/s^2}$ 加速。
(B)The block accelerates at $0.4~\mathrm{m/s^2}$.物块以 $0.4~\mathrm{m/s^2}$ 加速。
(C)The block does not move; static friction is $12~\mathrm{N}$ opposing the applied force.物块不动，静摩擦力为 $12~\mathrm{N}$，方向与施力相反。
(D)The block does not move; static friction is exactly $20~\mathrm{N}$.物块不动，静摩擦力恰好为 $20~\mathrm{N}$。
Two blocks of masses $m_1 = 2~\mathrm{kg}$ and $m_2 = 3~\mathrm{kg}$ sit in contact on a frictionless horizontal surface. A horizontal force $F = 10~\mathrm{N}$ is applied to $m_1$, pushing both blocks. The contact force between them is质量为 $m_1 = 2~\mathrm{kg}$ 和 $m_2 = 3~\mathrm{kg}$ 的两个物块相互接触，置于无摩擦水平面上。水平力 $F = 10~\mathrm{N}$ 施加于 $m_1$，推动两物块运动。两物块之间的接触力为
(A) $4~\mathrm{N}$
(B) $6~\mathrm{N}$
(C) $10~\mathrm{N}$
(D) $8.3~\mathrm{N}$
Q6MEDIUM2.6 Inclined Plane2.6 斜面No Calculator
A block slides down a frictionless $30^\circ$ incline. Take $g = 10~\mathrm{m/s^2}$. Its acceleration along the incline is一物块沿无摩擦 $30^\circ$ 斜面下滑，取 $g = 10~\mathrm{m/s^2}$。其沿斜面方向的加速度为
(A) $5.0~\mathrm{m/s^2}$
(B) $8.66~\mathrm{m/s^2}$
(C) $10~\mathrm{m/s^2}$
(D) $9.8~\mathrm{m/s^2}$
Q7MEDIUM2.4 Static vs. Kinetic Friction2.4 静摩擦与动摩擦Calculator
A $4.0~\mathrm{kg}$ block sits on a horizontal surface with $\mu_s = 0.30$ and $\mu_k = 0.20$. A horizontal force of $F = 16~\mathrm{N}$ is applied. The block's acceleration is closest to一个 $4.0~\mathrm{kg}$ 的物块置于水平面上，静摩擦系数 $\mu_s = 0.30$，动摩擦系数 $\mu_k = 0.20$。施加水平力 $F = 16~\mathrm{N}$。该物块的加速度最接近
An object falls through air under gravity and a velocity-dependent drag force opposing motion. At terminal velocity, the object's一物体在重力及与速度有关的阻力（方向与运动相反）共同作用下在空气中下落。在达到终端速度时，该物体的
(A)velocity is zero.速度为零。
(B)acceleration is zero.加速度为零。
(C)acceleration equals $g$.加速度等于 $g$。
(D)acceleration equals the drag force divided by its mass.加速度等于阻力除以质量。
Q9MEDIUM2.7 Spring Combinations2.7 弹簧组合No Calculator
Two identical springs, each with constant $k$, are connected end-to-end (in series). The equivalent spring constant of the combination is两个相同的弹簧，每个劲度系数均为 $k$，首尾相连（串联）。该组合的等效弹簧系数为
Two point masses separated by a distance $r$ feel a gravitational attraction of magnitude $F$. If the separation is doubled to $2r$ (and the masses are unchanged), the new gravitational force is两个质点相距 $r$，受到大小为 $F$ 的引力。若距离增大为 $2r$（质量不变），新的引力为
A $1200~\mathrm{kg}$ car rounds a flat (unbanked) curve of radius $r = 50~\mathrm{m}$. The static-friction coefficient between tires and road is $\mu_s = 0.50$. The maximum speed without slipping is closest to一辆 $1200~\mathrm{kg}$ 的汽车驶过半径 $r = 50~\mathrm{m}$ 的平坦（无倾斜）弯道，轮胎与路面间的静摩擦系数 $\mu_s = 0.50$。不发生侧滑的最大速度最接近
(A) $9.0~\mathrm{m/s}$
(B) $15.7~\mathrm{m/s}$
(C) $22.0~\mathrm{m/s}$
(D) $49.0~\mathrm{m/s}$
Q12MEDIUM2.6 Slip Threshold on Incline2.6 斜面滑动临界角No Calculator
A block sits at rest on an incline that is slowly tilted upward. The static-friction coefficient between block and incline is $\mu_s$. The block first begins to slide when the incline angle satisfies一物块静止在斜面上，斜面缓慢向上倾斜。物块与斜面之间的静摩擦系数为 $\mu_s$。当斜面角满足下列哪个条件时，物块开始滑动？
(A) $\tan\theta = \mu_s$
(B) $\sin\theta = \mu_s$
(C) $\cos\theta = \mu_s$
(D) $\theta = 45^\circ$ regardless of $\mu_s$
Q13MEDIUM2.10 Kepler's Third Law2.10 开普勒第三定律No Calculator
A satellite orbits Earth at radius $R$ with period $T$. A second satellite at radius $4R$ has period一颗卫星以半径 $R$ 绕地球运行，周期为 $T$。另一颗卫星运行半径为 $4R$，其周期为
(A) $T$
(B) $2T$
(C) $4T$
(D) $8T$
Q14HARD2.2 Atwood Machine2.2 阿特伍德机No Calculator
Two masses with $m_1 < m_2$ hang from opposite ends of a light inextensible string passing over a frictionless, massless pulley. The downward acceleration of $m_2$ is$m_1 < m_2$ 的两个质量分别悬挂在一根轻质不可伸长的绳子两端，绳子绕过无摩擦、无质量的滑轮。$m_2$ 向下的加速度为
(A) $\dfrac{(m_2 - m_1)\,g}{m_1 + m_2}$
(B) $\dfrac{(m_2 - m_1)\,g}{m_2}$
(C) $\dfrac{m_2\, g}{m_1 + m_2}$
(D) $g$
Q15HARD2.5 Linear Drag, Time Constant2.5 线性阻力，时间常数Calculator
A particle of mass $m$ falls from rest under gravity with linear drag $F_\mathrm{drag} = -bv$. Its velocity satisfies $v(t) = v_t\bigl(1 - e^{-t/\tau}\bigr)$ with terminal velocity $v_t = mg/b$ and time constant $\tau = m/b$. The time at which the speed reaches $0.9\,v_t$ is closest to质量为 $m$ 的质点从静止开始在重力和线性阻力 $F_\mathrm{drag} = -bv$ 作用下下落。其速度满足 $v(t) = v_t\bigl(1 - e^{-t/\tau}\bigr)$，终端速度 $v_t = mg/b$，时间常数 $\tau = m/b$。速度达到 $0.9\,v_t$ 所需时间最接近
(A) $0.5\,\tau$
(B) $1.5\,\tau$
(C) $2.3\,\tau$
(D) $9\,\tau$
Q16HARD2.9 Conical Pendulum2.9 圆锥摆No Calculator
A bob of mass $m$ swings in a horizontal circle at the end of a string of length $\ell$ that makes a fixed angle $\theta$ with the vertical (a conical pendulum). The angular speed satisfies质量为 $m$ 的摆球悬于长度为 $\ell$ 的绳端，绳与竖直方向成固定角 $\theta$，摆球在水平圆上运动（圆锥摆）。角速度满足
A $5.0~\mathrm{kg}$ block rests on a frictionless incline of angle $\theta = 25^\circ$. A force $F = 30~\mathrm{N}$ is applied parallel to the slope, directed up the slope. The block's acceleration along the incline is closest to一个 $5.0~\mathrm{kg}$ 的物块静止在 $\theta = 25^\circ$ 的无摩擦斜面上。沿斜面向上施加 $F = 30~\mathrm{N}$ 的力。该物块沿斜面方向的加速度最接近
(A)$1.86~\mathrm{m/s^2}$ up the slope$1.86~\mathrm{m/s^2}$，沿斜面向上
(B)$1.86~\mathrm{m/s^2}$ down the slope$1.86~\mathrm{m/s^2}$，沿斜面向下
(C)$4.14~\mathrm{m/s^2}$ up the slope$4.14~\mathrm{m/s^2}$，沿斜面向上
(D)$0$ (block remains at rest)$0$（物块保持静止）
Q18HARD2.8 Escape vs. Orbital Speed2.8 逃逸速度与轨道速度Calculator
The escape speed from a planet's surface is $v_e = \sqrt{2GM/R}$. Compared with the circular orbital speed $v_o = \sqrt{GM/R}$ at radius $R$, the escape speed satisfies从行星表面的逃逸速度为 $v_e = \sqrt{2GM/R}$。与半径 $R$ 处的圆轨道速度 $v_o = \sqrt{GM/R}$ 相比，逃逸速度满足
Show all work in the space provided. Partial credit is awarded for correct setup, units, and reasoning. Use $g = 9.8~\mathrm{m/s^2}$ unless otherwise stated. Begin every problem with a clearly labeled free-body diagram.请在所提供的空白处写出所有解题过程。正确的建模思路、单位及推理过程均可获得部分分数。除另有说明外，取 $g = 9.8~\mathrm{m/s^2}$。每道题须先画出标注清晰的受力图。
A block of mass $m_1 = 2.0~\mathrm{kg}$ sits on a horizontal table with kinetic-friction coefficient $\mu_k = 0.30$ between block and table. A light, inextensible string runs from $m_1$ over a frictionless, massless pulley at the edge of the table to a hanging block $m_2 = 4.0~\mathrm{kg}$. The system is released from rest.质量 $m_1 = 2.0~\mathrm{kg}$ 的物块置于水平桌面上，物块与桌面之间的动摩擦系数 $\mu_k = 0.30$。一根轻质不可伸长的绳子从 $m_1$ 经桌边无摩擦、无质量的滑轮连接到悬挂的物块 $m_2 = 4.0~\mathrm{kg}$。系统从静止开始释放。
(a)Draw a free-body diagram for each block, labeling every force. State Newton's second law for $m_1$ along the table and for $m_2$ along the vertical.为每个物块画受力图，标注所有力。分别写出 $m_1$ 沿桌面方向和 $m_2$ 沿竖直方向的牛顿第二定律方程。
(b)Determine the magnitude of the system's acceleration.求系统加速度的大小。
(c)Determine the tension in the string.求绳中的张力。
(d)After the system has been moving for $1.5~\mathrm{s}$, find the kinetic energy of the table block $m_1$.系统运动 $1.5~\mathrm{s}$ 后，求桌面物块 $m_1$ 的动能。
A block of mass $m = 4.0~\mathrm{kg}$ rests on an incline of angle $\theta = 30^\circ$. The kinetic-friction coefficient between block and incline is $\mu_k = 0.25$. An external force $F = 25~\mathrm{N}$ is applied parallel to the slope, directed up the slope.质量 $m = 4.0~\mathrm{kg}$ 的物块静止在倾角 $\theta = 30^\circ$ 的斜面上。物块与斜面间的动摩擦系数 $\mu_k = 0.25$。沿斜面向上施加外力 $F = 25~\mathrm{N}$。
(a)Draw a free-body diagram, with all forces resolved along axes parallel and perpendicular to the incline surface.画受力图，将所有力分解为沿斜面平行方向和垂直方向的分量。
(b)Find the normal force on the block.求物块所受的法向力。
(c)Determine the block's acceleration up the slope. State whether the block actually moves up the slope and explain how you decided.求物块沿斜面向上的加速度。判断物块是否真的沿斜面向上运动，并说明判断依据。
(d)Determine the value of $F$ for which the block instead slides up the slope at constant velocity.求使物块以匀速沿斜面向上滑动时 $F$ 的值。
FRQ 3HARD2.5 Linear Drag, Derivation2.5 线性阻力，推导No Calculator
A particle of mass $m$ falls from rest through a fluid that exerts a drag force $\vec F_\mathrm{drag} = -b\vec v$ (linear in speed, opposing motion). Take downward as positive.质量为 $m$ 的质点从静止开始在流体中下落，流体施加的阻力 $\vec F_\mathrm{drag} = -b\vec v$（与速度成正比，方向与运动相反）。取向下为正方向。
(a)Apply Newton's second law and write the differential equation governing $v(t)$.应用牛顿第二定律，写出描述 $v(t)$ 的微分方程。
(b)Identify the terminal velocity $v_t$ in terms of $m$, $g$, and $b$. Justify your answer using the ODE from (a).用 $m$、$g$ 和 $b$ 表示终端速度 $v_t$，并利用 (a) 中的微分方程对答案进行论证。
(c)Solve the ODE with the initial condition $v(0) = 0$ to show that $v(t) = v_t\bigl(1 - e^{-t/\tau}\bigr)$, identifying $\tau$ in terms of $m$ and $b$.以初始条件 $v(0) = 0$ 求解该微分方程，证明 $v(t) = v_t\bigl(1 - e^{-t/\tau}\bigr)$，并用 $m$ 和 $b$ 表示 $\tau$。
(d)On the axes provided, sketch $v(t)$ for $0 \le t \le 4\tau$. Mark the point $(\tau, 0.63\,v_t)$ and the asymptote at $v = v_t$.在所给坐标轴上，画出 $0 \le t \le 4\tau$ 范围内 $v(t)$ 的草图。标出点 $(\tau, 0.63\,v_t)$ 和渐近线 $v = v_t$。
A small block of mass $m$ slides without friction along a track that includes a vertical circular loop of radius $R$. The block enters the loop at the bottom moving horizontally with speed $v_0$. Use energy conservation and the centripetal-force condition.质量为 $m$ 的小物块沿无摩擦轨道滑行，轨道包含一个半径为 $R$ 的竖直圆环。物块以水平速度 $v_0$ 从圆环底部进入。利用能量守恒和向心力条件求解。
(a)Derive the minimum entry speed $v_{0,\min}$ such that the block maintains contact with the track at the very top of the loop.推导使物块在圆环顶部仍与轨道保持接触的最小入口速度 $v_{0,\min}$。
(b)For the case $v_0 = 2\,v_{0,\min}$, determine the block's speed at the top of the loop in terms of $g$ and $R$.当 $v_0 = 2\,v_{0,\min}$ 时，用 $g$ 和 $R$ 表示物块在圆环顶部的速度。
(c)For the same case, determine the magnitude of the normal force the track exerts on the block at the top of the loop, in terms of $m$ and $g$.对于同一情况，用 $m$ 和 $g$ 表示轨道在圆环顶部对物块法向力的大小。
(d)Without doing the algebra, briefly describe how the normal force at the bottom of the loop compares with the value at the top, and identify which two physical effects make it larger.无需进行代数推导，简要说明圆环底部法向力与顶部相比如何，并指出使底部法向力更大的两个物理因素。