Each item below restates the prompt and choices, marks the correct letter, and gives a one- to three-sentence justification keyed to the AP topic.以下每题均重述题干与选项，标出正确字母，并给出对应 AP 考点的一至三句解析。
Q1EASY4.1 Linear Momentum4.1 线动量No Calculator
A $0.50~\mathrm{kg}$ ball travels east at $4.0~\mathrm{m/s}$. The magnitude of its momentum is一个质量为 $0.50~\mathrm{kg}$ 的小球以 $4.0~\mathrm{m/s}$ 的速度向东运动。其动量大小为
(A) $0.5~\mathrm{kg \cdot m/s}$
(B) $2.0~\mathrm{kg \cdot m/s}$
(C) $4.0~\mathrm{kg \cdot m/s}$
(D) $8.0~\mathrm{kg \cdot m/s}$
Answer:答案：(B)
$$|\vec p| = m v = 0.50(4.0) = 2.0~\mathrm{kg \cdot m/s}$$
A constant force of $10~\mathrm{N}$ acts on an object for $3.0~\mathrm{s}$. The magnitude of the impulse delivered to the object is一个 $10~\mathrm{N}$ 的恒力作用在某物体上，持续时间为 $3.0~\mathrm{s}$。该力对物体施加的冲量大小为
(A) $3.3~\mathrm{N \cdot s}$
(B) $13~\mathrm{N \cdot s}$
(C) $30~\mathrm{N \cdot s}$
(D) $300~\mathrm{N \cdot s}$
Answer:答案：(C)
$$J = F\,\Delta t = 10(3.0) = 30~\mathrm{N \cdot s}$$
Trap (D) is $J g$ (extra factor of 10).干扰项 (D) 多乘了 $g$（额外的 10 倍因子）。
Q3EASY4.3 Conservation of Momentum4.3 动量守恒No Calculator
Two carts of equal mass sit at rest on a frictionless track, in contact with a compressed spring between them. After the spring is released, cart A moves left at $2.0~\mathrm{m/s}$. Cart B then moves两辆质量相等的小车静止在无摩擦轨道上，中间夹着一根压缩弹簧。弹簧释放后，小车 A 以 $2.0~\mathrm{m/s}$ 向左运动，则小车 B 的运动情况为
(A)right at $1.0~\mathrm{m/s}$以 $1.0~\mathrm{m/s}$ 向右
(B)right at $2.0~\mathrm{m/s}$以 $2.0~\mathrm{m/s}$ 向右
(C)left at $2.0~\mathrm{m/s}$以 $2.0~\mathrm{m/s}$ 向左
(D)right at $4.0~\mathrm{m/s}$以 $4.0~\mathrm{m/s}$ 向右
Answer:答案：(B)
Total $\vec p$ was zero at rest and stays zero (no external horizontal force). With equal masses, equal-and-opposite speeds:系统静止时总动量 $\vec p$ 为零，无水平外力故保持为零。质量相等，速度大小相等方向相反：
Q4EASY4.4 Inelastic vs. Elastic4.4 非弹性碰撞与弹性碰撞No Calculator
In which type of collision is the total kinetic energy of the system not conserved?在哪种碰撞类型中，系统的总动能不守恒？
(A)Elastic collision弹性碰撞
(B)Perfectly inelastic collision完全非弹性碰撞
(C)Two billiard balls colliding glancingly without spinning两个台球在无自旋情况下的斜碰
(D)Two pucks bouncing off a frictionless wall两个冰球从无摩擦墙壁反弹
Answer:答案：(B)
Perfectly inelastic collisions have the maximum possible KE loss (the two bodies share a common final velocity). Elastic collisions preserve KE by definition; idealized billiard and frictionless-wall scenarios are taken as elastic. Momentum is conserved in every case, only KE differs.完全非弹性碰撞的动能损失最大（两物体碰后共享同一末速度）。弹性碰撞根据定义保持动能守恒；理想化的台球碰撞和无摩擦墙壁反弹均视为弹性碰撞。所有情况下动量均守恒，差别仅在于动能。
Q5MEDIUM4.2 Impulse from $F$-$t$ Graph4.2 由 $F$-$t$ 图求冲量No Calculator
A force exerted on a $2.0~\mathrm{kg}$ object varies linearly from $0$ at $t = 0$ to $20~\mathrm{N}$ at $t = 0.40~\mathrm{s}$, then drops back to $0$ at $t = 0.80~\mathrm{s}$ (a triangular pulse). The object starts at rest. Its speed at $t = 0.80~\mathrm{s}$ is作用在 $2.0~\mathrm{kg}$ 物体上的力从 $t = 0$ 时的 $0$ 线性增大到 $t = 0.40~\mathrm{s}$ 时的 $20~\mathrm{N}$，再降回 $t = 0.80~\mathrm{s}$ 时的 $0$（三角脉冲）。物体由静止开始，$t = 0.80~\mathrm{s}$ 时的速度大小为
(A) $2.0~\mathrm{m/s}$
(B) $4.0~\mathrm{m/s}$
(C) $8.0~\mathrm{m/s}$
(D) $16~\mathrm{m/s}$
Answer:答案：(B)
Impulse is the area under the $F$-$t$ curve:冲量等于 $F$-$t$ 曲线下的面积：
Then $\Delta p = J \Rightarrow v = J/m = 8.0/2.0 = 4.0~\mathrm{m/s}$. Trap (C) uses a rectangular pulse (no half-factor).则 $\Delta p = J \Rightarrow v = J/m = 8.0/2.0 = 4.0~\mathrm{m/s}$。干扰项 (C) 按矩形脉冲计算（漏掉了 $\tfrac{1}{2}$ 因子）。
Q6MEDIUM4.2 Vector Change in Momentum4.2 动量的矢量变化No Calculator
A $0.20~\mathrm{kg}$ ball strikes a wall horizontally at $5.0~\mathrm{m/s}$ and rebounds straight back at $4.0~\mathrm{m/s}$. The magnitude of the impulse delivered by the wall to the ball is一个 $0.20~\mathrm{kg}$ 的小球以 $5.0~\mathrm{m/s}$ 水平撞墙，以 $4.0~\mathrm{m/s}$ 原路反弹。墙壁对小球施加的冲量大小为
(A) $0.20~\mathrm{N \cdot s}$
(B) $1.0~\mathrm{N \cdot s}$
(C) $1.8~\mathrm{N \cdot s}$
(D) $9.0~\mathrm{N \cdot s}$
Answer:答案：(C)
Treat incoming as $+x$. Then $v_i = +5$, $v_f = -4$ (rebound reverses direction).以入射方向为 $+x$，则 $v_i = +5$，$v_f = -4$（反弹方向相反）。
A $3.0~\mathrm{kg}$ cart moving at $4.0~\mathrm{m/s}$ collides with and sticks to a stationary $1.0~\mathrm{kg}$ cart on a frictionless track. The speed of the combined carts immediately after the collision is一辆质量为 $3.0~\mathrm{kg}$、速度为 $4.0~\mathrm{m/s}$ 的小车在无摩擦轨道上与静止的 $1.0~\mathrm{kg}$ 小车碰撞并粘在一起。碰后两车合体的速度大小为
A $60~\mathrm{kg}$ skater stands at rest on frictionless ice and throws a $3.0~\mathrm{kg}$ ball horizontally at $8.0~\mathrm{m/s}$ relative to the ground. The skater's recoil speed is一名 $60~\mathrm{kg}$ 的溜冰者静止站在无摩擦冰面上，水平抛出一个 $3.0~\mathrm{kg}$ 的球，球相对地面的速度为 $8.0~\mathrm{m/s}$。溜冰者的反冲速度大小为
In a one-dimensional elastic collision, a moving puck strikes an identical stationary puck. Immediately after the collision,在一维弹性碰撞中，一个运动的冰球撞击一个相同质量且静止的冰球。碰撞后瞬间，
(A)the moving puck stops and the target puck takes on its original velocity.运动的冰球静止，目标冰球获得原来的速度。
(B)both pucks move forward with half the original speed.两个冰球均以原速度的一半向前运动。
(C)the moving puck rebounds at full speed; the target puck stays at rest.运动的冰球以原速弹回；目标冰球保持静止。
(D)both pucks come to rest, since momentum and KE both balance to zero.两个冰球均静止，因为动量和动能均归零。
Answer:答案：(A)
For equal-mass 1-D elastic collisions, momentum and KE conservation force a full velocity exchange (use the standard $v_1' = ((m_1-m_2)/(m_1+m_2))v_1$ with $m_1 = m_2$ to get $v_1' = 0$, $v_2' = v_1$). Trap (B) would conserve momentum but lose KE; (C) would conserve KE but not momentum.对于等质量一维弹性碰撞，动量和动能守恒要求速度完全交换（代入标准公式 $v_1' = ((m_1-m_2)/(m_1+m_2))v_1$，当 $m_1 = m_2$ 时得 $v_1' = 0$，$v_2' = v_1$）。干扰项 (B) 动量守恒但动能不守恒；(C) 动能守恒但动量不守恒。
Two pucks of equal mass collide and stick together. Just before the collision, puck 1 moves east at $3.0~\mathrm{m/s}$ and puck 2 moves north at $4.0~\mathrm{m/s}$. Immediately after the collision, the speed of the combined object is两个质量相等的冰球碰撞后粘在一起。碰前，冰球 1 以 $3.0~\mathrm{m/s}$ 向东运动，冰球 2 以 $4.0~\mathrm{m/s}$ 向北运动。碰后合体的速度大小为
Trap (D) directly takes $\sqrt{3^2 + 4^2}$, that's the speed of the vector sum of velocities, but the combined mass dilutes each component by 2.干扰项 (D) 直接取 $\sqrt{3^2 + 4^2}$，那是速度矢量和的大小，但合体质量将每个分量稀释了 2 倍。
Q11MEDIUM4.2 Average Force4.2 平均力Calculator
A $0.15~\mathrm{kg}$ baseball arrives at home plate moving horizontally at $40~\mathrm{m/s}$ and is hit straight back at $50~\mathrm{m/s}$. The contact between bat and ball lasts $1.5~\mathrm{ms}$. The magnitude of the average force exerted by the bat on the ball is closest to一个 $0.15~\mathrm{kg}$ 的棒球以 $40~\mathrm{m/s}$ 水平飞向本垒板，被击后以 $50~\mathrm{m/s}$ 原路返回。球棒与球的接触时间为 $1.5~\mathrm{ms}$。球棒对球施加的平均力大小最接近
(A) $1{,}000~\mathrm{N}$
(B) $4{,}000~\mathrm{N}$
(C) $9{,}000~\mathrm{N}$
(D) $14{,}000~\mathrm{N}$
Answer:答案：(C)
Take incoming as $+$: $v_i = +40$, $v_f = -50$.以入射方向为正：$v_i = +40$，$v_f = -50$。
$$\bar F = \frac{|\Delta p|}{\Delta t} = \frac{13.5}{0.0015} = 9{,}000~\mathrm{N}$$
Trap (A) treats $\Delta v = 10$ (just the speed difference, sign-blind); (D) drops the time-to-ms conversion.干扰项 (A) 取 $\Delta v = 10$（仅计算速率差，忽略方向）；(D) 漏掉了毫秒换算。
Q12MEDIUM4.4 KE Before vs. After4.4 碰撞前后动能比较No Calculator
Two carts of equal mass collide head-on at the same speed and stick together. Compared with the total kinetic energy before the collision, the total kinetic energy immediately after the collision is两辆质量相等的小车以相同速度正面碰撞并粘在一起。与碰前总动能相比，碰后瞬间的总动能为
(A)the same相同
(B)half一半
(C)one-quarter四分之一
(D)zero零
Answer:答案：(D)
Total momentum before is $m v + m(-v) = 0$, so the combined object is at rest after sticking together. $K_f = 0$, $K_i = 2 \cdot \tfrac{1}{2} m v^2 = m v^2$, ratio $= 0$. All of the initial KE is dissipated. Trap (B) is the result for one cart moving and one at rest (max possible non-zero loss for perfect inelastic).碰前总动量为 $m v + m(-v) = 0$，故粘合后合体静止。$K_f = 0$，$K_i = 2 \cdot \tfrac{1}{2} m v^2 = m v^2$，比值为 $0$。所有初始动能均被耗散。干扰项 (B) 是一辆运动一辆静止情形的结果（完全非弹性碰撞的最大非零损失）。
Q13MEDIUM4.1 Momentum as a Vector4.1 动量的矢量性No Calculator
Object X has mass $2m$ and moves east at speed $v$. Object Y has mass $m$ and moves north at speed $2v$. The magnitudes of the two momentum vectors satisfy物体 X 质量为 $2m$，以速度 $v$ 向东运动；物体 Y 质量为 $m$，以速度 $2v$ 向北运动。两动量向量大小满足
(A) $|\vec p_X| > |\vec p_Y|$
(B) $|\vec p_X| < |\vec p_Y|$
(C) $|\vec p_X| = |\vec p_Y|$
(D)Cannot be determined without more information信息不足，无法确定
Answer:答案：(C)
$|\vec p_X| = 2 m v$ and $|\vec p_Y| = m(2v) = 2 m v$. Same magnitudes, different directions (perpendicular).$|\vec p_X| = 2 m v$，$|\vec p_Y| = m(2v) = 2 m v$。大小相等，方向不同（互相垂直）。
A puck of mass $M$ moves at speed $v$ and undergoes a one-dimensional elastic collision with a stationary puck of mass $3M$. Immediately after the collision, the velocities of the incoming and target pucks are, respectively,质量为 $M$ 的冰球以速度 $v$ 运动，与质量为 $3M$ 的静止冰球发生一维弹性碰撞。碰后瞬间，入射冰球和目标冰球的速度分别为
(A) $+v/2$ and $+v/2$
(B) $-v/2$ and $+v/2$
(C) $-v$ and $+2v$
(D) $0$ and $+v/3$
Answer:答案：(B)
Standard 1-D elastic-collision result with $m_2$ at rest:$m_2$ 静止的标准一维弹性碰撞公式：
The lighter incoming puck rebounds; the heavier target moves forward. Trap (A) would conserve momentum but not KE.质量较轻的入射冰球反弹，质量较重的目标冰球向前运动。干扰项 (A) 动量守恒但动能不守恒。
Q15HARD4.3 Ballistic Pendulum4.3 弹道摆Calculator
A $0.020~\mathrm{kg}$ bullet moving horizontally at $300~\mathrm{m/s}$ strikes and embeds itself in a $2.98~\mathrm{kg}$ block hanging at rest at the end of a long string. Take $g = 9.8~\mathrm{m/s^2}$. The maximum height (above the lowest point of the swing) reached by the block-plus-bullet is closest to一颗质量为 $0.020~\mathrm{kg}$ 的子弹以 $300~\mathrm{m/s}$ 水平飞行，嵌入悬挂在长绳末端静止的 $2.98~\mathrm{kg}$ 木块中。取 $g = 9.8~\mathrm{m/s^2}$，木块加子弹摆动到的最大高度（相对摆动最低点）最接近
Trap (D) skips the collision step and uses $v_b^2/(2g)$ for the bullet alone (treating it as if its KE all became PE), but most of the bullet's KE is dissipated in the collision.干扰项 (D) 跳过碰撞步骤，直接对子弹用 $v_b^2/(2g)$（假设子弹动能全部转化为势能），但实际上大部分子弹动能在碰撞中已被耗散。
Two pucks of equal mass $m$ undergo an elastic collision on a frictionless surface. Puck 1 moves east at speed $v$ and strikes puck 2, which is at rest. After the collision, puck 1 moves at $60^\circ$ north of east with speed $v/2$. The speed of puck 2 immediately after the collision is两个质量均为 $m$ 的冰球在无摩擦表面上发生弹性碰撞。冰球 1 以速度 $v$ 向东运动并撞击静止的冰球 2。碰后，冰球 1 以速度 $v/2$ 沿东偏北 $60^\circ$ 方向运动。碰后瞬间冰球 2 的速度大小为
(A) $v/4$
(B) $v/2$
(C) $\dfrac{\sqrt{3}}{2}\,v$
(D) $v$
Answer:答案：(C)
Equal masses + elastic + one target initially at rest forces $\vec v_1' \perp \vec v_2'$ (a standard 2-D elastic result). Then KE conservation gives the magnitude:等质量、弹性碰撞、目标初始静止，这三个条件决定 $\vec v_1' \perp \vec v_2'$（标准二维弹性结论）。由动能守恒求大小：
Cross-check the perpendicularity: puck 1 at $+60^\circ$ implies puck 2 at $-30^\circ$ from east; the angle between them is $90^\circ$. ✓验证垂直性：冰球 1 在东偏北 $+60^\circ$，则冰球 2 在东偏南 $30^\circ$（即 $-30^\circ$），两者夹角为 $90^\circ$。✓
A particle of mass $m$ initially at rest experiences a force $F(t) = \alpha t$ (with $\alpha$ a positive constant) along the $+x$ direction from $t = 0$ to $t = T$. The particle's speed at $t = T$ is质量为 $m$ 的质点由静止开始，在 $t = 0$ 到 $t = T$ 期间受到沿 $+x$ 方向的力 $F(t) = \alpha t$（$\alpha$ 为正常数）。$t = T$ 时质点的速度大小为
Trap (A) treats $F$ as if it were constant at $\alpha$ (units of N/s, not N); (B) skips the half-factor from integrating $t$.干扰项 (A) 把 $F$ 视为常数 $\alpha$（单位为 N/s，而非 N）；(B) 积分 $t$ 时遗漏了 $\tfrac{1}{2}$ 因子。
Q18HARD4.4 Fraction of KE Lost4.4 动能损失比例No Calculator
A projectile of mass $m$ moving at speed $v$ collides perfectly inelastically with a stationary target of mass $M$. The fraction of the projectile's original kinetic energy that is dissipated (converted to heat, sound, deformation, etc.) is质量为 $m$、速度为 $v$ 的抛射体与质量为 $M$ 的静止靶体发生完全非弹性碰撞。抛射体原有动能中被耗散（转化为热能、声能、形变能等）的比例为
(A) $\dfrac{m}{m+M}$
(B) $\dfrac{M}{m+M}$
(C) $\dfrac{1}{2}$
(D) $0$
Answer:答案：(B)
Final speed $v' = m v/(m+M)$. Energy fraction retained:末速度 $v' = m v/(m+M)$。保留的动能比例：
$$\frac{K_f}{K_i} = \frac{\tfrac{1}{2}(m+M)v'^2}{\tfrac{1}{2} m v^2} = \frac{m}{m+M}$$
Hence fraction lost is $1 - m/(m+M) = M/(m+M)$. Limits: $M \to 0$ (no target) gives no loss; $M \to \infty$ (immovable target) gives full loss.故损失比例为 $1 - m/(m+M) = M/(m+M)$。极限情况：$M \to 0$（无靶体）无损失；$M \to \infty$（不可移动靶体）全部损失。
Each FRQ walks every part in the canonical AP-style setup → execute → evaluate structure. Numbers use $g = 9.8~\mathrm{m/s^2}$.每道自由作答题均按标准 AP 解题结构（建模→计算→评估）逐步作答。数值计算取 $g = 9.8~\mathrm{m/s^2}$。
For a linear ramp from zero, $\bar F = F_\text{peak}/2$, consistent with the triangular-pulse identity (area $= \tfrac{1}{2}\,\text{base}\,\cdot\,\text{height}$).对于从零线性增大的力，$\bar F = F_\text{peak}/2$，与三角脉冲面积公式（面积 $= \tfrac{1}{2} \times$ 底 $\times$ 高）一致。
(d)Same impulse over $\Delta t = 0.10~\mathrm{s}$:在 $\Delta t = 0.10~\mathrm{s}$ 内施加相同冲量：
Safety implication.安全意义。The change in momentum (e.g., bringing a passenger to rest in a crash) is fixed by the speed change. Air bags and padded dashboards extend the contact time $\Delta t$, lowering the peak force $\bar F = \Delta p/\Delta t$, and hence the peak deceleration of the body, even though the total impulse $J$ is unchanged.动量的变化量（例如碰撞中使乘客减速至静止）由速度变化量决定。安全气囊和软质仪表板延长接触时间 $\Delta t$，从而降低峰值力 $\bar F = \Delta p/\Delta t$（即身体的峰值减速度），即使总冲量 $J$ 保持不变。
FRQ 2MEDIUM4.3 Recoil & Center of Mass4.3 反冲与质心Calculator
$60~\mathrm{kg}$ skater throws a $4.0~\mathrm{kg}$ medicine ball from rest on frictionless ice. Ball speed $6.0~\mathrm{m/s}$ (lab frame in parts a-c).$60~\mathrm{kg}$ 溜冰者在无摩擦冰面上由静止抛出 $4.0~\mathrm{kg}$ 实心球，球速 $6.0~\mathrm{m/s}$（a-c 部分均为实验室参考系）。
(a)Recoil. Total $\vec p = 0$ before and after:反冲。前后总动量 $\vec p = 0$：
Source: chemical/biological energy in the skater's muscles converted to mechanical KE during the throw. This is not a kinetic-energy-conserving process; it's a "controlled explosion" driven by an internal energy reservoir.来源：溜冰者肌肉中的化学/生物能量在抛球过程中转化为机械动能。这不是动能守恒过程，而是由内部能量库驱动的"受控爆炸"。
(d)Ball at $6.0~\mathrm{m/s}$ in the skater's frame. Let $v_s$ be the skater's lab velocity; the ball's lab velocity is $v_b = v_s + 6.0$. Momentum conservation:球在溜冰者参考系中速度为 $6.0~\mathrm{m/s}$。设溜冰者的实验室速度为 $v_s$，则球的实验室速度为 $v_b = v_s + 6.0$。动量守恒：
Compared with part (a), same nominal "throw speed", but $v_s$ is slightly smaller in magnitude because the ball didn't have to reach $6~\mathrm{m/s}$ in the lab frame.与 (a) 部分相比，名义"抛出速度"相同，但 $v_s$ 的绝对值略小，因为球在实验室参考系中不必达到 $6~\mathrm{m/s}$。
Unchanged by the collision because the collision forces are internal to the two-puck system; no external horizontal force acts during the brief contact, so total momentum (and therefore $v_\text{cm}$) is invariant.碰撞不改变质心速度，因为碰撞力是两冰球系统的内力；短暂接触期间无水平外力作用，故总动量（以及 $v_\text{cm}$）不变。
FRQ 4HARD4.4 2-D Collision4.4 二维碰撞Calculator
$2.0~\mathrm{kg}$ puck east at $5.0~\mathrm{m/s}$ collides with stationary $3.0~\mathrm{kg}$ puck. After: $2.0~\mathrm{kg}$ puck at $30^\circ$ N of E, speed $2.0~\mathrm{m/s}$.$2.0~\mathrm{kg}$ 冰球以 $5.0~\mathrm{m/s}$ 向东，与静止的 $3.0~\mathrm{kg}$ 冰球碰撞。碰后：$2.0~\mathrm{kg}$ 冰球以 $2.0~\mathrm{m/s}$ 沿东偏北 $30^\circ$ 运动。
(a)Component momentum conservation. Let east be $+x$, north $+y$. After-collision puck-1 components: $v_{1x}' = 2.0\cos 30^\circ \approx 1.732$, $v_{1y}' = 2.0\sin 30^\circ = 1.0$.分量动量守恒。取东为 $+x$，北为 $+y$。碰后冰球 1 的分量：$v_{1x}' = 2.0\cos 30^\circ \approx 1.732$，$v_{1y}' = 2.0\sin 30^\circ = 1.0$。
$\vec v_\text{cm}$ is preserved because the collision involves only internal forces, no external horizontal force on the two-puck system.$\vec v_\text{cm}$ 不变，因为碰撞仅涉及内力，两冰球系统无水平外力作用。