PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 18 marksAP 风格选择题 + 安/卑省考短答 · 共 18 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. For short-answer items, state units in every answer. Treat momentum as a vector and state your sign convention (positive direction) before calculating. No calculator on Q1-Q4; calculator permitted on Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。短答题每道都要写出单位。把动量视为矢量，计算前先说明正方向约定。Q1-Q4 不可使用计算器；Q5 可用计算器。
A $3.0\ \text{kg}$ object moves east at $8.0\ \text{m/s}$. What is the magnitude of its momentum?一个 $3.0\ \text{kg}$ 的物体向东以 $8.0\ \text{m/s}$ 运动。其动量的大小是多少？
A constant force of $200\ \text{N}$ acts on a ball for $0.050\ \text{s}$. What is the magnitude of the impulse delivered to the ball?一个 $200\ \text{N}$ 的恒力对一个球作用 $0.050\ \text{s}$。施加给球的冲量大小是多少？
A $0.42\ \text{kg}$ soccer ball is kicked and leaves the foot at $25\ \text{m/s}$.一只 $0.42\ \text{kg}$ 的足球被踢出，离脚时速度为 $25\ \text{m/s}$。
(a)Compute the momentum of the ball as it leaves the foot, with units.计算球离脚时的动量，并写出单位。[2]
(b)The ball started at rest. State the impulse the foot delivered to the ball, and explain why $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$.球起初静止。写出脚施加给球的冲量，并解释为何 $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$。[1]
(c)State the direction of the impulse relative to the ball's motion.写出冲量相对于球运动方向的方向。[1]
A $3.0\ \text{kg}$ rifle, initially at rest, fires a $0.015\ \text{kg}$ bullet forward at $500\ \text{m/s}$.一支 $3.0\ \text{kg}$、最初静止的步枪向前射出一颗 $0.015\ \text{kg}$、速度 $500\ \text{m/s}$ 的子弹。
(a)Find the recoil velocity of the rifle.求步枪的反冲速度。[3]
(b)State the total momentum of the rifle-bullet system before and after firing.写出发射前后枪-子弹系统的总动量。[1]
(c)State the condition under which momentum is conserved here.写出此处动量守恒所需满足的条件。[1]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. State the conservation law (momentum, and where relevant kinetic energy) before substituting values. State units in every final answer. Choose and state your sign convention (positive direction) at the start of each question. Calculator permitted on Q6-Q9.每一步推理都要写出。在代入数值前先注明所用守恒律（动量，必要时含动能）。每个最终答案都要写单位。每题开头请选择并说明正方向约定。Q6-Q9 可用计算器。
A $0.058\ \text{kg}$ tennis ball arrives at a racket at $30\ \text{m/s}$ and is returned in the opposite direction at $40\ \text{m/s}$. The contact lasts $4.0\ \text{ms}$. Take the outgoing direction as positive.一只 $0.058\ \text{kg}$ 的网球以 $30\ \text{m/s}$ 飞向球拍，并以 $40\ \text{m/s}$ 反向弹回。接触持续 $4.0\ \text{ms}$。取弹回方向为正。
(a)Find the change in momentum of the ball.求球的动量变化。[3]
(b)Find the average force the racket exerts on the ball.求球拍对球施加的平均力。[2]
(c)A player wants to lower the peak force on the ball without changing the incoming and outgoing speeds. Explain what they should change and why.某球员想在不改变来球与回球速率的前提下降低对球的峰值力。说明应改变什么以及原因。[2]
(d)State why the change in momentum is larger than it would be if the ball had simply stopped.说明为何此处的动量变化大于球仅仅停下时的情形。[1]
A $1200\ \text{kg}$ car moving east at $15\ \text{m/s}$ rear-ends a stationary $800\ \text{kg}$ car. The two lock together and move off as one.一辆 $1200\ \text{kg}$、向东以 $15\ \text{m/s}$ 行驶的车追尾一辆静止的 $800\ \text{kg}$ 车。两车卡在一起作为一个整体离开。
(a)Find the common velocity of the wreckage just after impact.求碰后残骸的共同速度。[3]
(b)Find the kinetic energy before and after the collision.求碰撞前后的动能。[3]
(c)State how much kinetic energy was lost and where it went.写出损失了多少动能及其去向。[1]
(d)State whether momentum was conserved, and justify your answer.说明动量是否守恒，并说明理由。[1]
A $2.0\ \text{kg}$ cart moving right at $3.0\ \text{m/s}$ collides head-on and elastically with a stationary $1.0\ \text{kg}$ cart. Take rightward as positive.一辆 $2.0\ \text{kg}$、向右以 $3.0\ \text{m/s}$ 运动的小车与一辆静止的 $1.0\ \text{kg}$ 小车发生正面弹性碰撞。取向右为正。
(a)Write the momentum-conservation equation for this collision.写出本次碰撞的动量守恒方程。[2]
(b)Write the kinetic-energy-conservation equation for this collision.写出本次碰撞的动能守恒方程。[2]
(c)Solve for the final velocity of each cart.解出每辆小车的末速度。[4]
(d)Verify that the relative speed of separation equals the relative speed of approach.验证分离的相对速率等于接近的相对速率。[1]
On a frictionless surface, a $3.0\ \text{kg}$ puck moving east at $4.0\ \text{m/s}$ strikes a $4.0\ \text{kg}$ puck moving north at $3.0\ \text{m/s}$. The two stick together. Take east as $+x$ and north as $+y$.在无摩擦表面上，一个 $3.0\ \text{kg}$、向东以 $4.0\ \text{m/s}$ 运动的冰球撞上一个 $4.0\ \text{kg}$、向北以 $3.0\ \text{m/s}$ 运动的冰球。两者粘在一起。取向东为 $+x$，向北为 $+y$。
(a)Find the total momentum in the $x$-direction before the collision.求碰撞前 $x$ 方向的总动量。[2]
(b)Find the total momentum in the $y$-direction before the collision.求碰撞前 $y$ 方向的总动量。[2]
(c)Find the speed of the combined pucks after the collision.求碰后合并冰球的速率。[3]
(d)Find the direction of motion after the collision (angle north of east).求碰后运动的方向（北偏东的角度）。[2]
(e)State whether this collision is elastic, and how you can tell without computing the kinetic energies.说明这次碰撞是否弹性，以及不计算动能也能判断的依据。[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 28 marks阿省毕业考 + 通用题型 · 共 28 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define symbols (with units) at the start of each question. State the conservation law used before substituting. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义符号（含单位）。代入数值前先写出所用守恒律。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
In a crash test, a $75\ \text{kg}$ dummy travelling at $14\ \text{m/s}$ is brought to rest. A safety engineer compares stopping it against a rigid dashboard versus into an airbag.在一次碰撞测试中，一个以 $14\ \text{m/s}$ 行驶的 $75\ \text{kg}$ 假人被制止。一位安全工程师比较让它撞向刚性仪表台与撞入安全气囊两种情形。
(a)Find the magnitude of the change in momentum of the dummy.求假人动量变化的大小。[3]
(b)Against the rigid dashboard the stop takes $0.020\ \text{s}$. Find the average force on the dummy.撞向刚性仪表台时停止历时 $0.020\ \text{s}$。求假人所受的平均力。[3]
(c)With an airbag the stop takes $0.30\ \text{s}$. Find the average force on the dummy.使用安全气囊时停止历时 $0.30\ \text{s}$。求假人所受的平均力。[2]
(d)State the factor by which the airbag reduces the average force, and explain the principle.写出安全气囊将平均力降低的倍数，并解释其原理。[1]
A $4.0\ \text{kg}$ firework shell at rest bursts into two fragments. A $1.0\ \text{kg}$ fragment flies off at $6.0\ \text{m/s}$ to the right. Take rightward as positive.一枚 $4.0\ \text{kg}$、静止的烟花弹炸裂成两块碎片。一块 $1.0\ \text{kg}$ 的碎片以 $6.0\ \text{m/s}$ 向右飞出。取向右为正。
(a)Find the velocity of the other ($3.0\ \text{kg}$) fragment.求另一块（$3.0\ \text{kg}$）碎片的速度。[3]
(b)Find the total kinetic energy of the fragments after the burst.求爆裂后碎片的总动能。[3]
(c)The shell had zero kinetic energy before the burst. State where the kinetic energy of the fragments came from.爆裂前烟花弹的动能为零。写出碎片动能的来源。[2]
(d)State why the total momentum stays zero even though kinetic energy increased.说明为何总动量保持为零，即使动能增加了。[1]
A $0.020\ \text{kg}$ bullet travelling at $300\ \text{m/s}$ embeds itself in a $2.0\ \text{kg}$ wooden block resting on a frictionless surface. The block (with the bullet inside) then moves off.一颗 $0.020\ \text{kg}$、以 $300\ \text{m/s}$ 飞行的子弹嵌入一块静置于无摩擦表面、质量 $2.0\ \text{kg}$ 的木块中。木块（连同内部的子弹）随后开始运动。
(a)Find the speed of the block-and-bullet immediately after the bullet embeds.求子弹嵌入后木块与子弹的共同速率。[3]
(b)Find the kinetic energy before and after the bullet embeds.求子弹嵌入前后的动能。[3]
(c)Find the percentage of the original kinetic energy that is lost.求损失的动能占原动能的百分比。[2]
(d)Explain why so much kinetic energy is lost here while momentum is fully conserved.解释为何此处损失了如此多的动能，而动量却完全守恒。[2]
🇨🇦 Alberta阿尔伯塔Physics 30 Unit A · 30-A1.1k · 30-A1.2k · 30-A1.3k · 30-A1.4k · 30-A1.5k
Full Syllabus Map lives in ../Study Guides/Unit_4_Momentum_and_Collisions.html. Note: NGSS HS-PS2-2 is assessed for two bodies in one dimension; the two-dimensional collision (Q9) sits above the NGSS-assessed floor but is core for ON SPH4U / BC Physics 12 / AB Physics 30.完整大纲对照表见 ../Study Guides/Unit_4_Momentum_and_Collisions.html。注：NGSS HS-PS2-2 仅评估一维两体系统；二维碰撞题（Q9）超出 NGSS 考查范围，但为安大略 SPH4U / 卑诗物理 12 / 阿省物理 30 的核心内容。