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Momentum and Collisions · Solutions动量与碰撞 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 18 marksAP 选择题 + 安/卑省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Momentum动量 · HS-PS2-2 [3 marks][3 分]

A $3.0$ kg object moves east at $8.0$ m/s. Magnitude of its momentum?$3.0$ kg 物体向东以 $8.0$ m/s 运动，其动量大小为多少？

Answer:答案：  (A)  $24\ \text{kg}\,\text{m/s}$

(a) Apply $p = mv$ with units套用 $p = mv$ 并带单位 M1·A1·A1

$$ p \;=\; m v \;=\; (3.0)(8.0) \;=\; 24 \;\text{kg}\,\text{m/s} \;\; (\text{east}). $$ Magnitude is $24\ \text{kg}\,\text{m/s}$, option (A).大小为 $24\ \text{kg}\,\text{m/s}$，选 (A)。
Why the distractors fail.干扰项分析。
(B) $2.7\ \text{kg}\,\text{m/s}$: divides mass by speed, $3.0/8.0 \times$ a factor, instead of multiplying.把质量与速率相除而非相乘所得。
(C) $11\ \text{kg}\,\text{m/s}$: adds mass and speed, $3.0 + 8.0$.将质量与速率相加，$3.0 + 8.0$。
(D) $0.38\ \text{kg}\,\text{m/s}$: inverts the product, $3.0/8.0$.把乘积颠倒，算成 $3.0/8.0$。
Momentum is mass times velocity, a vector pointing the way the object moves.动量是质量乘以速度，方向与物体运动方向相同的矢量。 $\vec p = m\vec v$ has units $\text{kg}\,\text{m/s}$ and always points along the velocity. The number $24$ alone is incomplete: a full momentum answer states the direction (east here). Because momentum scales with both mass and speed, a heavy slow object can carry the same momentum as a light fast one. This vector bookkeeping, with a sign in one dimension, is the discipline the whole unit rests on, so naming the positive direction before substituting is the safest habit.$\vec p = m\vec v$ 的单位是 $\text{kg}\,\text{m/s}$，方向始终沿速度。仅写 $24$ 并不完整：完整的动量答案要写明方向（此处向东）。由于动量同时随质量与速率增长，又重又慢的物体可与又轻又快的物体携带相同动量。这种带符号的矢量记账（一维中体现为正负号）是整个单元所依赖的纪律，故代入前先指明正方向是最稳妥的习惯。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Impulse冲量 · HS-PS2-3 [3 marks][3 分]

A constant $200$ N force acts on a ball for $0.050$ s. Magnitude of the impulse?$200$ N 的恒力对球作用 $0.050$ s。冲量大小为多少？

Answer:答案：  (B)  $10\ \text{N}\,\text{s}$

(a) Apply $J = F\,\Delta t$套用 $J = F\,\Delta t$ M1·A1·A1

$$ J \;=\; F\,\Delta t \;=\; (200)(0.050) \;=\; 10 \;\text{N}\,\text{s}. $$ Option (B). Equivalently $10\ \text{kg}\,\text{m/s}$, since impulse and momentum share units.选 (B)。等价于 $10\ \text{kg}\,\text{m/s}$，因为冲量与动量单位相同。
Why the distractors fail.干扰项分析。
(A) $4000\ \text{N}\,\text{s}$: divides $200$ by $0.050$ instead of multiplying.把 $200$ 除以 $0.050$ 而非相乘。
(C) $200\ \text{N}\,\text{s}$: reports the force alone, forgetting the time factor.只报力本身，忘了乘时间。
(D) $0.25\ \text{N}\,\text{s}$: divides time by force, $0.050/200$.把时间除以力，$0.050/200$。
Impulse is force times the time interval it acts over.冲量是力乘以它作用的时间间隔。 $\vec J = \vec F\,\Delta t$ measures the cumulative effect of a force acting for a time, with units $\text{N}\,\text{s}$. That $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$ is no coincidence: $\text{N} = \text{kg}\,\text{m/s}^2$, so $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$. This unit identity foreshadows the impulse-momentum theorem, where impulse equals the change in momentum. When the force varies in time, the impulse is the area under the force-time graph; here the force is constant, so the area is simply a rectangle, $F\,\Delta t$.$\vec J = \vec F\,\Delta t$ 度量一个力作用一段时间的累积效果，单位为 $\text{N}\,\text{s}$。$\text{N}\,\text{s} = \text{kg}\,\text{m/s}$ 并非巧合：$\text{N} = \text{kg}\,\text{m/s}^2$，故 $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$。这一单位等式预示了冲量-动量定理，其中冲量等于动量的变化。当力随时间变化时，冲量是力-时间图下的面积；此处力恒定，面积就是一个矩形 $F\,\Delta t$。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §1 · §2 Momentum & impulse动量与冲量 · SPH4U C2 [4 marks][4 分]

$0.42$ kg ball kicked, leaves foot at $25$ m/s from rest. (a) Momentum. (b) Impulse and unit identity. (c) Direction of impulse.$0.42$ kg 的球被踢，自静止以 $25$ m/s 离脚。(a) 动量。(b) 冲量及单位等式。(c) 冲量方向。

Answer:答案：  (a) $p = 10.5\ \text{kg}\,\text{m/s}$  ·  (b) $J = 10.5\ \text{N}\,\text{s}$  ·  (c) same direction as the ball's motion与球运动方向相同

(a) Momentum of the ball as it leaves the foot球离脚时的动量 M1·A1

$$ p \;=\; m v \;=\; (0.42)(25) \;=\; 10.5 \;\text{kg}\,\text{m/s} \;\; (\text{in the kick direction}). $$

(b) Impulse delivered, and why $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$所施加的冲量，以及为何 $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$ A1

The ball started at rest, so the impulse equals the change in momentum: $J = \Delta p = 10.5 - 0 = 10.5\ \text{N}\,\text{s}$. The units match because $\text{N} = \text{kg}\,\text{m/s}^2$, so $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$.球起初静止，故冲量等于动量的变化：$J = \Delta p = 10.5 - 0 = 10.5\ \text{N}\,\text{s}$。单位相符是因为 $\text{N} = \text{kg}\,\text{m/s}^2$，故 $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$。

(c) Direction of the impulse冲量的方向 A1

Impulse is a vector equal to $\Delta \vec p$. Since the ball gains momentum in the direction it is kicked, the impulse points the same way as the ball's motion.冲量是矢量，等于 $\Delta \vec p$。由于球在被踢的方向上获得动量，冲量方向与球的运动方向相同。
For an object starting from rest, the impulse delivered equals its final momentum.对于从静止出发的物体，所受冲量等于其末动量。 The impulse-momentum theorem says $J = \Delta p = p_f - p_i$. When $p_i = 0$, the impulse is simply the final momentum, numerically equal but carrying the units $\text{N}\,\text{s}$ instead of $\text{kg}\,\text{m/s}$. Recognising the unit identity $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$ is a frequently tested checkpoint on Ontario SPH4U: it confirms that impulse and momentum are the same physical quantity viewed two ways, one through force-and-time, the other through mass-and-velocity.冲量-动量定理为 $J = \Delta p = p_f - p_i$。当 $p_i = 0$ 时，冲量就等于末动量，数值相等，但单位用 $\text{N}\,\text{s}$ 而非 $\text{kg}\,\text{m/s}$。认识单位等式 $\text{N}\,\text{s} = \text{kg}\,\text{m/s}$ 是安大略 SPH4U 常考的要点：它确认了冲量与动量是同一物理量的两种视角，一是力与时间，二是质量与速度。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §4 · §5 Elastic vs inelastic弹性与非弹性 · HS-PS2-2 [3 marks][3 分]

Two carts collide and stick together. Which statement is correct?两辆小车碰撞后粘在一起。哪项陈述正确？

Answer:答案：  (A)  Momentum conserved, kinetic energy not动量守恒，动能不守恒

(a) Identify the collision type and what each law says判别碰撞类型并明确各守恒律 M1·A1·A1

Sticking together is a perfectly inelastic collision. Momentum is conserved in every collision (no external net force), so it is conserved here. Kinetic energy is not conserved: some is converted to heat, sound, and deformation. Option (A).粘在一起是完全非弹性碰撞。每次碰撞动量都守恒（无外部净力），故此处守恒。动能不守恒：部分转化为热、声与形变。选 (A)。
Why the distractors fail.干扰项分析。
(B): reverses the truth: momentum is always conserved, kinetic energy is what is lost.把事实颠倒：动量始终守恒，损失的是动能。
(C): this describes an elastic collision; sticking together is the opposite extreme.这描述的是弹性碰撞；粘合恰是相反的极端。
(D): momentum is never lost in a collision when no external net force acts.无外部净力时，碰撞中动量绝不会损失。
Momentum is conserved in every collision; kinetic energy is the test that separates elastic from inelastic.每次碰撞动量都守恒；动能才是区分弹性与非弹性的检验。 This is the single most tested conceptual point in the unit. Momentum conservation follows from Newton's third law and holds for elastic, inelastic, and perfectly inelastic collisions alike, as long as no external net force acts. Kinetic energy is the discriminator: conserved in an elastic collision, partly lost in an inelastic one, and maximally lost in a perfectly inelastic ("stick-together") collision. The lost kinetic energy is not destroyed; it becomes heat, sound, and permanent deformation, with total energy still conserved.这是本单元最常考的概念要点。动量守恒源自牛顿第三定律，只要无外部净力，对弹性、非弹性与完全非弹性碰撞都成立。动能才是区分器：弹性碰撞中守恒，非弹性碰撞中部分损失，完全非弹性（"粘合"）碰撞中损失最大。损失的动能并未消失；它化为热、声与永久形变，总能量仍守恒。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Conservation (recoil)守恒（反冲） · Physics 12 [5 marks][5 分]

$3.0$ kg rifle at rest fires a $0.015$ kg bullet forward at $500$ m/s. (a) Recoil velocity. (b) Total momentum before/after. (c) Conservation condition.$3.0$ kg 步枪静止，向前射出 $0.015$ kg 子弹，速度 $500$ m/s。(a) 反冲速度。(b) 前后总动量。(c) 守恒条件。

Answer:答案：  (a) $v_r = -2.5\ \text{m/s}$ (backward)（向后）  ·  (b) zero before and after前后均为零  ·  (c) no external net force无外部净力

(a) Recoil velocity from conservation of momentum由动量守恒求反冲速度 M1·A1·A1

Take forward as positive. Total momentum before firing is zero (everything at rest):取向前为正。发射前总动量为零（一切静止）： $$ 0 \;=\; m_{\text{b}} v_{\text{b}} + m_{\text{r}} v_{\text{r}} \;=\; (0.015)(500) + (3.0)v_{\text{r}}. $$ $$ v_{\text{r}} \;=\; -\frac{(0.015)(500)}{3.0} \;=\; -\frac{7.5}{3.0} \;=\; -2.5 \;\text{m/s}. $$ The minus sign means the rifle recoils backward at $2.5$ m/s.负号表示步枪以 $2.5$ m/s 向后反冲。

(b) Total momentum before and after发射前后的总动量 A1

Before: $0$. After: $(0.015)(500) + (3.0)(-2.5) = 7.5 - 7.5 = 0$. The total is conserved at zero.之前：$0$。之后：$(0.015)(500) + (3.0)(-2.5) = 7.5 - 7.5 = 0$。总动量守恒为零。

(c) Condition for conservation守恒条件 A1

Momentum is conserved because no external net force acts on the rifle-bullet system during firing. The firing forces are internal and cancel in equal-and-opposite pairs.动量守恒是因为发射过程中枪-子弹系统不受外部净力。发射的力是内力，以等大反向的成对方式抵消。
Recoil is conservation of momentum from a zero start: equal and opposite momenta sum to zero.反冲是从零起的动量守恒：等大反向的动量相加为零。 Because the system begins at rest, its total momentum is zero and must stay zero. The light bullet's large forward momentum is exactly balanced by the heavy rifle's small backward momentum: $m_{\text{b}} v_{\text{b}} = -m_{\text{r}} v_{\text{r}}$. The mass ratio ($3.0 / 0.015 = 200$) is why the rifle moves $200$ times slower than the bullet. This same reasoning explains rocket propulsion, ice-skater push-offs, and the astronaut-throws-a-tool scenario: in every case, a closed system starting from rest splits into equal-and-opposite momenta.由于系统从静止开始，其总动量为零且必须保持为零。轻子弹的大向前动量恰被重步枪的小向后动量平衡：$m_{\text{b}} v_{\text{b}} = -m_{\text{r}} v_{\text{r}}$。质量比（$3.0 / 0.015 = 200$）正是步枪速率比子弹慢 $200$ 倍的原因。同样的推理解释了火箭推进、滑冰者互推以及宇航员抛工具的情形：每一例中，从静止出发的封闭系统都分裂为等大反向的动量。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Impulse-momentum theorem冲量-动量定理 · HS-PS2-3 [8 marks][8 分]

$0.058$ kg tennis ball: in at $30$ m/s, returned at $40$ m/s (opposite), contact $4.0$ ms. Outgoing positive. (a) $\Delta p$. (b) Average force. (c) Lower peak force. (d) Why $\Delta p$ is large.$0.058$ kg 网球：来球 $30$ m/s，回球 $40$ m/s（反向），接触 $4.0$ ms。取回球方向为正。(a) $\Delta p$。(b) 平均力。(c) 降低峰值力。(d) 为何 $\Delta p$ 大。

Answer:答案：  (a) $\Delta p = 4.06\ \text{kg}\,\text{m/s}$  ·  (b) $F = 1015\ \text{N}$  ·  (c) lengthen the contact time延长接触时间  ·  (d) the velocity reverses速度反向

(a) Change in momentum动量变化 M1·A1·A1

Outgoing positive, so $v_f = +40$ m/s and the incoming ball $v_i = -30$ m/s:回球为正，故 $v_f = +40$ m/s，来球 $v_i = -30$ m/s： $$ \Delta p \;=\; m(v_f - v_i) \;=\; 0.058\,(40 - (-30)) \;=\; 0.058 \times 70 \;=\; 4.06 \;\text{kg}\,\text{m/s.} $$

(b) Average force (impulse-momentum theorem)平均力（冲量-动量定理） M1·A1

With $\Delta t = 4.0\ \text{ms} = 0.0040$ s:取 $\Delta t = 4.0\ \text{ms} = 0.0040$ s： $$ F \;=\; \frac{\Delta p}{\Delta t} \;=\; \frac{4.06}{0.0040} \;=\; 1015 \;\text{N.} $$

(c) How to lower the peak force如何降低峰值力 A1·A1

Keeping the speeds fixed keeps $\Delta p$ fixed. Since $F = \Delta p / \Delta t$, the player should lengthen the contact time (a softer, longer follow-through with the strings), which reduces the average force for the same change in momentum.保持速率不变即保持 $\Delta p$ 不变。由于 $F = \Delta p / \Delta t$，球员应延长接触时间（用球弦更柔、更长的随挥），从而在相同动量变化下减小平均力。

(d) Why the change in momentum is large为何动量变化很大 A1

The ball reverses direction, so $\Delta p$ adds the incoming and outgoing speeds ($70$ m/s of velocity change), far larger than the $30$ m/s change if the ball had merely stopped.球反向，故 $\Delta p$ 把来球与回球速率相加（速度变化 $70$ m/s），远大于球仅停下时 $30$ m/s 的变化。
For a fixed change in momentum, force and contact time trade off: $F = \Delta p / \Delta t$.对固定的动量变化，力与接触时间此消彼长：$F = \Delta p / \Delta t$。 The impulse-momentum theorem $\vec F\,\Delta t = \Delta \vec p$ is the engine behind every padding, airbag, and follow-through. Because a reversal counts the incoming and outgoing speeds both, a bounced ball experiences a much larger impulse than one that simply stops, which is why a struck ball feels a force of over a kilonewton in just $4$ ms. The sign convention is essential: writing $v_i = -30$ (not $+30$) is what produces the correct $70$ m/s velocity change. Forgetting the sign and using $40 - 30 = 10$ is the single most common error here.冲量-动量定理 $\vec F\,\Delta t = \Delta \vec p$ 是一切护垫、气囊与随挥背后的引擎。由于反向把来球与回球速率都计入，弹回的球比仅停下的球受到大得多的冲量，这正是被击球在短短 $4$ ms 内受力超过一千牛的原因。符号约定至关重要：写 $v_i = -30$（而非 $+30$）才得出正确的 $70$ m/s 速度变化。忘记符号、用 $40 - 30 = 10$ 是此处最常见的错误。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 · §5 Inelastic collision非弹性碰撞 · SPH4U C3 [8 marks][8 分]

$1200$ kg car at $15$ m/s rear-ends a stationary $800$ kg car; they lock together. (a) Common velocity. (b) KE before and after. (c) KE lost. (d) Momentum conserved?$1200$ kg 车以 $15$ m/s 追尾静止的 $800$ kg 车；两车卡在一起。(a) 共同速度。(b) 碰前后动能。(c) 损失的动能。(d) 动量是否守恒？

Answer:答案：  (a) $v = 9.0\ \text{m/s}$  ·  (b) $KE_i = 135\,000\ \text{J},\ KE_f = 81\,000\ \text{J}$  ·  (c) $54\,000\ \text{J}$ lost损失  ·  (d) yes是

(a) Common velocity (conserve momentum, stick together)共同速度（守恒动量，粘合） M1·A1·A1

East positive; the cars stick, sharing one velocity $v$:向东为正；两车粘合，共享一个速度 $v$： $$ (1200)(15) + 0 \;=\; (1200 + 800)v \;\Longrightarrow\; v \;=\; \frac{18\,000}{2000} \;=\; 9.0 \;\text{m/s.} $$

(b) Kinetic energy before and after碰前后的动能 M1·A1·A1

$$ KE_i \;=\; \tfrac{1}{2}(1200)(15)^2 \;=\; 135\,000 \;\text{J.} $$ $$ KE_f \;=\; \tfrac{1}{2}(2000)(9.0)^2 \;=\; \tfrac{1}{2}(2000)(81) \;=\; 81\,000 \;\text{J.} $$

(c) Kinetic energy lost损失的动能 A1

$\Delta KE = 135\,000 - 81\,000 = 54\,000$ J, converted to crumpling metal, heat, and sound.$\Delta KE = 135\,000 - 81\,000 = 54\,000$ J，转化为金属溃缩、热与声。

(d) Was momentum conserved?动量是否守恒？ A1

Yes. Before: $18\,000\ \text{kg}\,\text{m/s}$; after: $(2000)(9.0) = 18\,000\ \text{kg}\,\text{m/s}$. Momentum is conserved because no external net force acts during the brief impact, even though kinetic energy is not.是。之前：$18\,000\ \text{kg}\,\text{m/s}$；之后：$(2000)(9.0) = 18\,000\ \text{kg}\,\text{m/s}$。动量守恒是因为短暂碰撞中无外部净力，尽管动能不守恒。
In an inelastic collision momentum is conserved while kinetic energy is lost; the two are independent bookkeeping.非弹性碰撞中动量守恒而动能损失；二者是相互独立的记账。 Here $40\%$ of the kinetic energy disappears ($54\,000$ of $135\,000$ J), yet momentum is conserved to the kilogram-metre-per-second. This is the heart of crash safety: engineers deliberately design cars to be more inelastic, channelling that lost kinetic energy into crumple zones rather than the occupants. The standard trap is to assume that "energy lost" means "momentum lost." They are governed by different laws: momentum conservation comes from Newton's third law (always holds with no external force), while kinetic-energy conservation is an extra condition that only elastic collisions satisfy.此处 $40\%$ 的动能消失（$135\,000$ J 中的 $54\,000$ J），然而动量守恒精确到千克·米每秒。这是撞击安全的核心：工程师刻意把车设计得更非弹性，把损失的动能导入溃缩区而非乘员。标准陷阱是误以为"能量损失"意味着"动量损失"。二者受不同定律支配：动量守恒源自牛顿第三定律（无外力时恒成立），而动能守恒是仅弹性碰撞才满足的额外条件。
Q8HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 Elastic collision (unequal mass)弹性碰撞（质量不等） · Physics 12 [9 marks][9 分]

$2.0$ kg cart at $3.0$ m/s hits a stationary $1.0$ kg cart, elastic, head-on. Rightward positive. (a) Momentum equation. (b) KE equation. (c) Final velocities. (d) Verify relative-speed reversal.$2.0$ kg 小车以 $3.0$ m/s 撞静止的 $1.0$ kg 小车，弹性正碰。向右为正。(a) 动量方程。(b) 动能方程。(c) 末速度。(d) 验证相对速度反向。

Answer:答案：  (c) $v_1 = 1.0\ \text{m/s},\ v_2 = 4.0\ \text{m/s}$  ·  (d) approach $= 3.0$ m/s $=$ separation接近 $= 3.0$ m/s $=$ 分离

(a) Conservation of momentum动量守恒 M1·A1

$$ (2.0)(3.0) + (1.0)(0) \;=\; (2.0)v_1 + (1.0)v_2 \;\Longrightarrow\; 2v_1 + v_2 \;=\; 6. $$

(b) Conservation of kinetic energy动能守恒 M1·A1

$$ \tfrac{1}{2}(2.0)(3.0)^2 \;=\; \tfrac{1}{2}(2.0)v_1^2 + \tfrac{1}{2}(1.0)v_2^2 \;\Longrightarrow\; 2v_1^2 + v_2^2 \;=\; 18. $$

(c) Solve for the final velocities解出末速度 M1·A1·A1·A1

From the momentum equation, $v_2 = 6 - 2v_1$. Substitute into the energy equation:由动量方程得 $v_2 = 6 - 2v_1$。代入能量方程： $$ 2v_1^2 + (6 - 2v_1)^2 \;=\; 18 \;\Longrightarrow\; 6v_1^2 - 24v_1 \;=\; 0 \;\Longrightarrow\; 6v_1(v_1 - 4) \;=\; 0. $$ So $v_1 = 0$ (the trivial "no collision" root) or $v_1 = 4$. The physical root is the one where the carts have actually interacted. Using the standard elastic formulas confirms which: $v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 = \frac{1}{3}(3.0) = 1.0$ m/s, and then $v_2 = 6 - 2(1.0) = 4.0$ m/s.故 $v_1 = 0$（平凡的"未碰撞"根）或 $v_1 = 4$。物理根是两车确已相互作用的那个。用标准弹性公式可确认：$v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 = \frac{1}{3}(3.0) = 1.0$ m/s，进而 $v_2 = 6 - 2(1.0) = 4.0$ m/s。 Check momentum: $2(1.0) + 1(4.0) = 6\ \checkmark$. Check energy: $2(1.0)^2 + (4.0)^2 = 2 + 16 = 18\ \checkmark$.验证动量：$2(1.0) + 1(4.0) = 6\ \checkmark$。验证能量：$2(1.0)^2 + (4.0)^2 = 2 + 16 = 18\ \checkmark$。

(d) Relative-speed reversal相对速度反向 A1

Approach: $u_1 - u_2 = 3.0 - 0 = 3.0$ m/s. Separation: $v_2 - v_1 = 4.0 - 1.0 = 3.0$ m/s. They are equal, as required for an elastic collision.接近：$u_1 - u_2 = 3.0 - 0 = 3.0$ m/s。分离：$v_2 - v_1 = 4.0 - 1.0 = 3.0$ m/s。两者相等，符合弹性碰撞的要求。
In a 1D elastic collision the relative speed of separation equals the relative speed of approach.一维弹性碰撞中，分离的相对速率等于接近的相对速率。 Solving the quadratic always yields two roots: the trivial $v_1 = u_1$ ("they pass through untouched") and the physical post-collision root. Knowing this lets you reject the spurious root immediately. The relative-velocity relation $u_1 - u_2 = -(v_1 - v_2)$ is a faster route than the quadratic and is the cleanest check: any 1D elastic collision conserves the relative speed and merely reverses its sign. When the masses were equal, this relation forces a velocity swap; here, with $m_1 = 2m_2$, the heavier cart keeps moving forward (slower) while the lighter one shoots ahead.求解二次方程总得两个根：平凡根 $v_1 = u_1$（"二者无接触地穿过"）与碰后的物理根。知道这一点便可立即排除虚假根。相对速度关系 $u_1 - u_2 = -(v_1 - v_2)$ 比二次方程更快，也是最简洁的检验：任何一维弹性碰撞都守恒相对速率而仅反转其符号。质量相等时该关系强制速度交换；此处 $m_1 = 2m_2$，较重的小车继续向前（变慢），较轻的小车则冲到前面。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 2D collision二维碰撞 · HS-PS2-2 (above 1D floor)（超出一维基准） [10 marks][10 分]

$3.0$ kg puck east at $4.0$ m/s strikes $4.0$ kg puck north at $3.0$ m/s; they stick. East $= +x$, north $= +y$. (a) $p_x$. (b) $p_y$. (c) Final speed. (d) Direction. (e) Elastic?$3.0$ kg 冰球向东 $4.0$ m/s 撞 $4.0$ kg 冰球向北 $3.0$ m/s；二者粘合。东 $= +x$，北 $= +y$。(a) $p_x$。(b) $p_y$。(c) 末速率。(d) 方向。(e) 是否弹性？

Answer:答案：  (a) $p_x = 12\ \text{kg}\,\text{m/s}$  ·  (b) $p_y = 12\ \text{kg}\,\text{m/s}$  ·  (c) $v = 2.42\ \text{m/s}$  ·  (d) $45°$ N of E北偏东  ·  (e) inelastic非弹性

(a) Total $x$-momentum before碰前 $x$ 方向总动量 M1·A1

Only the $3.0$ kg puck moves along $x$:只有 $3.0$ kg 冰球沿 $x$ 运动： $$ p_x \;=\; (3.0)(4.0) + (4.0)(0) \;=\; 12 \;\text{kg}\,\text{m/s.} $$

(b) Total $y$-momentum before碰前 $y$ 方向总动量 M1·A1

Only the $4.0$ kg puck moves along $y$:只有 $4.0$ kg 冰球沿 $y$ 运动： $$ p_y \;=\; (3.0)(0) + (4.0)(3.0) \;=\; 12 \;\text{kg}\,\text{m/s.} $$

(c) Final speed of the combined pucks合并冰球的末速率 M1·A1·A1

Momentum is conserved on each axis. Combined mass $M = 7.0$ kg, total momentum $p = \sqrt{p_x^2 + p_y^2} = \sqrt{12^2 + 12^2} = \sqrt{288} = 16.97\ \text{kg}\,\text{m/s}$. Then:动量在每个轴上守恒。合并质量 $M = 7.0$ kg，总动量 $p = \sqrt{p_x^2 + p_y^2} = \sqrt{12^2 + 12^2} = \sqrt{288} = 16.97\ \text{kg}\,\text{m/s}$。于是： $$ v \;=\; \frac{p}{M} \;=\; \frac{16.97}{7.0} \;\approx\; 2.42 \;\text{m/s.} $$

(d) Direction of motion (angle north of east)运动方向（北偏东角度） M1·A1

$$ \theta \;=\; \arctan\!\left(\frac{p_y}{p_x}\right) \;=\; \arctan\!\left(\frac{12}{12}\right) \;=\; 45°. $$

(e) Elastic or not?是否弹性？ A1

The pucks stick together, so the collision is perfectly inelastic. You can tell without computing kinetic energies: any "stick-together" collision loses kinetic energy and so cannot be elastic.两冰球粘在一起，故为完全非弹性碰撞。无需计算动能即可判断：任何"粘合"碰撞都损失动能，故不可能是弹性的。
In two dimensions, momentum is conserved on each axis separately; resolve, conserve per axis, then recombine.在二维中，动量在每个轴上分别守恒；分解、按轴守恒，再合成。 A single vector conservation law splits into two scalar equations, one for $x$ and one for $y$. Here each incoming puck contributes to only one axis, which keeps the arithmetic clean. After conserving each component, the final velocity is reassembled with the Pythagorean theorem for magnitude and $\arctan$ for direction, exactly as with projectile components. The symmetry ($p_x = p_y = 12$) is what produces the tidy $45°$ result. Note the conceptual shortcut in (e): the "stick-together" outcome is the signature of a perfectly inelastic collision, so the kinetic-energy test is unnecessary, though it would confirm energy is lost.单个矢量守恒律拆成两个标量方程，一个对 $x$、一个对 $y$。此处每个来球只贡献一个轴，使运算清爽。按分量守恒后，用勾股定理求大小、用 $\arctan$ 求方向重新合成末速度，与抛体分量完全一致。对称性（$p_x = p_y = 12$）正是产生整齐 $45°$ 结果的原因。注意 (e) 的概念捷径："粘合"结果是完全非弹性碰撞的标志，故无需动能检验，尽管检验会确认能量有损失。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解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

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §2 Impulse (vehicle safety)冲量（车辆安全） · 30-A1.2k [9 marks][9 分]

$75$ kg dummy at $14$ m/s brought to rest. (a) $|\Delta p|$. (b) Force if stop takes $0.020$ s. (c) Force if airbag stop takes $0.30$ s. (d) Factor of reduction.$75$ kg 假人以 $14$ m/s 被制止。(a) $|\Delta p|$。(b) 停止历时 $0.020$ s 的力。(c) 气囊停止历时 $0.30$ s 的力。(d) 降低的倍数。

Answer:答案：  (a) $1050\ \text{kg}\,\text{m/s}$  ·  (b) $52\,500\ \text{N}$  ·  (c) $3500\ \text{N}$  ·  (d) factor of $15$降为 $1/15$

(a) Magnitude of the change in momentum动量变化的大小 M1·A1·A1

$$ |\Delta p| \;=\; m\,|\Delta v| \;=\; 75 \times (14 - 0) \;=\; 1050 \;\text{kg}\,\text{m/s.} $$

(b) Force against the rigid dashboard撞向刚性仪表台的力 M1·A1·A1

$$ F_{\text{dash}} \;=\; \frac{|\Delta p|}{\Delta t} \;=\; \frac{1050}{0.020} \;=\; 52\,500 \;\text{N.} $$

(c) Force with the airbag使用气囊时的力 M1·A1

$$ F_{\text{airbag}} \;=\; \frac{1050}{0.30} \;=\; 3500 \;\text{N.} $$

(d) Factor of reduction降低的倍数 A1

$F_{\text{dash}} / F_{\text{airbag}} = 52\,500 / 3500 = 15$. The airbag cuts the average force to one fifteenth, because for a fixed $\Delta p$ a $15\times$ longer stopping time gives a $15\times$ smaller force.$F_{\text{dash}} / F_{\text{airbag}} = 52\,500 / 3500 = 15$。气囊把平均力降到十五分之一，因为对固定的 $\Delta p$，停止时间延长 $15$ 倍则力减小为原来的 $1/15$。
An airbag does not change $\Delta p$; it lengthens $\Delta t$, and $F = \Delta p / \Delta t$ does the rest.气囊不改变 $\Delta p$；它延长 $\Delta t$，而 $F = \Delta p / \Delta t$ 完成其余的事。 The occupant's change in momentum is fixed by the crash: they go from $14$ m/s to rest no matter what they hit. The only lever a safety device has is the stopping time. A rigid dashboard stops the occupant in milliseconds, producing a force of over $50$ kilonewtons, many times the force that breaks bone. The airbag stretches that same momentum change over $0.30$ s, slashing the force by the same factor as the time was lengthened. This is the explicit engineering of NGSS HS-PS2-3 and AB 30-A1.2k: crumple zones, helmets, and landing by bending the knees all work by the identical principle of increasing contact time.乘员的动量变化由碰撞决定：无论撞到什么，都从 $14$ m/s 变为静止。安全装置唯一能调的杠杆是停止时间。刚性仪表台在毫秒内使乘员停下，产生超过 $50$ 千牛的力，是折断骨头所需力的许多倍。气囊把同样的动量变化拉伸到 $0.30$ s，把力按时间延长的相同倍数削减。这就是 NGSS HS-PS2-3 与 AB 30-A1.2k 所明确的工程：溃缩区、头盔与屈膝落地都遵循增大接触时间这一相同原理。
Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §3 Explosion / recoil爆炸 / 反冲 · 30-A1.3k [9 marks][9 分]

$4.0$ kg shell at rest bursts into two fragments; a $1.0$ kg fragment flies right at $6.0$ m/s. Rightward positive. (a) Velocity of the $3.0$ kg fragment. (b) Total KE after the burst. (c) Source of that KE. (d) Why total momentum stays zero.$4.0$ kg 烟花弹静止炸裂成两块；一块 $1.0$ kg 碎片以 $6.0$ m/s 向右飞出。取向右为正。(a) $3.0$ kg 碎片的速度。(b) 爆裂后总动能。(c) 该动能的来源。(d) 为何总动量保持为零。

Answer:答案：  (a) $v = -2.0\ \text{m/s}$ (left)（向左）  ·  (b) $24\ \text{J}$  ·  (c) stored chemical energy储存的化学能  ·  (d) burst forces are internal爆裂力为内力

(a) Velocity of the $3.0$ kg fragment (conserve momentum from rest)$3.0$ kg 碎片的速度（由静止守恒动量） M1·A1·A1

Rightward positive. The shell is at rest, so total momentum before and after is zero:取向右为正。烟花弹静止，故前后总动量均为零： $$ 0 \;=\; (1.0)(6.0) + (3.0)v \;\Longrightarrow\; v \;=\; -\frac{6.0}{3.0} \;=\; -2.0 \;\text{m/s.} $$ The minus sign means the $3.0$ kg fragment moves left at $2.0$ m/s.负号表示 $3.0$ kg 碎片以 $2.0$ m/s 向左运动。

(b) Total kinetic energy after the burst爆裂后的总动能 M1·A1·A1

$$ KE \;=\; \tfrac{1}{2}(1.0)(6.0)^2 + \tfrac{1}{2}(3.0)(2.0)^2 \;=\; 18 + 6.0 \;=\; 24 \;\text{J.} $$

(c) Where the kinetic energy came from动能的来源 A1·A1

The shell had zero kinetic energy before the burst. The $24$ J came from chemical energy stored in the explosive, released and converted into the kinetic energy of the fragments.爆裂前烟花弹的动能为零。这 $24$ J 来自炸药中储存的化学能，被释放并转化为碎片的动能。

(d) Why total momentum stays zero为何总动量保持为零 A1

The burst forces are internal and act in equal-and-opposite pairs (Newton's third law), so they cannot change the total momentum. The system started at rest with zero momentum, so it must remain zero, even though kinetic energy increased.爆裂力是内力，以等大反向的成对方式作用（牛顿第三定律），故不能改变总动量。系统从静止出发、动量为零，因此必须保持为零，即使动能增加了。
An explosion is recoil in reverse: internal stored energy creates kinetic energy while total momentum is conserved at zero.爆炸是反过来的反冲：内部储存的能量产生动能，而总动量守恒为零。 This is the mirror image of the perfectly inelastic collision. There, two objects merge and kinetic energy is lost; here, one object splits and kinetic energy is gained. Both obey momentum conservation because the forces are internal, but kinetic energy is not conserved in either case. The key insight is that momentum and energy are bookkept separately: a closed system starting from rest must end with zero total momentum, so the $1.0$ kg and $3.0$ kg fragments carry equal-and-opposite momenta ($6.0$ and $-6.0\ \text{kg}\,\text{m/s}$). The lighter fragment moves three times faster, exactly as the rifle-bullet recoil, while the chemical energy released sets the total kinetic energy.这是完全非弹性碰撞的镜像。那里两物体合并、动能损失；此处一物体分裂、动能增加。两者都遵守动量守恒，因为力是内力，但两种情形动能都不守恒。关键洞见是动量与能量分开记账：从静止出发的封闭系统末态总动量必为零，故 $1.0$ kg 与 $3.0$ kg 碎片携带等大反向的动量（$6.0$ 与 $-6.0\ \text{kg}\,\text{m/s}$）。较轻的碎片速率快三倍，恰如枪-子弹反冲，而释放的化学能决定了总动能。
Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3 · §5 Ballistic embedding子弹嵌入 · HS-PS2-2 [10 marks][10 分]

$0.020$ kg bullet at $300$ m/s embeds in a $2.0$ kg block on a frictionless surface; they move off together. (a) Combined speed. (b) KE before and after. (c) Percentage of KE lost. (d) Why so much KE is lost while momentum is conserved.$0.020$ kg 子弹以 $300$ m/s 嵌入无摩擦表面上的 $2.0$ kg 木块；二者一起运动。(a) 共同速率。(b) 嵌入前后的动能。(c) 损失动能的百分比。(d) 为何动能损失如此之大而动量守恒。

Answer:答案：  (a) $v = 2.97\ \text{m/s}$  ·  (b) $KE_i = 900\ \text{J},\ KE_f = 8.91\ \text{J}$  ·  (c) $99\%$ lost损失 $99\%$  ·  (d) perfectly inelastic完全非弹性

(a) Combined speed (conserve momentum, embed)共同速率（守恒动量，嵌入） M1·A1·A1

The bullet embeds, so block and bullet share one velocity $v$:子弹嵌入，故木块与子弹共享一个速度 $v$： $$ (0.020)(300) + 0 \;=\; (0.020 + 2.0)v \;\Longrightarrow\; v \;=\; \frac{6.0}{2.020} \;=\; 2.97 \;\text{m/s.} $$

(b) Kinetic energy before and after嵌入前后的动能 M1·A1·A1

$$ KE_i \;=\; \tfrac{1}{2}(0.020)(300)^2 \;=\; \tfrac{1}{2}(0.020)(90\,000) \;=\; 900 \;\text{J.} $$ $$ KE_f \;=\; \tfrac{1}{2}(2.020)(2.97)^2 \;=\; \tfrac{1}{2}(2.020)(8.82) \;=\; 8.91 \;\text{J.} $$

(c) Percentage of kinetic energy lost损失动能的百分比 M1·A1

$$ \frac{KE_i - KE_f}{KE_i} \times 100\% \;=\; \frac{900 - 8.91}{900} \times 100\% \;=\; 99.0\%. $$

(d) Why so much KE is lost while momentum is conserved为何动能损失如此之大而动量守恒 A1·A1

Momentum is conserved because the forces between bullet and block are internal and equal-and-opposite, with no external net force on the frictionless surface. Kinetic energy is not conserved because this is a perfectly inelastic collision: the friction of the bullet ploughing into the wood converts almost all the kinetic energy into heat and deformation. The loss is extreme because the tiny bullet must drag the much heavier block up to speed, and a large mass mismatch in a stick-together collision dumps most of the kinetic energy.动量守恒是因为子弹与木块间的力是内力、等大反向，无摩擦表面上无外部净力。动能不守恒是因为这是完全非弹性碰撞：子弹钻入木头的摩擦把几乎全部动能转化为热与形变。损失之所以极大，是因为微小的子弹必须把重得多的木块拖到同一速度，而粘合碰撞中巨大的质量差会耗散绝大部分动能。
A bullet embedding in a heavy block is the extreme inelastic case: momentum is fully conserved, yet almost all kinetic energy is lost.子弹嵌入重木块是非弹性碰撞的极端情形：动量完全守恒，而几乎全部动能损失。 The ballistic pendulum and the embedded-bullet block are classic AP-feeder problems precisely because they force the two laws apart. Momentum is conserved across the impact (internal forces only), so the common speed follows from $m_b u_b = (m_b + m_{\text{block}})v$. But fraction of kinetic energy retained equals the mass ratio $m_b /(m_b + m_{\text{block}})$, which here is $0.020 / 2.020 \approx 1\%$. That is why $99\%$ of the energy vanishes into heat and splintered wood. The standard trap is to use energy conservation to find $v$; energy is not conserved here, so only momentum conservation gives the correct common speed.弹道摆与嵌入子弹的木块是经典的 AP 衔接题，正因为它们把两条定律分开。碰撞过程动量守恒（仅内力），故共同速率由 $m_b u_b = (m_b + m_{\text{block}})v$ 给出。但保留的动能比例等于质量比 $m_b /(m_b + m_{\text{block}})$，此处为 $0.020 / 2.020 \approx 1\%$。这就是为何 $99\%$ 的能量化为热与木屑。标准陷阱是用能量守恒求 $v$；此处能量不守恒，故只有动量守恒才能给出正确的共同速率。