📚 Explosions and Crash Landings: Applying Momentum and Impulse | 爆炸与坠毁着陆：动量与冲量的应用
Explosions and crash landings are dramatic events that showcase the power of momentum and impulse. In A-Level Physics, these phenomena are not just spectacular; they illustrate fundamental principles of conservation of momentum, impulse, and force. Whether a rocket blasting off or a car crumpling against a barrier, the same physics governs the outcome. This article unpacks the core concepts, walks through worked examples, and explains how engineers design safety systems to protect lives.
爆炸和坠毁着陆是展现动量与冲量威力的戏剧性事件。在A-Level物理中，这些现象不仅壮观，更生动地阐释了动量守恒、冲量和力的基本原理。无论是火箭发射还是汽车撞向护栏，起作用的都是相同的物理规律。本文将剖析核心概念，带你完成例题解析，并说明工程师如何设计安全系统来保护生命。
1. Momentum and Impulse Basics | 动量与冲量基础
Momentum p is the product of an object' object when a force F acts over a time interval Δt: J = F Δt = Δp. The unit of momentum is kg m s⁻l and impulse has the same unit, or equivalently N s.
动量p是物体质量m与其速度v的乘积，写作p = mv。它是一个矢量，方向至关重要。冲量J定义为力F在一段时间Δt内作用于物体时动量的变化：J = F Δt = Δp。动量的单位是kg m s⁻l，冲量的单位与此相同，或者等效为N s。
p = mv
J = F Δt = Δp = mv - mu
The area under a force–time graph equals the impulse delivered. For a constant force, it is simply the product F Δt; for a varying force, integration or area estimation is needed.
力–时间图像下的面积等于传递的冲量。对于恒力，它就是乘积F Δt；对于变力，则需要积分或面积估算。
2. Conservation of Momentum | 动量守恒
The total momentum of an isolated system remains constant in the absence of external resultant forces. This principle applies to both collisions and explosions. In an explosion, the system has zero total momentum before the event, so after the event the vector sum of momenta of all fragments must be zero. In a crash landing, the initial momentum of the vehicle is shared with the Earth, but because Earth's mass is enormous, its velocity change is negligible.
在没有外部合外力的情况下，孤立系统的总动量保持不变。这一原理既适用于碰撞，也适用于爆炸。在爆炸中，系统在爆炸前的总动量为零，因此爆炸后所有碎片的动量矢量之和必须为零。在坠毁着陆中，车辆的初始动量会与地球共享，但由于地球质量极其巨大，其速度变化可以忽略不计。
presents initial velocities and v represents final velocities. The equation is applied vectorially.
数学上，对于两体系统：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂，其中u代表初速度，v代表末速度。需要以矢量形式运用此方程。
3. Explosions: The Big Bang of Momentum | 爆炸：动量的“大爆发”
. The fragments move in opposite directions with speeds inversely proportional to their masses. Kinetic energy increases – al potential energy into kinetic energy; momentum is conserved, but kinetic energy is not.
设想一个静止的炸弹爆炸成两碎片。初始总动量为零。根据守恒，m₁v₁ + m₂v₂ = 0，因此v₂ = - (m₁/m₂) v₁。碎片向相反方向运动，速率与质量成反比。动能增加——爆炸将化学势能转化为动能；动量守恒，但动能并不守恒。
When an explosion ejects three or more fragments, vector addition is required. The sum of all fragment momenta must equal the initial momentum (usually zero). This idea is used in analysing firearm recoil or rocket propulsion.
当爆炸产生三块或更多碎片时，需要进行矢量加法。所有碎片动量的总和必须等于初始动量（通常为零）。这一思路用于分析枪械反冲或火箭推进。
4. One-Dimensional Explosions | 一维爆炸
In a one-dimensional scenario, all velocities lie along the same straight line. A common exam problem: a stationary firework rocket explodes into two pieces of masses 0.40 kg and 0.60 kg. The heavier piece travels at 12 m s⁻l to the right. Find the velocity of the lighter piece.
在一维情形中，所有速度都沿同一直线。常见的考题：一个静止的烟花火箭爆炸成0.40 kg和0.60 kg的两块。较重的碎片以12 m s⁻l的速度向右运动。求较轻碎片的速度。
0 = (0.40 kg)v₁ + (0.60 kg)(12 m s⁻l) ⇒ v₁ = -18 m s⁻l
The negative sign indicates motion to the left. The lighter piece has a higher speed, consistent with the inverse mass relationship.
负号表示向左运动。较轻的碎片具有更高的速率，这与质量反比关系一致。
5. Two-Dimensional Explosions | 二维爆炸
When fragments scatter in a plane, we resolve momentum into perpendicular components (usually x and y). Suppose a stationary object explodes into three pieces with momenta shown: piece A (0.50 kg, 4.0 m s⁻l east), piece B (0.30 kg, 6.0 m s⁻l north). Find the momentum vector of the third piece (mass 0.20 kg).
当碎片在平面内飞散时，我们将动量分解为正交分量（通常为x和y方向）。假设一个静止物体爆炸成三块，已知动量：A块（0.50 kg，4.0 m s⁻l 向东），B块（0.30 kg，6.0 m s⁻l 向北）。求第三块（质量0.20 kg）的动量矢量。
Initial momentum = 0. Total momentum after: pₐ = 0.50×4.0 = 2.0 kg m s⁻l east, pB = 0.30×6.0 = 1.8 kg m s⁻l north. For the total to be zero, pC must be 2.0 kg m s⁻l west and 1.8 kg m s⁻l south. Magnitude = √2.0² + 1.8² and velocity of C = 2.69/0.20 = 13.5 m s⁻l at an angle tan⁻l(1.8/2.0) = 42° south of west.
初始动量为零。爆炸后总动量：pₐ = 0.50×4.0 = 2.0 kg m s⁻l 向东，pB = 0.30×6.0 = 1.8 kg m s⁻l 向北。要使总动量为零，pC必须为2.0 kg m s⁻l 向西、1.8 kg m s⁻l 向南。大小为√2.0²+1.8² ≈ 2.69 kg m s⁻l，C的速度为2.69/0.20 = 13.5 m s⁻l，方向为西偏南tan⁻l(1.8/2.0)=42°。
6. Crash Landings: Bringing Vehicles to Rest | 坠毁着陆：让车辆停下
A crash landing, such as an aircraft hitting the ground or a car colliding with a barrier, involves rapid deceleration. The vehicle's momentum is reduced to near zero in a very short time, generating huge forces. The same impulse principle applies: the change in momentum (impulse) is fixed by the initial speed and mass, so a shorter impact time leads to a larger average force.
坠毁着陆，例如飞机撞击地面或汽车撞上护栏，涉及急速减速。车辆的动量在极短时间内减至接近零，产生巨大的力。同样的冲量原理在起作用：动量的变化（冲量）由初速度和初质量决定，因此撞击时间越短，平均力越大。
m s⁻l. Average force F = Δp/Δt = -30000/0.10 = -300000 N. That is over 30 tonnes of force!
如果一辆1500 kg的汽车以20 m s⁻l行驶，撞墙后在0.10 s内停下，Δp = 1500 × (0-20) = -30000 kg m s⁻l。平均力F = Δp/Δt = -30000/0.10 = -300000 N。这相当于30多吨的力！
7. Impulse and Impact Force | 冲量与撞击力
hat to reduce the force on an occupant, we must increase the impact time Δt for a given momentum change. This is the physics behind all passive safety systems in vehicles. The force experienced is inversely proportional to the duration over which the velocity change occurs.
冲量方程F = Δp/Δt表明，对于给定的动量变化，要减小乘员承受的力，必须增加撞击时间Δt。这是车辆上所有被动安全系统背后的物理原理。人体感受到的力与速度变化发生的时间成反比。
Force–time graphs for rigid and padded impacts illustrate this: a rigid collision shows a tall, narrow spike, whereas a padded or crumpling collision spreads the impulse over a longer interval, giving a lower maximum force. The area under both graphs (the impulse) is identical.
刚性和缓冲撞击的力–时间图能说明这一点：刚性碰撞表现为高而窄的尖峰，而带有缓冲或溃缩的碰撞将冲量分布在更长的时间间隔内，从而使最大力降低。两个图形下的面积（冲量）是相同的。
8. Crumple Zones and Force Reduction | 碰撞吸能区与力衰减
ntrolled manner, they increase the stopping distance and thus the impact time, reducing the average force transmitted to the passenger compartment. The momentum change remains the same, but Δt is larger, so F = Δp/Δt is smaller.
吸能区是车辆前后部分设计的结构区域，在碰撞时会发生塑性变形。通过受控溃缩，它们增加了停车距离，从而延长了撞击时间，降低了传递到乘员舱的平均力。动量变化保持不变，但Δt更大，因此F = Δp/Δt更小。
In a crash test, a car without a crumple zone stops in 0.05 s, producing a force of 400 kN on a dummy. With a crumple zone that extends the stopping time to 0.15 s, the average force drops to about 133 kN for the same speed and mass. This drastic reduction saves lives.
在碰撞测试中，没有吸能区的汽车在0.05 s内停下，对假人产生400 kN的力。有了吸能区，停车时间延长至0.15 s，相同速度和质量下平均力降至约133 kN。这一显著下降足以挽救生命。
9. Airbags and Seatbelts | 安全气囊与安全带
Airbags further increase the time over which a passenger’s momentum is brought to zero. Without restraints, a passenger continues forward at the original speed until striking the dashboard or windscreen in a few milliseconds. An airbag deploys in about 0.05 s and deflates gradually, providing a longer deceleration interval and spreading the force over a larger area of the body.
安全气囊进一步延长了乘客动量减至零的时间。如果没有约束装置，乘客会以原速继续向前，在几毫秒内撞上仪表盘或挡风玻璃。安全气囊在约0.05 s内展开，并逐渐泄气，提供了更长的减速时间，并将力分散到身体的更大面积上。
Seatbelts also play a dual role: they immediately engage to prevent ejection and they are slightly stretchy, adding precious milliseconds to the deceleration. The combination of seatbelt pretensioners, load limiters, and airbags works together to keep the occupant’s deceleration below critical thresholds.
安全带也起着双重作用：它们立即收紧以防止乘客飞出，并且略微具有延展性，为减速过程增添了宝贵的几毫秒。预紧式安全带、限力器与安全气囊的组合协同作用，将乘员的减速度控制在临界阈值以下。
10. Energy Dissipation in Collisions | 碰撞中的能量耗散
While momentum is always conserved in isolated systems, kinetic energy may or may not be conserved. Explosions and many crash landings are inelastic; kinetic energy decreases as it is converted into sound, heat, and deformation work. In a perfectly inelastic collision, objects stick together and kinetic energy loss is maximum.
虽然孤立系统中的动量总是守恒，但动能不一定守恒。爆炸和许多坠毁着陆都是非弹性的；动能因转化为声能、热能和变形功而减少。在完全非弹性碰撞中，物体粘在一起，动能损失最大。
ergy is dissipated in deforming metal and generating heat. Crumple zones are designed to absorb as much energy as possible in a controlled manner.
在汽车发生溃缩的碰撞中，初始动能KE = ½mv²。若m=1200 kg, v=15 m s⁻l，动能=135 kJ。几乎全部能量都在金属变形和生热过程中耗散。吸能区被设计为以受控方式尽可能多地吸收能量。
11. Worked Example: Explosion Problem | 例题：爆炸问题
Problem: A 2.0 kg firework at rest explodes into three fragments. Fragment 1 (0.60 kg) moves north at 8.0 m s⁻l. Fragment 2 (0.40 kg) moves east at 10 m s⁻l. Calculate the velocity (magnitude and direction) of the third fragment (mass 1.0 kg).
问题：一个静止的2.0 kg烟花爆炸成三块碎片。碎片1（0.60 kg）以8.0 m s⁻l向北运动；碎片2（0.40 kg）以10 m s⁻l向东运动。计算第三块碎片（质量1.0 kg）的速度大小和方向。
.40×10 = 4.0 kg m s⁻l east. For zero total, p₃ must be 4.8 south and 4.0 west. Magnitude = √4.8²+4.0² 6.25/1.0 = 6.25 m s⁻l. Direction θ = tan⁻l(4.8/4.0) = 50° south of west (or bearing 230°).
解答：初始动量为零。p₁ = 0.60×8.0 = 4.8 kg m s⁻l 向北；p₂ = 0.40×10 = 4.0 kg m s⁻l 向东。要使总动量为零，p₃须为4.8向南、4.0向西。大小=√4.8²+4.0²=6.25 kg m s⁻l。碎片3的速度 = p₃/m₃ = 6.25/1.0 = 6.25 m s⁻l。方向θ = tan⁻l(4.8/4.0)=50°，即西偏南50°（或方位角230°）。
12. Worked Example: Crash Landing | 例题：坠毁着陆
Problem: A 2500 kg SUV crashes head-on into a solid concrete wall at 18 m s⁻l and stops in 0.12 s. Calculate the average force exerted on the vehicle. If a passenger of mass 75 kg was not wearing a seatbelt and struck the dashboard, stopping in 0.03 s, estimate the force on the passenger.
问题：一辆2500 kg的SUV以18 m s⁻l的速度正面撞上一堵坚固的混凝土墙，在0.12秒内停下。计算车辆受到的平均力。如果一名75 kg的乘客未系安全带，撞上仪表盘并在0.03秒内停下，估算乘客所受的力。
te to motion). For the passenger, initial velocity same as SUV (18 m s⁻l). Δp = 75×(-18) = -1350 kg m s⁻l, Δt = 0.03 s, so F = -1350/0.03 = -45000 N. This is over 4.5 tonnes of force on the passenger's body, highlighting the need for seatbelts and airbags.
解答：对SUV，Δp = 2500×(0-18) = -45000 kg m s⁻l。平均力F = Δp/Δt = -45000/0.12 = -375000 N（负号表示方向与运动相反）。对乘客而言，初速度与SUV相同（18 m s⁻l）。Δp = 75×(-18) = -1350 kg m s⁻l，Δt = 0.03 s，因此F = -1350/0.03 = -45000 N。这相当于乘客身体承受超过4.5吨的力，突显了安全气囊与安全带的必要性。
Published by TutorHao | 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