线动量

Catch a raw egg with stiff hands and it breaks; draw your hands back as it lands and it survives. The egg stops either way, so its change in momentum is the same. What changes is how long the stopping takes. This unit is about that trade, and about the one quantity a collision can never create or destroy.

Linear momentum 动量 is mass times velocity – a vector pointing the same way as the velocity:

It measures "how hard it is to stop" a moving object. A heavy slow truck and a light fast ball can have the same momentum. Its unit is $\text{kg m/s}$, the same as $\text{N s}$.

A net force acting over time changes momentum. Using a constant force, or its average over the interval, the impulse 冲量 delivered is

This is the impulse–momentum theorem 冲量–动量定理: impulse equals the change in momentum. On a force–time graph 力–时间图, impulse is the area 面积 under the curve. It explains why airbags 安全气囊 and follow-through help – spreading the same momentum change over a longer time reduces the force.

Worked example. A $0.15\ \text{kg}$ ball hits a wall at $20\ \text{m/s}$ and bounces straight back at $15\ \text{m/s}$. The contact lasts $0.020\ \text{s}$. Find the average force on the ball. Take the rebound direction as positive, so $u=-20\ \text{m/s}$ and $v=+15\ \text{m/s}$:

The sign work matters: forgetting that the velocity reverses is the most common mistake here.

Worked example (reading a force–time graph). A $0.050\ \text{kg}$ golf ball at rest is struck by a club. The force rises steadily from zero to $400\ \text{N}$ and falls back to zero, a triangle lasting $0.010\ \text{s}$. The impulse is the triangle's area:

The average force 平均力 is $J/\Delta t=200\ \text{N}$, half the peak 峰值, as it must be for a triangle.

If the net external force 合外力 on a system is zero, its total momentum is conserved 守恒:

Internal forces (like the push between two colliding carts) come in third-law pairs 第三定律力对 and cancel, so they cannot change the total momentum. This is the key tool for collisions 碰撞 and explosions 爆炸 – apply it separately in the $x$ and $y$ directions.

Worked example (recoil 反冲). A $60\ \text{kg}$ skater, initially at rest on frictionless ice, throws a $2.0\ \text{kg}$ ball at $8.0\ \text{m/s}$. Find her recoil speed. The total momentum starts at zero and stays zero:

so she moves at $0.27\ \text{m/s}$ in the opposite direction to the ball – the principle behind rockets 火箭 and guns.

A whole collection of objects can be described by a single center-of-mass velocity 质心速度:

Since total momentum is conserved when no net external force acts, $\vec{v}_{\text{cm}}$ then stays constant – a $v_{\text{cm}}$-versus-time graph is flat unless an outside force acts. An internal collision or explosion never changes it, however violently the parts fly apart.

Momentum is conserved in every collision (with no external force). Kinetic energy 动能 is not:

• In an elastic collision 弹性碰撞, kinetic energy is also conserved (objects bounce apart cleanly).
• In an inelastic collision 非弹性碰撞, some kinetic energy becomes heat or deformation 形变. In a perfectly inelastic 完全非弹性 collision the objects stick together 粘在一起 and move with one common velocity afterward.

Strategy: always write momentum conservation; add energy conservation only if the collision is stated to be elastic. One elastic fact worth memorising: in a 1D elastic collision between equal masses, the two objects simply swap velocities (a moving ball striking an identical stationary 静止的 one stops dead, and the target flies off at the incoming speed).

Worked example. A $1000\ \text{kg}$ car moving at $20\ \text{m/s}$ runs into a stationary $1500\ \text{kg}$ car and they lock together. Find their common speed, and the kinetic energy lost. Momentum conservation gives

Kinetic energy before is $\tfrac12(1000)(20^2)=2.0\times10^{5}\ \text{J}$; after is $\tfrac12(2500)(8.0^2)=8.0\times10^{4}\ \text{J}$. So $1.2\times10^{5}\ \text{J}$ (about $60\%$) is lost to crumpling and heat – momentum is still conserved, but kinetic energy is not.

Worked example (is it elastic?). A $0.50\ \text{kg}$ cart at $2.0\ \text{m/s}$ hits a $1.5\ \text{kg}$ cart at rest. Afterward the first cart moves back at $1.0\ \text{m/s}$ and the second moves forward at $1.0\ \text{m/s}$. Momentum: $0.50(2.0)=1.0$ before and $0.50(-1.0)+1.5(1.0)=1.0\ \text{kg m/s}$ after. Kinetic energy: $\tfrac12(0.50)(2.0^2)=1.0\ \text{J}$ before and $\tfrac12(0.50)(1.0^2)+\tfrac12(1.5)(1.0^2)=0.25+0.75=1.0\ \text{J}$ after. Both are conserved, so this collision is elastic. Test kinetic energy only after momentum checks out.

Every momentum answer starts with one decision: which objects are in the system 系统. Momentum is conserved only for a system on which the net external impulse 外部冲量 is zero, or small enough to ignore.

• Internal forces 内力 act between objects inside the system. They come in Newton's-third-law pairs, give equal and opposite impulses, and cannot change the total.
• External forces 外力 come from outside it. A ball falling alone is not a conserved system, because gravity is external. Ball and Earth together is: the Earth gains exactly the momentum the ball does, in the opposite direction.
• A short collision usually lets you ignore friction 摩擦力 and gravity 重力. The collision forces are huge, but they act for so little time that the outside forces add almost no impulse. So "conserved during the collision" is safe; "conserved for the whole trip" usually is not.

Momentum bar charts: the system total is the same before and after the collision

Momentum bar charts 动量柱状图 are the AP way to show this: one bar per object before and after, plus the system total. The figure shows the car collision above. The system bars match, so the claim "momentum is conserved" is visible, not just asserted.

Worked example (two stages, two laws). A $0.010\ \text{kg}$ bullet at $300\ \text{m/s}$ embeds in a $2.0\ \text{kg}$ block hanging on a string, a ballistic pendulum 冲击摆. How high does the block swing?

• Stage 1, the collision. It is over in a fraction of a second, so the string and gravity add negligible 可忽略的 impulse: momentum is conserved. $0.010(300)=(2.010)\,v$, so $v=1.49\ \text{m/s}$. Kinetic energy is not conserved here: it falls from $450\ \text{J}$ to about $2.2\ \text{J}$.
• Stage 2, the swing. Now gravity acts for a long time, so momentum is no longer conserved, but no energy is lost: $\tfrac12v^2=gh$, so $h=\dfrac{1.49^2}{2(9.8)}=0.11\ \text{m}$.

Using energy in stage 1, or momentum in stage 2, gives a wrong answer. Name the stage before you choose the law.

Momentum is a vector 矢量, so it is conserved separately along each axis 轴.

Worked example. A $2.0\ \text{kg}$ puck moving east at $4.0\ \text{m/s}$ strikes a $3.0\ \text{kg}$ puck at rest. Afterward the $2.0\ \text{kg}$ puck moves at $1.5\ \text{m/s}$ at $60°$ north of east. Find the velocity of the $3.0\ \text{kg}$ puck.

• $x$ (east): $2.0(4.0)=2.0(1.5\cos60°)+3.0\,v_x$, so $8.0=1.5+3.0\,v_x$ and $v_x=2.17\ \text{m/s}$.
• $y$ (north): $0=2.0(1.5\sin60°)+3.0\,v_y$, so $v_y=-0.87\ \text{m/s}$, south.
• Speed $\sqrt{2.17^2+0.87^2}=2.33\ \text{m/s}$, at $\tan^{-1}(0.87/2.17)=22°$ south of east.

Check the energy: $16\ \text{J}$ before and $2.25+8.2=10.4\ \text{J}$ after. Kinetic energy fell, as it must in a real collision. If it had risen, an arithmetic slip would be the first suspect.

Free-response questions often ask you to justify 论证 a conservation claim without numbers. A full-credit answer names the system, the forces and the law. For two carts 小车 colliding on a level, low-friction track 轨道:

Yes, the total momentum of the two-cart system is conserved. During the collision, the only horizontal forces on each cart are the forces the carts exert on each other. These are internal to the system and, by Newton's third law, equal and opposite, so their impulses cancel. Friction from the track is negligible, so the net external impulse on the system is zero.

"Momentum is always conserved" earns nothing: it is false for the wrong system, and it names no reason.

Answer each before reading the answers below. Take $g=9.8\ \text{m/s}^2$.

1. A $0.40\ \text{kg}$ ball moving at $5.0\ \text{m/s}$ is caught and stopped in $0.10\ \text{s}$. Find the average force on it.
2. The same ball is stopped in $0.020\ \text{s}$ instead. Find the force, and explain why a padded glove is used.
3. A $3.0\ \text{kg}$ cart at $4.0\ \text{m/s}$ hits a $1.0\ \text{kg}$ cart at rest and they stick. Find their common speed and the kinetic energy lost.
4. A $50\ \text{kg}$ astronaut floating at rest pushes off a $150\ \text{kg}$ satellite, also at rest, and moves away at $0.60\ \text{m/s}$. Find the satellite's velocity. What happens to the center of mass of the pair?
5. A student says: "In question 3 kinetic energy was lost, so some momentum must have been lost too." Correct the student.

1. $\Delta p=0.40(5.0)=2.0\ \text{kg m/s}$, so $F=2.0/0.10=20\ \text{N}$.
2. $F=2.0/0.020=100\ \text{N}$. The momentum change is fixed; padding makes the stopping last longer, which lowers the force on the hand.
3. $3.0(4.0)=(4.0)v$, so $v=3.0\ \text{m/s}$. Kinetic energy falls from $\tfrac12(3.0)(4.0^2)=24\ \text{J}$ to $\tfrac12(4.0)(3.0^2)=18\ \text{J}$: $6.0\ \text{J}$ is lost.
4. $0=50(0.60)+150\,v$, so $v=-0.20\ \text{m/s}$: $0.20\ \text{m/s}$ in the opposite direction. The total momentum stays zero, so the center of mass stays at rest.
5. Momentum and kinetic energy obey separate laws. With no net external impulse, momentum is conserved in every collision. Kinetic energy can turn into internal energy (heat, sound, deformation), so it may fall while momentum stays exactly the same.

• Momentum is a vector — assign $+$/$-$ signs before adding; a ball that rebounds reverses its velocity, giving a large $\Delta p$.
• Momentum is conserved in every collision (no external force); kinetic energy is conserved only if the collision is stated to be elastic.
• In a perfectly inelastic collision the objects stick and move with one common velocity.
• Impulse $=F\,\Delta t=\Delta p$ = the area under a force–time graph; spreading a collision over a longer time reduces the force (airbags, bending knees).
• For recoil/explosions, set the total momentum equal before and after (often zero before).

Common mistakes

• Subtracting speeds for a rebound. A ball that hits a wall at $20\ \text{m/s}$ and returns at $15\ \text{m/s}$ changes velocity by $35\ \text{m/s}$, not $5\ \text{m/s}$.
• Claiming momentum is conserved for a single object acted on by gravity or friction. Conservation needs a system with zero net external impulse.
• Using kinetic energy conservation for a collision that is not stated (or shown) to be elastic. Write momentum first; test kinetic energy only if the question asks.
• Saying a larger force means a larger impulse. Impulse is force times time: a soft catch has a smaller force but the same impulse.
• Adding the magnitudes of momenta in a 2D collision. Conserve the $x$ and $y$ components separately, then combine.
• Using momentum conservation for the swing of a ballistic pendulum, or energy conservation for the impact. Each stage has its own law.