线动量

Linear momentum 动量 is mass 质量 times velocity 速度:

It is a vector 矢量 pointing along the velocity, and it is the quantity for analysing collisions 碰撞 and explosions 爆炸. A collection of objects can be treated as one system 系统 moving with the velocity of its center of mass 质心:

so the system's total momentum is its total mass times $\vec{v}_{\text{cm}}$ – one object's worth of bookkeeping for any number of parts.

Newton's second law 牛顿第二定律 for a system containing the same particles is:

This reduces to $m\vec{a}$ for constant total mass. If matter crosses the chosen boundary, include the momentum it carries. Sand arriving with zero horizontal speed and leaving on a conveyor belt 传送带 at speed $v$ needs horizontal force $F=\dot{m}v$, where $\dot{m}$ is the mass arriving per second. A rocket 火箭 gains momentum as exhaust carries momentum away; applying $d(m\vec{v})/dt$ to the rocket alone without this momentum flow is incorrect. The impulse 冲量 delivered by a force is its time integral 对时间的积分, and the impulse–momentum theorem 冲量-动量定理 says it equals the change in momentum:

For a constant net force 合力, the impulse points in its direction. Read it off graphs both ways:

• On a force–time graph 力–时间图, impulse is the area 面积 under the curve.
• On a momentum–time graph, the net force is the slope 斜率 at each instant 时刻.

Spreading the same $\Delta p$ over a longer time lowers the force – that is why airbags 安全气囊, crumple zones 溃缩区, and soft landings work, and why you bend your knees when you land.

Worked example. A time-varying force $F(t)=10t\ \text{N}$ acts on a $2.0\ \text{kg}$ object for $2.0\ \text{s}$. The impulse is $J=\displaystyle\int_0^2 10t\,dt=\big[5t^2\big]_0^2=20\ \text{N}\cdot\text{s}$, so the speed changes by $\Delta v=J/m=10\ \text{m/s}$.

Internal forces 内力 come in Newton's-third-law pairs, so they cancel inside any system: they can shuffle momentum between parts but never change the total. For a fixed set of particles, total momentum changes only through a net external force ($\vec{J}=\Delta\vec{p}$). So, with zero net external force 合外力, total momentum is conserved 守恒:

Momentum is conserved over a collision when the net external impulse is negligible; choose a system containing both colliding objects. Apply conservation separately along each axis 坐标轴; AP asks for quantitative work in one or two dimensions.

Worked example (explosion). A $6.0\ \text{kg}$ shell at rest splits into a $2.0\ \text{kg}$ piece moving at $9.0\ \text{m/s}$ east and a $4.0\ \text{kg}$ piece. Total momentum stays zero, so the heavy piece moves west at $v=\dfrac{2.0(9.0)}{4.0}=4.5\ \text{m/s}$. The kinetic energy came from stored (chemical or spring) energy – momentum conservation does not require kinetic-energy conservation.

When the net external impulse is negligible, collisions conserve momentum; they differ in what happens to the kinetic energy 动能:

Strategy: always write momentum conservation first; add the kinetic-energy equation only when the problem says "elastic". Two useful elastic facts: equal masses in a 1D elastic collision simply exchange velocities, and in the center-of-mass frame 质心参考系 each object just reverses its velocity.

Worked example. A $1000\ \text{kg}$ car at $20\ \text{m/s}$ strikes a stationary $1500\ \text{kg}$ car and they lock together – perfectly inelastic. Momentum: $v=\dfrac{1000(20)}{2500}=8.0\ \text{m/s}$. Kinetic energy falls from $2.0\times10^5\ \text{J}$ to $\tfrac12(2500)(8.0)^2=8.0\times10^4\ \text{J}$: about $60\%$ is lost, even though momentum is exactly conserved.

Worked example (2D). A puck moving east at $4.0\ \text{m/s}$ strikes an identical puck at rest; after the glancing hit, one moves at $2.0\ \text{m/s}$ at $60^\circ$ north of east. Conserve each axis: east–west, $m(4.0)=m(2.0)\cos60^\circ+mv_x$, so $v_x=3.0\ \text{m/s}$; north–south, $0=m(2.0)\sin60^\circ-mv_y$, so $v_y=1.7\ \text{m/s}$. The second puck moves at $\sqrt{3.0^2+1.7^2}=3.5\ \text{m/s}$, about $30^\circ$ south of east.

Exam skill. On FRQs, justify with the condition, not the slogan: "the net external force on the two-puck system is zero during the collision, so its total momentum is constant." If asked whether the collision is elastic, compute the kinetic energy before and after and compare – never assume.

• Impulse equals the momentum change: $\vec J=\int \vec F\,dt=\Delta\vec p$, and it is the area under a force–time graph.
• Momentum is conserved whenever the net external force is zero — always the go-to for collisions.
• Distinguish elastic (kinetic energy conserved) from inelastic (kinetic energy decreases) collisions; in a perfectly inelastic collision the objects stick and move with one velocity.
• Apply conservation to each component 分量 (x and y) separately in 2-D.
• Connect to center of mass: total momentum $=M\vec v_{cm}$.

Common mistakes

• Conserving momentum while an external impulse is large, for example during a long push from the ground. Conservation needs a negligible net external impulse over the interval.
• Calling a collision elastic because momentum is conserved. Check kinetic energy too.
• Reading impulse as the peak force times the time. Impulse is the area under the $F$–$t$ graph.
• Adding momenta in two dimensions as magnitudes. Conserve the $x$ and $y$ components separately.
• Forgetting that internal forces cannot move the centre of mass: an explosion leaves $\vec v_{cm}$ unchanged.