Conservation of momentum
| English | Chinese | Pinyin |
|---|---|---|
| external force | 外力 | wài lì |
| conservation of momentum | 动量守恒 | dòng liàng shǒu héng |
| collision | 碰撞 | pèng zhuàng |
| explosion | 爆炸 | bào zhà |
| recoil | 反冲 | fǎn chōng |
| elastic | 弹性 | tán xìng |
| inelastic | 非弹性 | fēi tán xìng |
| thrust | 推力 | tuī lì |
Push off and drift apart
- Two ice skaters stand still, then push on each other and glide apart.
- One goes left, one goes right — the total momentum stays zero.
- Nothing pushed from outside, so momentum was conserved.
The principle
- If there is no resultant external force 外力, the total momentum of a system stays constant.
- This is conservation of momentum 动量守恒.
Free-body diagram of a book being pulled on a table
Conservation of momentum
m₁u₁ + m₂u₂ = (m₁+m₂)v
In a collision the total momentum is conserved.
Total momentum stays constant when the resultant ____ force on the system is zero.
With no resultant external force, the system's total momentum is conserved (internal forces come in third-law pairs and cancel).
When it holds
- It works for collisions 碰撞, explosions 爆炸 and recoil 反冲 — any time outside forces cancel.
- In two dimensions, momentum is conserved along each direction on its own.
Forces on a falling object in a fluid
Elastic 弹性 vs inelastic 非弹性
- In every collision, momentum is conserved.
- Elastic: kinetic energy is also conserved (relative speed of approach = of separation).
- Inelastic: some KE becomes heat/sound/deformation. If they stick, it is perfectly inelastic.
Newton's third law in an isolated two-particle system: equal and opposite forces
In an inelastic collision, momentum is still conserved.
Momentum is conserved in every collision (no external force). In an inelastic one it is the kinetic energy that is not conserved.
Compared with collisions in general, what is special about an elastic collision?
All collisions conserve momentum; an elastic collision also conserves kinetic energy.
Solving a head-on collision
- Use signed velocities: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$.
- If it is elastic, add $u_1 - u_2 = -(v_1 - v_2)$ to get a second equation.
A $2.0\ \text{kg}$ trolley at $3.0\ \dfrac{\text{m}}{\text{s}}$ hits a stationary $1.0\ \text{kg}$ trolley and they stick. Find their common velocity.
Momentum: $2.0 \times 3.0 = (2.0 + 1.0)\,v$, so $v = \dfrac{6.0}{3.0} = 2.0\ \dfrac{\text{m}}{\text{s}}$.
Collisions in two dimensions
- Split each velocity into perpendicular components.
- Apply conservation of momentum along each axis separately.
In two dimensions, resolve along two axes and conserve momentum on each one separately
In two dimensions, momentum is conserved separately along each perpendicular direction.
Yes — resolve into components and apply conservation of momentum to each axis on its own.
Rockets and recoil
- A rocket throws gas one way and is pushed the other way (Newton's third law).
- Thrust 推力 $F = \dot{m}\,u$ — mass thrown per second times its speed.
A rocket ejects gas at $100\ \dfrac{\text{m}}{\text{s}}$ with a mass-flow rate of $2.0\ \dfrac{\text{kg}}{\text{s}}$. What thrust does it feel?
Thrust $F = \dot{m}\,u = 2.0 \times 100 = 200\ \text{N}$.
You've got it
- no external resultant force → total momentum is conserved
- momentum is conserved in all collisions; elastic ones also conserve KE
- solve 1-D collisions with $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ (use signs)