Momentum (Further Mechanics)
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| momentum/məʊˈmentəm/ | động lượng |
| collision/kəˈlɪʒn/ | va chạm |
| conservation of momentum/ˌkɒnsəˈveɪʃn ɒv məʊˈmentəm/ | bảo toàn động lượng |
| coefficient of restitution/ˌkəʊɪˈfɪʃənt ɒv rɪstɪˈtjuːʃn/ | hệ số phục hồi |
| perfectly elastic/ˈpɜːfektlɪ ɪˈlæstɪk/ | đàn hồi hoàn toàn |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | động năng |
| perfectly inelastic/ˈpɜːfektlɪ ɪnɪˈlæstɪk/ | không đàn hồi hoàn toàn |
Newton's cradle and the perfect bounce
- Drop one ball of a Newton's cradle and exactly one flies off the far end. Why one, not two?
- The answer is two conservation ideas working together: momentum 动量, and how much bounce a collision 碰撞 has.
Conservation of momentum 动量守恒
- In any collision (or explosion), with no external force, total momentum is conserved:
Momentum before equals momentum after — the single most useful equation in collisions.
Conservation of momentum
m₁u₁ + m₂u₂ = (m₁+m₂)v
Two bodies collide — the total momentum before equals the total after.
A 2 kg body at 3 m/s hits a stationary 1 kg body and they stick together. By momentum conservation 2(3) = (2+1)v, find v (m/s).
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6 = 3v, so v = 2 m/s.
The coefficient of restitution 恢复系数
- How bouncy is the collision? The coefficient of restitution $e$ compares the speeds:
- $e = 1$: perfectly elastic 完全弹性 (no kinetic energy 动能 lost). $e = 0$: perfectly inelastic 完全非弹性 (bodies stick together).
The coefficient of restitution $e$ compares how fast the bodies separate with how fast they approached
A Newton's cradle demonstrates conservation of momentum in collisions
The coefficient of restitution e always lies in which range?
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e runs from 0 (perfectly inelastic) to 1 (perfectly elastic).
The coefficient of restitution e = speed of separation ÷ speed of ______.
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e = separation speed / approach speed.
If two bodies coalesce (stick together) on impact, the coefficient of restitution is:
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They move off together (zero separation speed), so e = 0 — perfectly inelastic.
Using both equations
- Two unknowns (the final speeds) need two equations: momentum conservation and Newton's restitution law ($\text{separation} = e \times \text{approach}$).
- Solve them together to find both final velocities.
Kinetic energy is only conserved when $e = 1$. For any $e < 1$ some kinetic energy is lost (to heat and sound) — so never assume energy is conserved in a collision unless told it is perfectly elastic.
- In a direct or oblique impact, total linear momentum is conserved (conservation of linear momentum) and Newton's experimental law gives the coefficient of restitution.
Kinetic energy is conserved in every collision where momentum is conserved.
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False. Momentum is conserved in all collisions, but kinetic energy is conserved only when e = 1 (perfectly elastic); otherwise some is lost.
You've got it
- total momentum is conserved: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
- $e = \dfrac{\text{separation speed}}{\text{approach speed}}$, with $0 \leq e \leq 1$
- combine momentum + restitution to find both final speeds; KE conserved only if $e = 1$
Write from a common speed of 3 metres per second
- A 3 kg body at 4 metres per second hits a stationary 1 kg body and they stick together.
- Find the common speed. Kinetic energy is conserved only when the coefficient of restitution is 1.
A 3 kg body at 4 metres per second hits a stationary 1 kg body and they stick together. What is their common speed, and why is kinetic energy not conserved?
Momentum is conserved in the impact. An inelastic collision keeps momentum and loses kinetic energy.
What is the common speed?
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3 times 4 is 12, and 12 divided by 4 is 3 metres per second. Other options — 2 metres per second (the lesson speed for a 2 kg body at 3 metres per second); 12 (the momentum before impact).