冲量与动量：冲击档案
引入| English | 释义 | 拼音 |
|---|---|---|
| impulse/ˈɪmpʌls/ | 冲量 | chōng liàng |
| linear momentum/ˈlɪnɪə məʊˈmentəm/ | 动量 | dòng liàng |
| coefficient of restitution/ˌkəʊɪˈfɪʃənt ɒv rɪstɪˈtjuːʃn/ | 恢复系数 | huī fù xì shù |
| steady flow | 稳态流 | wěn tài liú |
Two frozen pictures
- The certified impact lasts 8.0 ms - too fast to integrate forces, exactly right for their time integral: impulse 冲量 $\mathbf{J} = \Delta\mathbf{p}$, the change of linear momentum 动量.
- The till arrives at 5.943 m/s and leaves at $e \times 5.943 = 2.080$ m/s with coefficient of restitution 恢复系数 $e = 0.35$.
*Two pictures and one arrow of impulse between them
Rebound speed (m/s, three decimals)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
e v.
Rebound height (m, four decimals)?
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v² / 2g.
The coefficient of restitution is the ratio of:
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Speeds at the contact, not energies.
The numbers that size the welds
- $J = m(v + ev) = 2.5 \times 1.35 \times 5.943 = 20.06$ N·s.
- Average force $= J/\Delta t = 2507$ N over 8.0 ms; gravity's impulse in the same window is 1.0% of J.
A 2.5 kg till hits at 5.943 m/s with e = 0.35: impulse magnitude (N s, two decimals)?
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J = m(1+e)v.
Same impact over 8.0 ms: average force (N, nearest integer)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
J / 0.008.
Momentum keeps, energy does not
- The marshalling test: 2.5 kg at 3.00 m/s into 4.0 kg at rest, $e = 0.60$ → 0.046 and 1.846 m/s; momentum 7.500 kept to the digit, 4.431 J of kinetic energy gone.
- Steady flow: the rig's fan thrusts $\dot{m}v = 0.120 \times 8.50 = 1.020$ N.
Kept, lost, or transferred?
Classify each quantity through the certified impact.
2.5 kg at 3.00 m/s into 4.0 kg at rest, e = 0.60: the 4.0 kg cart leaves at (m/s, three decimals)?
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Momentum + separation equations.
Same collision: kinetic energy lost (J, three decimals)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
11.25 - 6.82.
In the certified impact, kinetic energy is conserved because momentum is.
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
Only e = 1 conserves energy; here 87.8% was absorbed.
Three momentum slips
- Losing the sign. Down-in and up-out subtract: $J = m(v + ev)$.
- Applying e to velocities. It applies to the separation/approach pair.
- Concluding energy is conserved. Only e = 1 does that; 87.8% of arrival energy was absorbed here.
Recap
- $\sum \mathbf{J} = \Delta\mathbf{p}$; $e$ = separation over approach.
- $J = 20.06$ N·s, average force 2507 N over 8 ms; rebound height 0.2205 m.
- Carts: 0.046 and 1.846 m/s, 4.431 J lost; fan thrust 1.020 N.