Mechanics11/Seite13/Loesung4.16

m₁ = 1700kg

v₁ = 15 m/s

m₂ = 650kg

v₂ = 10 m/s

a)

p₁ = m₁ * v₁ = 1700kg * 15 m/s = 25500Ns

p₂ = m₂ * v₂ = 650kg * 10 m/s = 6500Ns

p'= p₁ + p₂ = 25500Ns + 6500Ns = 32000Ns

p'= (m₁ + m₂) * u => u= p' / (m₁ + m₂) = 13,6m/s

b) F_kin = 1/2 m * v²

F_busVorher = 0.5 * 1700kg * (15 m/s)² = 190000N

F_busNachher = 0.5 * 1700kg * (13.6 m/s)² = 160000N

F_busVerloren = 190000N - 160000N = 30000N

c) Kinetische Energie wird umgewandelt in Deformationsarbeit (enthält Erwärmung)

E_kinBus = 30kJ

E_kinSmart = 60kJ

=> Delta E = 30kJ

  -> WD = 3kJ

Zweiter Lösungsweg Kinetische energie vor dem unfall: Ekin_bus + Ekin_smart - Ekin_{smart und bus nach unfall} = Edeformation

d) p' = p₁ - p₂ = 19000Ns

u = p' / (m₁ + m₂) = 8.1 m/s

F_busNachher = 0.5 * 1700kg * (8.1 m/s)² = 56000N

F_busVerloren = 190000N - 56000N = 134000N

von CarO, Chantal, Kenny 08:11, 18 April 2008 (UTC)

FatF's Lösung

geg.:

• m₁ = 1700 kg
• m₂ = 650 kg
• v₁ = 54 km/h = 15 m/s
• v₂ = 36 km/h = 10 m/s

Impulsverhalten

a)

p = p'

p₁ + p₂ = p'

m₁v₁ + m₂v₂ = (m₁ + m₂) * u

[math]u = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \frac{m}{s}[/math]

[math]u = \frac{1700*15 + 650*10}{1700 + 650} \frac{m}{s}[/math]

[math]u = \frac{32000}{2350} \frac{m}{s}[/math]

[math]u = 14 \frac{m}{s}[/math]

b)

[math]E_{kin} = \frac{1}{2} m_1 v_1^2 = 191 kJ[/math]

[math]E'_{kin} = \frac{1}{2} m_2 u^2 = 158 kJ[/math]

>>> [math]\Delta E = 33 kJ \frac{}{}[/math]

-- FatF 07:57, 18 April 2008 (UTC)