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)
Fkin = 1/2 m * v2
FbusVorher = 0.5 * 1700kg * (15 m/s)2 = 190000N
FbusNachher = 0.5 * 1700kg * (13.6 m/s)2 = 160000N
FbusVerloren = 190000N - 160000N = 30000N
c)
Kinetische Energie wird umgewandelt in Deformationsarbeit (enthält Erwärmung)
EkinBus = 30kJ
EkinSmart = 60kJ
=> Delta E = 30kJ
-> WD = 3kJ
Zweiter Lösungsweg Kinetische energie vor dem unfall: Ekinbus + Ekinsmart - Ekinsmart und bus nach unfall = Edeformation
d)
p' = p1 - p2 = 19000Ns
u = p' / (m1 + m2) = 8.1 m/s
FbusNachher = 0.5 * 1700kg * (8.1 m/s)2 = 56000N
FbusVerloren = 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)