# 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

$u = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \frac{m}{s}$

$u = \frac{1700*15 + 650*10}{1700 + 650} \frac{m}{s}$

$u = \frac{32000}{2350} \frac{m}{s}$

$u = 14 \frac{m}{s}$

b)

$E_{kin} = \frac{1}{2} m_1 v_1^2 = 191 kJ$

$E'_{kin} = \frac{1}{2} m_2 u^2 = 158 kJ$

>>> $\Delta E = 33 kJ \frac{}{}$

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