# Mechanics11/Seite13/Loesung4.17

4.17 Autoscooter: Lösung

a.)
$m_1$ = 160kg . . . . . . . $v_1$ = 5,0 $\frac{m}{s}$

$m_2$ = 90kg . . . . . . . . $v_2$ = 1,0 $\frac{m}{s}$

$u_1$ = $\frac{(m_1 - m_2)* v_1 + 2m_2v_2}{m_1 + m_2}$ = $\frac{(160kg - 90kg)* 5,0m/s + 2*90kg * 1,0m/s}{160kg + 90kg}$ = 2,12$\frac{m}{s}$

$u_2$ = $\frac{(m_2 - m_1)* v_2 + 2m_1v_1}{m_1 + m_2}$ = $\frac{(90kg - 160kg)* 1,0m/s + 2*160kg * 5,0m/s}{160kg + 90kg}$ = 6,12$\frac{m}{s}$

b.)
$E_Kin(davor)$ = $E_Kin(danach)$
$\frac{1}{2}$$m_1$$v^2_1$ + $\frac{1}{2}$$m_2$$v^2_2$ = $\frac{1}{2}$$m_1$$u^2_1$ + $\frac{1}{2}$$m_2$$u^2_2$

$\frac{1}{2}$ * 160kg * (5,0 m/s)² + $\frac{1}{2}$ * 90kg * (1,0m/s)² = $\frac{1}{2}$ * 160kg * (2,12 m/s)² + $\frac{1}{2}$ * 90kg * (6,12m/s)²

2000J + 45J = 359,6J + 1685,4J

2045J = 2045J

c.)
$m_1$ = 160kg . . . . . . . $v_1$ = 1,0 $\frac{m}{s}$

$m_2$ = 90kg . . . . . . . . $v_2$ = -5,0 $\frac{m}{s}$

$u_1$ = $\frac{(m_1 - m_2)* v_1 + 2m_2v_2}{m_1 + m_2}$ = $\frac{(160kg - 90kg)* 1,0m/s + 2*90kg * (-5,0m/s)}{160kg + 90kg}$ = -3,32$\frac{m}{s}$

$u_2$ = $\frac{(m_2 - m_1)* v_2 + 2m_1v_1}{m_1 + m_2}$ = $\frac{(90kg - 160kg)* (-5,0m/s) + 2*160kg * 1,0m/s}{160kg + 90kg}$ = 2,68$\frac{m}{s}$

--Ida.lein 08:11, 18 April 2008 (UTC)

d)

Impuls davor:

m = 45kg v = 1 m/s

p = v2 * m

p = 45J

Impuls danach:

m = 45kg v =