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 =