Umstellen einer Formel

Question

Aufgabe:

1.) \( \left(v_{1}+\frac{m_{2}}{m_{1}}\left(v_{2}-u_{2}\right)\right)^{2}=v_{1}^{2}+\frac{m_{2}}{m_{1}}\left(v_{2}^{2}-u_{2}^{2}\right) \) \( \downarrow_{\text {nach }} u_{2} \) umstellen
2.) \( u_{2}=\frac{\left(m_{2}-m_{1}\right) v_{2}+2 m_{2} \cdot v_{2}}{m_{1}+m_{2}} \)

Problem/Ansatz:

brauche die Rechnungsschritte wie man von der ersten Formel zu der zweiten kommt

Answer 1

\( \left(v_{1}+\frac{m_{2}}{m_{1}}\left(v_{2}-u_{2}\right)\right)^{2}=v_{1}^{2}+\frac{m_{2}}{m_{1}}\left(v_{2}^{2}-u_{2}^{2}\right) \)

Erst mal links binomische Formel :

\( v_{1}^2+2v_1\frac{m_{2}}{m_{1}}\left(v_{2}-u_{2}\right)+\frac{m_{2}^2}{m_{1}^2}\left(v_{2}-u_{2}\right)^2=v_{1}^{2}+\frac{m_{2}}{m_{1}}\left(v_{2}^{2}-u_{2}^{2}\right) \)

\( 2v_1\frac{m_{2}}{m_{1}}\left(v_{2}-u_{2}\right)+\frac{m_{2}^2}{m_{1}^2}\left(v_{2}-u_{2}\right)^2=\frac{m_{2}}{m_{1}}\left(v_{2}^{2}-u_{2}^{2}\right) \)

rechts die 3. binomi. Formel.

\( 2v_1\frac{m_{2}}{m_{1}}\left(v_{2}-u_{2}\right)+\frac{m_{2}^2}{m_{1}^2}\left(v_{2}-u_{2}\right)^2=\frac{m_{2}}{m_{1}}\left(v_{2}-u_{2}\right)\left(v_{2}+u_{2}\right) \)

Falls klar ist, dass v2 - u2 nicht 0 ist, dadurch dividieren

\( 2v_1\frac{m_{2}}{m_{1}}+\frac{m_{2}^2}{m_{1}^2}\left(v_{2}-u_{2}\right)=\frac{m_{2}}{m_{1}}\left(v_{2}+u_{2}\right) \)

Und noch durch m2 / m1

\( 2v_1+\frac{m_{2}}{m_{1}}\left(v_{2}-u_{2}\right)=\left(v_{2}+u_{2}\right) \)

mal m1

\( 2v_1m_1+m_{2}\left(v_{2}-u_{2}\right)=m_1\left(v_{2}+u_{2}\right) \)

\( 2v_1m_1+m_{2}v_{2}-m_{2}u_{2}=m_1v_{2}+m_{1}u_{2} \)

\( 2v_1m_1+m_{2}v_{2}-m_1v_{2}=m_{2}u_{2}+m_{1}u_{2}=u_{2}( m_{2}+m_1)\)

und dann durch die Klammer teilen.

Comments

Wie würde dann die Lösung heißen, wenn man durch die Klammer teilt?

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\( 2v_1m_1+m_{2}v_{2}-m_1v_{2}=u_{2}( m_{2}+m_1)\)

==>

\( \frac{2v_1m_1+m_{2}v_{2}-m_1v_{2}}{ m_{2}+m_1}=u_{2}\)

und dann im Zähler noch v2 ausklammern

\( \frac{2v_1m_1+v_2(m_{2}-m_1)}{ m_{2}+m_1}=u_{2}\)

Answer 2

\( \left(v_{1}+\frac{m_{2}}{m_{1}}\left(v_{2}-u_{2}\right)\right)^{2}=v_{1}^{2}+\frac{m_{2}}{m_{1}}\left(v_{2}^{2}-u_{2}^{2}\right) \)

v₁=a  m₂=d   m₁=c  v₂=b  u₂=e

\( \left(a+\frac{d}{c}\left(b-e\right)\right)^{2}=a^{2}+\frac{d}{c}\left(b^{2}-e^{2}\right) \)

\( a^{2}+\frac{2 a d}{c} \cdot(b-e)+\frac{d^{2}}{c^{2}} \cdot(b-e)^{2}=a^{2}+\frac{d}{c} \cdot\left(b^{2}-e^{2}\right) \)

\( \frac{2 a d}{c} \cdot(b-e)+\frac{d^{2}}{c^{2}} \cdot(b-e)^{2}=\frac{d}{c} \cdot\left(b^{2}-e^{2}\right) \mid \cdot c^{2} \)

\( 2 a c d \cdot(b-e)+d^{2} \cdot(b-e)^{2}=d c \cdot\left(b^{2}-e^{2}\right) \mid: d \)

\( 2 a c \cdot(b-e)+d \cdot(b-e)^{2}=c \cdot\left(b^{2}-e^{2}\right) \)

\( 2 a c \cdot(b-e)+d \cdot(b-e)^{2}=c \cdot(b+e) \cdot(b-e) \mid:(b-e) \) wobei \( (b-e) \neq 0 \)

\( 2 a c+d \cdot(b-e)=c \cdot(b+e) \)

\( 2 a c+b d-d e=c \cdot b+c e \mid+d e-c b \)

\( c e+d e=2 a c+b \cdot d-c \cdot b \)

\( e=\frac{2 a c+b \cdot d-b \cdot c}{c+d} \)

Und nun wieder zurück.