Ian: New page: Using ${{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}$ $\begin{align} & 3\times 4+5\times (-2)=3\times 0+5v \\ & 12-10=5v \\ & v=\frac{2}{5...$

2010-09-23T16:21:46Z

New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> <math>\begin{align} & 3\times 4+5\times (-2)=3\times 0+5v \\ & 12-10=5v \\ & v=\frac{2}{5...

New page

Using

<math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math>

<math>\begin{align}
& 3\times 4+5\times (-2)=3\times 0+5v \\
& 12-10=5v \\
& v=\frac{2}{5}=0\textrm{.}4\text{ m}{{\text{s}}^{\text{-1}}} \\
\end{align}</math>

Continues to move in same direction at a speed 0.4 <math>\text{m}{{\text{s}}^{\text{-1}}}</math>.

Solution 16.3 - Revision history

Ian: New page: Using {{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}} \begin{align} & 3\times 4+5\times (-2)=3\times 0+5v \\ & 12-10=5v \\ & v=\frac{2}{5...

Ian: New page: Using ${{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}$ $\begin{align} & 3\times 4+5\times (-2)=3\times 0+5v \\ & 12-10=5v \\ & v=\frac{2}{5...$