Ian: New page: Using ${{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}$ and ${{u}_{A}}={{u}_{B}}$ $\begin{align} & 1000\times 20+4000\times 10=(1...$

Ian — Thu, 23 Sep 2010 16:06:38 GMT

New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> and <math>{{u}_{A}}={{u}_{B}}</math> <math>\begin{align} & 1000\times 20+4000\times 10=(1...

New page

Using

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

and

<math>{{u}_{A}}={{u}_{B}}</math>

<math>\begin{align}
& 1000\times 20+4000\times 10=(1000+4000)v \\
& v=\frac{60000}{5000}=12\text{ m}{{\text{s}}^{\text{-1}}} \\
\end{align}</math>

Solution 16.1 - Revision history

Ian: New page: Using {{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}} and {{u}_{A}}={{u}_{B}} \begin{align} & 1000\times 20+4000\times 10=(1...

Ian: New page: Using ${{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}$ and ${{u}_{A}}={{u}_{B}}$ $\begin{align} & 1000\times 20+4000\times 10=(1...$