Ian: New page: Using ${{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}$ and ${{\mathbf{u}}_{A}}={{\mathbf{u}}_{B}}$
2010-09-23T18:24:36Z

New page: Using <math>{{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}</math> and <math> {{\mathbf{u}}_{A}}={{\mathbf{u}}_{B}}</mat...

New page
Using

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

and

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

Assume the particles have mass <math>m</math>

<math>\begin{align}
& m(5\mathbf{i})+m(8\mathbf{j})=2m\mathbf{v} \\
& \mathbf{v}=2\textrm{.}5\mathbf{i}+4\mathbf{j} \\
& v=\sqrt{{{2\textrm{.}5}^{2}}+{{4}^{2}}}=4\textrm{.}72\text{ m}{{\text{s}}^{\text{-1}}} \\
\end{align}</math>

Solution 16.7a - Revision history