Solution 16.6
From Mechanics
(Difference between revisions)
(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>...) |
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& 2(9\mathbf{i}+5\mathbf{j})+3(-3\mathbf{i}+4\mathbf{j})=(2+3)\mathbf{v} \\ | & 2(9\mathbf{i}+5\mathbf{j})+3(-3\mathbf{i}+4\mathbf{j})=(2+3)\mathbf{v} \\ | ||
& 18\mathbf{i}+10\mathbf{j}-9\mathbf{i}+12\mathbf{j}=5\mathbf{v} \\ | & 18\mathbf{i}+10\mathbf{j}-9\mathbf{i}+12\mathbf{j}=5\mathbf{v} \\ | ||
| - | & \mathbf{v}=\frac{9\mathbf{i}+22\mathbf{j}}{5}=(1\textrm{.}8\mathbf{i}+4\textrm{.}4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}} \\ | + | & \mathbf{v}=\frac{\smash{9\mathbf{i}+22\mathbf{j}}}{5}=(1\textrm{.}8\mathbf{i}+4\textrm{.}4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}} \\ |
\end{align}</math> | \end{align}</math> | ||
Current revision
Using
\displaystyle {{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}
and
\displaystyle {{\mathbf{u}}_{A}}={{\mathbf{u}}_{B}}
\displaystyle \begin{align} & 2(9\mathbf{i}+5\mathbf{j})+3(-3\mathbf{i}+4\mathbf{j})=(2+3)\mathbf{v} \\ & 18\mathbf{i}+10\mathbf{j}-9\mathbf{i}+12\mathbf{j}=5\mathbf{v} \\ & \mathbf{v}=\frac{\smash{9\mathbf{i}+22\mathbf{j}}}{5}=(1\textrm{.}8\mathbf{i}+4\textrm{.}4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}} \\ \end{align}
