Ian: New page: Using ${{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}$ and ${{v}_{A}}={{v}_{B}}=0$ $\begin{align} & 0=40\times 3+2v \\ & v=-...$

Ian — Thu, 23 Sep 2010 16:25:33 GMT

New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> and <math>{{v}_{A}}={{v}_{B}}=0</math> <math>\begin{align} & 0=40\times 3+2v \\ & v=-...

New page

Using

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

and

<math>{{v}_{A}}={{v}_{B}}=0</math>

<math>\begin{align}
& 0=40\times 3+2v \\
& v=-\frac{120}{2}=-60\text{ m}{{\text{s}}^{\text{-1}}} \\
\end{align}</math>

Skate board moves in the opposite direction to the child at <math>\text{60 }\ \text{m}{{\text{s}}^{\text{-1}}}</math>.

Solution 16.2 - Revision history

Ian: New page: Using {{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}} and {{v}_{A}}={{v}_{B}}=0 \begin{align} & 0=40\times 3+2v \\ & v=-...

Ian: New page: Using ${{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}$ and ${{v}_{A}}={{v}_{B}}=0$ $\begin{align} & 0=40\times 3+2v \\ & v=-...$