Solution 16.5

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(New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> Assume the 2 kg particle is moving in the positive direction, <math>\begin{align} & 2\time...)
Current revision (16:36, 27 March 2011) (edit) (undo)
 
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<math>\begin{align}
<math>\begin{align}
& 2\times 5+3\times (-6)=2\times (-2)+3v \\
& 2\times 5+3\times (-6)=2\times (-2)+3v \\
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& 10-12=-4+3v \\
+
& 10-18=-4+3v \\
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& 2=3v \\
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& -4=3v \\
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& v=\frac{2}{3}\text{ m}{{\text{s}}^{\text{-1}}} \\
+
& v=\frac{-4}{3}\text{ m}{{\text{s}}^{\text{-1}}} \\
\end{align}</math>
\end{align}</math>
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The 3 kg particle moves in the opposite direction to its original velocity.
+
The 3 kg particle moves in the same direction to its original velocity.

Current revision

Using

\displaystyle {{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}

Assume the 2 kg particle is moving in the positive direction,

\displaystyle \begin{align} & 2\times 5+3\times (-6)=2\times (-2)+3v \\ & 10-18=-4+3v \\ & -4=3v \\ & v=\frac{-4}{3}\text{ m}{{\text{s}}^{\text{-1}}} \\ \end{align}

The 3 kg particle moves in the same direction to its original velocity.