Ian at 16:36, 27 March 2011

Ian — Sun, 27 Mar 2011 16:36:59 GMT

←Older revision		Revision as of 16:36, 27 March 2011
Line 7:		Line 7:
	<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 \\
-	& 10-12=-4+3v \\	+	& 10-18=-4+3v \\
-	& 2=3v \\	+	& -4=3v \\
-	& 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>

-	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.

Ian: New page: Using ${{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}$ Assume the 2 kg particle is moving in the positive direction, $\begin{align} & 2\time...$

Ian — Thu, 23 Sep 2010 17:07:42 GMT

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...

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\times 5+3\times (-6)=2\times (-2)+3v \\
& 10-12=-4+3v \\
& 2=3v \\
& v=\frac{2}{3}\text{ m}{{\text{s}}^{\text{-1}}} \\
\end{align}</math>

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

Solution 16.5 - Revision history

Ian at 16:36, 27 March 2011

Ian: New page: Using {{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, \begin{align} & 2\time...

Ian: New page: Using ${{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}$ Assume the 2 kg particle is moving in the positive direction, $\begin{align} & 2\time...$