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	<title>Conservation of Momentum EX 2 - Revision history</title>
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	<updated>2026-04-20T19:36:55Z</updated>
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	<entry>
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		<title>imported&gt;Patrick at 16:50, 27 August 2011</title>
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		<updated>2011-08-27T16:50:12Z</updated>

		<summary type="html">&lt;p&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;Newton&amp;#039;s Second Law can be axpressed as&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F} = m \vec{a} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;quot;During an interval of time a constant force acting on an object causes a change of momentum.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F} = {\Delta \vec{p} \over \Delta t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\Delta \vec{p} = \vec{F} \Delta t&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Given&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F} = 1000 \cos(127^{\circ}) \hat{i} + 1000 \sin(127^{\circ}) \hat{j} = -600 \hat{i} + 800 \hat{j} N&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
we can compute&lt;br /&gt;
&lt;br /&gt;
 &amp;lt;math&amp;gt;\Delta \vec{p} = (-600 \hat{i} + 800 \hat{j})(0.001) = -0.600 \hat{i} + 0.800 \hat{j} N&amp;lt;/math&amp;gt;&lt;/div&gt;</summary>
		<author><name>imported&gt;Patrick</name></author>
	</entry>
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