16. Conservation of momentum - Revision history

Ian at 15:33, 10 April 2011

2011-04-10T15:33:00Z

←Older revision		Revision as of 15:33, 10 April 2011
Line 135:		Line 135:
	& 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\		& 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\
	& 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\		& 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\
-	& \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}}	+	& \mathbf{v}=\frac{\smash{14\mathbf{i}-8\mathbf{j}}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}}
	\end{align}</math>		\end{align}</math>

Ian at 14:06, 8 March 2010

2010-03-08T14:06:08Z

←Older revision		Revision as of 14:06, 8 March 2010
Line 11:		Line 11:
	== '''Key Points''' ==		== '''Key Points''' ==

-	In all collisions, where no external forces act, momentum will be conserved and ~~we can apply~~	+	In all collisions, where no external forces act, momentum will be conserved and thus the total momentum just after the collision will be the same as the total momentum just before the collision, which gives,

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

Ian at 18:00, 18 February 2010

2010-02-18T18:00:36Z

←Older revision		Revision as of 18:00, 18 February 2010
Line 29:		Line 29:
	Before the collision:		Before the collision:

-	<math>{{u}_{B}}=250</math>	+	<math>{{u}_{B}}=250\text{ m}{{\text{s}}^{\text{-1}}}</math>
	and		and
	<math>{{u}_{T}}=0</math>		<math>{{u}_{T}}=0</math>
Line 35:		Line 35:
	After the collision:		After the collision:

-	<math>{{v}_{B}}={{v}_{T}}=10</math>	+	<math>{{v}_{B}}={{v}_{T}}=10\text{ m}{{\text{s}}^{\text{-1}}}</math>

	Also the mass of the bullet should be converted to kg:		Also the mass of the bullet should be converted to kg:
Line 58:		Line 58:

	Before the collision:		Before the collision:
-	<math>{{u}_{V}}=12</math>	+	<math>{{u}_{V}}=12\text{ m}{{\text{s}}^{\text{-1}}}</math>
	and		and
	<math>{{u}_{C}}=0</math>		<math>{{u}_{C}}=0</math>
Line 116:		Line 116:

	Before the collision:		Before the collision:
-	<math>{{\mathbf{u}}_{A}}=4\mathbf{i}+2\mathbf{j}</math>	+	<math>{{\mathbf{u}}_{A}}=4\mathbf{i}+2\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>
	and		and
-	<math>{{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}</math>	+	<math>{{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>

	After the collision:		After the collision:
	<math>{{\mathbf{v}}_{A}}={{\mathbf{v}}_{B}}=\mathbf{v}</math>		<math>{{\mathbf{v}}_{A}}={{\mathbf{v}}_{B}}=\mathbf{v}</math>

-	The masses are defined:	+	The masses are defined (in kg):
	<math>{{m}_{A}}=2</math>		<math>{{m}_{A}}=2</math>
	and		and
Line 135:		Line 135:
	& 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\		& 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\
	& 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\		& 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\
-	& \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j}	+	& \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}}
	\end{align}</math>		\end{align}</math>

Line 149:		Line 149:
	[[Image:E16.4fig1.GIF]]		[[Image:E16.4fig1.GIF]]

-	<math>{{\mathbf{u}}_{C}}=15\mathbf{i}</math>	+	<math>{{\mathbf{u}}_{C}}=15\mathbf{i}\text{ m}{{\text{s}}^{\text{-1}}}</math>
	and		and
-	<math>{{\mathbf{u}}_{V}}=U\mathbf{j}</math>	+	<math>{{\mathbf{u}}_{V}}=U\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>

	This diagram shows the velocity after the collision.		This diagram shows the velocity after the collision.
Line 157:		Line 157:
	[[Image:E16.4fig2.GIF]]		[[Image:E16.4fig2.GIF]]

-	<math>{{\mathbf{v}}_{C}}={{\mathbf{v}}_{V}}=V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}</math>	+	<math>{{\mathbf{v}}_{C}}={{\mathbf{v}}_{V}}=V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>

	Using conservation of momentum gives:		Using conservation of momentum gives:

Ian at 17:21, 18 February 2010

2010-02-18T17:21:41Z

←Older revision		Revision as of 17:21, 18 February 2010
Line 7:		Line 7:
	\|}		\|}

-	Conservation of Momentum

-	Key ~~Results~~	+
		+	== '''Key Points''' ==

	In all collisions, where no external forces act, momentum will be conserved and we can apply		In all collisions, where no external forces act, momentum will be conserved and we can apply

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

	or		or

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


Line 30:		Line 24:
	A bullet of mass 40 grams is travelling horizontally at 250 <math>\text{m}{{\text{s}}^{-1}}</math>. It hits a wooden trolley that is at rest. The bullet and trolley then move together at		A bullet of mass 40 grams is travelling horizontally at 250 <math>\text{m}{{\text{s}}^{-1}}</math>. It hits a wooden trolley that is at rest. The bullet and trolley then move together at
	10 <math>\text{m}{{\text{s}}^{-1}}</math>.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.		10 <math>\text{m}{{\text{s}}^{-1}}</math>.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.
-

	'''Solution'''		'''Solution'''

	Before the collision:		Before the collision:
-

	<math>{{u}_{B}}=250</math>		<math>{{u}_{B}}=250</math>
	and		and
	<math>{{u}_{T}}=0</math>		<math>{{u}_{T}}=0</math>
-

	After the collision:		After the collision:

-
	<math>{{v}_{B}}={{v}_{T}}=10</math>		<math>{{v}_{B}}={{v}_{T}}=10</math>
-

	Also the mass of the bullet should be converted to kg:		Also the mass of the bullet should be converted to kg:

-
	<math>{{m}_{B}}=0\textrm{.}04</math>		<math>{{m}_{B}}=0\textrm{.}04</math>
-

	Using conservation of momentum gives:		Using conservation of momentum gives:

-
	<math>\begin{align}		<math>\begin{align}
	& {{m}_{B}}{{u}_{B}}+{{m}_{T}}{{u}_{T}}={{m}_{B}}{{v}_{B}}+{{m}_{T}}{{v}_{T}} \\		& {{m}_{B}}{{u}_{B}}+{{m}_{T}}{{u}_{T}}={{m}_{B}}{{v}_{B}}+{{m}_{T}}{{v}_{T}} \\
Line 63:		Line 49:
	& {{m}_{T}}=\frac{10-0\textrm{.}4}{10}=0\textrm{.}96\text{ kg}		& {{m}_{T}}=\frac{10-0\textrm{.}4}{10}=0\textrm{.}96\text{ kg}
	\end{align}</math>		\end{align}</math>
-


Line 69:		Line 54:

	A van, of mass 2.5 tonnes, drives directly into the back of a stationary car, of mass 1.5 tonnes. The van was travelling at 12 <math>\text{m}{{\text{s}}^{-1}}</math> and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision.		A van, of mass 2.5 tonnes, drives directly into the back of a stationary car, of mass 1.5 tonnes. The van was travelling at 12 <math>\text{m}{{\text{s}}^{-1}}</math> and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision.
-

	'''Solution'''		'''Solution'''
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	and		and
	<math>{{u}_{C}}=0</math>		<math>{{u}_{C}}=0</math>
-

	After the collision:		After the collision:
	<math>{{v}_{V}}={{v}_{C}}=v</math>		<math>{{v}_{V}}={{v}_{C}}=v</math>
-

	The masses should be converted to kilograms:		The masses should be converted to kilograms:

-
	<math>{{m}_{V}}=2500</math>		<math>{{m}_{V}}=2500</math>
	and		and
	<math>{{m}_{C}}=1500</math>		<math>{{m}_{C}}=1500</math>
-

	Using conservation of momentum gives:		Using conservation of momentum gives:

-
	<math>\begin{align}		<math>\begin{align}
	& {{m}_{V}}{{u}_{V}}+{{m}_{C}}{{u}_{C}}={{m}_{V}}{{v}_{V}}+{{m}_{C}}{{v}_{C}} \\		& {{m}_{V}}{{u}_{V}}+{{m}_{C}}{{u}_{C}}={{m}_{V}}{{v}_{V}}+{{m}_{C}}{{v}_{C}} \\
Line 100:		Line 79:
	& v=\frac{30000}{4000}=7\textrm{.}5\text{ m}{{\text{s}}^{\text{-1}}}		& v=\frac{30000}{4000}=7\textrm{.}5\text{ m}{{\text{s}}^{\text{-1}}}
	\end{align}</math>		\end{align}</math>
-


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	Two particles, A and B of mass m and 3m are moving towards each other with speeds of 4u and u respectively along a straight line. They collide and coalesce. Describe how the motion of each particle changes during the collision.		Two particles, A and B of mass m and 3m are moving towards each other with speeds of 4u and u respectively along a straight line. They collide and coalesce. Describe how the motion of each particle changes during the collision.
-

	'''Solution'''		'''Solution'''
Line 114:		Line 91:
	and		and
	<math>{{u}_{B}}=-u</math>		<math>{{u}_{B}}=-u</math>
-

	After the collision:		After the collision:
	<math>{{v}_{A}}={{v}_{B}}=v</math>		<math>{{v}_{A}}={{v}_{B}}=v</math>
-

	Using conservation of momentum gives:		Using conservation of momentum gives:

-
	<math>\begin{align}		<math>\begin{align}
	& {{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}={{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}} \\		& {{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}={{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}} \\
Line 138:		Line 112:
	<math>(2\mathbf{i}-4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}}</math>		<math>(2\mathbf{i}-4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}}</math>
	. If the particles coalesce during the collision, find their final velocity.		. If the particles coalesce during the collision, find their final velocity.
-

	'''Solution'''		'''Solution'''
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	and		and
	<math>{{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}</math>		<math>{{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}</math>
-

	After the collision:		After the collision:
	<math>{{\mathbf{v}}_{A}}={{\mathbf{v}}_{B}}=\mathbf{v}</math>		<math>{{\mathbf{v}}_{A}}={{\mathbf{v}}_{B}}=\mathbf{v}</math>
-

	The masses are defined:		The masses are defined:
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	and		and
	<math>{{m}_{B}}=3</math>		<math>{{m}_{B}}=3</math>
-

	Using conservation of momentum gives:		Using conservation of momentum gives:
-

	<math>\begin{align}		<math>\begin{align}
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	& \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j}		& \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j}
	\end{align}</math>		\end{align}</math>
-


Line 180:		Line 148:

	[[Image:E16.4fig1.GIF]]		[[Image:E16.4fig1.GIF]]
-

	<math>{{\mathbf{u}}_{C}}=15\mathbf{i}</math>		<math>{{\mathbf{u}}_{C}}=15\mathbf{i}</math>
	and		and
	<math>{{\mathbf{u}}_{V}}=U\mathbf{j}</math>		<math>{{\mathbf{u}}_{V}}=U\mathbf{j}</math>
-

	This diagram shows the velocity after the collision.		This diagram shows the velocity after the collision.
Line 192:		Line 158:

	<math>{{\mathbf{v}}_{C}}={{\mathbf{v}}_{V}}=V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}</math>		<math>{{\mathbf{v}}_{C}}={{\mathbf{v}}_{V}}=V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}</math>
-

	Using conservation of momentum gives:		Using conservation of momentum gives:
-

	<math>\begin{align}		<math>\begin{align}
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	& 1200\times 15\mathbf{i}+1400\times U\mathbf{j}=2600(V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j})		& 1200\times 15\mathbf{i}+1400\times U\mathbf{j}=2600(V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j})
	\end{align}</math>		\end{align}</math>
-

	Considering the <math>\mathbf{i}</math> component gives:		Considering the <math>\mathbf{i}</math> component gives:
-

	<math>\begin{align}		<math>\begin{align}
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	& V=\frac{1200\times 15}{2600\cos 20{}^\circ }=\frac{180}{26\cos 20{}^\circ }=7\textrm{.}36\text{ m}{{\text{s}}^{\text{-1}}}		& V=\frac{1200\times 15}{2600\cos 20{}^\circ }=\frac{180}{26\cos 20{}^\circ }=7\textrm{.}36\text{ m}{{\text{s}}^{\text{-1}}}
	\end{align}</math>		\end{align}</math>
-

	Considering the <math>\mathbf{j}</math> component gives:		Considering the <math>\mathbf{j}</math> component gives:
-

	<math>\begin{align}		<math>\begin{align}

Ian at 16:08, 29 September 2009

2009-09-29T16:08:34Z

←Older revision		Revision as of 16:08, 29 September 2009
Line 179:		Line 179:
	This diagram shows the velocities before the collision.		This diagram shows the velocities before the collision.

-		+	[[Image:E16.4fig1.GIF]]
-		+
-		+
-		+
-		+


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	This diagram shows the velocity after the collision.		This diagram shows the velocity after the collision.

-		+	[[Image:E16.4fig2.GIF]]
-		+
-		+

	<math>{{\mathbf{v}}_{C}}={{\mathbf{v}}_{V}}=V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}</math>		<math>{{\mathbf{v}}_{C}}={{\mathbf{v}}_{V}}=V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}</math>

Ian at 16:00, 29 September 2009

2009-09-29T16:00:42Z

←Older revision		Revision as of 16:00, 29 September 2009
Line 26:		Line 26:


-	Example 16.1	+	'''[[Example 16.1]]'''
		+
	A bullet of mass 40 grams is travelling horizontally at 250 <math>\text{m}{{\text{s}}^{-1}}</math>. It hits a wooden trolley that is at rest. The bullet and trolley then move together at		A bullet of mass 40 grams is travelling horizontally at 250 <math>\text{m}{{\text{s}}^{-1}}</math>. It hits a wooden trolley that is at rest. The bullet and trolley then move together at
	10 <math>\text{m}{{\text{s}}^{-1}}</math>.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.		10 <math>\text{m}{{\text{s}}^{-1}}</math>.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.

-	Solution	+
		+	'''Solution'''
		+
	Before the collision:		Before the collision:

Line 62:		Line 65:


-	Example 16.2	+
		+	'''[[Example 16.2]]'''
		+
	A van, of mass 2.5 tonnes, drives directly into the back of a stationary car, of mass 1.5 tonnes. The van was travelling at 12 <math>\text{m}{{\text{s}}^{-1}}</math> and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision.		A van, of mass 2.5 tonnes, drives directly into the back of a stationary car, of mass 1.5 tonnes. The van was travelling at 12 <math>\text{m}{{\text{s}}^{-1}}</math> and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision.

-	Solution	+
		+	'''Solution'''
		+
	Before the collision:		Before the collision:
	<math>{{u}_{V}}=12</math>		<math>{{u}_{V}}=12</math>
Line 95:		Line 102:


-	Example 16.3	+
		+	'''[[Example 16.3]]'''
		+
	Two particles, A and B of mass m and 3m are moving towards each other with speeds of 4u and u respectively along a straight line. They collide and coalesce. Describe how the motion of each particle changes during the collision.		Two particles, A and B of mass m and 3m are moving towards each other with speeds of 4u and u respectively along a straight line. They collide and coalesce. Describe how the motion of each particle changes during the collision.

-	Solution	+
		+	'''Solution'''
		+
	Before the collision:		Before the collision:
	<math>{{u}_{A}}=4u</math>		<math>{{u}_{A}}=4u</math>
Line 120:		Line 131:


-	Example 16.4	+	'''[[Example 16.4]]'''

	A particle, A, of mass 2 kg has velocity		A particle, A, of mass 2 kg has velocity
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	. If the particles coalesce during the collision, find their final velocity.		. If the particles coalesce during the collision, find their final velocity.

-	Solution	+
		+	'''Solution'''
		+
	Before the collision:		Before the collision:
	<math>{{\mathbf{u}}_{A}}=4\mathbf{i}+2\mathbf{j}</math>		<math>{{\mathbf{u}}_{A}}=4\mathbf{i}+2\mathbf{j}</math>
Line 157:		Line 170:


-	Example 16.5	+
		+	'''[[Example 16.5]]'''
		+
	A car, of mass 1.2 tonnes, is travelling at 15 <math>\text{m}{{\text{s}}^{-1}}</math>, when it is hit by a van, of mass 1.4 tonnes, travelling at right angles to the path of the first car. After the collision the two vehicles move together at an angle of 20<math>{}^\circ </math> to the original motion of the car. Find the speed of the heavier van just before the collision.		A car, of mass 1.2 tonnes, is travelling at 15 <math>\text{m}{{\text{s}}^{-1}}</math>, when it is hit by a van, of mass 1.4 tonnes, travelling at right angles to the path of the first car. After the collision the two vehicles move together at an angle of 20<math>{}^\circ </math> to the original motion of the car. Find the speed of the heavier van just before the collision.

-	Solution	+	'''Solution'''
		+
	This diagram shows the velocities before the collision.		This diagram shows the velocities before the collision.

Ian at 15:46, 29 September 2009

2009-09-29T15:46:04Z

←Older revision		Revision as of 15:46, 29 September 2009
Line 48:		Line 48:


-	<math>{{m}_{B}}=0.04</math>	+	<math>{{m}_{B}}=0\textrm{.}04</math>


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	<math>\begin{align}		<math>\begin{align}
	& {{m}_{B}}{{u}_{B}}+{{m}_{T}}{{u}_{T}}={{m}_{B}}{{v}_{B}}+{{m}_{T}}{{v}_{T}} \\		& {{m}_{B}}{{u}_{B}}+{{m}_{T}}{{u}_{T}}={{m}_{B}}{{v}_{B}}+{{m}_{T}}{{v}_{T}} \\
-	& 0.04\times 250+{{m}_{T}}\times 0=0.04\times 10+{{m}_{T}}\times 10 \\	+	& 0\textrm{.}04\times 250+{{m}_{T}}\times 0=0\textrm{.}04\times 10+{{m}_{T}}\times 10 \\
-	& 10=0.4+10{{m}_{T}} \\	+	& 10=0\textrm{.}4+10{{m}_{T}} \\
-	& {{m}_{T}}=\frac{10-0.4}{10}=0.96\text{ kg}	+	& {{m}_{T}}=\frac{10-0\textrm{.}4}{10}=0\textrm{.}96\text{ kg}
	\end{align}</math>		\end{align}</math>

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	& 2500\times 12+1500\times 0=2500v+1500v \\		& 2500\times 12+1500\times 0=2500v+1500v \\
	& 30000=4000v \\		& 30000=4000v \\
-	& v=\frac{30000}{4000}=7.5\text{ m}{{\text{s}}^{\text{-1}}}	+	& v=\frac{30000}{4000}=7\textrm{.}5\text{ m}{{\text{s}}^{\text{-1}}}
	\end{align}</math>		\end{align}</math>

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	& 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\		& 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\
	& 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\		& 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\
-	& \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2.8\mathbf{i}-1.6\mathbf{j}	+	& \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j}
	\end{align}</math>		\end{align}</math>

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	<math>\begin{align}		<math>\begin{align}
	& 1200\times 15=2600V\cos 20{}^\circ \\		& 1200\times 15=2600V\cos 20{}^\circ \\
-	& V=\frac{1200\times 15}{2600\cos 20{}^\circ }=\frac{180}{26\cos 20{}^\circ }=7.36\text{ m}{{\text{s}}^{\text{-1}}}	+	& V=\frac{1200\times 15}{2600\cos 20{}^\circ }=\frac{180}{26\cos 20{}^\circ }=7\textrm{.}36\text{ m}{{\text{s}}^{\text{-1}}}
	\end{align}</math>		\end{align}</math>

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	<math>\begin{align}		<math>\begin{align}
	& 1400U=2600\times \frac{180}{26\cos 20{}^\circ }\times \sin 20{}^\circ \\		& 1400U=2600\times \frac{180}{26\cos 20{}^\circ }\times \sin 20{}^\circ \\
-	& U=\frac{2600\times 180}{1400\times 26}\tan 20{}^\circ =4.68\text{ m}{{\text{s}}^{\text{-1}}}	+	& U=\frac{2600\times 180}{1400\times 26}\tan 20{}^\circ =4\textrm{.}68\text{ m}{{\text{s}}^{\text{-1}}}
	\end{align}</math>		\end{align}</math>

Ian at 15:39, 29 September 2009

2009-09-29T15:39:40Z

←Older revision		Revision as of 15:39, 29 September 2009
Line 15:		Line 15:

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

	or		or

-	<math>{{m}_{A}}{{\mathbf{v}}_~~{\mathbf~~{A}}}+{{m}_{B}}{{\mathbf{v}}_~~{\mathbf~~{B}}}={{m}_{A}}{{\mathbf{u}}_~~{\mathbf~~{A}}}+{{m}_{B}}{{\mathbf{u}}_~~{\mathbf~~{B}}}</math>	+
		+	<math>{{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}</math>
		+
		+



	Example 16.1		Example 16.1
-	A bullet of mass 40 grams is travelling horizontally at 250 ms-1. It hits a wooden trolley that is at rest. The bullet and trolley then move together at	+	A bullet of mass 40 grams is travelling horizontally at 250 <math>\text{m}{{\text{s}}^{-1}}</math>. It hits a wooden trolley that is at rest. The bullet and trolley then move together at
-	10 ms-1.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.	+	10 <math>\text{m}{{\text{s}}^{-1}}</math>.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.

	Solution		Solution
Line 59:		Line 63:

	Example 16.2		Example 16.2
-	A van, of mass 2.5 tonnes, drives directly into the back of a stationary car, of mass 1.5 tonnes. The van was travelling at 12 ms-1 and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision.	+	A van, of mass 2.5 tonnes, drives directly into the back of a stationary car, of mass 1.5 tonnes. The van was travelling at 12 <math>\text{m}{{\text{s}}^{-1}}</math> and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision.

	Solution		Solution
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	Example 16.5		Example 16.5
-	A car, of mass 1.2 tonnes, is travelling at 15 ms-1, when it is hit by a van, of mass 1.4 tonnes, travelling at right angles to the path of the first car. After the collision the two vehicles move together at an angle of ~~20~~ to the original motion of the car. Find the speed of the heavier van just before the collision.	+	A car, of mass 1.2 tonnes, is travelling at 15 <math>\text{m}{{\text{s}}^{-1}}</math>, when it is hit by a van, of mass 1.4 tonnes, travelling at right angles to the path of the first car. After the collision the two vehicles move together at an angle of 20<math>{}^\circ </math> to the original motion of the car. Find the speed of the heavier van just before the collision.

	Solution		Solution
Line 189:		Line 193:


-	Considering the i component gives:	+	Considering the <math>\mathbf{i}</math> component gives:


Line 198:		Line 202:


-	Considering the j component gives:	+	Considering the <math>\mathbf{j}</math> component gives:

Ian at 15:18, 29 September 2009

2009-09-29T15:18:05Z

←Older revision		Revision as of 15:18, 29 September 2009
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	\|}		\|}
		+
		+	Conservation of Momentum
		+
		+	Key Results
		+
		+	In all collisions, where no external forces act, momentum will be conserved and we can apply
		+
		+
		+	<math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math>
		+
		+	or
		+
		+	<math>{{m}_{A}}{{\mathbf{v}}_{\mathbf{A}}}+{{m}_{B}}{{\mathbf{v}}_{\mathbf{B}}}={{m}_{A}}{{\mathbf{u}}_{\mathbf{A}}}+{{m}_{B}}{{\mathbf{u}}_{\mathbf{B}}}</math>
		+
		+
		+
		+	Example 16.1
		+	A bullet of mass 40 grams is travelling horizontally at 250 ms-1. It hits a wooden trolley that is at rest. The bullet and trolley then move together at
		+	10 ms-1.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.
		+
		+	Solution
		+	Before the collision:
		+
		+
		+	<math>{{u}_{B}}=250</math>
		+	and
		+	<math>{{u}_{T}}=0</math>
		+
		+
		+	After the collision:
		+
		+
		+	<math>{{v}_{B}}={{v}_{T}}=10</math>
		+
		+
		+	Also the mass of the bullet should be converted to kg:
		+
		+
		+	<math>{{m}_{B}}=0.04</math>
		+
		+
		+	Using conservation of momentum gives:
		+
		+
		+	<math>\begin{align}
		+	& {{m}_{B}}{{u}_{B}}+{{m}_{T}}{{u}_{T}}={{m}_{B}}{{v}_{B}}+{{m}_{T}}{{v}_{T}} \\
		+	& 0.04\times 250+{{m}_{T}}\times 0=0.04\times 10+{{m}_{T}}\times 10 \\
		+	& 10=0.4+10{{m}_{T}} \\
		+	& {{m}_{T}}=\frac{10-0.4}{10}=0.96\text{ kg}
		+	\end{align}</math>
		+
		+
		+	Example 16.2
		+	A van, of mass 2.5 tonnes, drives directly into the back of a stationary car, of mass 1.5 tonnes. The van was travelling at 12 ms-1 and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision.
		+
		+	Solution
		+	Before the collision:
		+	<math>{{u}_{V}}=12</math>
		+	and
		+	<math>{{u}_{C}}=0</math>
		+
		+
		+	After the collision:
		+	<math>{{v}_{V}}={{v}_{C}}=v</math>
		+
		+
		+	The masses should be converted to kilograms:
		+
		+
		+	<math>{{m}_{V}}=2500</math>
		+	and
		+	<math>{{m}_{C}}=1500</math>
		+
		+
		+	Using conservation of momentum gives:
		+
		+
		+	<math>\begin{align}
		+	& {{m}_{V}}{{u}_{V}}+{{m}_{C}}{{u}_{C}}={{m}_{V}}{{v}_{V}}+{{m}_{C}}{{v}_{C}} \\
		+	& 2500\times 12+1500\times 0=2500v+1500v \\
		+	& 30000=4000v \\
		+	& v=\frac{30000}{4000}=7.5\text{ m}{{\text{s}}^{\text{-1}}}
		+	\end{align}</math>
		+
		+
		+	Example 16.3
		+	Two particles, A and B of mass m and 3m are moving towards each other with speeds of 4u and u respectively along a straight line. They collide and coalesce. Describe how the motion of each particle changes during the collision.
		+
		+	Solution
		+	Before the collision:
		+	<math>{{u}_{A}}=4u</math>
		+	and
		+	<math>{{u}_{B}}=-u</math>
		+
		+
		+	After the collision:
		+	<math>{{v}_{A}}={{v}_{B}}=v</math>
		+
		+
		+	Using conservation of momentum gives:
		+
		+
		+	<math>\begin{align}
		+	& {{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}={{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}} \\
		+	& m\times 4u+3m\times (-u)=mv+3mv \\
		+	& mu=4mv \\
		+	& v=\frac{mu}{4mu}=\frac{u}{4}
		+	\end{align}</math>
		+
		+
		+	Example 16.4
		+
		+	A particle, A, of mass 2 kg has velocity
		+	<math>(4\mathbf{i}+2\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}}</math>
		+	. It collides with a second particle, B, of mass 3 kg and velocity
		+	<math>(2\mathbf{i}-4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}}</math>
		+	. If the particles coalesce during the collision, find their final velocity.
		+
		+	Solution
		+	Before the collision:
		+	<math>{{\mathbf{u}}_{A}}=4\mathbf{i}+2\mathbf{j}</math>
		+	and
		+	<math>{{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}</math>
		+
		+
		+	After the collision:
		+	<math>{{\mathbf{v}}_{A}}={{\mathbf{v}}_{B}}=\mathbf{v}</math>
		+
		+
		+	The masses are defined:
		+	<math>{{m}_{A}}=2</math>
		+	and
		+	<math>{{m}_{B}}=3</math>
		+
		+
		+	Using conservation of momentum gives:
		+
		+
		+	<math>\begin{align}
		+	& {{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}={{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}} \\
		+	& 2\times (4\mathbf{i}+2\mathbf{j})+3\times (2\mathbf{i}-4\mathbf{j})=2\mathbf{v}+3\mathbf{v} \\
		+	& 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\
		+	& 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\
		+	& \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2.8\mathbf{i}-1.6\mathbf{j}
		+	\end{align}</math>
		+
		+
		+	Example 16.5
		+	A car, of mass 1.2 tonnes, is travelling at 15 ms-1, when it is hit by a van, of mass 1.4 tonnes, travelling at right angles to the path of the first car. After the collision the two vehicles move together at an angle of 20 to the original motion of the car. Find the speed of the heavier van just before the collision.
		+
		+	Solution
		+	This diagram shows the velocities before the collision.
		+
		+
		+
		+
		+
		+
		+
		+
		+	<math>{{\mathbf{u}}_{C}}=15\mathbf{i}</math>
		+	and
		+	<math>{{\mathbf{u}}_{V}}=U\mathbf{j}</math>
		+
		+
		+	This diagram shows the velocity after the collision.
		+
		+
		+
		+
		+
		+	<math>{{\mathbf{v}}_{C}}={{\mathbf{v}}_{V}}=V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}</math>
		+
		+
		+	Using conservation of momentum gives:
		+
		+
		+	<math>\begin{align}
		+	& {{m}_{C}}{{\mathbf{u}}_{C}}+{{m}_{V}}{{\mathbf{u}}_{V}}={{m}_{C}}{{\mathbf{v}}_{C}}+{{m}_{V}}{{\mathbf{v}}_{V}} \\
		+	& 1200\times 15\mathbf{i}+1400\times U\mathbf{j}=2600(V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j})
		+	\end{align}</math>
		+
		+
		+	Considering the i component gives:
		+
		+
		+	<math>\begin{align}
		+	& 1200\times 15=2600V\cos 20{}^\circ \\
		+	& V=\frac{1200\times 15}{2600\cos 20{}^\circ }=\frac{180}{26\cos 20{}^\circ }=7.36\text{ m}{{\text{s}}^{\text{-1}}}
		+	\end{align}</math>
		+
		+
		+	Considering the j component gives:
		+
		+
		+	<math>\begin{align}
		+	& 1400U=2600\times \frac{180}{26\cos 20{}^\circ }\times \sin 20{}^\circ \\
		+	& U=\frac{2600\times 180}{1400\times 26}\tan 20{}^\circ =4.68\text{ m}{{\text{s}}^{\text{-1}}}
		+	\end{align}</math>

Ian: New page: NOTOC {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #797979" width="5px" | {{Selected tab|[[16. Conservation of momentu...

2009-09-29T14:53:33Z

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{{Selected tab|[[16. Conservation of momentum|Theory]]}}
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