15. Moment och impuls - Versionshistorik

Tek: flyttade 15. Momentum and impulse till 15. Moment och impuls

Tek — Mon, 01 Mar 2010 14:11:17 GMT

flyttade 15. Momentum and impulse till 15. Moment och impuls

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Versionen från 1 mars 2010 kl. 14.11

Ian den 17 februari 2010 kl. 19.37

Ian — Wed, 17 Feb 2010 19:37:48 GMT

← Äldre version		Versionen från 17 februari 2010 kl. 19.37
Rad 128:		Rad 128:
	& \text{Initial Momentum }=\text{ m}\mathbf{u} \\		& \text{Initial Momentum }=\text{ m}\mathbf{u} \\
	& =0\textrm{.}2\times 10\mathbf{i} \\		& =0\textrm{.}2\times 10\mathbf{i} \\
-	& =2\mathbf{i}	+	& =2\mathbf{i} \text{ Ns}
	\end{align}</math>		\end{align}</math>

Rad 134:		Rad 134:
	& \text{Final Momentum }=m\mathbf{v} \\		& \text{Final Momentum }=m\mathbf{v} \\
	& =0\textrm{.}2(16\cos 60{}^\circ \mathbf{i}+16\sin 60{}^\circ \mathbf{j}) \\		& =0\textrm{.}2(16\cos 60{}^\circ \mathbf{i}+16\sin 60{}^\circ \mathbf{j}) \\
-	& =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j}	+	& =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j} \text{ Ns}
	\end{align}</math>		\end{align}</math>

Rad 140:		Rad 140:
	& \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\		& \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\
	& =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j}-2\mathbf{i} \\		& =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j}-2\mathbf{i} \\
-	& =-0\textrm{.}4\mathbf{i}+2\textrm{.}77\mathbf{j}	+	& =-0\textrm{.}4\mathbf{i}+2\textrm{.}77\mathbf{j} \text{ Ns}
	\end{align}</math>		\end{align}</math>

Ian den 17 februari 2010 kl. 18.58

Ian — Wed, 17 Feb 2010 18:58:56 GMT

← Äldre version		Versionen från 17 februari 2010 kl. 18.58
Rad 8:		Rad 8:


-	15. Momentum and Impulse

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

	Momentum is defined as the product of the mass and velocity of a body.		Momentum is defined as the product of the mass and velocity of a body.
Rad 18:		Rad 18:
	or Momentum =		or Momentum =
	<math>m\mathbf{v}</math>		<math>m\mathbf{v}</math>
-

	If the velocity of a body, of mass		If the velocity of a body, of mass
Rad 50:		Rad 49:
	<math>F</math>		<math>F</math>
	is assumed to be constant and represents the average force.		is assumed to be constant and represents the average force.
-


	'''[[Example 15.1]]'''		'''[[Example 15.1]]'''
-

	Find the momentum of a car, of mass 1.2 tonnes, travelling in a straight line		Find the momentum of a car, of mass 1.2 tonnes, travelling in a straight line
Rad 63:		Rad 60:
	Note that 1.2 tonnes is 1200 kg. Using Momentum = mv gives:		Note that 1.2 tonnes is 1200 kg. Using Momentum = mv gives:

-
	<math>\begin{align}		<math>\begin{align}
	& \text{Momentum}=1200\times 30 \\		& \text{Momentum}=1200\times 30 \\
Rad 83:		Rad 79:
	& =-6600\text{ Ns}		& =-6600\text{ Ns}
	\end{align}</math>		\end{align}</math>
-

	Use the formula		Use the formula
	<math>\text{Impulse }=Ft</math>		<math>\text{Impulse }=Ft</math>

-
-
	<math>\begin{align}		<math>\begin{align}
	& -6600=F\times 2\textrm{.}2 \\		& -6600=F\times 2\textrm{.}2 \\
	& F=-3000\text{ N}		& F=-3000\text{ N}
	\end{align}</math>		\end{align}</math>
-

	The average force has magnitude 3000 N		The average force has magnitude 3000 N
Rad 100:		Rad 92:

	'''[[Example 15.3]]'''		'''[[Example 15.3]]'''
-

	The speed of a car, travelling in a straight line, is reduced from 20 <math>\text{m}{{\text{s}}^{-1}}</math> to		The speed of a car, travelling in a straight line, is reduced from 20 <math>\text{m}{{\text{s}}^{-1}}</math> to
Rad 108:		Rad 99:

	First find the impulse (change in momentum).		First find the impulse (change in momentum).
-

	<math>\begin{align}		<math>\begin{align}
Rad 115:		Rad 105:
	& =-6000\text{ Ns}		& =-6000\text{ Ns}
	\end{align}</math>		\end{align}</math>
-

	Use the formula		Use the formula
	<math>\text{Impulse }=Ft</math>		<math>\text{Impulse }=Ft</math>
-
-

	<math>\begin{align}		<math>\begin{align}
Rad 137:		Rad 124:

	[[Image:E15.4.GIF]]		[[Image:E15.4.GIF]]
-

	<math>\begin{align}		<math>\begin{align}
Rad 144:		Rad 130:
	& =2\mathbf{i}		& =2\mathbf{i}
	\end{align}</math>		\end{align}</math>
-
-

	<math>\begin{align}		<math>\begin{align}
Rad 153:		Rad 137:
	\end{align}</math>		\end{align}</math>

-
-
	<math>\begin{align}		<math>\begin{align}
	& \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\		& \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\
Rad 160:		Rad 142:
	& =-0\textrm{.}4\mathbf{i}+2\textrm{.}77\mathbf{j}		& =-0\textrm{.}4\mathbf{i}+2\textrm{.}77\mathbf{j}
	\end{align}</math>		\end{align}</math>
		+

	'''[[Example 15.5]]'''		'''[[Example 15.5]]'''
Rad 176:		Rad 159:
	& =2\textrm{.}4\mathbf{i}		& =2\textrm{.}4\mathbf{i}
	\end{align}</math>		\end{align}</math>
-

	<math>\begin{align}		<math>\begin{align}
Rad 184:		Rad 166:
	\end{align}</math>		\end{align}</math>

-
-
	<math>\begin{align}		<math>\begin{align}
	& \text{Impulse }=m\mathbf{v}-m\mathbf{u} \\		& \text{Impulse }=m\mathbf{v}-m\mathbf{u} \\
	& =1\textrm{.}8\mathbf{j}-2\textrm{.}4\mathbf{i}		& =1\textrm{.}8\mathbf{j}-2\textrm{.}4\mathbf{i}
	\end{align}</math>		\end{align}</math>
-

	The diagram shows the impulse.		The diagram shows the impulse.
Rad 199:		Rad 178:
	using Pythagoras:		using Pythagoras:

-
	<math>I=\sqrt{{{1\textrm{.}8}^{2}}+{{2\textrm{.}4}^{2}}}=3\text{ Ns}</math>		<math>I=\sqrt{{{1\textrm{.}8}^{2}}+{{2\textrm{.}4}^{2}}}=3\text{ Ns}</math>
-

	The angle, <math>\alpha </math>, can be found using trigonometry:		The angle, <math>\alpha </math>, can be found using trigonometry:

Ian den 26 september 2009 kl. 18.04

Ian — Sat, 26 Sep 2009 18:04:55 GMT

← Äldre version		Versionen från 26 september 2009 kl. 18.04
Rad 3:		Rad 3:
	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
	{{Selected tab\|[[15. Momentum and impulse\|Theory]]}}		{{Selected tab\|[[15. Momentum and impulse\|Theory]]}}
-	{{Not selected tab\|[[2. Exercises\|Exercises]]}}	+	{{Not selected tab\|[[15. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}

Ian den 26 september 2009 kl. 18.04

Ian — Sat, 26 Sep 2009 18:04:15 GMT

← Äldre version		Versionen från 26 september 2009 kl. 18.04
Rad 2:		Rad 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Selected tab\|[[15. ~~Forces~~ and ~~vectors~~\|Theory]]}}	+	{{Selected tab\|[[15. Momentum and impulse\|Theory]]}}
-	{{Not selected tab\|[[15. Exercises\|Exercises]]}}	+	{{Not selected tab\|[[2. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}

Ian den 24 september 2009 kl. 14.38

Ian — Thu, 24 Sep 2009 14:38:59 GMT

← Äldre version		Versionen från 24 september 2009 kl. 14.38
Rad 91:		Rad 91:

	<math>\begin{align}		<math>\begin{align}
-	& -6600=F\times 2.2 \\	+	& -6600=F\times 2\textrm{.}2 \\
	& F=-3000\text{ N}		& F=-3000\text{ N}
	\end{align}</math>		\end{align}</math>
Rad 136:		Rad 136:
	The diagrams show the initial and final velocities of the ball and the unit vectors <math>\mathbf{i}</math> and <math>\mathbf{j}</math>.		The diagrams show the initial and final velocities of the ball and the unit vectors <math>\mathbf{i}</math> and <math>\mathbf{j}</math>.

-		+	[[Image:E15.4.GIF]]
-		+
-		+
-		+
-		+


	<math>\begin{align}		<math>\begin{align}
	& \text{Initial Momentum }=\text{ m}\mathbf{u} \\		& \text{Initial Momentum }=\text{ m}\mathbf{u} \\
-	& =0.2\times 10\mathbf{i} \\	+	& =0\textrm{.}2\times 10\mathbf{i} \\
	& =2\mathbf{i}		& =2\mathbf{i}
	\end{align}</math>		\end{align}</math>
Rad 153:		Rad 149:
	<math>\begin{align}		<math>\begin{align}
	& \text{Final Momentum }=m\mathbf{v} \\		& \text{Final Momentum }=m\mathbf{v} \\
-	& =0.2(16\cos 60{}^\circ \mathbf{i}+16\sin 60{}^\circ \mathbf{j}) \\	+	& =0\textrm{.}2(16\cos 60{}^\circ \mathbf{i}+16\sin 60{}^\circ \mathbf{j}) \\
-	& =3.2\cos 60{}^\circ \mathbf{i}+3.2\sin 60{}^\circ \mathbf{j}	+	& =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j}
	\end{align}</math>		\end{align}</math>

Rad 161:		Rad 157:
	<math>\begin{align}		<math>\begin{align}
	& \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\		& \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\
-	& =3.2\cos 60{}^\circ \mathbf{i}+3.2\sin 60{}^\circ \mathbf{j}-2\mathbf{i} \\	+	& =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j}-2\mathbf{i} \\
-	& =-0.4\mathbf{i}+2.77\mathbf{j}	+	& =-0\textrm{.}4\mathbf{i}+2\textrm{.}77\mathbf{j}
	\end{align}</math>		\end{align}</math>

Rad 173:		Rad 169:
	The diagrams show the initial and final velocities of the ball and the unit vectors <math>\mathbf{i}</math> and <math>\mathbf{j}</math>.		The diagrams show the initial and final velocities of the ball and the unit vectors <math>\mathbf{i}</math> and <math>\mathbf{j}</math>.

-		+	[[Image:E15.5fig1.GIF]]
-		+
-		+
-		+

	<math>\begin{align}		<math>\begin{align}
	& \text{Initial Momentum }=\text{ }m\mathbf{u} \\		& \text{Initial Momentum }=\text{ }m\mathbf{u} \\
-	& =0.3\times 8\mathbf{i} \\	+	& =0\textrm{.}3\times 8\mathbf{i} \\
-	& =2.4\mathbf{i}	+	& =2\textrm{.}4\mathbf{i}
	\end{align}</math>		\end{align}</math>

Rad 187:		Rad 180:
	<math>\begin{align}		<math>\begin{align}
	& \text{Final Momentum }=m\mathbf{v} \\		& \text{Final Momentum }=m\mathbf{v} \\
-	& =0.3\times 6\mathbf{j} \\	+	& =0\textrm{.}3\times 6\mathbf{j} \\
-	& =1.8\mathbf{j}	+	& =1\textrm{.}8\mathbf{j}
	\end{align}</math>		\end{align}</math>

Rad 195:		Rad 188:
	<math>\begin{align}		<math>\begin{align}
	& \text{Impulse }=m\mathbf{v}-m\mathbf{u} \\		& \text{Impulse }=m\mathbf{v}-m\mathbf{u} \\
-	& =1.8\mathbf{j}-2.4\mathbf{i}	+	& =1\textrm{.}8\mathbf{j}-2\textrm{.}4\mathbf{i}
	\end{align}</math>		\end{align}</math>


	The diagram shows the impulse.		The diagram shows the impulse.
		+
		+	[[Image:E15.5fig2.GIF]]

	The magnitude of the impulse, <math>I</math>, is found		The magnitude of the impulse, <math>I</math>, is found
Rad 205:		Rad 200:


-	<math>I=\sqrt{{{1.8}^{2}}+{{2.4}^{2}}}=3\text{ Ns}</math>	+	<math>I=\sqrt{{{1\textrm{.}8}^{2}}+{{2\textrm{.}4}^{2}}}=3\text{ Ns}</math>


	The angle, <math>\alpha </math>, can be found using trigonometry:		The angle, <math>\alpha </math>, can be found using trigonometry:

Ian den 24 september 2009 kl. 14.06

Ian — Thu, 24 Sep 2009 14:06:19 GMT

← Äldre version		Versionen från 24 september 2009 kl. 14.06
Rad 14:		Rad 14:
	Momentum is defined as the product of the mass and velocity of a body.		Momentum is defined as the product of the mass and velocity of a body.

-	Momentum = mv or Momentum = mv	+	Momentum =
		+	<math>mv</math>
		+	or Momentum =
		+	<math>m\mathbf{v}</math>

-	If the velocity of a body, of mass m, changes from u to v, then

-	Change in momentum or impulse = mv - mu.	+	If the velocity of a body, of mass
		+	<math>m</math>
		+	, changes from
		+	<math>u</math>
		+	to
		+	<math>v</math>
		+	, then
		+
		+	Change in momentum or impulse =
		+	<math>mv</math>
		+	-
		+	<math>mu</math>
		+

	We often use		We often use
Rad 31:		Rad 45:
	or		or
	<math>\mathbf{I}=\mathbf{F}t</math>		<math>\mathbf{I}=\mathbf{F}t</math>
-	can be used where F is constant or where F is assumed to be constant and represents the average force.	+	can be used where
		+	<math>F</math>
		+	is constant or where
		+	<math>F</math>
		+	is assumed to be constant and represents the average force.
		+
		+
		+
		+	'''[[Example 15.1]]'''


-	Example 15.1
	Find the momentum of a car, of mass 1.2 tonnes, travelling in a straight line		Find the momentum of a car, of mass 1.2 tonnes, travelling in a straight line
-	at 30 ms-1.	+	at 30 <math>\text{m}{{\text{s}}^{-1}}</math>.
		+
		+	'''Solution'''

-	Solution
	Note that 1.2 tonnes is 1200 kg. Using Momentum = mv gives:		Note that 1.2 tonnes is 1200 kg. Using Momentum = mv gives:

Rad 48:		Rad 70:


-	Example 15.2	+	'''[[Example 15.2]]'''
-	A car, of mass 1.1 tonnes, which was initially travelling at 6 ms-1 is brought to rest in 2.2 seconds by a wall. Find the average force exerted by the wall on the car.	+
		+	A car, of mass 1.1 tonnes, which was initially travelling at 6 <math>\text{m}{{\text{s}}^{-1}}</math> is brought to rest in 2.2 seconds by a wall. Find the average force exerted by the wall on the car.
		+
		+	'''Solution'''

-	Solution
	First find the impulse (change in momentum).		First find the impulse (change in momentum).

Rad 73:		Rad 97:

	The average force has magnitude 3000 N		The average force has magnitude 3000 N
-	Example 15.3
-	The speed of a car, travelling in a straight line, is reduced from 20 ms-1 to
-	15 ms-1 in 20 seconds. The mass of the car is 1200 kg. Find the average force acting on the car.

-	Solution	+
		+	'''[[Example 15.3]]'''
		+
		+
		+	The speed of a car, travelling in a straight line, is reduced from 20 <math>\text{m}{{\text{s}}^{-1}}</math> to
		+	15 <math>\text{m}{{\text{s}}^{-1}}</math> in 20 seconds. The mass of the car is 1200 kg. Find the average force acting on the car.
		+
		+	'''Solution'''
		+
	First find the impulse (change in momentum).		First find the impulse (change in momentum).

Rad 99:		Rad 128:


-	Example 15.4	+	'''[[Example 15.4]]'''
-	A ball has a mass of 200 grams. It initially travels horizontally at 10 ms-1. After being hit it travels at 16 ms-1 at an angle of 60 above the horizontal. Find the initial and final momentum of the ball and the change in momentum.	+

-	~~Solution~~	+	A ball has a mass of 200 grams. It initially travels horizontally at 10 <math>\text{m}{{\text{s}}^{-1}}</math>. After being hit it travels at 16 <math>\text{m}{{\text{s}}^{-1}}</math> at an angle of 60<math>{}^\circ </math> above the horizontal. Find the initial and final momentum of the ball and the change in momentum.

-	The diagrams show the initial and final velocities of the ball and the unit vectors i and j.	+	'''Solution'''
		+
		+	The diagrams show the initial and final velocities of the ball and the unit vectors <math>\mathbf{i}</math> and <math>\mathbf{j}</math>.


Rad 135:		Rad 165:
	\end{align}</math>		\end{align}</math>

-	Example 15.5	+	'''[[Example 15.5]]'''
-	A ball is travelling horizontally at 8 ms-1, when it is hit. After being hit it initially travels upwards at 6 ms-1. The mass of the ball is 300 grams. Find the magnitude and direction of the impulse on the ball.	+
		+	A ball is travelling horizontally at 8 <math>\text{m}{{\text{s}}^{-1}}</math>, when it is hit. After being hit it initially travels upwards at 6 <math>\text{m}{{\text{s}}^{-1}}</math>. The mass of the ball is 300 grams. Find the magnitude and direction of the impulse on the ball.
		+
		+	'''Solution'''

-	~~Solution~~	+	The diagrams show the initial and final velocities of the ball and the unit vectors <math>\mathbf{i}</math> and <math>\mathbf{j}</math>.
-	The diagrams show the initial and final velocities of the ball and the unit vectors i and j.	+


Rad 169:		Rad 201:
	The diagram shows the impulse.		The diagram shows the impulse.

-	The magnitude of the impulse, I, is found	+	The magnitude of the impulse, <math>I</math>, is found
	using Pythagoras:		using Pythagoras:

Rad 176:		Rad 208:


-	The angle, α, can be found using trigonometry:	+	The angle, <math>\alpha </math>, can be found using trigonometry:

Ian den 24 september 2009 kl. 13.42

Ian — Thu, 24 Sep 2009 13:42:05 GMT

← Äldre version		Versionen från 24 september 2009 kl. 13.42
Rad 6:		Rad 6:
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}
		+
		+
		+	15. Momentum and Impulse
		+
		+	Key Results
		+
		+	Momentum is defined as the product of the mass and velocity of a body.
		+
		+	Momentum = mv or Momentum = mv
		+
		+	If the velocity of a body, of mass m, changes from u to v, then
		+
		+	Change in momentum or impulse = mv - mu.
		+
		+	We often use
		+
		+	<math>I=mv-mu</math>
		+	or
		+	<math>\mathbf{I}=m\mathbf{v}-m\mathbf{u}</math>
		+
		+
		+	The relationships
		+	<math>I=Ft</math>
		+	or
		+	<math>\mathbf{I}=\mathbf{F}t</math>
		+	can be used where F is constant or where F is assumed to be constant and represents the average force.
		+
		+
		+	Example 15.1
		+	Find the momentum of a car, of mass 1.2 tonnes, travelling in a straight line
		+	at 30 ms-1.
		+
		+	Solution
		+	Note that 1.2 tonnes is 1200 kg. Using Momentum = mv gives:
		+
		+
		+	<math>\begin{align}
		+	& \text{Momentum}=1200\times 30 \\
		+	& =36000\text{ Ns}
		+	\end{align}</math>
		+
		+
		+	Example 15.2
		+	A car, of mass 1.1 tonnes, which was initially travelling at 6 ms-1 is brought to rest in 2.2 seconds by a wall. Find the average force exerted by the wall on the car.
		+
		+	Solution
		+	First find the impulse (change in momentum).
		+
		+	<math>\begin{align}
		+	& \text{Impulse }=mv-mu \\
		+	& =1100\times 0-1100\times 6 \\
		+	& =-6600\text{ Ns}
		+	\end{align}</math>
		+
		+
		+	Use the formula
		+	<math>\text{Impulse }=Ft</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& -6600=F\times 2.2 \\
		+	& F=-3000\text{ N}
		+	\end{align}</math>
		+
		+
		+	The average force has magnitude 3000 N
		+	Example 15.3
		+	The speed of a car, travelling in a straight line, is reduced from 20 ms-1 to
		+	15 ms-1 in 20 seconds. The mass of the car is 1200 kg. Find the average force acting on the car.
		+
		+	Solution
		+	First find the impulse (change in momentum).
		+
		+
		+	<math>\begin{align}
		+	& \text{Impulse }=mv-mu \\
		+	& 1200\times 15-1200\times 20 \\
		+	& =-6000\text{ Ns}
		+	\end{align}</math>
		+
		+
		+	Use the formula
		+	<math>\text{Impulse }=Ft</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& -6000=F\times 20 \\
		+	& F=-300\text{ N} \\
		+	\end{align}</math>
		+
		+
		+	Example 15.4
		+	A ball has a mass of 200 grams. It initially travels horizontally at 10 ms-1. After being hit it travels at 16 ms-1 at an angle of 60 above the horizontal. Find the initial and final momentum of the ball and the change in momentum.
		+
		+	Solution
		+
		+	The diagrams show the initial and final velocities of the ball and the unit vectors i and j.
		+
		+
		+
		+
		+
		+
		+
		+
		+	<math>\begin{align}
		+	& \text{Initial Momentum }=\text{ m}\mathbf{u} \\
		+	& =0.2\times 10\mathbf{i} \\
		+	& =2\mathbf{i}
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& \text{Final Momentum }=m\mathbf{v} \\
		+	& =0.2(16\cos 60{}^\circ \mathbf{i}+16\sin 60{}^\circ \mathbf{j}) \\
		+	& =3.2\cos 60{}^\circ \mathbf{i}+3.2\sin 60{}^\circ \mathbf{j}
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\
		+	& =3.2\cos 60{}^\circ \mathbf{i}+3.2\sin 60{}^\circ \mathbf{j}-2\mathbf{i} \\
		+	& =-0.4\mathbf{i}+2.77\mathbf{j}
		+	\end{align}</math>
		+
		+	Example 15.5
		+	A ball is travelling horizontally at 8 ms-1, when it is hit. After being hit it initially travels upwards at 6 ms-1. The mass of the ball is 300 grams. Find the magnitude and direction of the impulse on the ball.
		+
		+	Solution
		+	The diagrams show the initial and final velocities of the ball and the unit vectors i and j.
		+
		+
		+
		+
		+
		+
		+	<math>\begin{align}
		+	& \text{Initial Momentum }=\text{ }m\mathbf{u} \\
		+	& =0.3\times 8\mathbf{i} \\
		+	& =2.4\mathbf{i}
		+	\end{align}</math>
		+
		+
		+	<math>\begin{align}
		+	& \text{Final Momentum }=m\mathbf{v} \\
		+	& =0.3\times 6\mathbf{j} \\
		+	& =1.8\mathbf{j}
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& \text{Impulse }=m\mathbf{v}-m\mathbf{u} \\
		+	& =1.8\mathbf{j}-2.4\mathbf{i}
		+	\end{align}</math>
		+
		+
		+	The diagram shows the impulse.
		+
		+	The magnitude of the impulse, I, is found
		+	using Pythagoras:
		+
		+
		+	<math>I=\sqrt{{{1.8}^{2}}+{{2.4}^{2}}}=3\text{ Ns}</math>
		+
		+
		+	The angle, α, can be found using trigonometry:

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Ian — Thu, 24 Sep 2009 13:21:05 GMT

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15. Moment och impuls - Versionshistorik

Tek: flyttade 15. Momentum and impulse till 15. Moment och impuls

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