Relative Velocity - Revision history

YorickAndeweg: /* History */

2019-08-11T17:00:01Z

History

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	==History==		==History==

	The equation to convert between relative velocities was the natural result of the Galilean transformation equations. Although they are named after [[Galileo Galilei]], those equations are in turn the natural result of Newton's perception of space and time as he described them in <i>Principia Mathematica</i>. Although no one in particular is credited with the Galilean transformation equations, both Newton and Galileo (and many other physicists) used them. They remained unchanged until the early 20th century, at which time [[Albert Einstein]] rewrote them to accommodate his theories of relativity. Einstein also famously said that it is impossible to tell the velocity of one's reference frame by performing experiments locally, popularizing the idea that no frame was more legitimate or correct than any other frame.		The equation to convert between relative velocities was the natural result of the Galilean transformation equations. Although they are named after [[Galileo Galilei]], those equations are in turn the natural result of Newton's perception of space and time as he described them in <i>Principia Mathematica</i><sup>[[#References\|3]]</sup>. Although no one in particular is credited with the Galilean transformation equations, both Newton and Galileo (and many other physicists) used them. They remained unchanged until the early 20th century, at which time [[Albert Einstein]] rewrote them to accommodate his theories of relativity. Einstein also famously said that it is impossible to tell the velocity of one's reference frame by performing experiments locally, popularizing the idea that no frame was more legitimate or correct than any other frame.

	==See also==		==See also==

YorickAndeweg: /* References */

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References

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	#Chabay, Ruth W., and Bruce A. Sherwood. <i>Matter and Interactions</i>. Hoboken, NJ: Wiley, 2011. Print.		#Chabay, Ruth W., and Bruce A. Sherwood. <i>Matter and Interactions</i>. Hoboken, NJ: Wiley, 2011. Print.
	#Wikipedia contributors. (2019, June 7). Relative velocity. In Wikipedia, The Free Encyclopedia. Retrieved 16:06, August 11, 2019, from https://en.wikipedia.org/w/index.php?title=Relative_velocity&oldid=900770643		#Wikipedia contributors. (2019, June 7). Relative velocity. In Wikipedia, The Free Encyclopedia. Retrieved 16:06, August 11, 2019, from https://en.wikipedia.org/w/index.php?title=Relative_velocity&oldid=900770643
			#Wikipedia contributors. (2019, July 18). Galilean transformation. In Wikipedia, The Free Encyclopedia. Retrieved 16:57, August 11, 2019, from https://en.wikipedia.org/w/index.php?title=Galilean_transformation&oldid=906797474
	#[http://hyperphysics.phy-astr.gsu.edu/hbase/relmot.html#c1 HyperPhysics: Relative Motion]		#[http://hyperphysics.phy-astr.gsu.edu/hbase/relmot.html#c1 HyperPhysics: Relative Motion]
	#[https://commons.wikimedia.org/wiki/File:Wikipage_pic.PNG Example from Frame of Reference]		#[https://commons.wikimedia.org/wiki/File:Wikipage_pic.PNG Example from Frame of Reference]

YorickAndeweg: /* See also */

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YorickAndeweg: /* References */

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References

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	==References==		==References==
	~~[http://hyperphysics.phy-astr.gsu.edu/hbase/relmot.html#c1 HyperPhysics: Relative Motion]~~

	[https://commons.wikimedia.org/wiki/File:Wikipage_pic.PNG Example from Frame of Reference]		#Chabay, Ruth W., and Bruce A. Sherwood. <i>Matter and Interactions</i>. Hoboken, NJ: Wiley, 2011. Print.
			#Wikipedia contributors. (2019, June 7). Relative velocity. In Wikipedia, The Free Encyclopedia. Retrieved 16:06, August 11, 2019, from https://en.wikipedia.org/w/index.php?title=Relative_velocity&oldid=900770643
			#[http://hyperphysics.phy-astr.gsu.edu/hbase/relmot.html#c1 HyperPhysics: Relative Motion]
			#[https://commons.wikimedia.org/wiki/File:Wikipage_pic.PNG Example from Frame of Reference]


	[[Category:Velocity and Momentum]]		[[Category:Velocity and Momentum]]

YorickAndeweg: /* See also */

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YorickAndeweg: /* 3. (Difficult) */

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3. (Difficult)

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	Alice, Bob, and Charlie, go ice skating together. Alice and Bob are skating next to each other in the same direction at 4m/s (relative to the ice skating rink) when Bob accidentally collides with Charlie, who is initially at rest (relative to the ice skating rink). Bob and Charlie grab onto each other as they collide to stabilize themselves, so their collision is inelastic and Bob and Charlie end up at the same speed. Bob weighs 60kg and Charlie weighs 80kg. What is the final speed of Bob and Charlie? Assume that there is no friction between the skates and the ice. <b>Solve this problem using conservation of momentum in two different reference frames: the ice skating rink and Alice. Show that both solutions agree with each other.</b>		Alice, Bob, and Charlie, go ice skating together. Alice and Bob are skating next to each other in the same direction at 4m/s (relative to the ice skating rink) when Bob accidentally collides with Charlie, who is initially at rest (relative to the ice skating rink). Bob and Charlie grab onto each other as they collide to stabilize themselves, so their collision is inelastic and Bob and Charlie end up at the same speed. Bob weighs 60kg and Charlie weighs 80kg. What is the final speed of Bob and Charlie? Assume that there is no friction between the skates and the ice. <b>Solve this problem using conservation of momentum in two different reference frames: the ice skating rink and Alice. Show that both solutions agree with each other.</b>

	[[File:Relativeveliceskating]]		[[File:Relativeveliceskating.png]]

	Solution:		Solution:

YorickAndeweg: /* 3. (Difficult) */

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3. (Difficult)

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	This is a one-dimensional problem because the trajectories of both Bob and Charlie are constrained to the same axis. Let us define the positive direction as the direction in which Bob is initially traveling.		This is a one-dimensional problem because the trajectories of both Bob and Charlie are constrained to the same axis. Let us define the positive direction as the direction in which Bob is initially traveling.

	First let us solve the problem from the reference frame of the ice skating rink. Relative to the ice skating rink, the initial velocities of Bob and Charlie are 4m/s and 0m/s respectively. This means their momenta are 240kgm/s and 0kgm/s respectively. The initial total momentum of the Bob-Charlie system is therefore 240kgm/s. Since this ~~is a~~ collision ~~that~~ occurs in isolation of any external forces, the final momentum of the system is also 240kgm/s. Bob and Charlie are moving together at the end of the collision. Their combined mass is 140kg, so their velocity must be <math>\frac{240}{140} = 1.71</math>m/s.		First let us solve the problem from the reference frame of the ice skating rink. Relative to the ice skating rink, the initial velocities of Bob and Charlie are 4m/s and 0m/s respectively. This means their momenta are 240kgm/s and 0kgm/s respectively. The initial total momentum of the Bob-Charlie system is therefore 240kgm/s. Since this collision occurs in isolation of any external forces, the final momentum of the system is also 240kgm/s. Bob and Charlie are moving together at the end of the collision. Their combined mass is 140kg, so their velocity must be <math>\frac{240}{140} = 1.71</math>m/s.

	Now let us solve the problem from the reference frame of Alice. Relative to the ice skating rink, the initial velocities of Bob and Charlie are 0m/s and -4m/s respectively. (If you cannot tell this intuitively, you can calculate it by using the relative velocity transformation equation ~~asdfasdfasd~~.)		Now let us solve the problem from the reference frame of Alice. Relative to the ice skating rink, the initial velocities of Bob and Charlie are 0m/s and -4m/s respectively. (If you cannot tell this intuitively, you can calculate it by using the relative velocity transformation equation <math>\vec{u} = \vec{u}' + \vec{v}</math>.) This means their momenta are 0kgm/s and -320kgm/s respectively. The initial total momentum of the Bob-Charlie system is therefore -320kgm/s. Since this collision occurs in isolation of any external forces, the final momentum of the system is also -320kgm/s. Bob and Charlie are moving together at the end of the collision. Their combined mass is 140kg, so their velocity must be <math>\frac{-320}{140} = -2.29</math>m/s.

			Now let us show that our two answers are consistent with each other despite being numerically different. Relative to Alice, we found that the final velocity of Bob and Charlie is -2.29m/s. However, Alice is herself moving relative to the ice skating rink at a velocity of 4m/s. Using the equation <math>\vec{u} = \vec{u}' + \vec{v}</math>, the final velocity of Bob and Charlie relative to the ice skating rink is -2.29 + 4 = 1.71m/s. This numerically matches the calculations we did from the reference frame of the ice skating rink.

			This problem shows that the relative nature of velocity translates to other properties that depend on velocity, such as kinetic energy and momentum. Calculations involving these properties can be correctly performed from a variety of reference frames.

	== See also ==		== See also ==

YorickAndeweg: /* Examples */

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Examples

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	<math>\theta = \sin^{-1}(\frac{0.6}{1.5}) = 23.6</math>		<math>\theta = \sin^{-1}(\frac{0.6}{1.5}) = 23.6</math>

			===3. (Difficult)===

			Note: this problem requires knowledge of [[Collisions\|collisions]].

			Alice, Bob, and Charlie, go ice skating together. Alice and Bob are skating next to each other in the same direction at 4m/s (relative to the ice skating rink) when Bob accidentally collides with Charlie, who is initially at rest (relative to the ice skating rink). Bob and Charlie grab onto each other as they collide to stabilize themselves, so their collision is inelastic and Bob and Charlie end up at the same speed. Bob weighs 60kg and Charlie weighs 80kg. What is the final speed of Bob and Charlie? Assume that there is no friction between the skates and the ice. <b>Solve this problem using conservation of momentum in two different reference frames: the ice skating rink and Alice. Show that both solutions agree with each other.</b>

			[[File:Relativeveliceskating]]

			Solution:

			This is a one-dimensional problem because the trajectories of both Bob and Charlie are constrained to the same axis. Let us define the positive direction as the direction in which Bob is initially traveling.

			First let us solve the problem from the reference frame of the ice skating rink. Relative to the ice skating rink, the initial velocities of Bob and Charlie are 4m/s and 0m/s respectively. This means their momenta are 240kgm/s and 0kgm/s respectively. The initial total momentum of the Bob-Charlie system is therefore 240kgm/s. Since this is a collision that occurs in isolation of any external forces, the final momentum of the system is also 240kgm/s. Bob and Charlie are moving together at the end of the collision. Their combined mass is 140kg, so their velocity must be <math>\frac{240}{140} = 1.71</math>m/s.

			Now let us solve the problem from the reference frame of Alice. Relative to the ice skating rink, the initial velocities of Bob and Charlie are 0m/s and -4m/s respectively. (If you cannot tell this intuitively, you can calculate it by using the relative velocity transformation equation asdfasdfasd.)

	== See also ==		== See also ==

YorickAndeweg: /* 2. (Middling) */

2019-08-09T20:00:51Z

2. (Middling)

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	<math>\sin\theta = \frac{0.6}{1.5}</math>		<math>\sin\theta = \frac{0.6}{1.5}</math>

	<math>\theta = \sin^{-1}(\frac{0.6}{1.5})</math>		<math>\theta = \sin^{-1}(\frac{0.6}{1.5}) = 23.6</math>

	== See also ==		== See also ==

YorickAndeweg: /* 2. (Middling) */

2019-08-09T19:59:54Z

2. (Middling)

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	===2. (Middling)===		===2. (Middling)===

	A rower can row their rowboat at a speed of 1.5m/s relative to the water. The rower comes to a straight river flowing to the east at a speed of 0.6m/s. The rower wants to cross the river perpendicularly from south to north. At what angle <math>\theta</math> west of north should the rower row so that the boat travels in a straight perpendicular line across the river?		A rower can row their rowboat at a speed of 1.5m/s relative to the water. The rower comes to a straight river flowing to the east at a speed of 0.6m/s. The rower wants to cross the river perpendicularly from south to north. At what angle <math>\theta</math> west of north should the rower row so that the boat travels in a straight perpendicular line across the river from the point of view of the riverbank?

	[[File:Relativevelrowboat.png\|600px]]		[[File:Relativevelrowboat.png\|600px]]
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	Let us call east and north the +x and +y directions respectively.		Let us call east and north the +x and +y directions respectively.

	In this problem, there are two frames of reference: the water and the riverbank. Using trigonometry, we can find the velocity of the rowboat relative to the water: <math>\left \langle {-1.5\sin\theta,1.5\cos\theta} \right \rangle</math>.		In this problem, there are two frames of reference: the water and the riverbank. Using trigonometry, we can find the velocity of the rowboat relative to the water: <math>\left \langle {-1.5\sin\theta,1.5\cos\theta} \right \rangle</math>m/s.

	~~Now~~, by the ~~formula~~		By the formula <math>\vec{u} = \vec{u}' + \vec{v}</math>, the velocity of the rowboat relative to the riverbank equals the velocity of the rowboat relative to the water plus the velocity of the water relative to the riverbank:

			<math>\vec{u}_{rowboat/riverbank} = \left \langle {-1.5\sin\theta,1.5\cos\theta}m/s \right \rangle + \left \langle {0.6,0} \right \rangle</math>m/s

			<math>\vec{u}_{rowboat/riverbank} = \left \langle {-1.5\sin\theta+0.6,1.5\cos\theta}m/s \right \rangle</math>m/s

			The rower's goal is for the boat to cross the river perpendicularly from the point of view of the riverbank. That is, <math>\vec{u}_{rowboat/riverbank}</math> must have only a y component; the x component must be equal to 0, or the boat's path would not be perpendicular to the river.

			<math>-1.5\sin\theta+0.6 = 0</math>

			<math>-1.5\sin\theta = -0.6</math>

			<math>\sin\theta = \frac{0.6}{1.5}</math>

			<math>\theta = \sin^{-1}(\frac{0.6}{1.5})</math>

	== See also ==		== See also ==

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	This problem shows that the relative nature of velocity translates to other properties that depend on velocity, such as kinetic energy and momentum. Calculations involving these properties can be correctly performed from a variety of reference frames.		This problem shows that the relative nature of velocity translates to other properties that depend on velocity, such as kinetic energy and momentum. Calculations involving these properties can be correctly performed from a variety of reference frames.

	== See also ==		==Connectedness==

			Since the goal of the field of mechanics is to predict the motion of systems, velocity is critical to this branch of physics. In addition to being a value of interest in and of itself, properties such as Linear Momentum, Angular Momentum, and Magnetic Force depend on the velocities of objects.

			Because velocity is relative, properties that depend on it are also relative. That is, properties such as momentum and energy might have different values when measured relative to different reference frames. This fits into the larger theme in physics that a problem can be solved from different, equally valid perspectives.

			==History==

			The equation to convert between relative velocities was the natural result of the Galilean transformation equations. Although they are named after [[Galileo Galilei]], those equations are in turn the natural result of Newton's perception of space and time as he described them in <i>Principia Mathematica</i>. Although no one in particular is credited with the Galilean transformation equations, both Newton and Galileo (and many other physicists) used them. They remained unchanged until the early 20th century, at which time [[Albert Einstein]] rewrote them to accommodate his theories of relativity. Einstein also famously said that it is impossible to tell the velocity of one's reference frame by performing experiments locally, popularizing the idea that no frame was more legitimate or correct than any other frame.

			==See also==
	*[[Velocity]]		*[[Velocity]]

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	== See also ==		== See also ==
	*[[Velocity]]		*[[Velocity]]
	*[[Speed vs Velocity]]
	*[[Derivation of Average Velocity]]
	*[[2-Dimensional Motion]]

	===External links===		===External links===