Inelastic Collisions - Revision history

Gmyers30 at 01:54, 2 December 2025

2025-12-02T01:54:20Z

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	Edited/Claimed by Gillian Myers, Fall 2025		Edited/Claimed by Gillian Myers, Fall 2025

	~~Edited/Claimed by Jessica Ouyang, Fall 2018~~

	~~Edited/Claimed by Vansh Kareer, Spring 2017~~

	~~Edited/Claimed by Abigail Brady, Fall 2017~~


	This topic covers the concept of inelastic collisions and the results of the collisions on the bodies involved. Included are the methods that can be applied to solve problems related to these types of collisions, along with worked examples.		This topic covers the concept of inelastic collisions and the results of the collisions on the bodies involved. Included are the methods that can be applied to solve problems related to these types of collisions, along with worked examples.
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	[[File:Pat_Devlin_(American_football).jpg]]		[[File:Pat_Devlin_(American_football).jpg]]


			The science of explosions is not just used for action movies, but to prevent accidents in industrial settings as well. When factories process sugar (grinding it down), it releases large amounts of sugar dust into the air. Other instances also include handling large packages of sugar and refineries which finely mill sugar. Grinding sugar down increases its surface area while the chemical composition of sugar actually makes it extremely prone to combustion when ignited. Due to a factories confined space and the constant milling and handling of sugar, this becomes a significant hazard. Any spark from a machine can instantly ignite the air and other sugar around it. This was the case of the Imperial Sugar Refinery Accident in 2008, where improper maitenance of machinery resulted in an explosion in the factories basement, which traveled to the main buildings in a devastating chain reaction.


	== History ==		== History ==

Gmyers30 at 01:24, 2 December 2025

2025-12-02T01:24:05Z

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	2)Label all initial states in both X and Y directions. Most likely, the problem will start out with a stationary object that has not exploded, which will make your initial conditions = 0. If your object is not initially stationary, make sure to adjust for any initial forces or velocities that are not 0.		2)Label all initial states in both X and Y directions. Most likely, the problem will start out with a stationary object that has not exploded, which will make your initial conditions = 0. If your object is not initially stationary, make sure to adjust for any initial forces or velocities that are not 0.

	3)Label all final states in both X and Y directions. If there are objects flying at different angles, make sure that the Y coordinate = <math>mvsin(~~theta~~)</math> and the X coordinate = <math>mvcos(~~theta~~)</math>.		3)Label all final states in both X and Y directions. If there are objects flying at different angles, make sure that the Y coordinate = <math>mvsin(θ)</math> and the X coordinate = <math>mvcos(θ)</math>.

	4)Set the final and initial states for each coordinate equal to eachother, and solve.		4)Set the final and initial states for each coordinate equal to eachother, and solve.

	Special Note) If you need to find the velocity of the an object that is thrown at an angle, leave the object labeled as <math>P_x</math> and <math>P_y</math> in your equation. DO NOT write it out as <math>mv(sin/cos(~~θ~~))</math>.		Special Note) If you need to find the velocity of the an object that is thrown at an angle, leave the object labeled as <math>P_x</math> and <math>P_y</math> in your equation. DO NOT write it out as <math>mv(sin/cos(θ))</math>.

	===Simple===		===Simple===

Gmyers30 at 01:23, 2 December 2025

2025-12-02T01:23:04Z

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	4)Set the final and initial states for each coordinate equal to eachother, and solve.		4)Set the final and initial states for each coordinate equal to eachother, and solve.

	Special Note) If you need to find the velocity of the an object that is thrown at an angle, leave the object labeled as <math>~~P_#x~~</math> and <math>~~P_#y~~</math> in your equation. DO NOT write it out as <math>~~mv_3~~(sin/cos(theta))</math>.		Special Note) If you need to find the velocity of the an object that is thrown at an angle, leave the object labeled as <math>P_x</math> and <math>P_y</math> in your equation. DO NOT write it out as <math>mv(sin/cos(θ))</math>.

	===Simple===		===Simple===

Gmyers30 at 01:20, 2 December 2025

2025-12-02T01:20:28Z

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	'''Types of Inelastic Collisions'''		'''Types of Inelastic Collisions'''

			===Maximally Inelastic===
	One type of inelastic collisions is the maximally inelastic collision, also known as the perfectly inelastic collision, in which two objects collide and then remain connected for the duration of the observable interaction. This does not indicate an end to movement, as momentum must be conserved by the end of the interaction. See problem 1 for an example. More information about such problems can be found in the page [[Maximally Inelastic Collision]].		One type of inelastic collisions is the maximally inelastic collision, also known as the perfectly inelastic collision, in which two objects collide and then remain connected for the duration of the observable interaction. This does not indicate an end to movement, as momentum must be conserved by the end of the interaction. See problem 1 for an example. More information about such problems can be found in the page [[Maximally Inelastic Collision]].

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	In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf. An example of a typically inelastic collision is a car crash.		In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf. An example of a typically inelastic collision is a car crash.

			===Explosions===
	Explosions are a somewhat special case of inelastic collisions. In these types of problems, an object bursts apart, and some of its internal energy is converted into kinetic energy. As a result, ΔK is actually positive and ΔEint is actually negative, in that internal energy is used up in the course of the explosion and causes an increase in the speed of the particles involved. This can be seen in the relation Ki < Kf, wherein the change in internal energy of the lends itself to the increase in kinetic energy from Ki to Kf, which directly correlates to an increase in particle speed in the explosion's aftermath.		Explosions are a somewhat special case of inelastic collisions. Its like a collision in reverse. In these types of problems, an object bursts apart, and some of its internal energy is converted into kinetic energy. As a result, ΔK is actually positive and ΔEint is actually negative, in that internal energy is used up in the course of the explosion and causes an increase in the speed of the particles involved. This can be seen in the relation Ki < Kf, wherein the change in internal energy of the lends itself to the increase in kinetic energy from Ki to Kf, which directly correlates to an increase in particle speed in the explosion's aftermath.

	[[File:Wiki Image 5.jpg\|thumb\|center\|1020x620px]]		[[File:Wiki Image 5.jpg\|thumb\|center\|1020x620px]]
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	6) Next, you can find the change in kinetic energy, or ΔK, and this will be equal to the negative change in the internal energy (whether it be heat, rotation, etc.) gained by the objects in the course of the collision. In other words <math>ΔE_{int} = -ΔK</math>.		6) Next, you can find the change in kinetic energy, or ΔK, and this will be equal to the negative change in the internal energy (whether it be heat, rotation, etc.) gained by the objects in the course of the collision. In other words <math>ΔE_{int} = -ΔK</math>.

			'''Typical Process for solving Explosions'''

			In Explosions, momentum is still conserved. Additionally, each part of the object are usually thrown in multiple directions, making it more challenging due to angles and coordinate changes. However, because momentum is conserved, <math>P_f = P_i</math> still applies.

			1)Remember the conservation of momentum <math>P_f = P_i</math>, which can be further parsed into X and Y coordinates: <math>P_fx = P_ix</math> & <math>P_fy = P_iy</math>.

			2)Label all initial states in both X and Y directions. Most likely, the problem will start out with a stationary object that has not exploded, which will make your initial conditions = 0. If your object is not initially stationary, make sure to adjust for any initial forces or velocities that are not 0.

			3)Label all final states in both X and Y directions. If there are objects flying at different angles, make sure that the Y coordinate = <math>mvsin(theta)</math> and the X coordinate = <math>mvcos(theta)</math>.

			4)Set the final and initial states for each coordinate equal to eachother, and solve.

			Special Note) If you need to find the velocity of the an object that is thrown at an angle, leave the object labeled as <math>P_#x</math> and <math>P_#y</math> in your equation. DO NOT write it out as <math>mv_3(sin/cos(theta))</math>.

	===Simple===		===Simple===

Gmyers30 at 00:51, 2 December 2025

2025-12-02T00:51:54Z

← Older revision		Revision as of 20:51, 1 December 2025
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	Created by Csorensen6		Created by Csorensen6

			Edited/Claimed by Gillian Myers, Fall 2025

	Edited/Claimed by Jessica Ouyang, Fall 2018		Edited/Claimed by Jessica Ouyang, Fall 2018
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	Edited/Claimed by Abigail Brady, Fall 2017		Edited/Claimed by Abigail Brady, Fall 2017


	This topic covers the concept of inelastic collisions and the results of the collisions on the bodies involved. Included are the methods that can be applied to solve problems related to these types of collisions, along with worked examples.		This topic covers the concept of inelastic collisions and the results of the collisions on the bodies involved. Included are the methods that can be applied to solve problems related to these types of collisions, along with worked examples.

Sydney: /* Examples */

2019-07-08T20:15:29Z

Examples

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	1) Draw a picture of the initial and final states.		1) Draw a picture of the initial and final states.

	2) Given that collisions involve extremely large forces acting over short time intervals, it is accurate to say that ~~Fnet~~,ext = 0, because the external forces are typically much, much smaller than the internal forces involved in the collision.		2) Given that collisions involve extremely large forces acting over short time intervals, it is accurate to say that <math>F_{net,ext} = 0</math>, because the external forces are typically much, much smaller than the internal forces involved in the collision.

	3) Knowing that ~~Fnet~~,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ~~ΔPsystem~~ = ~~Fnet~~,ext*ΔT, and if ~~Fnet~~,ext is zero, then the right side of that equation is also zero.		3) Knowing that <math>F_{net,ext} = 0</math>, this means that momentum is conserved and that <math>P_f = P_i</math>. This comes from the momentum principle, in that <math>ΔP_{system} = F_{net,ext}ΔT</math>, and if <math>F_{net,ext}</math> is zero, then the right side of that equation is also zero.

	4) Next, knowing that Pf = Pi, you can solve for any unknown velocities of the objects involved via the momentum principle.		4) Next, knowing that <math>P_f = P_i</math>, you can solve for any unknown velocities of the objects involved via the momentum principle.

	5) Having determined the velocity of each object, you can then determine the initial and final kinetic energies of the system using K = (1/2)m\|v\|^2 for each object.		5) Having determined the velocity of each object, you can then determine the initial and final kinetic energies of the system using <math>K = (1/2)m\|v\|^2</math> for each object.

	6) Next, you can find the change in kinetic energy, or ΔK, and this will be equal to the negative change in the internal energy (whether it be heat, rotation, etc.) gained by the objects in the course of the collision. In other words ~~ΔEint~~ = -ΔK.		6) Next, you can find the change in kinetic energy, or ΔK, and this will be equal to the negative change in the internal energy (whether it be heat, rotation, etc.) gained by the objects in the course of the collision. In other words <math>ΔE_{int} = -ΔK</math>.

	===Simple===		===Simple===
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	<math> m_{A}v_{A,i} + m_{B}v_{B,i} = m_{f}v_{f} </math>		<math> m_{A}v_{A,i} + m_{B}v_{B,i} = m_{f}v_{f} </math>

	(2)(5) + (1.5)(2) = (2 + 1.5)(vf)		<math>(2)(5) + (1.5)(2) = (2 + 1.5)(v_f)</math>

	10 + 3 = 3.~~5vf~~		<math>10 + 3 = 3.5v_f</math>

	vf = 3.714 m/s		<math>v_f = 3.714 m/s</math>

	===Middling===		===Middling===
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	(c) Calculate the kinetic energy of the bullet plus the block after the collision:		(c) Calculate the kinetic energy of the bullet plus the block after the collision:

	d) Was this collision elastic or inelastic?		(d) Was this collision elastic or inelastic?

	(e) Calculate the rise in thermal energy of the bullet plus block as a result of the collision:		(e) Calculate the rise in thermal energy of the bullet plus block as a result of the collision:
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	d)		d)
	Due to the fact that the final kinetic energy is less than the initial kinetic energy (change in kinetic energy), while the momentum of the interaction is conserved, the interaction can be said to be inelastic. The result of the change in kinetic energy is a rise in internal/thermal energy.		Due to the fact that the final kinetic energy is less than the initial kinetic energy (change in kinetic energy), while the
			momentum of the interaction is conserved, the interaction can be said to be inelastic. The result of the change in kinetic
			energy is a rise in internal/thermal energy.


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	e)		e)
	As the collision takes place in a closed system, and the collision considers forces that render frictional forces by surroundings negligible, there is no transfer of energy Q between the surroundings and the system.		As the collision takes place in a closed system, and the collision considers forces that render frictional forces by surroundings
			negligible, there is no transfer of energy Q between the surroundings and the system.
	Q=0 J		Q=0 J

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	ΔEinternal = Ki - Kf		ΔEinternal = Ki - Kf

	ΔEinternal = (1/2)(15)(10^2 + 17^2 + 0^2) - (1/2)(3)(8^2 + 7^2 + 0^2) - (1/2)(4)(15^2 + 0^2 + 0^2) - (1/2)(3.5)(12^2 + (-5)^ + 0^2) - (1/2)(4.5)(5.33^2 + 55.89^2 + 0^2)		ΔEinternal = (1/2)(15)(10^2 + 17^2 + 0^2) - (1/2)(3)(8^2 + 7^2 + 0^2) - (1/2)(4)(15^2 + 0^2 + 0^2) - (1/2)(3.5)(12^2 + (-5)^2 + 0^2)
			- (1/2)(4.5)(5.33^2 + 55.89^2 + 0^2)
	ΔEinternal = 2971.5 - 169.5 - 450 - 295.75 - 7092		ΔEinternal = 2971.5 - 169.5 - 450 - 295.75 - 7092

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	d)		d)
	The final kinetic energy is greater than the initial kinetic energy, and this is because, by exploding, some of the internal energy stored in the firework was released and then harnessed as kinetic energy.		The final kinetic energy is greater than the initial kinetic energy, and this is because by exploding, some of the internal energy
			stored in the firework was released and then harnessed as kinetic energy.
	Consequently, the change in internal energy is negative.		Consequently, the change in internal energy is negative.

Sydney: /* Mathematical Model */

2019-07-08T20:04:32Z

Mathematical Model

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	A very simple way to approach many inelastic collision problems is using the conservation of momentum principle. Since the two objects would not bounce and thus stick together- they become one object in terms of calculations. So the final equation would be: <math>		A very simple way to approach many inelastic collision problems is using the conservation of momentum principle. Since the two objects would not bounce and thus stick together- they become one object in terms of calculations. So the final equation would be: <math>
	m_1v_1 + m_2v_2 = (m_1 + m_2)v_f.		m_1v_1 + m_2v_2 = (m_1 + m_2)v_f</math>.

	===Computational Model===		===Computational Model===

Sydney: /* The Main Idea */

2019-07-08T20:03:42Z

The Main Idea

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	Because the final kinetic energy differs from the initial kinetic energy, inelastic collisions witness a change in the internal energy of the objects involved in the collision. A change in internal energy often manifests itself in the form of deformation, rotation, heat, vibration, explosions, or energy state excitation.		Because the final kinetic energy differs from the initial kinetic energy, inelastic collisions witness a change in the internal energy of the objects involved in the collision. A change in internal energy often manifests itself in the form of deformation, rotation, heat, vibration, explosions, or energy state excitation.


	'''Types of Inelastic Collisions'''		'''Types of Inelastic Collisions'''

			One type of inelastic collisions is the maximally inelastic collision, also known as the perfectly inelastic collision, in which two objects collide and then remain connected for the duration of the observable interaction. This does not indicate an end to movement, as momentum must be conserved by the end of the interaction. See problem 1 for an example. More information about such problems can be found in the page [[Maximally Inelastic Collision]].
	One type of inelastic collisions is the maximally inelastic collision, also known as the perfectly inelastic collision, in which two objects collide and then remain connected for the duration of the observable interaction. This does not indicate an end to movement, as momentum must be conserved by the end of the interaction. See problem 1 for an example. More information about such problems can be found in the page [Maximally Inelastic Collision].


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	In the car crash depicted above, there is a loss of kinetic energy, which is converted into internal energy in the form of produced heat and deformation of the car and the tree.		In the car crash depicted above, there is a loss of kinetic energy, which is converted into internal energy in the form of produced heat and deformation of the car and the tree.

	This is in accordance with the relation ~~ΔEinternal~~ = -~~ΔKtrans~~=-0.5mass(Δvelocity)^2.		This is in accordance with the relation <math>ΔE_{internal} = -ΔK_{trans}=-0.5(mass)(Δvelocity)^2</math>.

	A very simple way to approach many inelastic collision problems is using the conservation of momentum principle. Since the two objects would not bounce and thus stick together- they become one object in terms of calculations. So the final equation would be:		A very simple way to approach many inelastic collision problems is using the conservation of momentum principle. Since the two objects would not bounce and thus stick together- they become one object in terms of calculations. So the final equation would be: <math>
	~~m1v1~~ + ~~m2v2~~ = (m1 + m2)vf.		m_1v_1 + m_2v_2 = (m_1 + m_2)v_f.

	===Computational Model===		===Computational Model===

	Using the Collision Lab [[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html]], model different scenarios for the inelastic collision. Such as when the two bodies have the same mass, or when one is moving at a higher speed than the other. This simulation can help visualize inelastic collisions under different circumstances.		Using the Collision Lab [https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html], you can model different scenarios for the inelastic collision. Such as when the two bodies have the same mass, or when one is moving at a higher speed than the other. This simulation can help visualize inelastic collisions under different circumstances.

	==Examples==		==Examples==

Sydney: /* The Main Idea */

2019-07-08T19:55:59Z

The Main Idea

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	==The Main Idea==		==The Main Idea==

	There are two types of collisions: inelastic and elastic. The big identifying characteristics of inelastic collisions that distinguish them from elastic collisions is that in inelastic collisions, the momentum of the interacting bodies are conserved, but the kinetic energy is not. This change in kinetic energy from initial to final states is what differentiates inelastic collisions from elastic collisions. Generally, the forces involved in an inelastic collisions are so large relative to external forces that those external forces can be ignored. Consequently, given that <math>ΔP_{system} = F_{net,ext}ΔT, if we approximate F{net,ext} as zero, then ΔP_{system} = 0, and P_i = P_f.		There are two types of collisions: inelastic and elastic. The big identifying characteristics of inelastic collisions that distinguish them from elastic collisions is that in inelastic collisions, the momentum of the interacting bodies are conserved, but the kinetic energy is not. This change in kinetic energy from initial to final states is what differentiates inelastic collisions from elastic collisions. Generally, the forces involved in an inelastic collisions are so large relative to external forces that those external forces can be ignored. Consequently, given that <math>ΔP_{system} = F_{net,ext}ΔT</math>, if we approximate <math>F_{net,ext} </math>as zero, then <math>ΔP_{system} = 0</math>, and <math>P_i = P_f</math>.

Sydney: /* The Main Idea */

2019-07-08T18:33:18Z

The Main Idea

← Older revision		Revision as of 14:33, 8 July 2019
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	==The Main Idea==		==The Main Idea==

	There are two types of collisions: inelastic and elastic. The big identifying characteristics of inelastic collisions that distinguish them from elastic collisions is that in inelastic collisions, the momentum of the interacting bodies are conserved, but the kinetic energy is not. This change in kinetic energy from initial to final states is what differentiates inelastic collisions from elastic collisions. Generally, the forces involved in an inelastic collisions are so large relative to external forces that those external forces can be ignored. Consequently, given that ~~ΔPsystem~~ = ~~Fnet~~,ext*ΔT, if we approximate ~~Fnet~~,ext as zero, then ~~ΔPsystem~~ = 0, and Pi = Pf.		There are two types of collisions: inelastic and elastic. The big identifying characteristics of inelastic collisions that distinguish them from elastic collisions is that in inelastic collisions, the momentum of the interacting bodies are conserved, but the kinetic energy is not. This change in kinetic energy from initial to final states is what differentiates inelastic collisions from elastic collisions. Generally, the forces involved in an inelastic collisions are so large relative to external forces that those external forces can be ignored. Consequently, given that <math>ΔP_{system} = F_{net,ext}ΔT, if we approximate F{net,ext} as zero, then ΔP_{system} = 0, and P_i = P_f.