Physics Book - User contributions [en]

Inelastic Collisions

2016-04-15T17:48:19Z

Csorensen6:

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides examples with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext*ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0, and Pi = Pf.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates some characteristics of a typical, inelastic collision: a car crash.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are two main equations to consider, with the third being derived from the second.:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. See problem 1 for an example. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deform, get hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle.

The steps for solving inelastic collision problems are often as follows:

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.

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.

4) Next, knowing that Pf = Pi, 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.

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.

== Examples ==

'''Simple'''

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

'''Middling'''

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

'''Difficult'''

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,0> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.
One piece has a mass of 3 kg and a velocity of <8,7,0> m/s.
Another has a mass of 4 kg and a velocity of <15,0,0> m/s.
The third has a mass of 3.5 kg and a velocity of <12,-5,0> m/s.

Reference the picture below.

[[File:Wiki Image 7.jpg]]

a) What is the mass of the fourth piece?

b) What is the momentum and velocity of that piece?

c) What is the change in internal energy?

d) Justify the sign of the answer found in part c).

Answer

a)
m1 = m2 + m3 + m4 + m5
15 kg = 3 kg + 4 kg + 3.5 kg + m5
15 kg = 10.5 kg + m5

m5 = 4.5 kg

b)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

15 kg <10,17,0> m/s = 3 kg <8,7,0> m/s + 4 kg <15,0,0> m/s + 3.5 kg <12,-5,0> m/s + p5
<150,255,0> kg*m/s = <24,21,0> kg*m/s + <60,0,0> kg*m/s + <42,-17.5,0> kg*m/s + p5
<150,225,0> kg*m/s = <126,3.5,0> kg*m/s + p5

p5 = <24,251.5,0> kg*m/s

p = mv
p5 = m5v5
<24,251.5,0> kg*m/s = 4.5kg*v5

v5 = <5.33,55.89,0> m/s

c)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest = 0
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal
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 = 2971.5 - 169.5 - 450 - 295.75 - 7092

ΔEinternal = -5036 J

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.
Consequently, the change in internal energy is negative.

== Connectedness ==

The science of collisions and the study of momentum are especially prominent when it comes to the sport of football, and particularly in regards to concussions, or traumatic brain injuries sustained due to hard blows to the head. Significant research within the biomedical field is being conducted into how to reduce concussion frequency and severity, and this often entails specifically designing helmets to minimize the impact of high-speed, hard-hitting tackles. This research has come to greater prominence with the relatively-recent discovery of the role football plays in the acquisition of CTE, a degenerative disease which slowly causes the loss of cognitive function, similar to dementia, and which is especially prominent among football players and other athletes who suffer repeated head trauma.

[[File:Wiki Image 8.jpg]]

== See Also ==

For more information about the momentum principle, check out this page!

[[http://www.physicsbook.gatech.edu/Momentum_Principle]]

For more on the energy principle, check here!

[[http://www.physicsbook.gatech.edu/The_Energy_Principle]]

If you want to learn more about elastic or maximally inelastic collisions, click here or here, respectively!

[[http://www.physicsbook.gatech.edu/Elastic_Collisions]]

[[http://www.physicsbook.gatech.edu/Maximally_Inelastic_Collision]]

== References ==

Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. Hoboken, NJ: Wiley, 2011. Print

Inelastic Collisions

2016-04-15T17:45:58Z

Csorensen6: /* Connectedness */

Inelastic Collisions

2016-04-15T17:42:26Z

Csorensen6: /* Connectedness */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides examples with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext*ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0, and Pi = Pf.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates some characteristics of a typical, inelastic collision: a car crash.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are two main equations to consider, with the third being derived from the second.:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. See problem 1 for an example. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deform, get hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle.

The steps for solving inelastic collision problems are often as follows:

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.

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.

4) Next, knowing that Pf = Pi, 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.

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.

== Examples ==

'''Simple'''

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

'''Middling'''

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

'''Difficult'''

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,0> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.
One piece has a mass of 3 kg and a velocity of <8,7,0> m/s.
Another has a mass of 4 kg and a velocity of <15,0,0> m/s.
The third has a mass of 3.5 kg and a velocity of <12,-5,0> m/s.

Reference the picture below.

[[File:Wiki Image 7.jpg]]

a) What is the mass of the fourth piece?

b) What is the momentum and velocity of that piece?

c) What is the change in internal energy?

d) Justify the sign of the answer found in part c).

Answer

a)
m1 = m2 + m3 + m4 + m5
15 kg = 3 kg + 4 kg + 3.5 kg + m5
15 kg = 10.5 kg + m5

m5 = 4.5 kg

b)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

15 kg <10,17,0> m/s = 3 kg <8,7,0> m/s + 4 kg <15,0,0> m/s + 3.5 kg <12,-5,0> m/s + p5
<150,255,0> kg*m/s = <24,21,0> kg*m/s + <60,0,0> kg*m/s + <42,-17.5,0> kg*m/s + p5
<150,225,0> kg*m/s = <126,3.5,0> kg*m/s + p5

p5 = <24,251.5,0> kg*m/s

p = mv
p5 = m5v5
<24,251.5,0> kg*m/s = 4.5kg*v5

v5 = <5.33,55.89,0> m/s

c)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest = 0
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal
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 = 2971.5 - 169.5 - 450 - 295.75 - 7092

ΔEinternal = -5036 J

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.
Consequently, the change in internal energy is negative.

== Connectedness ==

The science of collisions and the study of momentum are especially prominent when it comes to the sport of football, and particularly in regards to concussions, or traumatic brain injuries sustained due to hard blows to the head. Significant research within the biomedical field is being conducted into how to reduce concussion frequency and severity, and this often entails specifically designing helmets to minimize the impact of high-speed, hard-hitting tackles.

[[File:Wiki Image 8.jpg]]

== See Also ==

For more information about the momentum principle, check out this page!

[[http://www.physicsbook.gatech.edu/Momentum_Principle]]

For more on the energy principle, check here!

[[http://www.physicsbook.gatech.edu/The_Energy_Principle]]

If you want to learn more about elastic or maximally inelastic collisions, click here or here, respectively!

[[http://www.physicsbook.gatech.edu/Elastic_Collisions]]

[[http://www.physicsbook.gatech.edu/Maximally_Inelastic_Collision]]

File:Wiki Image 8.jpg

2016-04-15T17:42:12Z

Csorensen6:

Inelastic Collisions

2016-04-15T17:41:02Z

Csorensen6:

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides examples with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext*ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0, and Pi = Pf.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates some characteristics of a typical, inelastic collision: a car crash.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are two main equations to consider, with the third being derived from the second.:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. See problem 1 for an example. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deform, get hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle.

The steps for solving inelastic collision problems are often as follows:

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.

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.

4) Next, knowing that Pf = Pi, 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.

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.

== Examples ==

'''Simple'''

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

'''Middling'''

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

'''Difficult'''

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,0> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.
One piece has a mass of 3 kg and a velocity of <8,7,0> m/s.
Another has a mass of 4 kg and a velocity of <15,0,0> m/s.
The third has a mass of 3.5 kg and a velocity of <12,-5,0> m/s.

Reference the picture below.

[[File:Wiki Image 7.jpg]]

a) What is the mass of the fourth piece?

b) What is the momentum and velocity of that piece?

c) What is the change in internal energy?

d) Justify the sign of the answer found in part c).

Answer

a)
m1 = m2 + m3 + m4 + m5
15 kg = 3 kg + 4 kg + 3.5 kg + m5
15 kg = 10.5 kg + m5

m5 = 4.5 kg

b)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

15 kg <10,17,0> m/s = 3 kg <8,7,0> m/s + 4 kg <15,0,0> m/s + 3.5 kg <12,-5,0> m/s + p5
<150,255,0> kg*m/s = <24,21,0> kg*m/s + <60,0,0> kg*m/s + <42,-17.5,0> kg*m/s + p5
<150,225,0> kg*m/s = <126,3.5,0> kg*m/s + p5

p5 = <24,251.5,0> kg*m/s

p = mv
p5 = m5v5
<24,251.5,0> kg*m/s = 4.5kg*v5

v5 = <5.33,55.89,0> m/s

c)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest = 0
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal
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 = 2971.5 - 169.5 - 450 - 295.75 - 7092

ΔEinternal = -5036 J

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.
Consequently, the change in internal energy is negative.

== Connectedness ==

The science of collisions and the study of momentum are especially prominent when it comes to the sport of football, and particularly in regards to concussions, or traumatic brain injuries sustained due to hard blows to the head. Significant research within the biomedical field is being conducted into how to reduce concussion frequency and severity, and this often entails specifically designing helmets to minimize the impact of high-speed, hard-hitting tackles.
[[File:Wiki Image 8.jpg]]

== See Also ==

For more information about the momentum principle, check out this page!

[[http://www.physicsbook.gatech.edu/Momentum_Principle]]

For more on the energy principle, check here!

[[http://www.physicsbook.gatech.edu/The_Energy_Principle]]

If you want to learn more about elastic or maximally inelastic collisions, click here or here, respectively!

[[http://www.physicsbook.gatech.edu/Elastic_Collisions]]

[[http://www.physicsbook.gatech.edu/Maximally_Inelastic_Collision]]

Inelastic Collisions

2016-04-15T17:24:21Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

'''Simple'''

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

'''Middling'''

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

'''Difficult'''

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,0> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.
One piece has a mass of 3 kg and a velocity of <8,7,0> m/s.
Another has a mass of 4 kg and a velocity of <15,0,0> m/s.
The third has a mass of 3.5 kg and a velocity of <12,-5,0> m/s.

Reference the picture below.

[[File:Wiki Image 7.jpg]]

a) What is the mass of the fourth piece?

b) What is the momentum and velocity of that piece?

c) What is the change in internal energy?

d) Justify the sign of the answer found in part c).

Answer

a)
m1 = m2 + m3 + m4 + m5
15 kg = 3 kg + 4 kg + 3.5 kg + m5
15 kg = 10.5 kg + m5

m5 = 4.5 kg

b)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

15 kg <10,17,0> m/s = 3 kg <8,7,0> m/s + 4 kg <15,0,0> m/s + 3.5 kg <12,-5,0> m/s + p5
<150,255,0> kg*m/s = <24,21,0> kg*m/s + <60,0,0> kg*m/s + <42,-17.5,0> kg*m/s + p5
<150,225,0> kg*m/s = <126,3.5,0> kg*m/s + p5

p5 = <24,251.5,0> kg*m/s

p = mv
p5 = m5v5
<24,251.5,0> kg*m/s = 4.5kg*v5

v5 = <5.33,55.89,0> m/s

c)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest = 0
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal
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 = 2971.5 - 169.5 - 450 - 295.75 - 7092

ΔEinternal = -5036 J

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.
Consequently, the change in internal energy is negative.

Inelastic Collisions

2016-04-15T17:18:08Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

'''Simple'''

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

'''Middling'''

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

'''Difficult'''

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,0> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.
One piece has a mass of 3 kg and a velocity of <8,7,0> m/s.
Another has a mass of 4 kg and a velocity of <15,0,0> m/s.
The third has a mass of 3.5 kg and a velocity of <12,-5,0> m/s.

Reference the picture below.

[[File:Wiki Image 7.jpg]]

a) What is the mass of the fourth piece?

b) What is the momentum and velocity of that piece?

c) What is the change in internal energy?

d) Justify the sign of the answer found in part c).

Answer

a)
m1 = m2 + m3 + m4 + m5
15 kg = 3 kg + 4 kg + 3.5 kg + m5
15 kg = 10.5 kg + m5

m5 = 4.5 kg

b)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

15 kg <10,17,0> m/s = 3 kg <8,7,0> m/s + 4 kg <15,0,0> m/s + 3.5 kg <12,-5,0> m/s + p5
<150,255,0> kg*m/s = <24,21,0> kg*m/s + <60,0,0> kg*m/s + <42,-17.5,0> kg*m/s + p5
<150,225,0> kg*m/s = <126,3.5,0> kg*m/s + p5

p5 = <24,221.5,0> kg*m/s

p = mv
p5 = m5v5

<24,221.5,0> kg*m/s = 4.5kg v5

v5 = <5.33,49.22,0> m/s

c)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest = 0
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal
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 + 49.22^2 + 0^2)
ΔEinternal = 2971.5 - 169.5 - 450 - 295.75 - 5514.8

ΔEinternal = -3459 J

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.
Consequently, the change in internal energy is negative.

Inelastic Collisions

2016-04-15T17:17:38Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

'''Simple'''

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

'''Middling'''

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

'''Difficult'''

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,0> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.
One piece has a mass of 3 kg and a velocity of <8,7,0> m/s.
Another has a mass of 4 kg and a velocity of <15,0,0> m/s.
The third has a mass of 3.5 kg and a velocity of <12,-5,0> m/s.

Reference the picture below.

[[File:Wiki Image 7.jpg]]

a) What is the mass of the fourth piece?

b) What is the momentum and velocity of that piece?

c) What is the change in internal energy?

d) Justify the sign of the answer found in part c).

Answer

a)
m1 = m2 + m3 + m4 + m5
15 kg = 3 kg + 4 kg + 3.5 kg + m5
15 kg = 10.5 kg + m5

m5 = 4.5 kg

b)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

15 kg <10,17,0> m/s = 3 kg <8,7,0> m/s + 4 kg <15,0,0> m/s + 3.5 kg <12,-5,0> m/s + p5
<150,255,0> kg*m/s = <24,21,0> kg*m/s + <60,0,0> kg*m/s + <42,-17.5,0> kg*m/s + p5
<150,225,0> kg*m/s = <126,3.5,0> kg*m/s + p5

p5 = <24,221.5,0> kg*m/s

p = mv
p5 = m5v5

<24,221.5,0> kg*m/s = 4.5kg v5

v5 = <5.33,49.22,0> m/s

c)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest = 0
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal
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 + 49.22^2 + 0^2)
ΔEinternal = 2971.5 - 169.5 - 450 - 295.75 - 5514.8

ΔEinternal = -3459 J

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. Consequently, the change in internal energy is negative.

Inelastic Collisions

2016-04-15T17:17:17Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

'''Simple'''

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

'''Middling'''

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

'''Difficult'''

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,0> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.
One piece has a mass of 3 kg and a velocity of <8,7,0> m/s.
Another has a mass of 4 kg and a velocity of <15,0,0> m/s.
The third has a mass of 3.5 kg and a velocity of <12,-5,0> m/s.

Reference the picture below.

[[File:Wiki Image 7.jpg]]

a) What is the mass of the fourth piece?

b) What is the momentum and velocity of that piece?

c) What is the change in internal energy?

d) Justify the sign of the answer found in part c).

Answer

a)
m1 = m2 + m3 + m4 + m5
15 kg = 3 kg + 4 kg + 3.5 kg + m5
15 kg = 10.5 kg + m5

m5 = 4.5 kg

b)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

15 kg <10,17,0> m/s = 3 kg <8,7,0> m/s + 4 kg <15,0,0> m/s + 3.5 kg <12,-5,0> m/s + p5
<150,255,0> kg*m/s = <24,21,0> kg*m/s + <60,0,0> kg*m/s + <42,-17.5,0> kg*m/s + p5
<150,225,0> kg*m/s = <126,3.5,0> kg*m/s + p5

p5 = <24,221.5,0> kg*m/s

p = mv
p5 = m5v5

<24,221.5,0> kg*m/s = 4.5kg v5

v5 = <5.33,49.22,0> m/s

c)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest = 0
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal
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 + 49.22^2 + 0^2)
ΔEinternal = 2971.5 - 169.5 - 450 - 295.75 - 5514.8

ΔEinternal = -3459 J

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. Consequently, the change in internal energy is negative.

Inelastic Collisions

2016-04-15T17:16:34Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

'''Simple'''

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

'''Middling'''

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

'''Difficult'''

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,0> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.
One piece has a mass of 3 kg and a velocity of <8,7,0> m/s.
Another has a mass of 4 kg and a velocity of <15,0,0> m/s.
The third has a mass of 3.5 kg and a velocity of <12,-5,0> m/s.

Reference the picture below.

[[File:Wiki Image 7.jpg]]

a) What is the mass of the fourth piece?

b) What is the momentum and velocity of that piece?

c) What is the change in internal energy?

d) Justify the sign of the answer found in part c).

Answer

a)
m1 = m2 + m3 + m4 + m5
15 kg = 3 kg + 4 kg + 3.5 kg + m5
15 kg = 10.5 kg + m5

m5 = 4.5 kg

b)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

15 kg <10,17,0> m/s = 3 kg <8,7,0> m/s + 4 kg <15,0,0> m/s + 3.5 kg <12,-5,0> m/s + p5
<150,255,0> kg*m/s = <24,21,0> kg*m/s + <60,0,0> kg*m/s + <42,-17.5,0> kg*m/s + p5
<150,225,0> kg*m/s = <126,3.5,0> kg*m/s + p5

p5 = <24,221.5,0> kg*m/s

p = mv
p5 = m5v5

<24,221.5,0> kg*m/s = 4.5kg v5

v5 = <5.33,49.22,0> m/s

c)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest = 0
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal
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 + 49.22^2 + 0^2)
ΔEinternal = 2971.5 - 169.5 - 450 - 295.75 - 5514.8

ΔEinternal = -3459 J

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.
Consequently, the change in internal energy is negative.

Inelastic Collisions

2016-04-15T17:04:52Z

Csorensen6: /* Typical Process for Solving */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

'''Simple'''

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

'''Middling'''

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

'''Difficult'''

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,0> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.
One piece has a mass of 3 kg and a velocity of <8,7,0> m/s.
Another has a mass of 4 kg and a velocity of <15,0,0> m/s.
The third has a mass of 3.5 kg and a velocity of <12,-5,0> m/s.

Reference the picture below.
[[File:Wiki Image 7.jpg]]

a) What is the mass of the fourth piece?

b) What is the momentum of that piece?

c) What is the change in internal energy?

d) Justify the sign of the answer found in part c).

Answer

a)

b)

c)

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.
Consequently, the change in internal energy is negative.

File:Wiki Image 7.jpg

2016-04-15T17:04:16Z

Csorensen6:

Inelastic Collisions

2016-04-15T17:03:44Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

'''Simple'''

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

'''Middling'''

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

'''Difficult'''

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,0> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.
One piece has a mass of 3 kg and a velocity of <8,7,0> m/s.
Another has a mass of 4 kg and a velocity of <15,0,0> m/s.
The third has a mass of 3.5 kg and a velocity of <12,-5,0> m/s.

Reference the picture below.
[[File:Wiki Image 7.jpg]]

a) What is the mass of the fourth piece?

b) What is the momentum of that piece?

c) What is the change in internal energy?

d) Justify the sign of the answer found in part c).

Answer

a)

b)

c)

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.
Consequently, the change in internal energy is negative.

Inelastic Collisions

2016-04-15T16:51:18Z

Csorensen6: /* Types of Inelastic Collisions */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

'''Maximally Inelastic Collisions'''

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

'''Explosions'''

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

'''Typical Inelastic Collisions'''

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

Difficult

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,-5> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.

Inelastic Collisions

2016-04-15T16:50:02Z

Csorensen6: /* The Main Idea */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

'''Mathematical Model'''

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

Difficult

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,-5> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.

Inelastic Collisions

2016-04-15T15:57:19Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

Difficult

A lit firework of mass 15 kg is launched into the sky with an initial velocity of <10,17,-5> m/s.
Eventually, its fuse burns up and the firework explodes, bursting into four parts.

Inelastic Collisions

2016-04-15T15:55:15Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf

vf = <-3.33, 0, 0> m/s

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v

v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J

Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J

Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0

ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:54:19Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

Just do part A.

[[File:Wiki Image 4.jpg]]

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> + M<10,0,0> = (2M + M)*vf
<-10M,0,0> = 3M * vf
vf = <-3.33, 0, 0> m/s

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a)
ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c)
K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d)
ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:36:19Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.

[[File:Wiki Image 4.jpg]]

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> m/s + m<10,0,0> m/s =

b) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

2M <-10,0,0> m/s + M<10,0,0> m/s = (2M + M)*vf
<-10M,0,0> m/s = 3M * vf
vf = <-3.33, 0, 0> m/s

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

File:Wiki Image 5.jpg

2016-04-15T15:31:13Z

Csorensen6: Csorensen6 uploaded a new version of "File:Wiki Image 5.jpg"

File:Wiki Image 5.jpg

2016-04-15T15:30:33Z

Csorensen6:

Inelastic Collisions

2016-04-15T15:29:38Z

Csorensen6: /* Types of Inelastic Collisions */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.

[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:26:45Z

Csorensen6: /* Types of Inelastic Collisions */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

[[Media:Wiki Image 5.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.

[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:26:05Z

Csorensen6: /* Types of Inelastic Collisions */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

[[File:Wiki Image 5.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.

[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:25:47Z

Csorensen6: /* Types of Inelastic Collisions */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.
[[File:Wiki Image 3.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.

[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:25:29Z

Csorensen6: /* Types of Inelastic Collisions */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.

[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:24:44Z

Csorensen6: /* Types of Inelastic Collisions */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.
[[File:Wiki Image 2.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.

[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:23:42Z

Csorensen6: /* The Main Idea */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[File:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.

[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:23:19Z

Csorensen6: /* The Main Idea */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[Media:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.

[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:22:55Z

Csorensen6: /* Examples */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[Media:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple

This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.

[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:22:02Z

Csorensen6: /* The Main Idea */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[Media:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple
[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

File:Wiki Image 4.jpg

2016-04-15T15:20:44Z

Csorensen6:

Inelastic Collisions

2016-04-15T15:19:56Z

Csorensen6: /* Example Problem */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[Media:WikiImage3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Examples ==

Simple
[[File:Wiki Image 4.jpg]]

Middling

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Difficult

Inelastic Collisions

2016-04-15T15:17:26Z

Csorensen6: /* Typical Process for Solving */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[Media:WikiImage3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Inelastic Collisions

2016-04-15T15:17:10Z

Csorensen6: /* Types of Inelastic Collisions */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[Media:WikiImage3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

Explosions

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. In other words, Ki < Kf.

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Inelastic Collisions

2016-04-15T15:15:12Z

Csorensen6: /* The Main Idea */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[Media:WikiImage3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

http://images.tutorvista.com/cms/images/83/inelastic-collision-image.gif

Explosions

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. In other words, Ki < Kf.

https://en.wikipedia.org/wiki/Explosion#/media/File:NTS_-_BEEF_-_WATUSI.jpg

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

File:Wiki Image 3.jpg

2016-04-15T15:14:43Z

Csorensen6:

Inelastic Collisions

2016-04-15T15:14:27Z

Csorensen6: /* The Main Idea */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

The picture below illustrates this.

[[Media:Wiki Image 3.jpg]]

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

http://images.tutorvista.com/cms/images/83/inelastic-collision-image.gif

Explosions

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. In other words, Ki < Kf.

https://en.wikipedia.org/wiki/Explosion#/media/File:NTS_-_BEEF_-_WATUSI.jpg

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Inelastic Collisions

2016-04-15T15:12:51Z

Csorensen6: /* The Main Idea */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.
The picture below illustrates this.

In these types of problems, there are three main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

http://images.tutorvista.com/cms/images/83/inelastic-collision-image.gif

Explosions

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. In other words, Ki < Kf.

https://en.wikipedia.org/wiki/Explosion#/media/File:NTS_-_BEEF_-_WATUSI.jpg

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Inelastic Collisions

2016-04-15T15:08:36Z

Csorensen6: /* The Main Idea */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

In these types of problems, there are two main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0

-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

http://images.tutorvista.com/cms/images/83/inelastic-collision-image.gif

Explosions

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. In other words, Ki < Kf.

https://en.wikipedia.org/wiki/Explosion#/media/File:NTS_-_BEEF_-_WATUSI.jpg

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Inelastic Collisions

2016-04-15T15:08:21Z

Csorensen6: /* The Main Idea */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

In these types of problems, there are two main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0
-ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

http://images.tutorvista.com/cms/images/83/inelastic-collision-image.gif

Explosions

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. In other words, Ki < Kf.

https://en.wikipedia.org/wiki/Explosion#/media/File:NTS_-_BEEF_-_WATUSI.jpg

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Inelastic Collisions

2016-04-15T15:08:04Z

Csorensen6: /* The Main Idea */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

In these types of problems, there are two main equations to consider:

ΔPsystem = Fnet,ext * ΔT = 0.

ΔE = ΔKtrans + ΔEinternal = 0 -ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

http://images.tutorvista.com/cms/images/83/inelastic-collision-image.gif

Explosions

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. In other words, Ki < Kf.

https://en.wikipedia.org/wiki/Explosion#/media/File:NTS_-_BEEF_-_WATUSI.jpg

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Inelastic Collisions

2016-04-15T15:07:41Z

Csorensen6: /* Definition */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== The Main Idea ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi. This is because, in all collisions, the forces involved in the collision are so large that external forces can be neglected. Consequently, given that ΔPsystem = Fnet,ext * ΔT, if we approximate Fnet,ext as zero, then ΔPsystem = 0.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state. In these collisions, the change in internal energy is equal to the negative change in kinetic energy, i.e. ΔEinternal = -ΔKtrans.

In these types of problems, there are two main equations to consider:
ΔPsystem = Fnet,ext * ΔT = 0.
ΔE = ΔKtrans + ΔEinternal = 0 -ΔKtrans = ΔEinternal

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

http://images.tutorvista.com/cms/images/83/inelastic-collision-image.gif

Explosions

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. In other words, Ki < Kf.

https://en.wikipedia.org/wiki/Explosion#/media/File:NTS_-_BEEF_-_WATUSI.jpg

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Inelastic Collisions

2016-04-15T15:01:37Z

Csorensen6: /* Definition */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== Definition ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is not the same as the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi.

Moreover, 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 objects deforming, rotating, getting hot, vibrating, exploding, or being excited to a higher energy state.

The video linked below explores in further detail the differences between elastic and inelastic collisions, and it also provides some examples of each.

In these types of problems, there are two main equations to consider:

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

http://images.tutorvista.com/cms/images/83/inelastic-collision-image.gif

Explosions

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. In other words, Ki < Kf.

https://en.wikipedia.org/wiki/Explosion#/media/File:NTS_-_BEEF_-_WATUSI.jpg

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Inelastic Collisions

2016-04-12T18:23:17Z

Csorensen6: /* Types of Inelastic Collisions */

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== Definition ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is greater than the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi.

Moreover, because the final kinetic energy is less than 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 objects deforming, rotating, getting hot, vibrating, or being excited to a higher energy state.

The video linked below explores in further detail the differences between elastic and inelastic collisions, and it also provides some examples of each.

[https://www.youtube.com/watch?v=Xe2r6wey26E]

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

http://images.tutorvista.com/cms/images/83/inelastic-collision-image.gif

Explosions

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. In other words, Ki < Kf.

https://en.wikipedia.org/wiki/Explosion#/media/File:NTS_-_BEEF_-_WATUSI.jpg

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

File:Wiki Image 2.jpg

2016-04-12T18:19:36Z

Csorensen6:

Inelastic Collisions

2016-04-12T18:19:18Z

Csorensen6:

This topic covers what inelastic collisions are, discusses a series of steps that can be undertaken to solve problems involving them, and then provides an example with the answers worked out.
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)

== Definition ==

Inelastic collisions differ from elastic collisions in one primary aspect: the initial kinetic energy is greater than the final kinetic energy. In other words, kinetic energy is NOT conserved, and Kf does not equal Ki.
Momentum, however, is still conserved, so Pf = Pi.

Moreover, because the final kinetic energy is less than 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 objects deforming, rotating, getting hot, vibrating, or being excited to a higher energy state.

The video linked below explores in further detail the differences between elastic and inelastic collisions, and it also provides some examples of each.

[https://www.youtube.com/watch?v=Xe2r6wey26E]

== Types of Inelastic Collisions ==

Maximally Inelastic Collisions

One of the most commonly-seen situations is the maximally inelastic collision, in which two objects collide and then stick together. The below image illustrates a maximally inelastic collision. More information about such problems can be found in the page about maximally inelastic collisions.

http://images.tutorvista.com/cms/images/83/inelastic-collision-image.gif

Explosions

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. In other words, Ki < Kf.
[[File:Wiki Image 2.jpg]]

Typical Inelastic Collisions

In these types, two or more objects collide but do not stick together, but as a result of their collision, one or both begins to rotate, begins to vibrate, deforms, gets hot, etc.
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf.
Most real-world collisions are inelastic like this, but to varying degrees.

== Typical Process for Solving ==

In such problems, it is often necessary to apply both the momentum principle and the energy principle. The image below provides an example of and the equations involved in solving a maximally inelastic collision problem.
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/inecol.gif

The steps for solving inelastic collision problems are often as follows:

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.

3) Knowing that Fnet,ext = 0, this means that momentum is conserved and that Pf = Pi. This comes from the momentum principle, in that ΔP = Fnet,ext*ΔT, and if Fnet,ext 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.

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.

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.

== Example Problem ==

Question

There are two hockey pucks traveling across the surface of a frozen pond. The first puck has a mass of 0.15 kg and is moving with a velocity of <2.5, 3.4, 0> m/s. The second puck has a mass of 0.13 kg and is moving with a velocity of <-3.1, 1.7, 0> m/s. After colliding, the first puck then has a velocity of <-1.9, 3.0, 0> m/s. (Reference the image below).

[[File:Wiki Image.jpg]]

a) What is the velocity of the second puck after the collision?

b) What is the initial kinetic energy?

c) What is the final kinetic energy?

d) What is the change of the internal energy of the two pucks?

Answer

a) ΔP = Pf - Pi = Fnet,ext*ΔT = 0
Pi = Pf

0.15kg<2.5, 3.4, 0> m/s + 0.13kg<-3.1, 1.7, 0> m/s = 0.15kg<-1.9, 3.0, 0> m/s + 0.13kg*v
<-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*m/s + 0.13kg*v
<0.257,0.281,0> kg*m/s = 0.13kg*v
v = <1.977, 2.162, 0> m/s

b) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Ki = (1/2)*0.15kg*((2.5 m/s)^2 + (3.4 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((-3.1 m/s)^2 + (1.7 m/s)^2 + (0 m/s)^2)
Ki = 1.336 J + 0.813 J
Ki = 2.149 J

c) K = (1/2)*m*|v|^2
|v| = (vx^2 + vy^2 + vz^2)^(1/2)
|v|^2 = vx^2 + vy^2 + vz^2

Kf = (1/2)*0.15kg*((-1.9 m/s)^2 + (3.0 m/s)^2 + (0 m/s)^2) + (1/2)*0.13kg*((1.977 m/s)^2 + (2.162 m/s)^2 + (0 m/s)^2)
Kf = 0.946 J + 0.558 J
Kf = 1.504 J

d) ΔE = Q + W = ΔK + ΔEinternal + ΔU + ΔErest
ΔE = ΔK + ΔEinternal = 0 = Kf - Ki + ΔEinternal

1.504 J - 2.149 J + ΔEinternal = 0
-0.645J + ΔEinternal = 0
ΔEinternal = 0.645 J

Inelastic Collisions

2016-04-05T17:59:41Z

Csorensen6:

Inelastic Collisions

2016-04-05T17:59:13Z

Csorensen6:

Inelastic Collisions

2016-04-05T01:06:06Z

Csorensen6: