Inelastic Collisions: Difference between revisions

From Physics Book
Jump to navigation Jump to search
 
(95 intermediate revisions by 5 users not shown)
Line 1: Line 1:
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.
Created by Csorensen6
--[[User:Csorensen6|Csorensen6]] ([[User talk:Csorensen6|talk]]) 21:06, 4 April 2016 (EDT)


== The Main Idea ==
Edited/Claimed by Jessica Ouyang, Fall 2018


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.
Edited/Claimed by Vansh Kareer, Spring 2017
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.


Edited/Claimed by Abigail Brady, Fall 2017


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


The picture below illustrates this.


[[Media:Wiki Image 3.jpg]]
==The Main Idea==


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


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


ΔPsystem = Fnet,ext * ΔT = 0.
Because the final kinetic energy differs from the initial kinetic energy, inelastic collisions witness a change in the internal energy of the objects involved in the collision. A change in internal energy often manifests itself in the form of deformation, rotation, heat, vibration, explosions, or energy state excitation.  


ΔE = ΔKtrans + ΔEinternal = 0 
'''Types of Inelastic Collisions'''
         
-ΔKtrans = ΔEinternal


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


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.
Most inelastic collisions are considered partially inelastic collisions. In these types, two or more objects collide but do not stick together (unlike maximally inelastic collisions), but as a result of their collision, one or both begin to rotate, vibrate, deform, heat up, etc.  
In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf. An example of a typically inelastic collision is a car crash.




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


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|thumb|center|1020x620px]]




===Mathematical Model===


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 inelastic collisions, the change in internal energy is equal to the negative change in kinetic energy. Based on these characteristics, and applying the momentum principle, a few basic equations can be set to outline an inelastic collision.
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 picture below illustrates some characteristics of a typical, inelastic collision: a car crash.


[[File:Car_crash_1.jpg|500px]]
In the car crash depicted above, there is a loss of kinetic energy, which is converted into internal energy in the form of produced heat and deformation of the car and the tree.
This is in accordance with the relation <math>ΔE_{internal} = -ΔK_{trans}=-0.5(mass)(Δvelocity)^2</math>.
A very simple way to approach many inelastic collision problems is using the conservation of momentum principle. Since the two objects would not bounce and thus stick together- they become one object in terms of calculations. So the final equation would be: <math>
m_1v_1 + m_2v_2 = (m_1 + m_2)v_f</math>.
===Computational Model===
Using the Collision Lab [https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html], you can model different scenarios for the inelastic collision. Such as when the two bodies have the same mass, or when one is moving at a higher speed than the other. This simulation can help visualize inelastic collisions under different circumstances.
==Examples==
'''Typical Process for Solving'''
In such problems, it is often necessary to apply both the momentum principle and the energy principle.
Momentum Principle: <math> p_{1f} + p_{2f} = p_{1i} </math>
Energy Principle: <math> K_{1f} + K_{2f} + E_{int,f} = K_{1i} + E_{int,i} </math>
As shown above, the change in internal energy is directly related to the change in kinetic energy, in the Energy Principle.


The steps for solving inelastic collision problems are often as follows:
The steps for solving inelastic collision problems are often as follows:
Line 52: Line 70:
1) Draw a picture of the initial and final states.  
1) Draw a picture of the initial and final states.  


2) Given that collisions involve extremely large forces acting over short time intervals, it is accurate to say that Fnet,ext = 0, because the external forces are typically much, much smaller than the internal forces involved in the collision.  
2) Given that collisions involve extremely large forces acting over short time intervals, it is accurate to say that <math>F_{net,ext} = 0</math>, because the external forces are typically much, much smaller than the internal forces involved in the collision.
 
3) Knowing that <math>F_{net,ext} = 0</math>, this means that momentum is conserved and that <math>P_f = P_i</math>. This comes from the momentum principle, in that <math>ΔP_{system} = F_{net,ext}ΔT</math>, and if <math>F_{net,ext}</math> is zero, then the right side of that equation is also zero.
 
4) Next, knowing that <math>P_f = P_i</math>, you can solve for any unknown velocities of the objects involved via the momentum principle.


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


4) Next, knowing that Pf = Pi, you can solve for any unknown velocities of the objects involved via the momentum principle.  
6) Next, you can find the change in kinetic energy, or ΔK, and this will be equal to the negative change in the internal energy (whether it be heat, rotation, etc.) gained by the objects in the course of the collision. In other words <math>ΔE_{int} = -ΔK</math>.


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.
===Simple===


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.
'''Question'''
  In other words ΔEint = -ΔK.


== Examples ==
Block A moves on a friction-less surface at a speed of 5 m/s towards block B. Block B is moving towards Block A at a speed of 2 m/s. Block A has a mass of 2 kg and block B has a mass of 1.5 kg. Assume the blocks are moving directly toward each other in the x direction and stick together on colliding. What is the speed of the blocks after the collision?


Simple
'''Answer'''


This problem illustrates well the difference between elastic and inelastic collisions, as well as providing practice with symbolic questions.
Using conservation of momentum, we can say that the momentum of the two blocks before the collision is equal to the momentum of the blocks after the collision.


[[File:Wiki Image 4.jpg]]
<math> p_{1i} + p_{2i} = p_{f} </math>


<math> m_{A}v_{A,i} + m_{B}v_{B,i} = m_{f}v_{f} </math>
<math>(2)(5) + (1.5)(2) = (2 + 1.5)(v_f)</math>
<math>10 + 3 = 3.5v_f</math>
<math>v_f = 3.714 m/s</math>
===Middling===
'''Question'''


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).
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]]
[[File:Wiki Image.jpg |thumb|center|1020x600px]]


a) What is the velocity of the second puck after the collision?
a) What is the velocity of the second puck after the collision?
Line 86: Line 116:
d) What is the change of the internal energy of the two pucks?  
d) What is the change of the internal energy of the two pucks?  


'''Answer'''


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


Line 95: Line 125:
   <-0.028, 0.731, 0> kg*m/s = <-0.285, 0.45, 0> kg*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
   <0.257,0.281,0> kg*m/s = 0.13kg*v
   v = <1.977, 2.162, 0> m/s
   v = <1.977, 2.162, 0> m/s


b) K = (1/2)*m*|v|^2     
b)  
  K = (1/2)*m*|v|^2     
   |v| = (vx^2 + vy^2 + vz^2)^(1/2)       
   |v| = (vx^2 + vy^2 + vz^2)^(1/2)       
   |v|^2 = vx^2 + vy^2 + vz^2
   |v|^2 = vx^2 + vy^2 + vz^2
Line 103: Line 135:
   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 = (1/2)*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 = 1.336 J + 0.813 J
   Ki = 2.149 J
   Ki = 2.149 J


c) K = (1/2)*m*|v|^2     
c)
  K = (1/2)*m*|v|^2     
   |v| = (vx^2 + vy^2 + vz^2)^(1/2)       
   |v| = (vx^2 + vy^2 + vz^2)^(1/2)       
   |v|^2 = vx^2 + vy^2 + vz^2   
   |v|^2 = vx^2 + vy^2 + vz^2   
Line 111: Line 145:
   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 = (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 = 0.946 J + 0.558 J
   Kf = 1.504 J
   Kf = 1.504 J


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


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






Difficult
'''Question'''
 
A bullet of mass 0.071 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 2 kg that is sitting at rest on a nearly frictionless surface.
 
(a) What is the speed of the block after the bullet embeds itself in the block?
 
(b) Calculate the kinetic energy of the bullet plus the block before the collision:
 
(c) Calculate the kinetic energy of the bullet plus the block after the collision:
 
(d) Was this collision elastic or inelastic?
 
(e) Calculate the rise in thermal energy of the bullet plus block as a result of the collision:
 
(f) What was the transfer of energy Q (microscopic work) from the surroundings into the block+bullet system during the collision? (Remember that Q represents energy transfer due to a temperature difference between a system and its surroundings.)
 
'''Answer'''
 
a)
  ΔP = Pf - Pi = Fnet,ext*ΔT = 0
  pi1+pi2=pf
  m1*v1+m2*v2=mf*vf
  vf=(m1*v1+m2*v2)/mf
  vf=(0.071 kg*100 m/s+2 kg*0 m/s)/(2.071 kg)
  vf=3.43 m/s
 
 
b)
  Ki=0.5*m1*v1i^2+0.5*m2*v2i^2
  Ki=0.5*0.071 kg*(100 m/s)^2+0.5*(2 kg)*(0 m/s)^2
  Ki=355 J
 
c)
  Kf=0.5*mf*vf^2
  Ki=0.5*2.071 kg*(3.43 m/s)^2
  Ki=12.18 J
 
 
d)
  Due to the fact that the final kinetic energy is less than the initial kinetic energy (change in kinetic energy), while the
  momentum of the interaction is conserved, the interaction can be said to be inelastic. The result of the change in kinetic
  energy is a rise in internal/thermal energy.
 
 
e)
  ΔEthermal,bullet + ΔEthermal,block = ΔK=Ki-Kf
  ΔEthermal,bullet + ΔEthermal,block = 355 J-12.18 J
  ΔEthermal,bullet + ΔEthermal,block = 342.8 J
 
 
e)
  As the collision takes place in a closed system, and the collision considers forces that render frictional forces by surroundings
  negligible, there is no transfer of energy Q between the surroundings and the system.
  Q=0 J
 
===Difficult===
 
'''Question'''
 
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|thumb|center|1020x620px]]
 
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)^2 + 0^2)
  - (1/2)(4.5)(5.33^2 + 55.89^2 + 0^2)
  ΔEinternal = 2971.5 - 169.5 - 450 - 295.75 - 7092
 
  ΔEinternal = -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 is especially prominent among football players and other athletes who suffer repeated head trauma.
 
The concepts presented through collisions are also the founding physics principles for studying car crashes, or car safety, as well as other real world events that contain collisions. By using the mathematical models introduced earlier, studies can be done to improve the user experience and overall safety of many things including vehicles.
 
[[File:Pat_Devlin_(American_football).jpg]]
 
== History ==
The idea of inelastic collisions is rooted in the conservation of linear momentum, which is implied by Newton's Laws. Newton published his book, the Philosophiae Naturalis Principia Mathematica, in 1687. This book, commonly known as Principia, described Newton’s findings on collisions and laid a foundation for the different types of collisions we study today.
 
The history of collisions in modern physics starts with Rutherford’s Gold Foil experiment. He found that while they did not touch, the alpha particles and the gold nucleus collided elastically. It is essential that the collision was elastic, in that it gives us the famous scatting pattern and led to the discovery of the positively charged nucleus.
 
== See Also ==
 
=== Further reading ===
 
For more information about the momentum principle, check out this page!
 
[[Momentum Principle]]
 
For more on the energy principle, check here:
 
[[Energy Principle]]
 
If you want to learn more about elastic or maximally inelastic collisions, check out these pages:
 
[[Elastic Collisions]]
 
[[Maximally Inelastic Collision]]
 
=== External links ===
 
A different explanation of [http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html inelastic collisions]
 
A Khan Academy video on [https://www.khanacademy.org/science/ap-physics-1/ap-linear-momentum/inelastic-collisions-and-2d-collisions-ap/v/elastic-and-inelastic-collisions elastic and inelastic collisions]
 
== References ==
 
Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. Hoboken, NJ: Wiley, 2011. Print
 
[[Category:Collisions]]

Latest revision as of 15:15, 8 July 2019

Created by Csorensen6

Edited/Claimed by Jessica Ouyang, Fall 2018

Edited/Claimed by Vansh Kareer, Spring 2017

Edited/Claimed by Abigail Brady, Fall 2017

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


The Main Idea

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


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

Types of Inelastic Collisions

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


Most inelastic collisions are considered partially inelastic collisions. In these types, two or more objects collide but do not stick together (unlike maximally inelastic collisions), but as a result of their collision, one or both begin to rotate, vibrate, deform, heat up, etc. In these problems, there is a positive ΔEint and a negative ΔK, and Ki > Kf. An example of a typically inelastic collision is a car crash.


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


Mathematical Model

In inelastic collisions, the change in internal energy is equal to the negative change in kinetic energy. Based on these characteristics, and applying the momentum principle, a few basic equations can be set to outline an inelastic collision.


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


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

This is in accordance with the relation [math]\displaystyle{ ΔE_{internal} = -ΔK_{trans}=-0.5(mass)(Δvelocity)^2 }[/math].

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

Computational Model

Using the Collision Lab [1], you can model different scenarios for the inelastic collision. Such as when the two bodies have the same mass, or when one is moving at a higher speed than the other. This simulation can help visualize inelastic collisions under different circumstances.

Examples

Typical Process for Solving

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

Momentum Principle: [math]\displaystyle{ p_{1f} + p_{2f} = p_{1i} }[/math]

Energy Principle: [math]\displaystyle{ K_{1f} + K_{2f} + E_{int,f} = K_{1i} + E_{int,i} }[/math]

As shown above, the change in internal energy is directly related to the change in kinetic energy, in 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 [math]\displaystyle{ F_{net,ext} = 0 }[/math], because the external forces are typically much, much smaller than the internal forces involved in the collision.

3) Knowing that [math]\displaystyle{ F_{net,ext} = 0 }[/math], this means that momentum is conserved and that [math]\displaystyle{ P_f = P_i }[/math]. This comes from the momentum principle, in that [math]\displaystyle{ ΔP_{system} = F_{net,ext}ΔT }[/math], and if [math]\displaystyle{ F_{net,ext} }[/math] is zero, then the right side of that equation is also zero.

4) Next, knowing that [math]\displaystyle{ P_f = P_i }[/math], you can solve for any unknown velocities of the objects involved via the momentum principle.

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

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

Simple

Question

Block A moves on a friction-less surface at a speed of 5 m/s towards block B. Block B is moving towards Block A at a speed of 2 m/s. Block A has a mass of 2 kg and block B has a mass of 1.5 kg. Assume the blocks are moving directly toward each other in the x direction and stick together on colliding. What is the speed of the blocks after the collision?

Answer

Using conservation of momentum, we can say that the momentum of the two blocks before the collision is equal to the momentum of the blocks after the collision.

[math]\displaystyle{ p_{1i} + p_{2i} = p_{f} }[/math]

[math]\displaystyle{ m_{A}v_{A,i} + m_{B}v_{B,i} = m_{f}v_{f} }[/math]

[math]\displaystyle{ (2)(5) + (1.5)(2) = (2 + 1.5)(v_f) }[/math]

[math]\displaystyle{ 10 + 3 = 3.5v_f }[/math]

[math]\displaystyle{ v_f = 3.714 m/s }[/math]

Middling

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).

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


Question

A bullet of mass 0.071 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 2 kg that is sitting at rest on a nearly frictionless surface.

(a) What is the speed of the block after the bullet embeds itself in the block?

(b) Calculate the kinetic energy of the bullet plus the block before the collision:

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

(d) Was this collision elastic or inelastic?

(e) Calculate the rise in thermal energy of the bullet plus block as a result of the collision:

(f) What was the transfer of energy Q (microscopic work) from the surroundings into the block+bullet system during the collision? (Remember that Q represents energy transfer due to a temperature difference between a system and its surroundings.)

Answer

a) 

  ΔP = Pf - Pi = Fnet,ext*ΔT = 0
  pi1+pi2=pf
  m1*v1+m2*v2=mf*vf
  vf=(m1*v1+m2*v2)/mf
  vf=(0.071 kg*100 m/s+2 kg*0 m/s)/(2.071 kg)
  vf=3.43 m/s


b) 

  Ki=0.5*m1*v1i^2+0.5*m2*v2i^2
  Ki=0.5*0.071 kg*(100 m/s)^2+0.5*(2 kg)*(0 m/s)^2
  Ki=355 J

c) 

  Kf=0.5*mf*vf^2
  Ki=0.5*2.071 kg*(3.43 m/s)^2
  Ki=12.18 J


d) 

  Due to the fact that the final kinetic energy is less than the initial kinetic energy (change in kinetic energy), while the 
  momentum of the interaction is conserved, the interaction can be said to be inelastic. The result of the change in kinetic 
  energy is a rise in internal/thermal energy.


e) 

  ΔEthermal,bullet + ΔEthermal,block = ΔK=Ki-Kf
  ΔEthermal,bullet + ΔEthermal,block = 355 J-12.18 J
  ΔEthermal,bullet + ΔEthermal,block = 342.8 J


e) 

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

Difficult

Question

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.

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)^2 + 0^2)
  - (1/2)(4.5)(5.33^2 + 55.89^2 + 0^2)
  ΔEinternal = 2971.5 - 169.5 - 450 - 295.75 - 7092
 
  ΔEinternal = -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 is especially prominent among football players and other athletes who suffer repeated head trauma.

The concepts presented through collisions are also the founding physics principles for studying car crashes, or car safety, as well as other real world events that contain collisions. By using the mathematical models introduced earlier, studies can be done to improve the user experience and overall safety of many things including vehicles.

History

The idea of inelastic collisions is rooted in the conservation of linear momentum, which is implied by Newton's Laws. Newton published his book, the Philosophiae Naturalis Principia Mathematica, in 1687. This book, commonly known as Principia, described Newton’s findings on collisions and laid a foundation for the different types of collisions we study today.

The history of collisions in modern physics starts with Rutherford’s Gold Foil experiment. He found that while they did not touch, the alpha particles and the gold nucleus collided elastically. It is essential that the collision was elastic, in that it gives us the famous scatting pattern and led to the discovery of the positively charged nucleus.

See Also

Further reading

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

Momentum Principle

For more on the energy principle, check here:

Energy Principle

If you want to learn more about elastic or maximally inelastic collisions, check out these pages:

Elastic Collisions

Maximally Inelastic Collision

External links

A different explanation of inelastic collisions

A Khan Academy video on elastic and inelastic collisions

References

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

Retrieved from "http://www.physicsbook.gatech.edu/index.php?title=Inelastic_Collisions&oldid=35674"