Talk:Elastic collision

message: This is my first contribution, so I hope I've done it right. I forgot to put in a message for the edit log, so here's what I did:
The article was looking promising until it was wiped and later replaced with a copy of 'Elastic', so I reverted it to the 07:50, 2 May 2005 version and merged a couple of the more recent changes. Squashed sultana 07:30, 5 August 2005 (UTC)
thread_title:  Contribution 

message: The equations were not displaying correctly, but I was able to fix the problem. All in all, you did a nice job.
Bmaganti 04:08, 20 April 2006 (UTC)
thread_title:  Equations 

message: (u)+(v)=(u')+(v')  where () is an apsolute value
u-v=v'-u'
u+u'=v+v'  —Preceding unsigned comment added by 91.150.113.93 (talk) 13:39, 8 September 2007 (UTC)
replies: text: relativ velcity is ewual before and after collision  —Preceding unsigned comment added by 91.150.113.93 (talk) 13:42, 8 September 2007 (UTC)
thread_title:  Non sequitur 

message: Removed the sentence "Here, room-temperature helium atoms are slowed down two trillion fold." from the caption for the picture because it was absurd.--Tigerthink (talk) 09:13, 21 October 2009 (UTC)
thread_title:  Two trillion fold? 

message: "The collisions of atoms are elastic collisions".  Is this always true? Can't atoms collide and for example ionize one of them? Lea phys (talk) 14:41, 31 January 2010 (UTC)
thread_title:  Collision of atoms 

message: The velocity formulae are wrong, simply because they do not have dimensions of velocity. Probably the correction would be to multiply the given expressions by the original velocity of the moving body/ 132.68.73.50 (talk) 16:01, 23 May 2010 (UTC)
thread_title:  The Two-dimensional Deflation Formulae 

message: I would suggest switching to a procedural language such as C. I haven't fully tested the code to replace it but I think if we use C we can then easily verify the correctness whereas with the current example we can't. If noone objects I'll fully check the correctness of the above, supply a testcase in a comment and add it to the article instead of the C# IRWolfie- (talk) 22:23, 10 March 2012 (UTC)
replies: text: C# is also a procedural language and there is nothing in this example that is not completely isomorphic to C/C++. I see no compelling reason to replace it. Qartar (talk) 21:15, 29 March 2012 (UTC)
replies: text: More to C++ than C. the current version can confuse since most of the class isn't actually defined. IRWolfie- (talk) 20:18, 19 April 2012 (UTC)
thread_title:  Switch to a procedural language for the example? 

message: Is it really necessary to state the "trivial solution" in the One-dimensional Newtonian Equations subsection?
"OR ...
The latter is the trivial solution, corresponding to the case that no collision has taken place (yet)."
Qartar (talk) 21:06, 29 March 2012 (UTC)
thread_title:  Trivial solution? 

message: Hi. I don't know anything about physics. However, I was trying to implement the code example in some real code. I was only able to get it to work after changing
ball.velocity = v2t + Dn * ((m2 - m1) / M * v2n.Length() + 2 * m1 / M * v1n.Length());
to
ball.velocity = v2t + Dn * ((m2 - m1) / M * v2n.Length() + (-2) * m1 / M * v1n.Length());
Otherwise, the ball bounces in the wrong (reverse) direction. Perhaps there was something else wrong in my code, but I checked pretty carefully and couldn't find anything. Thought I would point it out in case it was some obvious mistake.
Also, this seemed a little strange:
       // avoid division by zero
        if (delta == 0)
        {
            ball.Center = new Vector2(0.01f,0);
            return;
        }'''
I'm not sure how moving the ball to a completely new part of the map would help.
Controller143 (talk) 04:39, 19 April 2012 (UTC)
thread_title:  2d collision code example 

message: As an earlier user mentioned, atomic collisions are not necessarily elastic. Atoms in excited states can undergo collisional de-excitation. This would lead to a inelastic collision.
This is why the "forbidden" lines of emission nebulae are not seen under lab conditions. The excited states in question de-excite collisionally in the lab instead of by emission (as they do at extremely low densities in the nebula). I propose that the line in the introduction that states that atomic collisions are elastic be removed.
Kepler314 (talk) 13:42, 27 June 2012 (UTC)
thread_title:   Inelastic collision of atoms 

message: This article is already quite long, so maybe another article should be started.  I note that rotational momentum is not taken into account in the article at all.  Jan Hakenberg's web page (referenced) does treat rotational momentum, but does not take friction into account.  It seems to me that the physics of collision resolution between objects of arbitrary shape and initial state would be of interest to computer game writers.
I have a solution to these collisions which does not appear here or elsewhere.  I am reticent to include it here, because I am not sure here is where it belongs. LNeergaard (talk) 22:47, 30 June 2012 (UTC)
replies: text: While I see no issue with adding more content, it can always be broken out into a sub article, there is likely to be a problem with adding a solution that does not appear elsewhere. That sounds like it would be a perfect example of original research. You'd be better of submitting that for publication first. -AndrewDressel (talk) 16:44, 16 July 2012 (UTC)

text: Such a web page would be a mere wp:primary source. WP needs wp:secondary sources and wp:notability. If this work is indeed recent (original) research, then that might take a decade or so. So, creating that page in order to get it in here would be a waste of your —any anyone's— time. - DVdm (talk) 21:26, 19 July 2012 (UTC)
thread_title:  What about rotational momentum? 

message: An alternate solution method is to use the expression for coefficient of restitution as the second equation instead of kinetic energy. Has this been left out on purpose? I don't see it discussed anywhere. It is the method I've seen it Dynamics textbooks, such as the ones by Hibbeler and Beer & Johnston. Any objects to adding it? -AndrewDressel (talk) 16:36, 16 July 2012 (UTC)
thread_title:  What about coefficient of restitution? 

message: The multiple ball image shows a multitude of presumably equal balls moving in different directions at different velocities. However the explanatory images do not show a change in velocity of a ball due to an encounter with an equal mass ball. Is there a mathematical description of how an encounter between equal mass balls results in one of the balls achieving an increased velocity of motion?WFPM (talk) 03:35, 1 October 2012 (UTC)
thread_title: velocity variations

message: For some reason the term plastic collision leads to this article, which did not at all mention it.
The phenomena actually appears under the term Perfectly inelastic collision in the inelastic collision article.
Please change the referrals to avoid further confusion.  — Preceding unsigned comment added by 109.160.242.51 (talk) 02:31, 26 September 2013 (UTC)
replies: text: I have redirected it to Inelastic collision#Perfectly inelastic collision. Thanks for pointing this out. RockMagnetist (talk) 04:47, 26 September 2013 (UTC)
thread_title:  Plastic collision badly reffered 

message: Hi, i implemented the equation for 2D collision, there is an error in there
Following the link [2] for the source you'll see that the source is correct
Basically you have to consider the module of the speed in the equation (V1 and V2), not the single component (V1x, V1y, V2x,V2y)
otherwise you lose a lot of energy in the collision (biggest loss for 45° phase between velocity and collision angle  — Preceding unsigned comment added by 188.153.40.250 (talk) 12:34, 23 April 2014 (UTC)
replies: text: I believe this is right. It was right before as well, but for some reason Lutzl changed the equation. I used the old version in a computer game with collision physics and it was working fine. Now with the new version strange things happen. Also, as you mentioned, the source uses the norms of the velocities. 88.153.242.165 (talk) 12:56, 11 May 2014 (UTC)

text: Back to the previous state. -- This happens with uneven expectations. I expect speed to be a vector, especially since the result of the computation is given as a vector; the original author expected speed to be a scalar and that this to be implicitly understandable by the discussion of angles after that. --  I still think that this whole angle business, while nice to demonstrate the geometry of the process, is an unnecessary complication in actual implementations.--LutzL (talk) 00:28, 12 May 2014 (UTC)

text: I ran into issues with the vector form of the 2D collision equation. I actually calculated the energy, and there is a rather large amount of energy lost after about a 3 minute run time: the total kinetic energy went from about 800 kJ to less than 1 kJ. Can someone actually give a corrected version of the equation, either on the page or on this talk page (preferably on the page, of course)?Kiranganeshan14 (talk) 05:12, 28 November 2015 (UTC)

text: The vectorized 2D equation has an error.  The last factor (x_1 - x_2) must also be normalized to unit length.  We are projecting the velocity change magnitude on the axis of collision onto the collision axis itself.  If you think about it, it simply does not make sense that the updated speed should have anything to do with the raw position difference (x_1 - x_2) and, moreover, the units of this position difference measurement.  Just implement the existing version and you will see that it is incorrect, though it only needs the small fix as noted above to work correctly.  184.147.91.137 (talk) 03:03, 7 December 2020 (UTC)
thread_title:  Check the equation for 2D collision 

message: Why the article states that a elastic collision is a collision between two objects? shouldnt it say 'two or more'?
Also there is not only a clue on how to solve this kind of collision. I suppose that you treat the bodies on two by two but im only supposing.— Preceding unsigned comment added by 167.57.87.129 (talk) 19:23, 7 March 2018 (UTC)
replies: text: In order to have a collision between more that two objects, you need them to all simultaneously come into contact at exactly the same time (otherwise it is simply a collection of two by two collisions). Therefore, this kind of collision is a bit artificial in that the system essentially needs to be constructed in such a way that it occurs. The most common example is the break in billiards, where there are many billiard balls initially at rest in contact with many others when one of them is hit by the moving cue ball. Jdmiller97 (talk) 20:17, 5 September 2020 (UTC)
thread_title:  collision of more than two objects? 

message: The statement "A useful special case of elastic collision is when the two bodies have equal mass, in which case they will simply exchange their momenta" is only true in one dimension (and head-on collisions in higher dimensions).
In the introductory figure, the statement "On average, two atoms rebound from each other with the same kinetic energy as before a collision" in the caption is inaccurate (if it were true, then gases would never equilibriate). For hard spheres in two dimensions with equal masses, an average collision equalizes the kinetic energies of the two atoms. For hard spheres in more than two dimensions or with unequal masses, an average collision reduces the difference between the kinetic energies of the two atoms (the reduction is multiplicative and depends on the dimension and ratio of masses). The original statement is correct in general with the caveat that a thermodynamic equilibrium has been reached, but I believe this takes the characterization away from mechanics to statistical physics. Jdmiller97 (talk) 23:53, 16 July 2020 (UTC)
thread_title:  Inaccurate statements in the introduction 

message: See Wikipedia_talk:WikiProject_Physics#Large_addition_at_Elastic_collision. More comments welcome. - DVdm (talk) 08:37, 3 September 2020 (UTC)
thread_title:  Large, possibly undue or COI addition 

message: === Missing result for relativistic? ===
In the article section "One-dimensional relativistic", above the header "When u_1\ll c and u_2\ll c",   I miss both v_1 and v_2 as a function of only  u_1, u_2, m_1, m_2 and c.  Hence I find the statement "Therefore, the classical calculation holds true when the speed of both colliding bodies is much lower than the speed of light" hard to follow. See the section below.
=== Newtonian is a special case of the special relativity theory ===
    Note: this section must not apppear in a Wikipedia article, since it lacks a reference. I don't have the Serway book used in the article. (By the way, unfortunately, there is no reference in the article section about hyperbolic functions.)
Only at the end of the article section about hyperbolic functions, the relativistic result is given, with v_1 and v_2 both as a function of only  u_1, u_2, m_1, m_2 and c:
: Z=\sqrt{(1-u_1^2/c^2) (1-u_2^2/c^2)} 
: v_1=\frac{2 m_1 m_2 c^2 u_2 Z+2 m_2^2 c^2 u_2-(m_1^2+m_2^2) u_1 u_2^2+(m_1^2-m_2^2) c^2 u_1} {2 m_1 m_2 c^2 Z-2 m_2^2 u_1 u_2-(m_1^2-m_2^2) u_2^2+(m_1^2+m_2^2) c^2} 
: v_2=\frac{2 m_1 m_2 c^2 u_1 Z+2 m_1^2 c^2 u_1-(m_1^2+m_2^2) u_1^2 u_2+(m_2^2-m_1^2) c^2 u_2} {2 m_1 m_2 c^2 Z-2 m_1^2 u_1 u_2-(m_2^2-m_1^2) u_1^2+(m_1^2+m_2^2) c^2} .
Hence here is an opportunity to find out if the Newtonian ("classical") result is indeed a special case of the relativistic result, based directly on the relativistic result:
I start with c \to \infin. Then we have for the Lorentz factor: \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\to 1. And similarly Z \to 1.
We also have v_1 \to \infin/\infin and the same for v_2. This requires L'Hopital's rule: take the derivative of  both the numerator and the denominator with respect to c, until a computable result emerges. 
This has to be done two times here, transforming  c^2 into 2, thereby removing all occurencies of c. Terms without c^2 transform to zero. This gives for instance for v_1: 
: v_1=\frac{2}{2} \times 
   \frac{2 m_1 m_2  u_2 +2 m_2^2  u_2 - 0 + (m_1^2-m_2^2)  u_1} 
   {2 m_1 m_2  - 0 -0 +(m_1^2+m_2^2) }  
:    = \frac{2 m_1 m_2  u_2 +2 m_2^2  u_2  + (m_1^2-m_2^2)  u_1} 
   {2 m_1 m_2  +(m_1^2+m_2^2) }  
:     = \frac{(m_1-m_2)u_1 + 2m_2u_2}{m_1+m_2} 
This is the Newtonian result.  
Arie ten Cate (talk) 09:02, 9 December 2020 (UTC)
replies: text: Is there a source somewhere? This is nice, but effectively off-topic as far as the Wikipedia wp:talk page guidelines are concerned. Without a reliable source, this will not be taken on board in the article, and we are not allowed to even discuss it here. - DVdm (talk) 11:07, 14 December 2020 (UTC)
replies: text:  Isn't it simpler to put c infinite in the initial equations when relativistic goes over to Newtonian? One old accessible reference is to Whittaker who treats in some detail the relativistic case both in standard and hyperbolic notation. Whittaker E.T. Aether and Electricity, Harper NY 1953 pp.47 etc.  JFB80 (talk) 16:28, 16 December 2020 (UTC)
thread_title:  Newtonian versus relativistic 

message: Note: this section must not apppear in a Wikipedia article, since it lacks a reference. I don't have the Serway book used in the article. (By the way, unfortunately, there is no reference in the article section about hyperbolic functions.)
Only at the end of the article section about hyperbolic functions, the relativistic result is given, with v_1 and v_2 both as a function of only  u_1, u_2, m_1, m_2 and c:
: Z=\sqrt{(1-u_1^2/c^2) (1-u_2^2/c^2)} 
: v_1=\frac{2 m_1 m_2 c^2 u_2 Z+2 m_2^2 c^2 u_2-(m_1^2+m_2^2) u_1 u_2^2+(m_1^2-m_2^2) c^2 u_1} {2 m_1 m_2 c^2 Z-2 m_2^2 u_1 u_2-(m_1^2-m_2^2) u_2^2+(m_1^2+m_2^2) c^2} 
: v_2=\frac{2 m_1 m_2 c^2 u_1 Z+2 m_1^2 c^2 u_1-(m_1^2+m_2^2) u_1^2 u_2+(m_2^2-m_1^2) c^2 u_2} {2 m_1 m_2 c^2 Z-2 m_1^2 u_1 u_2-(m_2^2-m_1^2) u_1^2+(m_1^2+m_2^2) c^2} .
Hence here is an opportunity to find out if the Newtonian ("classical") result is indeed a special case of the relativistic result, based directly on the relativistic result:
I start with c \to \infin. Then we have for the Lorentz factor: \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\to 1. And similarly Z \to 1.
We also have v_1 \to \infin/\infin and the same for v_2. This requires L'Hopital's rule: take the derivative of  both the numerator and the denominator with respect to c, until a computable result emerges. 
This has to be done two times here, transforming  c^2 into 2, thereby removing all occurencies of c. Terms without c^2 transform to zero. This gives for instance for v_1: 
: v_1=\frac{2}{2} \times 
   \frac{2 m_1 m_2  u_2 +2 m_2^2  u_2 - 0 + (m_1^2-m_2^2)  u_1} 
   {2 m_1 m_2  - 0 -0 +(m_1^2+m_2^2) }  
:    = \frac{2 m_1 m_2  u_2 +2 m_2^2  u_2  + (m_1^2-m_2^2)  u_1} 
   {2 m_1 m_2  +(m_1^2+m_2^2) }  
:     = \frac{(m_1-m_2)u_1 + 2m_2u_2}{m_1+m_2} 
This is the Newtonian result.  
Arie ten Cate (talk) 09:02, 9 December 2020 (UTC)
replies: text: Is there a source somewhere? This is nice, but effectively off-topic as far as the Wikipedia wp:talk page guidelines are concerned. Without a reliable source, this will not be taken on board in the article, and we are not allowed to even discuss it here. - DVdm (talk) 11:07, 14 December 2020 (UTC)
replies: text:  Isn't it simpler to put c infinite in the initial equations when relativistic goes over to Newtonian? One old accessible reference is to Whittaker who treats in some detail the relativistic case both in standard and hyperbolic notation. Whittaker E.T. Aether and Electricity, Harper NY 1953 pp.47 etc.  JFB80 (talk) 16:28, 16 December 2020 (UTC)
thread_title:  Newtonian is a special case of the special relativity theory 

message: In the 1D elastic collision between two particles, it is obvious that if the final velocities of the two particles were equal to the initial velocities (v1 = u1 and v2 = u2), then they would give the correct values for the total momentum and total kinetic energy.
So at least a few words are necessary to explain why this trivial solution is not the physically correct solution. 2601:200:C000:1A0:3C4F:AE96:D771:5385 (talk) 01:18, 30 August 2022 (UTC)
thread_title:  Need to exclude initial velocities! 

message: The derivation of vectorial equation of the final velocities in the 2D case is quite simple, and I think is worth to include. I think is also interesting to note that the same equation is valid for the 3D case with hard spheres.
It would be something like this:
The impulse \mathbf J during the collision for each particle is:
:\mathbf{p'_1}-\mathbf{p_1}=\mathbf{J_1}
Conservation of Momentum implies \mathbf{J}\equiv\mathbf{J_1}=-\mathbf{J_2} 
Since the force during collision is perpendicular to both particles' surfaces at the contact point, the impulse is along the line parallel to \mathbf{x}_1-\mathbf{x}_2 \equiv\Delta \mathbf x , the relative vector between the particles' center at collision time:
: \mathbf J =\lambda\, \hat n, for some \lambda to be determined and \mathbf \hat n \equiv \frac{\Delta \mathbf x}{\|\Delta \mathbf x\|}
Then from ():
:\mathbf{v'_1} = \mathbf {v_1} + \frac{\lambda}{m_1} \mathbf \hat n
From above equations, conservation of kinetic energy now requires:
:\lambda^2\frac{m_1+m_2}{m_1 m_2}+2\lambda\,\langle \mathbf \hat n, \Delta \mathbf v\rangle = 0 ,\qquadwhith \Delta \mathbf v\equiv \mathbf{v}_1-\mathbf{v}_2
The both solutions of the this equation are \lambda = 0  and \lambda = -2 \frac{m_1 m_2}{m_1+m_2}\langle \mathbf \hat n, \Delta \mathbf v\rangle, where \lambda = 0  corresponds to the trivial case of no collision. Substituting the non trivial value of \lambda in () we get the desired result.
Since all equations are in vectorial form, this derivation is valid also for 3 dimensions with spheres. Wikon (talk) 12:39, 11 October 2022 (UTC)
thread_title:  Derivation of  equation of final velocities in 2D. 

message: Hi can someone make the five red atoms separate colours so they’re easier to track individually?? Ty Coloured balls (talk) 23:06, 3 September 2023 (UTC)
thread_title:  The balls bouncing against each other.