Talk:Elastic collision

Contribution

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)[reply]

Equations

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)[reply]

Equation in Section 1.1 - The way it is displayed in my browser is a bit confusing.
It looks like:
V1 = [AAA] x V2 = [BBB]
It's actually:
V1= [AAA] and V2 = [BBB].
Can we move the second equation to next line? The 'period' is confusing and almost seems like a multiplication operator. —Preceding unsigned comment added by 220.227.95.85 (talk) 04:47, 2 September 2009 (UTC)[reply]

Link contribution

According to policy, you shouldn't add external links to... A website that you own or maintain, even if the guidelines above imply that it should be linked to. This is because of neutrality and point-of-view concerns; neutrality is an important objective at Wikipedia, and a difficult one. If it is relevant and informative, mention it on the talk page and let other — neutral — Wikipedia editors decide whether to add the link.

I have never edited here before so I would rather have someone experienced decide if the following is appropriate content. I suggest someone add links to my pages on this topic with descriptions as follows:

"2-Dimensional Elastic Collisions Without Trigonometry" (http://www.geocities.com/vobarian/2dcollisions/) "Bouncescope - a simulator for 2-dimensional elastic collisions" (http://www.geocities.com/vobarian/bouncescope/)

Equation ?

How did you solve the equation, going from step 2 to step 3 ? You should explain how you removed v'2 in the v1=... equation and how you removed v'1 in the v2=... equation.

The equations for v1' and v2' are the result of basic but lengthy algebraic manipulations. The basic steps are: solving one equation for one unknown variable, substituting it into the other equation, resulting in an equation with one unknown, then proceeding to simplify, expand, collect like terms, move all quantities to one side of the equation, apply the quadratic formula, expand, collect like terms, take the square root, and simplify.

I didn't think it was necessary to include all these steps because the equations for v1' and v2' are already given in the Wikipedia entry without explanation, so I was actually just using those equations without having solved them myself.

There seems to be inconsistent use of symbols switching between u and v and v with and without prime to represent the velocities before and after the collisions 211.30.176.177 (talk) 12:37, 30 January 2009 (UTC)dnordon[reply]

I agree, I am extremely confused in the elastic collision section. I think there is at least one typo and this is not well explained. Unless I am missing something fundamental I'll make changes after a little more sleep. MystRivenExile (talk) 00:22, 14 September 2009 (UTC)[reply]

Ok, thanks for clarifying! This article helped me a lot.

It's a lot less messy if you temporarily change the frame of reference, I've added this to the article. —Preceding unsigned comment added by 131.111.213.36 (talk) 02:01, 25 November 2007 (UTC)[reply]

Care to explain how you do that? —Preceding unsigned comment added by 195.60.18.238 (talk) 21:37, 18 March 2008 (UTC)[reply]

Merge with Elastic scattering

It might make sense to add more particle physics stuff to this page, but I don't think there needs to be two pages. Flying fish 02:15, 6 March 2007 (UTC)[reply]

I'd have to disagree, along with those who commented on the elastic scattering page. The two terms are used in very different contexts. A single page dealing with both would have to be careful to avoid confusion, and it simply isn't necessary. --Starwed 11:47, 3 April 2007 (UTC)[reply]

I strongly disagree with the merge - others have already motivated why. I'm removing the merge tag. Sergio Ballestrero

Perhaps you got distracted? I removed the merge tag today. --Starwed 05:46, 3 August 2007 (UTC)[reply]

I personally see a reason behind keeping Elastic Collision and Elastic Scattering as distinct pages. While there definitely are paraellels between the two processes, elastic scattering is inherently a quantum process whereas elastic collision is within the realm of Newtonian mechanics. Rephrasing this distinction, elastic scattering is a sub-microscopic, relativistic process whereas elastic collision is primarily a macroscopic, sub-relativistic process. Approaching the one model from the vantage of the other often leads to incorrect description of phenomena and so-called paradoxical behavior. Thus separating the models by having two different pages should be pedagogically accurate. -Billyziege 21:26, 8 March 2007 (UTC)Billyziege[reply]

Non sequitur

(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)[reply]

relativ velcity is ewual before and after collision —Preceding unsigned comment added by 91.150.113.93 (talk) 13:42, 8 September 2007 (UTC)[reply]

Total kinetic energy is the same before and after the collision, hence:

this equation does not say what the numerical value of u is. I have no idea how to manipulate this equation because I don't know what u represents. —Preceding unsigned comment added by 71.198.202.203 (talk) 07:00, 6 November 2007 (UTC)[reply]

I suggest you actualy read the whole article before posting it says "let u be the velocity before collision" right before any equations are written

-John

If right about here you would put an image of the 2 m's rebounding from each other in elastic collisions, whereas they would cancel out their momentum in the case of an inelastic collision, I think that everyone would better understand the logic of the images as compared to understanding the math.WFPM (talk) 03:20, 1 October 2012 (UTC)[reply]

no, it do not remains

Total momentum remains constant throughout the collision:

\,\!m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2}.

Since if m=1kg, v=1 m/s before collision momentum equal to v*2m+v*m=1*2*1+1*1=3 and after collision 2m*v/3+5mv/3=2*1*1/3+5*1*1/3=2.666666666 so it is not equal. 3 and 2.7 is not the same. —Preceding unsigned comment added by Taupusensteinas (talk • contribs) 15:18, 15 October 2009 (UTC)[reply]

Before collision:

m1 = 2 m, v1 = v, m2 = m, v2 = -v

total momentum = m1 v1 + m2 v2 = 2 m v - m v = m v

total kinetic energy = 1/2 m1 v1 + 1/2 m2 v2^2 = m v^2 + 1/2 m v^2 = 3/2 m v^2

After collision:

m1 = 2 m, u1= -1/3 v, m2 = m, u2 = 5/3 v

total momentum = m1 u1 + m2 u2 = -2/3 m v + 5/3 m v = m v

total kinetic energy = 1/2 m1 u1^2 + 1/2 m2 u2^2 = 1/9 m v^2 + 25/18 m v^2 = 3/2 m v^2

Now please stop using this talk page for puzzles, challenges, homework? Thank you. DVdm (talk) 06:03, 16 October 2009 (UTC)[reply]

The article says that the collision results in a separating force between the colliding particles, but results in a separation velocity of 2, with 5/3 parts to the unit mass and only 1/3 part to to the 2 unit mass and a total momentum transfer of 8/3 units. How can the separation force between the masses be different?WFPM (talk) 23:12, 13 July 2012 (UTC)[reply]

Consider the problem as 1 particle colliding with 2 particles. If the 2 particle mass is moving forward with a velocity v and a momentum of 2mv and collides with a particle with momentum of 1mv, there is no way that its velocity direction will be reversed but only reduced by as much as 1mv as occurs in an inelastic collision. If such a energy transfer is possible, then it must be by an angular momentum transfer process where the the angular momentum is conserved in energy transfer processes.WFPM (talk) 17:43, 15 July 2012 (UTC)[reply]

Another exercise in mathematical illogic.WFPM (talk) 18:04, 15 July 2012 (UTC)[reply]

If the 2m particle were a spring that could store the collision angular momentum (8/3 units) then in the middle of the process we would have the unelastic condition of the combined 3m mass moving forward at 1/3 v. Then as the spring returned the 8/3 momentum to the associated masses, the force of expansion of the spring would result in the acceleration of the single mass being twice that of the spring mass, (assuming that all the mass of the spring was in the moving portion), and thus resulting in your mathematical equations saying that that the returned angular momentum would be evenly divided between the unit mass and the 2unit mass, with the unit mass achieving a forward velocity of 5/3 units and the 2 unit mass being given a reverse velocity of -2/3 units to a net negative velocity of - 1/3 units, and the math works out. But it still looks fishyWFPM (talk) 16:01, 16 July 2012 (UTC)[reply]

What angular momentum? About what point? Why introduce angular momentum into a 1D problem? -AndrewDressel (talk) 16:31, 16 July 2012 (UTC)[reply]

WFPM, see, for instance, this book on page 244. If you have a problem with such basic examples, perhaps you could try at our wp:reference desk/science. Good luck. - DVdm (talk) 16:39, 16 July 2012 (UTC)[reply]

Okay Okay!I was just offended by seeing an image of a 2m mass being apparently instantaneously reversed by a linear collision with a 1m mass when I know that couldn't happen. But evidently it does as part of a sequential operation. And your Math doesn't explain that sequence. I'm still trying to rationalize the parceling out the stored kinetic energy between the 2 different masses, where here you say they each get the same amount of returned linear momentum, and whereas in explosions each mass gets its proportion of the momentum. And in orbital mechanics I know that angular momentum would be conserved in any kind of angular motion. So so much for interested thinking!WFPM (talk) 17:42, 16 July 2012 (UTC)[reply]

As I said, try the reference desk. An article talk page —see wp:TPG— isn't supposed to be the place where the article subject (or aspects thereof) should be explained, rationalised or defended. Cheers - DVdm (talk) 17:53, 16 July 2012 (UTC)[reply]

It's nonsens, momentum should be summed up.

Erroneous section hidden - Click the Show button on the right to reveal content.

Particulary for over example I take such conditions: m_1=1000 kg, v_1=10 m/s; m_2=1 kg, v_2=0 m/s. And after colission with formula combination of kinetic energy and momentum from pdf or which used in this gif file, I get u_1=9.98 m/s, u_2=19.98 m/s. So Do you imagine like what nonsense it sounds? I kinda touch small ball and becouse my mass is big, this ball flying from me with insane speed. Like kinda I am superman or what? I have big mass, I with small speed touch some small 10 grams particle and it flying with ten or hundred times bigger speed than move my finger at that moment. You can use metal object instead finger if you don't like it softnes and nonsens will repeate itself one more time, isn't it? —Preceding unsigned comment added by Taupusensteinas (talk • contribs) 09:48, 16 October 2009 (UTC)[reply]

I suggest this my example solve in this way: Don't moving 1 kg mass m_2 object aquaring speed of heavier object m_1 and so now momentum of m_2 object is p_2=v_1*m_2=10*1=10; u_2=10 m/s. And Momentum of object m_1 becoming after colision m_1*v_1-v_1*m_2=1000*10-1*10=9990. So speed after collision u_1=9990/m_1=9990/1000=9.99 m/s. —Preceding unsigned comment added by Taupusensteinas (talk • contribs) 10:08, 16 October 2009 (UTC)[reply]

Here proposition how should behave two objects collision if m_1<m_2 and v_2=0 for object m_2. Example:

m_1=2 kg, v_1=100 m/s, r_1=m_1*v_1=2*100=200;

m_2=3 kg, v_2=0 m/s, r_2=m_2*v_2=3*0=0.

p_2=m_2*(v_1/(m_2/m_1))*(1/(m_2/m_1)=v_1*m_1*m_1/m_2=r_1*m_1/m_2=400/3=133.333333.

u_2=v_1/(m_2/m_1)*m_1/m_2=v_1*m_1*m_1/(m_2*m_2)=r_1*m_1/(m_2*m_2)=200*2/(3*3)=400/9=44.4444444 m/s.

u_1=v_1-(v_1/(m_2/m_1))=v_1-r_1/m_2=100-200/3=33.3333333 m/s.

p_1=m_1*u_1=2*100/3=66.666666.

And here we see conservation of momentum r_1+r_2=200+0=200 and p_1+p_2=66.6666+133.33333=200. First object getting oposit moving direction after collision.

And here proposition for objects when m_1>m_2 and v_2=0 of object m_2:

m_1=3 kg, v_1=100 m/s, r_1=3*100=300;

m_2=2 kg, v_2=0 m/s, r_2=2*0=0.

p_2=v_1*m_2=100*2=200.

u_2=100 m/s automaticly aquaring speed of m_1 object, because m_1 heavier than m_2.

p_1=r_1-u_2*m_2=r_1-p_2=300-200=100.

u_1=(r_1-p_2)/m_1=(300-200)/3=33.3333333 m/s.

So conservation of momentum remain the same: r_1+r_2=300 and p_1+p_2=100+200=300. —Preceding unsigned comment added by Taupusensteinas (talk • contribs) 14:15, 16 October 2009 (UTC)[reply]

Remarks:

"It's nonsens, momentum should be summed up." => Momentum is summed up - you just don't seem to understand that the v1, v2, u1 and u2 in the equations stand for velocities, and can be either be interpreted as vectors, or as projected components on some directed axis. In this case the u- and v-variables are components and are therefore positive in one direction and negative in the other direction.
Please note that this is not a place where you come for free lessons in elementary physics. If you want to propose a modification to the article, please provide a proper source.
Also please note the text on top of the edit window. It says:

This is a talk page. Please respect the talk page guidelines, and remember to sign your posts bytyping four tildes (~~~~).

Please take some time to read the warnings on your talk page? - Thank you. - DVdm (talk) 16:41, 16 October 2009 (UTC)[reply]

Two trillion fold?

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)[reply]

Collision of atoms

"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)[reply]

The Two-dimensional Deflation Formulae

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) There is no θ in the picture and also no υ1 and υ2。 Kidd lg M (talk) 12:54, 6 July 2010 (UTC)[reply]

Switch to a procedural language for the example?

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)[reply]

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)[reply]

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)[reply]

Trivial solution?

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)[reply]

I sense that this section dangerously assumes that the equal-mass particles are indistinguishable. If we have a 2 kg blue ball at 3 m/s and a 2 kg red ball at -5 m/s, the elastic collision will result in the blue ball at -5 m/s and the red ball at 3 m/s. Thus the equal-mass but distinguishable particles trade velocities, so the article erroneously claims that no collision took place. Is this correct? Gatorjakey (talk) 21:06, 23 August 2019 (UTC)[reply]

2d collision code example

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)[reply]

Inelastic collision of atoms

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)[reply]

What about rotational momentum?

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)[reply]

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)[reply]

OK. I have read the original research link and see your point. It cannot go here, and it is below the publication threshold of the standard publication sources. I know: I tried. It would appear the knowledge will remain hidden, unless I create an outside web page and then reference myself. I suspect this is not in the spirit of the prohibition on original work, but see no alternatives. LNeergaard (talk) 19:15, 19 July 2012 (UTC)[reply]

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)[reply]

In the introduction is written first just that "kinetic energy" should be conserved, and a few lines after "kinetic energy in their translational motions". This makes a huge difference, since the total kinetic energy (including rotational energy) in collisions of diatomic molecules is conserved, and therefore they would qualify as elastic collisions, but they would not if you only talk about translational motions. 129.206.28.122 (talk) 13:28, 9 October 2014 (UTC)Luca[reply]

What about coefficient of restitution?

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)[reply]

velocity variations

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) This is noted to NOT be the case in the 2 dimensional collision illustration.WFPM (talk) 06:55, 2 October 2012 (UTC)[reply]

Plastic collision badly reffered

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)[reply]

I have redirected it to Inelastic collision#Perfectly inelastic collision. Thanks for pointing this out. RockMagnetist (talk) 04:47, 26 September 2013 (UTC)[reply]

Check the equation for 2D collision

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)[reply]

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)[reply]

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)[reply]

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)kiranganeshan14[reply]

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)[reply]

possible issue in Two-dimensional collision section

It seems to me the this sentence: "Since the collision only imparts force along the line of collision, the velocities that are tangent to the point of collision do not change." has two issues: 1. firstly it is only true if there is no friction between two surfaces, which is not necessarily true all the time. 2. secondly, "tangent to the point" is a mathematically wrong sentence. you can not tangent anything to a point. it should be replaced by "tangent to the surface" — Preceding unsigned comment added by Foadsf (talk • contribs) 10:51, 14 February 2017 (UTC)[reply]

Please sign all your talk page messages with four tildes (~~~~). Thanks.

Yes, or "tangent to the surface at the point of collision". Good find! Feel free to correct, as in wp:sofixit. - DVdm (talk) 16:59, 14 February 2017 (UTC)[reply]

collision of more than two objects?

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)[reply]

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)[reply]

Inaccurate statements in the introduction

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)[reply]

Large, possibly undue or COI addition

See Wikipedia_talk:WikiProject_Physics#Large_addition_at_Elastic_collision. More comments welcome. - DVdm (talk) 08:37, 3 September 2020 (UTC)[reply]

Newtonian versus relativistic

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 {2m_{1}m_{2}c^{2}u_{2}Z+2m_{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}}{2m_{1}m_{2}c^{2}Z-2m_{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 {2m_{1}m_{2}c^{2}u_{1}Z+2m_{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}}{2m_{1}m_{2}c^{2}Z-2m_{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 $c\to \infty$ . 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 $v_{1}\to \infty /\infty$ 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 {2m_{1}m_{2}u_{2}+2m_{2}^{2}u_{2}-0+(m_{1}^{2}-m_{2}^{2})u_{1}}{2m_{1}m_{2}-0-0+(m_{1}^{2}+m_{2}^{2})}}

={\frac {2m_{1}m_{2}u_{2}+2m_{2}^{2}u_{2}+(m_{1}^{2}-m_{2}^{2})u_{1}}{2m_{1}m_{2}+(m_{1}^{2}+m_{2}^{2})}}

={\frac {(m_{1}-m_{2})u_{1}+2m_{2}u_{2}}{m_{1}+m_{2}}}

This is the Newtonian result. Arie ten Cate (talk) 09:02, 9 December 2020 (UTC)[reply]

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)[reply]

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.^[1] JFB80 (talk) 16:28, 16 December 2020 (UTC) JFB80 (talk) 17:17, 16 December 2020 (UTC)[reply]

References

^ Whittaker E.T. Aether and Electricity, Harper NY 1953 pp.47 etc.

Need to exclude initial velocities!

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)[reply]

Derivation of equation of final velocities in 2D.

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}}

\mathbf {p'_{2}} -\mathbf {p_{2}} =\mathbf {J_{2}}

(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 (1):

\mathbf {v'_{1}} =\mathbf {v_{1}} +{\frac {\lambda }{m_{1}}}\mathbf {\hat {n}}

\mathbf {v'_{2}} =\mathbf {v_{2}} -{\frac {\lambda }{m_{2}}}\mathbf {\hat {n}} ,

(2)

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,\qquad

whith

\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 (2) 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)[reply]

The balls bouncing against each other.

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)[reply]

[1] Whittaker E.T. Aether and Electricity, Harper NY 1953 pp.47 etc.

[1]