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Natural Science Forum / Physics / General Physics / July 2008Conservation laws
Two particles move in opposite direction and have an elastic collision Particle 1 has a mass of 1 kg and a velocity of 5 m/s. Particle 2 has a mass of 0,5 kg and a velocity of 10 m/s. What is the velocity after impact ? From the principle of conservation of linear momentum : m1*v1 - m2*v2 = m1*v3 + m2*v4 As 1*5 - 0,5*10 = 0 so m1*v3 +m2*v4 is also zero. Hence v3 and v4 are also 0. However, in de pre-impact fase the kinetic energy was : E pre = 0,5*m1*v1^2 + 0,5*m2*v2^2 = 0,5*1*5^2 + 0,5*0,5*10^2 = 37,5 Joule After impact the kinetic energy is E post = 0,5*m1*v3^2 + 0,5*m2*v4^2 = 0,5*1*0^2 + 0,5*0,5*0^2 = 0 Joule This violates the law of conservation of energy. Where did I went wrong? > Two particles move in opposite direction and have an elastic collision > [quoted text clipped - 22 lines] > > Where did I went wrong? Elastic collisions do not conserve energy. > > Two particles move in opposite direction and have an elastic collision > [quoted text clipped - 26 lines] > > - Tekst uit oorspronkelijk bericht weergeven - Sorry Eric but in an elasytic collision kinetic energy is conserved ! > > > Two particles move in opposite direction and have an elastic collision > [quoted text clipped - 28 lines] > > Sorry Eric but in an elasytic collision kinetic energy is conserved ! Whoops my bad, got them confused. The only way the particles will have no energy after the collision is if they stick together. You assume the velocities are equal to zero because the sum is zero - why? Eric Gisse <jowr.pi@gmail.com> wrote in message e908af84-71ac-459d-99db-bee3b55a7047@w39g2000prb.googlegroups.com >>>> Two particles move in opposite direction and have an elastic collision >> [quoted text clipped - 33 lines] > The only way the particles will have no energy after the collision is > if they stick together. HUH? Why shouldn't they have zero energy then??? > You assume the velocities are equal to zero > because the sum is zero - why? You're a bit confused this morning :-) see my reply :-) Dirk Vdm Dirk Van de moortel <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote in message zvFgk.3400$vp4.1466@newsfe28.ams2 > Eric Gisse <jowr.pi@gmail.com> wrote in message > e908af84-71ac-459d-99db-bee3b55a7047@w39g2000prb.googlegroups.com [quoted text clipped - 39 lines] > HUH? > Why shouldn't they have zero energy then??? Read: Why should they have zero energy then??? >> You assume the velocities are equal to zero >> because the sum is zero - why? [quoted text clipped - 3 lines] > > Dirk Vdm > Eric Gisse <jowr...@gmail.com> wrote in message > [quoted text clipped - 40 lines] > HUH? > Why shouldn't they have zero energy then??? Because they don't have zero energy - the particles are still moving. The only way for them to have no energy is for them to stop moving - if they stick together. > > You assume the velocities are equal to zero > > because the sum is zero - why? > > You're a bit confused this morning :-) Probably, but enough to spot the mistake. He assumes that the only solution to the momentum equation is zero for both velocities. > see my reply :-) > > Dirk Vdm Eric Gisse <jowr.pi@gmail.com> wrote in message c29df6ed-4b39-4718-b006-4ad4eed129f6@x36g2000prg.googlegroups.com >> Eric Gisse <jowr...@gmail.com> wrote in message >> [quoted text clipped - 42 lines] > > Because they don't have zero energy the particles are still moving. see my correction. You said: "The only way the particles will have no energy after the collision is if they stick together." But they do have energy, even if they stick together. If they move together, they still have energy. Thas was my point to your comment. > The only way for them to have no energy is for them to stop moving - > if they stick together. [quoted text clipped - 6 lines] > Probably, but enough to spot the mistake. He assumes that the only > solution to the momentum equation is zero for both velocities. Actually he didn't even have a momentum equation to begin with, since he assumed that the v's were speeds :-) Dirk Vdm > Eric Gisse <jowr...@gmail.com> wrote in message > [quoted text clipped - 54 lines] > If they move together, they still have energy. > Thas was my point to your comment. I think of this in terms of pool. I can't imagine that happening - the balls would have to stop to stick together in my mind. I do see your point though. > > The only way for them to have no energy is for them to stop moving - > > if they stick together. [quoted text clipped - 9 lines] > Actually he didn't even have a momentum equation to begin > with, since he assumed that the v's were speeds :-) m1*v1 - m2*v2 = m1*v3 + m2*v4 The only reason I disagree with that is because of the sign on the 2nd term. I think he wrote it right the first time, then got confused down the line. > Dirk Vdm > > > > Two particles move in opposite direction and have an elastic collision > [quoted text clipped - 36 lines] > > - Tekst uit oorspronkelijk bericht weergeven - Eric The total momentum Pt of the system before impact = 0 This total momentum after the collision has to be the same. So Pt = (m1 + m2)*v Solving this equation gives v = 0 HR > > Two particles move in opposite direction and have an elastic collision > > [quoted text clipped - 24 lines] > > Elastic collisions do not conserve energy. Any closed system locally conserves mass-energy.  SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 hubert.ruypers@skynet.be <hubert.ruypers@skynet.be> wrote in message c4090630-0ef3-4fd4-a35b-9f350679be0d@d77g2000hsb.googlegroups.com > Two particles move in opposite direction and have an elastic collision > [quoted text clipped - 5 lines] > From the principle of conservation of linear momentum : > m1*v1 - m2*v2 = m1*v3 + m2*v4 Try it with all velocities measured along the same axis. m1*v1 + m2*v2 = m1*v3 + m2*v4 with v1 = 5 v2 = -10 > As > 1*5 - 0,5*10 = 0 > so > m1*v3 +m2*v4 > is also zero. Hence v3 and v4 are also 0. No... hence v3 = -m2/m1 * v4 = -0.5 * v4 > However, in de pre-impact fase the kinetic energy was : > E pre = 0,5*m1*v1^2 + 0,5*m2*v2^2 = 0,5*1*5^2 + 0,5*0,5*10^2 = 37,5 > Joule > > After impact the kinetic energy is > E post = 0,5*m1*v3^2 + 0,5*m2*v4^2 = 0,5*1*0^2 + 0,5*0,5*0^2 = 0 Joule After impact the kinetic energy is 0,5*1*(-0.5* v4)^2 + 0,5*0,5*v4^2 = 37,5 giving with some algebra v4 = 5, since the second particle is assumed to reverse direction, and thus v3 = -2.5 > This violates the law of conservation of energy. > > Where did I went wrong? In confusing speeds with velocity projections onto an axis. Convervation of momentum is not valid for speeds. Cheers, Dirk Vdm > hubert.ruyp...@skynet.be <hubert.ruyp...@skynet.be> wrote in message > [quoted text clipped - 50 lines] > Cheers, > Dirk Vdm Dirk It is a one dimensional problem (head-on-collision). So the velocity = speed (or is this a wrong conclusion) HR hubert.ruypers@skynet.be <hubert.ruypers@skynet.be> wrote in message 5708ee1c-923e-4822-bdac-a0c0002896c6@d45g2000hsc.googlegroups.com >> hubert.ruyp...@skynet.be <hubert.ruyp...@skynet.be> wrote in message >> [quoted text clipped - 54 lines] > It is a one dimensional problem (head-on-collision). So the velocity = > speed (or is this a wrong conclusion) Yes, this is a very wrong conclusion :-) Speed is always positive, whereas in one-dim setups velocity can "be negative". Strictly speaking, the 'numeric velocities' in one dimension are projections of velocity vectors onto an axis, so they can be negative. Remember: speed = sqrt( velocity^2 ) This is always valid, i.e. in 1-dim with "velocity projections" where the sqauring (^2) denotes numeric multiplication, and in 1- or 2- or 3-dim with vectors, where the squaring denotes the vector dot-product. Convervation of momentum is not valid for speeds. Convervation of energy happens to be valid for speeds. Why? Dirk Vdm > hubert.ruyp...@skynet.be <hubert.ruyp...@skynet.be> wrote in message > [quoted text clipped - 72 lines] > Convervation of momentum is not valid for speeds. > Convervation of energy happens to be valid for speeds. Why? I have no idea. Possibly because energy is a dot product of a force and displacement vector ? > Dirk Vdm- Tekst uit oorspronkelijk bericht niet weergeven - > > - Tekst uit oorspronkelijk bericht weergeven - hubert.ruypers@skynet.be <hubert.ruypers@skynet.be> wrote in message bb40ae65-9a4c-4f2b-b2ff-fbb14834544f@2g2000hsn.googlegroups.com >> hubert.ruyp...@skynet.be <hubert.ruyp...@skynet.be> wrote in message >> [quoted text clipped - 75 lines] > I have no idea. Possibly because energy is a dot product of a force > and displacement vector ? That's just a bunch of words :-) Hint: speed^2 = velocity^2 See also speed = sqrt( velocity^2 ) Dirk Vdm > From the principle of conservation of linear momentum : > m1*v1 - m2*v2 = m1*v3 + m2*v4 [quoted text clipped - 4 lines] > m1*v3 +m2*v4 > is also zero. Hence v3 and v4 are also 0. No, v3 and v4 are not zero. Conservation of Momentum only says m1*v3 +m2*v4 = 0 Notice that you have 1 equation (m1*v1 + m2*v2 = m1*v3 + m2*v4) and 2 unknowns (v3 and v4). You need a second equation to exactly determine v3 and v4. That second equation is the Conservation of Energy which states: m1*v1^2 + m2*v2^2 = m1*v3^2 + m2*v4^2 So your 2 equations are v3 + (v4)/2 = 0 (momentum) v3^2 + (v4^2)/2 = 75 (energy) Let v3 = -(v4)/2 so v3^2 = v4^2/4 and substitute this into the energy equation to get (3/4) v4^2 = 75 or v4^2 = 100 or magnitude of v4 = 10 You have to fall back on the "law of obviousity" to conclude that v4 = +10 (it reverses direction) v3 = -5 Dwib > However, in de pre-impact fase the kinetic energy was : > E pre = 0,5*m1*v1^2 + 0,5*m2*v2^2 = 0,5*1*5^2 + 0,5*0,5*10^2 = 37,5 [quoted text clipped - 6 lines] > > Where did I went wrong? Dwib <dwibdwib@gmail.com> wrote in message 58765687-3d7b-4243-83d9-5af38de93de8@l42g2000hsc.googlegroups.com >> From the principle of conservation of linear momentum : >> m1*v1 - m2*v2 = m1*v3 + m2*v4 [quoted text clipped - 25 lines] > v4 = +10 (it reverses direction) > v3 = -5 Yes, I made a little mistake somewhere. v3 = ( m1 v1 - m2 v1 + 2 m2 v2 ) / (m1+m2) = -5 v4 = ( m2 v2 - m1 v2 + 2 m1 v1 ) / (m1+m2) = 10 Silly numerics :-) Dirk Vdm > Dwib <dwibd...@gmail.com> wrote in message > [quoted text clipped - 38 lines] > > - Tekst uit oorspronkelijk bericht weergeven - Thanks for the answers but now stil don't know why : ""Convervation of momentum is not valid for speeds. Convervation of energy happens to be valid for speeds. """ :-) HR hubert.ruypers@skynet.be <hubert.ruypers@skynet.be> wrote in message b3bf17df-aab8-40d4-9b48-212e0bfa2973@i76g2000hsf.googlegroups.com >> Dwib <dwibd...@gmail.com> wrote in message >> [quoted text clipped - 42 lines] > ""Convervation of momentum is not valid for speeds. > Convervation of energy happens to be valid for speeds. """ :-) I already explained. Conservation of energy happens to be valid for speeds because there is no difference between a squared velocity and a squared speed: velocity^2 = speed^2 where on the left hand side the squaring operator (^2) denotes dot-product of vectors in 1D, 2D, 3D, and on the right hand side it denotes numeric multiplication of numbers. In the 1D case of your problem the velocity vectors are silently identified with their component projections onto a directed axis whence they are signed numbers, whereas speeds are always positive numbers. Dirk Vdm > On 20 jul, 19:18, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO- > [quoted text clipped - 47 lines] > > HR Momentum is a vector. Energy is a scalar. Therein lies the difference. > Thanks for the answers but now stil don't know why : > ""Convervation of momentum is not valid for speeds. > Convervation of energy happens to be valid for speeds. """ :-) That's easy. Momentum is a vector (magnitude + direction). Energy is a number (no direction). Speed is a number while velocity is a vector. Dwight > Two particles move in opposite direction and have an elastic collision > [quoted text clipped - 11 lines] > m1*v3 +m2*v4 > is also zero. OK up to here. > Hence v3 and v4 are also 0. No! Why would you say this? Sure, v3 = 0 = v4 works, since m1*v3 + m2*v4 = 0, but it isn't the only solution. Just from the magic words, "elastic collision", I'd pick v3 = -5, v4 = +10 instead, which also gives m1*v3 + m2*v4 = 0. But numbers like v3 = -1,000,000 and v4 = +2,000,000 also work. Why not pick that? You have two unknown values, v3 and v4. Do you expect to find them from a single equation? The usual procedure is to have as many equations as you have unknown quantities. Where can you get a second equation from? You know where: > However, in de pre-impact fase the kinetic energy was : > E pre = 0,5*m1*v1^2 + 0,5*m2*v2^2 = 0,5*1*5^2 + 0,5*0,5*10^2 = 37,5 > Joule > > After impact the kinetic energy is > E post = 0,5*m1*v3^2 + 0,5*m2*v4^2 = 0,5*1*0^2 + 0,5*0,5*0^2 = 0 Joule Since the two quantities above must be equal (elastic collision), doesn't this give you a 2nd equation? Solve simultaneously, and done! SignatureTimo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/ E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html > Two particles move in opposite direction and have an elastic collision > [quoted text clipped - 11 lines] > m1*v3 +m2*v4 > is also zero. Hence v3 and v4 are also 0. Nonsense. If I told you that 2x + 3y = 0, would this demand that x and y are both 0? (Try, for example, x=6 and y=-4.) PD > However, in de pre-impact fase the kinetic energy was : > E pre = 0,5*m1*v1^2 + 0,5*m2*v2^2 = 0,5*1*5^2 + 0,5*0,5*10^2 = 37,5 [quoted text clipped - 6 lines] > > Where did I went wrong? > Two particles move in opposite direction and have an elastic collision One presumes they move toward each other along a common line. > Particle 1 has a mass of 1 kg and a velocity of 5 m/s. Particle 2 has > a mass of 0,5 kg and a velocity of 10 m/s. > > What is the velocity after impact ? > From the principle of conservation of linear momentum : > m1*v1 - m2*v2 = m1*v3 + m2*v4 [quoted text clipped - 15 lines] > > Where did I went wrong? Newton's cradle. SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 > Two particles move in opposite direction and have an elastic collision [snippers] From this thread we learn two important lessons about doing physics of 1-D elastic collisions. 1) Transform to the centre of momentum frame and the collision will be very simple. (Happily, your problem already has everything in the COM frame.) 2) Learning from a usenet news group is most often a losing proposition. Socks |
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