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B Inelastic collision and momentum

  1. Mar 17, 2016 #1
    Guys, I'm confused that how momentum is conserved when kinetic energy is lost?.....I know this question is asked by many people and i searched for it but I'm not getting the reason. Here i have sent two photos which shows that momentum is not conserved from www.simbucket.com. If im wrong at any point please correct me. Thankyou Screenshot_20160317-103325.png Screenshot_20160317-103318.png
     
  2. jcsd
  3. Mar 17, 2016 #2

    A.T.

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    The 3.3 m/s is rounded up. It's actually 3.333... m/s.
     
  4. Mar 17, 2016 #3
    But still its not equal right?I don't think that approximations should be taken
     
  5. Mar 17, 2016 #4

    A.T.

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    If you cannot interpret numerical results, do the symbolic math yourself.
     
  6. Mar 17, 2016 #5

    sophiecentaur

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    You have just confirmed my suspicion about using simulations to 'prove' ideas in Science. The difference you think you have found is just because of the inevitable rounding in the calculations.
    First of all, believe that Momentum is Conserved, which as been proved by measurement, time and time again (and to greater and greater accuracy). A simulation is not a measurement. Then stick with the 'algebra' to get answers - putting in the numbers at the very last moment.
     
  7. Mar 17, 2016 #6

    A.T.

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    I'm sure the issue here in not calculation accuracy, but merely screen space.
     
  8. Mar 17, 2016 #7

    ZapperZ

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    What EXACTLY are you asking in this thread?

    Are you asking if this simulation has a mistake in that the result doesn't EXACTLY show that momentum must be conserved, or are you trying to refute the principle of conservation of momentum in the case of inelastic collision?

    If you are asking for the former, then I believe there has been sufficient answers given already.

    If you are asking about the latter, then you need your head examined for using a made-up "simulation" as your evidence.

    Zz.
     
  9. Mar 17, 2016 #8

    jtbell

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    3.3 is the approximation displayed by your simulation. 3.33333333..... (or ##3 \frac{1}{3}##) is the exact value. Substitute that exact value into the calculation for the total momentum in your first diagram (the final state after collision) and you'll see that it does follow conservation of momentum.
     
  10. Mar 17, 2016 #9

    sophiecentaur

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    Yes and no. The display was rounded more than we might like but, otoh, that sort of confusion would never occur in a symbolic approach. The fact that the OP actually thought he might have found a 'loophole' due to the results of a simulation is worrying. (Not his fault but the fault of the common belief in simulations.)
     
  11. Mar 17, 2016 #10
    I'm asking about conservation of momentum in an inelastic collision..... How momentum is comserved when some of the kinetic energy is lost?
     
  12. Mar 17, 2016 #11

    ZapperZ

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    But what exactly do you mean by "how momentum is conserved"? I mean, how is momentum conserved in elastic collision? You appear to not have a problem with that!

    Conservation of momentum means that there are no external forces acting on the system. All the push/pull came ONLY from within the system, regardless if there is energy loss or not. That is HOW momentum is conserved.

    You could have easily asked this without trying to justify your question by using this simulation AND misinterpreting the result.

    Zz.
     
  13. Mar 17, 2016 #12

    A.T.

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    They are different quantities, so one can change while the other stays the same.
     
  14. Mar 17, 2016 #13

    sophiecentaur

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    You seem to be looking for something that just doesn't have to be there. Momentum and KE are very significantly different.
    Take identical two bullets, fired at each other at the same speeds (momentums would be +P and -P). The total momentum is zero but the KE could be many tens of Joules, and all that KE would go to zero and the total momentum would stay at zero. KE just doesn't have to be conserved in ANY model used in Physics. That simulation has done you no favours at all, I think.
     
  15. Mar 17, 2016 #14

    Nugatory

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    It's actually not quite right to say that the kinetic energy is "lost" - it's more that some of the kinetic energy goes somewhere else where we aren't counting it. In the example you started with, a 1 kg cart travelling to the right at 10 m/s collides inelastically with a 2 kg cart at rest and they both end up moving to the right at ##3\frac{1}{3}## m/s. Momentum is conserved (10 kg-meters/sec before and after, either as ##1\times{10}## or ##3\frac{1}{3}\times{3}##) but the kinetic energy before is 50 joules (by ##E_K=mv^2/2##) but only a bit more than 16 Joules after (same formula). The "lost" energy isn't really lost - it went into the work needed to smash and deform and bend the two carts, heat from friction in the collision, maybe sound waves from the noise of the collision, various other things.
     
  16. Mar 17, 2016 #15
    I'm fairly new to these forums but have noticed more than once where someone's honest question exposes him to a barrage of snarky comments. I was hoping for something better.

    Now to address the OP's question, keep in mind that momentum is a vector quantity, with magnitude and direction, whereas energy is a scalar. As an example, suppose there are two identical objects moving toward each other with the same speed. The initial total momentum of the system is therefore zero and there is some kinetic energy. If there is an inelastic collision, the resulting kinetic energy is lower than before the collision but the total momentum is still zero as the objects move away from each other with equal (reduced) speed. So some kinetic energy is lost but momentum is conserved.
     
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