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B Conservation of Velocity

  1. Dec 12, 2017 #1
    My question is, in an explosion, where 2 cars at rest are pushed apart by a spring between them, is velocity conserved? Does the total velocity for both cars before spring release equal the total velocity after release? Why or why not?
     
  2. jcsd
  3. Dec 12, 2017 #2

    fresh_42

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    If the cars are at rest, then the total velocity before is zero. This cannot be said of the velocities after release. Velocity isn't conserved and depends on many factors. One is, that the two cars don't even have a single velocity, since the spring applies an acceleration to the cars, not a velocity. I recommend to read about Newton's laws of motion. A good start would be the Wikipedia entry:
    https://en.wikipedia.org/wiki/Newton's_laws_of_motion
     
  4. Dec 12, 2017 #3

    PeroK

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    The quantity that is conserved is momentum, which is mass times velocity. Look up conservation of momentum.

    Note that if you have a number of particles all of the same mass, then conservation of momentum - in this special case - does indeed reduce to conservation of velocity!
     
  5. Dec 12, 2017 #4
    Thanks for the reply! I've been thinking about it for the last 30 mins or so, and I think I have a simpler explanation (at least for me). Please let me know if my thinking is correct.

    So, the reason velocity isn't conserved is because if the cars have different masses, the velocity is different. However, if the mass was the same, then velocity would be conserved in addition to momentum. Correct?
     
  6. Dec 12, 2017 #5
    Ah, I just realized that too! Thanks!
     
  7. Dec 12, 2017 #6

    fresh_42

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    Yes, if you take the directions into account. But both accelerate first and slow down again due to friction and air resistance (which also have to be the same), so it's not a single velocity, only velocities of equal amount in opposite directions, so the sum remains zero.
     
  8. Dec 12, 2017 #7

    Ibix

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    I wouldn't phrase it that way, because it's likely to lead to confusion: by this approach, velocity is sometimes "conserved" and sometimes isn't. Better to say momentum is conserved, and note that the masses cancel sometimes.
     
  9. Dec 12, 2017 #8
  10. Dec 13, 2017 #9

    Ibix

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    No. The conservation of momentum says that the sum of the initial momenta is equal to the sum of the final momenta. A "conservation of velocity" would say that the sum of the initial velocities is equal to the sum of the final velocities, which is not what the equation says. Rearrange slightly and you'll see that it says that, for a two-body elastic collision, the difference of the initial velocities is equal to the difference of the final velocities.

    Apparently the result is derived in another video. It's a consequence of conservation of energy.
     
    Last edited: Dec 13, 2017
  11. Dec 13, 2017 #10

    fresh_42

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    Where's the extra energy that has been stored in the spring? Also friction and air resistance is neglected. But the main difference is, that your proposed equation has been ##v_{1,initial}+v_{2,initial} = v_{1,final}+v_{2,final}## which is different and in general wrong. Here's the corresponding picture from the Wikipedia page:

    400px-Elastischer_sto%C3%9F3.gif

    It shows quite well, that the momentum is conserved, i.e. the masses are important. The equation of the elastic collision without masses results from the fact, that two equations with masses are simplified to a single equation in which the masses canceled out. What you have proposed was ##0 = v_{1,initial}+v_{2,initial} = v_{1,final}+v_{2,final} = \frac{4}{3}##.
     
  12. Dec 13, 2017 #11

    PeroK

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    You need to be careful not to confuse different with contrary. That statement - for an elastic collision - is something different.

    In fact, I would rewrite it as:

    ##v_{1i}-v_{2i} = -(v_{1f}-v_{2f})##

    Where ##v_1-v_2## is the relative velocity of the two objects. That equation says, therefore, that the relative velocity is reversed during an elastic collision.

    One example of this is a rubber ball bouncing on a hard floor, where the ball bounces (almost) as high as the point you dropped it from.
     
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