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Momentum Question

  1. Oct 17, 2011 #1
    How is that when:

    A wagon collides with the velocity (v) with another wagon (at rest, v=0) of the same mass (m) on a horizontal track, and when they collide they are connected together (think of wagons), that the velocity of the two objects are after the collision v/2?

    And now I exclude everything that has to do with friction (because in this text I just have written there was nothing said about friction).

    I hope you understand my question. And no, this ain't homework, I recently read it in a text but could never figure out how this was true.


    Thanks in Advance!
     
  2. jcsd
  3. Oct 17, 2011 #2
    Firstly the velocity is v/2 as can be seen by applying conservation of momentum principle (which principle holds since there are only internal forces between the two wagons). mv+0=(m+m)(v/2).

    There has to be some sort of internal force between the 2 wagons in order to become connected. This force is labeled as friction because when we do the energy calculation we found that some kinetic energy hass been lost and we have to assume this energy has been lost as heat due to this internal friction(between the two wagons, not between the wagons and the ground).
    Most of the lost energy becomes heat but in some cases of collisions a small fraction of this lost energy becomes chemical potential energy between the two colliding bodies (that move as one body after the collision that is the collision is called plastic).
     
  4. Oct 17, 2011 #3

    Ken G

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    Gold Member

    Or you could store the energy as potential energy, like if you had a spring and a coupling that held the two wagons together. There wouldn't need to be any friction or energy loss there, it could all go into the spring (in principle) and be re-extracted later when the wagons are uncoupled. What is the issue you are having with that, RobinSky? One could conserve both energy and momentum with no friction if one had an internal potential energy, or one could have friction and create internal heat to explain the lost kinetic energy as mentioned above.
     
  5. Oct 17, 2011 #4
    Because momentum, velocity times mass, is the thing that is conserved, not velocity itself. After the collision, the mass of the moving object has doubled, so the velocity must halve in order to keep the momentum constant. This is intuitive. Think about more extreme examples. What if a fly traveling at 60 mph hits a stationary truck on a flat surface with the brakes off? Will the truck be pushed by the fly to 60 mph, thus conserving the velocity? No, the fly has very little mass, and therefore very little momentum compared to the massive truck, and so does very little to truck.
     
  6. Oct 17, 2011 #5
    I think I'm thinking too much of newton's cradle and not knowing how it properly actually works... Let's say we have this cradle but just two balls. One ball goes down, hit's the other one, become to rest, the 2nd ball goes up, down, hit's the first ball again, first one goes up, down & motion is repeated again due to conservation of energy (or momentum in this case?).

    But I guess it ain't the same we don't have these other balls that usually is between the two swinging ones in newton's cradle?

    And sorry if I've made myself a fool but I'm still new to this :).

    Thanks.
     
  7. Oct 18, 2011 #6
    Newton's cradle is due to both conservation of energy and momentum.

    I guess what confuses you is when we have the intermediate balls between the two balls at two edges that swing. The energy and momentum is transfered from the first ball to the last ball through the intermediate balls.Its like having many successive collisions from one ball to the next one in which energy and momentum is passed from one ball to the next one. The last ball has nowhere to collide and thus it starts moving. There is no inelastic(plastic) collision here where two or more balls would move as one body, we have a series of sucessive elastic collisions. In each elastic collision full energy and momentum is passed from one ball to the next one.
     
  8. Oct 18, 2011 #7
    I think I'm trying to understand it now. Let's say the wagons were not connected right after the collision. Will the second wagon (that WAS at rest) then get the same velocity as the first wagon had before they collided? And the first wagon then be at velocity 0, at rest?

    But in my mind, if I want to think realistic (realistic might be a blurry word, but, at least how it would look like if I think about it)... The first wagon will also be knocked back a little with a small velocity due to the normal force (from the second wagon) pushing the first wagon back when they collide, or?

    Short summary, the second wagon get's the most of the momentum during the collision, and the first wagon get's just a little bit, because the second pushes it back. ?
     
  9. Oct 18, 2011 #8

    Low-Q

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    Correct

    No. It will rather go the same direction due to deformation and loss at the collision point. At no deformation and no loss, the actual collision point is not moving anywhere, and the first wagon will rest perfectly still, while the second wagon is pushed at the same velocity as the initial velocity of the first wagon (like an ideal Newtons cradle)

    No. See above. However, if it does, the velocity of the second wagon will be reduced accordingly.

    Vidar
     
  10. Oct 18, 2011 #9
    Ahh, thanks Vidar now some bells are ringing.

    Just an example, let's say they collide once again. The normal force from second wagon (that was at rest) is bigger, than, the force of colliding (first one) wagon. Will that result in a movement backwards for the first wagon, and the second one gaining the velocity that was transferred during the collision?
    For this to happen, a larger normal force is needed, I also guess the mass from the two objects will no longer be the same, or no... It won't ever happen (Normal force from wagon 2 > the force from wagon 1)? The normal force from wagon 2 will always be equal to the colliding force from the first wagon, or?

    Just like the normal force will be always be equal to force I'm applying to the ground while standing on a horizontal plane?

    Sorry if this post was a bit fuzzy, but at least I'm trying! :)
     
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