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B Is my solution to this question about a collision valid?

  1. Nov 26, 2018 #1
    QUESTION
    Ball A is moving with velocity of v, it collides elastically with five stationary balls. All six balls are of equal mass. What happens next?

    SOLUTION
    I want to find out how many balls move off from those that were stationary and with what velocity they move away.
    I apply the elastic collision equation ①, conservation of momentum ② and conservation of kinetic energy ③. After substituting into equation ② from ① and deriving an equation for the outgoing mass (mB) I then think about what possible outcomes there could be. This leads me to the conclusion that mB = m is the only valid solution. Then I feed that solution back into equation ② to give me vB = v.
    Please see attached PDF.

    I don't actually use equation ③ directly so it is superfluous, though equation ① is derived from it. My main concern is the use of equation ① in the absence of knowing exactly how much mass moves out of the problem - is my use of it correct? If my solution is not valid is there a way of using cons. of momentum and KE to solve this problem?*
    Any insight gratefully received.


    * I did try thinking about the problem with mA margining into the masses and having an unknown mC and mD moving away either left and right or right or stationary and right and using mA + mB = mC + mD, but even with that I couldn't derive enough equations to eliminate unknowns.
     

    Attached Files:

  2. jcsd
  3. Nov 26, 2018 #2

    sophiecentaur

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    Sorry but I cannot read that image. Did it start off as a scan? Make sure it's scanned at a reasonable resolution (300dpi) and try adding an attachment (button at the bottom of the answer window.)
    If it's about Newton's cradle, separate the balls by a microscopic amount. Consider the first collision, then move on to the next and the next. Write down the two conservation equations for the first collision and you find the second ball moves off with the same velocity as the first and the first ball ends up stationary. This happens all down the line and is no surprise because we have all seen the toy / demo.
     
  4. Nov 26, 2018 #3
    My apologies! I've converted it to a PDF and attached.
     
  5. Nov 26, 2018 #4

    Nugatory

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    That's Newton's cradle. Google is your friend here.
     
  6. Nov 26, 2018 #5
    Yeah that is a good way to tackle this. Would you say it is the only way to solve this at the simple level of conservation of momentum and KE?
     
  7. Nov 26, 2018 #6

    sophiecentaur

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    I think that at one point, you have described a situation where the five balls are welded together because you are assuming that they share the same exit velocity. (The KE equation) or you are making some other assumption about the remaining four / five being stationary. I don't think you can make any assumptions about what happens for all the collisions - if they are coincident in time. There may be a set of simultaneous equations but

    I'm sure it's valid to assume zero time for interactions and to allow a vanishingly small distance between the balls. After all - it gives the right answer!!
     
  8. Nov 26, 2018 #7

    gneill

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    Alternatively one could model the percussion as a sound wave traveling through steel at the speed of sound in steel. I presume that the symmetry of the sphere would focus the energy at the antipodal point from the point of percussion, then since there would be a good impedance match where the balls touch, the wave would pass through to the next ball in line without reflection (for the ideal case where all balls are perfectly identical). The last ball in line has nowhere to "dump" the wave at its far side except into the air, which does not present a good impedance match. So the momentum carried by the wave ends up trapped in the final ball as it begins to move.
     
  9. Nov 27, 2018 #8
    Ideally I would like to arrive at a solution that doesn't assume how many masses are involved in a collision so that it can be generalised. For example the method of separating out balls won't work for two balls incoming since in a 2m collision with stationary m the 2m ball needs to keep moving forward. We could separate out the balls into pairs but that is too much assumption for me and it wouldn't work if there were an odd number of balls. I'll keep looking around.
     
  10. Nov 27, 2018 #9

    sophiecentaur

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    It will work but it will be complicated. The first collision will, as you say, result in the incident ball be moving and it will collide with the first ball (which will be stationary after its first collision down the line) again. It's possible to write down what happens at every collision but you were 'lucky' in the simple case. For other cases you will get a sequence of equations which you could solve for n balls.
    I was wondering if there would be any future in taking the 2m ball and consider it hitting the first two balls in the row. That would correspond to a collision between equal masses so they would move off at v, hitting the next pair of balls etc..
    From what I remember, this is the sort ofd thing that happens when you play around with the normal Newton's Cradle. It sort of implies that, for an even number of balls on the cradle, the result would be 'trivial' and the final pair would leave with the initial v but for an odd number of balls, the row of balls would continue to move as the final ball departed.
    If you were motivated enough, it could be worth while googling Newton's Cradle. You can be sure that someone will have been through it all already. (Haha Do as I say and not as I do. :wink:)
     
    Last edited: Nov 27, 2018
  11. Nov 27, 2018 #10
    I like the sound of that - trying the sequence of collisions. I’ll give it a go.

    I did look for Newton’s cradle specifically (on DuckDuckGo not Google ; ) and came across a question like mine in Physics Stack Exchange. One answer says that 2 balls incoming is the maximum number that can be solved with conservation. Further solutions require more complicated physics - something to do with wave pulses that combine at the point the outgoing balls separate from the pack.
    Thanks for taking the time to respond.
     
  12. Nov 27, 2018 #11

    haruspex

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    You could try treating it as two massive linear springs of the same density but different lengths. Not easy, though.
     
  13. Nov 27, 2018 #12
    That is an interesting proposition! I’ll try the sequence of collisions first.
    Thanks
     
  14. Nov 27, 2018 #13

    robphy

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    This has been on my to-read list... but I've been too busy to actually do it.

    Possibly interesting issues on Newton's Cradle:

     
    Last edited: Nov 27, 2018
  15. Nov 28, 2018 #14
    Looks great. I’ll take a look. Thank you.
     
  16. Dec 2, 2018 #15
    Thanks, I came across those articles independently also. I guess they answer my question as to whether I can write a solution at A-Level standard and it seems as though the answer to that may be "no", but it'll be an interesting enquiry for myself.
     
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