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Newton Cradle ball height

  1. Nov 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Three small identical steel balls A,B and C are suspended by vertical threads of equal length from a horizontal support, with centres in a horizontal line and separated by a small gap. A is raised by a height 'h', with the thread taut, and released. All subsequent collisions are elastic.

    (i) When all the balls have mass 'M', determine the subsequent motion of the system and the height to which the centre of C is raised following its first collision.

    (ii) When A has mass 'M', B has mass '2M' and C has mass '3M', determine the height to which the centre of C is raised after its collision.
    3. The attempt at a solution
    This is an additional question that my Physics teacher gave me. I know PE=KE needs to be used here but I am not even getting close to a reasonable answer for (i) or (ii).

    Please help!
     
    Last edited: Nov 22, 2012
  2. jcsd
  3. Nov 22, 2012 #2

    Simon Bridge

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    Welcome to PF;
    If all the balls were touching, what is the height of the final ball in relation to the first one?

    What difference does the small gap make to the energy in the collision?
    (sketching the setup will help here. BTW: there are three balls.)
     
    Last edited: Nov 22, 2012
  4. Nov 22, 2012 #3
    It would be the same height, as there would be conservation of energy. The small gap means that some energy is lost in the collision.
    However the question says that all subsequent collisions are elastic.
     
  5. Nov 22, 2012 #4

    rcgldr

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    For a real Newtons cradle, small gaps don't have much effect, and virtually no effect on conservation of energy (other than aerodynamic drag). Also for a real Newton's cradle, there are complicating factors, such as the force that the frame exerts on the threads is an "external force" with a horizontal component, which will generate a horizontal force if the center ball(s) is (are) shifted (offset) during a cycle. Another factor is the rate of propagation of force through the balls (speed of sound within the balls) versus the collision interactions.
     
    Last edited: Nov 22, 2012
  6. Nov 22, 2012 #5

    Simon Bridge

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    You don't lose the energy in the collision then :)
    Where else can the energy go?

    Did you draw the picture?
    The ball has it's maximum kinetic energy at the bottom of the arc.
    Where is the ball when it impacts the next one?

    As rcgldr points out, this is a pretty idealized cradle.
    Though - for all the real-life ones I've used, small gaps (but define "small") do make a noticeable difference. Maybe I've just been lucky?
     
  7. Nov 22, 2012 #6
    My teacher explained when giving the question to me that the question was not as easy as it looked and that there was a hidden side to it. The point that rcgldr makes seems to be along the lines of what my teacher was referring to. Is there any way of calculating the height after collision just from the information given in the question and knowing the speed of sound? What equations would I use?
     
    Last edited: Nov 22, 2012
  8. Nov 22, 2012 #7

    rcgldr

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    I listed the complicating factors in the wrong order. I don't think you're supposed to consider the details like the speed of sound within the balls, versus the aspect of the collision reaction (how much and how quickly the balls deform during collisions). If not for this aspect, then the central part of the pack shouldn't start swinging, so the horizontal component of tension shouldn't be a factor.

    The center of mass of the pack of balls is moving back and forth, but for an idealized situation, you can assume the center ball does not move.

    For the initial collision, you can assume that all of the PE (m g h), is converted into KE (1/2 m v^2) at the moment of collision. Assume the center ball transmits forces instantaneously, so the same force that slows down ball #1 also accelerates ball #3, which are equal but opposing forces, so that the center ball does not move, and ball #3 ends up at the same height as ball #1.

    What is being ignored here is the mathematical possiblity that ball #1 could end up rebounding a bit backwards after the collision, which happens in real life, but not in an idealized Newton's cradle.

    I think your teacher is referring to the case with different masses, which isn't as straight forward.
     
    Last edited: Nov 22, 2012
  9. Nov 22, 2012 #8
    Thank you. How would I go about solving that question. Do you have any tips?
     
  10. Nov 22, 2012 #9

    Simon Bridge

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    OK, since you are sure, I'll leave you two to it then :)
     
  11. Nov 22, 2012 #10

    rcgldr

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    Pretend there are small gaps between the balls, and then treat each collsion as an independent event, the collision between ball #1 and ball #2, then the colllision between ball #2 and ball #3. Ignore how far the balls move between collisions.

    One issue I forgot to mention is compressability of the balls. During compression, the centers of masses of the balls move closer to each other, and which means the center of mass of ball #1 must be moving faster than the center of mass of ball #2, and the center of mass of ball #2 must be moving faser than the center of mass of ball #3. Likewise, during expainsion, ball #3 is moving the fastest and ball #1 the slowest. In an ideal situation, ball #1 and ball #2 end up at rest, with only ball #3 moving, but ball #2 has been shifted, which results in a horizontal component of tension in the threads that oppose the shifting.
     
  12. Nov 22, 2012 #11
    Thank you, I will give it a go and see what i obtain. Theoretically i should obtain a smaller fraction of the initial height. When 3 is raised the component of height is smaller due to 3 having a 3M mass (mgh). What equation can i use to take into consideration the shifting of 2?
     
  13. Nov 22, 2012 #12

    rcgldr

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    Just treat the problem as two separate elastic collision problems. ball #1 with mass M colliding into ball #2 with mass 2M (I think ball #1 may end up bouncing backwards). Then using the velocity you calculated for ball #2, use that as input for the elastic collision between ball #2 and ball #3 (again ball #2 may also end up moving backwards).

    You probaby should ignore the shifting effects by treating the balls as non-compressable.
     
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