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Homework Help: Question on Conservation of Momentum

  1. Dec 10, 2006 #1
    1. The problem statement, all variables and given/known data

    So I have one of these trippy toys
    http://www.spacehall.com/shop/images/AstroBlaster.JPG [Broken],[/URL] and have a question.

    Basically the way that it works, is that all the balls bounce, and the top one that isn't secured on rockets off very high while the others are secured in place on a rod. I know that this is because the combined mass is exerted upon the small top ball (Potential Energy = mass*gravity*height, thus the combined mass means that it has a huge potential energy).

    Anyways, my question is that if I have recorded all the data about it, know the elasticity of the top ball, and have the potential energy calculated. Is there any way to calculate how high the ball will bounce if I know the potential energy?

    (The potential energies for the entire thing and just the top ball are ~5000 and 200 J respectively)

    2. Relevant equations

    Potential Energy = MassGravityHeight , Kintetic Energy = 1/2 MassVelocity^2

    3. The attempt at a solution

    So far I have tried to divide the PE's and find a ratio, however the balls bounce at about 50cm and 3.8cm combined and just the top ball, but the ratios dont work out right.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Dec 10, 2006 #2


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    Did you try a Google search on this. I'm sure I have seen the whole thing worked out more than once.

    I'm not finding it now though. Do you know the masses of the balls?
    Last edited: Dec 10, 2006
  4. Dec 10, 2006 #3
    the masses are --

    3.88g for the top ball
    9.96 for the next bigger
    23.15 for the next bigger
    64.11 for the next bigger

    and 101.1g total

    Dropped from 5cm, the top ball bounced an average of 52cm

    Yeah, I tried googling it... haven't had any luck though. This SHOULD be really easy... I just can't find the right equation or am missing something!
  5. Dec 11, 2006 #4


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    Actually, I don't think it is easy. When you have that many objects involved in a process things get complicated in a hurry. I have come up with a solution based on an assumption that may not be completely valid, but it does lead to the conclusion that the top ball will rise to five times the initial height from which the blaster is dropped. The "five times" number is quoted in all the product descriptions I have seen.

    The assumption is that each ball deforms in proportion to its radius, and that the balls remain in contact with one another while they are compressed. That means the top ball moves farther than the bottom ball during the collision process. Imagine the balls all being compressed by some fraction of their radius at the lowest point in the collision, with the center of mass being half way between the top and bottom of each ball. Let's number the balls 1 through 4 from top to bottom and call the distance the bottom ball CM moves d4 = kR4. The CM of the next ball up moves d3 = k(2R4 + R3). The next will be d2 = k(2R4 + 2R3 + R2) and finally d1 = k(2R4 + 2R3 + 2R2 +R1).

    Assuming the balls all move together, the velocity of each ball at the end of collision will be proportional to the distance it moved up from the lowest point during the collision. To find those distances, you need the ratios of the radii. I calculated those assuming solid balls of the same density, which is not quite right because of the holes even if they are made of the same material. Using the velocity ratios from this calculation, I assumed all the mechanical energy was conserved and calculated the height each ball would reach if they were all free to move separately. (It looks to me from the picture that they can all separate a bit before the middle two get stopped by the stick) Then I took the total momentum of the bottom 3 balls and treated those as having a secondary collision that keeps them connected as one object.

    The result of this calculation is that the top ball bounces just over 5 times the drop height, while the bottom three ball combination bounces to about 0.61 times the drop height.
    Last edited: Dec 11, 2006
  6. Dec 11, 2007 #5
    Do you mind telling me where you got one of those trippy toys. Looks like a bit of fun
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