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Impact of solids vs non-solids.

  1. Oct 5, 2005 #1
    Hi everyone. I'm not a physicist but I need some help.

    In the hypothetical situation of a 1 kg block of concrete being dropped onto a concrete slab from 3 metres, the force of the impact is easy to calculate. However, if one were to pick up the pieces and re-drop them, what would be the force of the impact in relation to the previous impact? I.e. I'm under the impression that the 'impulse' would be less because the pieces would compress. Can anyone help me? I don't want to have to build a dangerous concrete-smasher in my yard...
     
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  3. Oct 6, 2005 #2

    berkeman

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    The "force" of the initial impact is not easy to calculate. You need to talk in terms of the impact for the initial drop as well as the later piecemeal (sp?) impacts. Real collisions involve some level of compression, which gives a duration to the impact and spreads the application of the force over time.
     
  4. Oct 6, 2005 #3

    PerennialII

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    Were you thinking about doing the 2nd drop with a re-assembled block of concrete (but not re-molded I'd think)? :biggrin:

    In the 2nd drop considering it 'realistically' (not as a rigid collision(s)) the energy would be consumed by further deformation, internal collisions within the block, internal friction within the the block and further cracking during the impact. Which would result in energy 'diverging' away from the actual impact, the impulse being smaller. Of course with the disclaimer that you're interested in what happens under these kinds of hypotheses?
     
  5. Oct 6, 2005 #4

    minger

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    Yes, trust me, falling force is 'not' easy to calculate at all. I had a similar problem that I posted here earlier this year. I had to find some very old textbook (handwriten actually) which had a relation using velocity and static displacement to get an approximation of the force of a falling object.

    Static displacement alone can be tricky unless you're dealing with a simple beam.
     
  6. Oct 6, 2005 #5
    OK... It makes sense to me that the same mass, broken up once already, would not have the same impact. So, what I need to know is what the force of the two impacts would be, relative to each other. Would the broken-up mass - say, 1kg of concrete broken up into ten gram pieces, average - hit for instance with half the force of the solid mass? Less? Or more?
     
  7. Oct 7, 2005 #6

    PerennialII

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    At this point it would be probably easier to build that concrete smasher :rofl: . Might get a some sort of an upper bound estimate if (extremely-) simplified to elastic deformation, simple friction and make some harsh approximations regarding the nature of the collisions (so that the result would have still at least some accuracy) ... could probably get something by considering 'multi-particle' contact generalized to this instance. I'll see what can come up with unless someone beats me to it.
     
  8. Oct 7, 2005 #7
    Cheers, this is getting complicated :uhh:

    My question concerns a hypothetical demolition, where I am interested in the efficiency of the event.

    If a weight is dropped on the uppermost floor that carries sufficient force to collapse that floor, can this process be continued in some kind of 'chain reaction' whereby the collapsing floors not only gain mass but also gain enough momentum to crush every floor down to ground level? You can see why I am trying to work out how energy transfers in an already-broken material.

    Can such a mass accelerate until it reaches terminal velocity or must it begin from zero at each floor?

    Any thoughts on this would be appreciated.
     
  9. Oct 8, 2005 #8

    PerennialII

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    About this part, yes, the chain reaction can be continued - actually not that far fetched of a scenario at all. The mass doesn't have to start from zero velocity wise at each floor, the processes of deformation, fracture, impact and the following collapse can be understood as being continuous (energy wise contributions to each occurring event to be exact) and the mass doesn't need to stop altogether.
     
  10. Oct 10, 2005 #9
    Thanks.

    So, is it possible to design a method of collapse whereby the falling mass can add sufficient energy in each impact to cause the overall speed to come close to that of a freefall drop? I mean, concrete has got to be thousands of times more resistant than air so is this a ridiculous idea?
     
  11. Oct 11, 2005 #10

    PerennialII

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    Intuitively would think you'd have to drop the thing from a space elevator, or attach a rocket or something to get it going, or work with extremely thin 'floors' compared to the mass & distance of descend of the falling object ... making it pretty impossible to accomplish (at least backyard wise :biggrin: .... don't know how close you could get by dropping a piece of concrete over thin layers whiches thickness would be around mms for example). Is there some particular reason to aim for conditions comparable to freefall?
     
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