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Hammer physics

  1. Mar 18, 2004 #1
    This came about on a test today (we have handed it in) and it just got to me. I didn't think it was hard until my teacher made an outrageous claim. Here goes nothin:

    A hammer strikes a nail, and the nail goes in .8 cm. On the second strike, it will go in .7 cm. It is a geometric sequence. How far will the nail go in an infinite amount of hammer strikes?

    Now, I said infinitely far. My teacher said that we should assume the nail could keep going even after it's flush with the wood. He also declared that there was an exact and real number that it would be (the answer). He compared it to the example he used earlier in the year: "I am walking towards that door. I will go halfway there each time. I won't ever reach the door."

    Now, I don't think that really applies here. My reasoning for saying 'infinitely far' is that each time, it goes further, therefore in infinitely many strike, it will go infinitely far.
  2. jcsd
  3. Mar 18, 2004 #2
    If it's going .1cm in each hit, then it will go infinitely far. On the other hand, if how far the nail goes in each time is a geometric sequence as you said, then it will only go in 1/8 whatever's left, i.e. 7/8 as much will be exposed after the hit each time.

    So the question becomes, what's

    [tex]\lim_{n\to \infty} 8\cdot\Big(\frac{7}{8}\Big)^n[/tex]


  4. Mar 18, 2004 #3

    matt grime

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    I add up the numbers 1, 1/2, 1/4, 1/8 etc. I add up an infinite number of them, the answer isn't infinity.
  5. Mar 18, 2004 #4


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    But how long will it take you to add them all up??
  6. Mar 18, 2004 #5

    matt grime

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    I do the first sum in a second, the second in half a second, the third in an eighth of a second. Time flies when you're having fun. Of course, if I well order a subset of time of ordinal type w^2 I can do an infinite number of infinite sums. I hope no one is taking this seriously.
  7. Mar 18, 2004 #6
    Right, in fact, it's whatever the last term is less than 2, right? You'd just have...let me see..
    8, 7, 49/8, 343/64...hmm, I dunno. I guess I just don't see where the asymptote is. Is it zero? Then I end up with an asymptote, and an area I'd like to calculate the are of (i think, to sum the terms), but the area looks infinite to me. I know it's not, of course..maybe I just have trouble with 7/8 more because it's less than half.

    anyhow, I'm still not sure what the answer is, because I havent used the calculation that cookiemonster dipsplayed before.

    edit: and before you go on, trust me, i understand how the "1/2 more each time" thing works...it's this 7/8 total thats giving me trouble.
    Last edited: Mar 18, 2004
  8. Mar 19, 2004 #7
    Wow, shoot me now. I just totally misread that question. 2 in a day.

  9. Mar 19, 2004 #8
    At first the nail goes in 0.8cm. In the second hit it goes in 0.7cm, which is 7/8 of the previous distance. Whenever you have a geometric series in which the ratio is smaller than 1 and bigger than -1, it converges and its sum is:

    [tex]S = a\frac {1}{1 - r}[/tex]

    This is because:

    [tex]S_n = a\frac {1 - r^{n+1}}{1 - r}[/tex]

    But [tex]n \rightarrow \infty[/tex] and |r| < 1, and any fraction to the power of infinity is 0.
  10. Mar 19, 2004 #9

    matt grime

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    A geometric series is one that goes


    a is the first term r is the common ratio.

    Let S(n) be the sum up to the term ar^{n-1}, the sum of the first n terms.

    What is rS(n)? It's ar+ar^2+..+ar^n

    so S(n)(r-1)=ar^n-a

    S(n) = a(r^n-1)/(r-1)

    Now let n tend to infinity. If r is less than 1 in absolute value then the sum converges to a/(1-r)

    Here the first term, a, is 0.8, and the ratio is 0.7/0.8 = 7/8

    so the nail travels

    (8/10)(1/(1-7/8)) = 64/10 or 6.4cm as we've been given decimals
  11. Mar 19, 2004 #10
    Thank you, Matt! Just to be clear: 6.4 will be an asymptote of the total distance, right?
  12. Mar 19, 2004 #11
    Yes, it will be a horizontal asymptote.
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