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Impulse problem about a medical biopsy needle.

  1. Oct 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Physicians perform needle biopsies to sample tissue from internal organs. A spring-loaded gun shoots a hollow needle into the tissue; extracting the needle brings out the tissue core. A particular device uses 8.4 needles that take 81 to stop in the tissue, which exerts a stopping force of 38 .
    How far into the tissue does the needle penetrate?

    2. Relevant equations

    Just F=ma and J=Ft, the same thing really.

    3. The attempt at a solution

    So I calculated J by multiplying the force through the time. I got 3.1x10^-3 m/s
    Then I did the following:
    Δp=J
    m(v2-v1)=J
    v1=-J/m ; assuming v2 equals zero since the needle stops
    v1=-2.604x10^-5 m/s

    Then I used a kinematic to solve for the acceleration:
    v2=v1+aΔt
    0=v1+aΔt ; once again assuming v2 equals zero since the needle stops
    a=v1/Δt
    a=3.2148x10^-4 m/s^s

    Using another kinematic now that i know the acceleration:
    Δd=v1Δt+(1/2)a(Δt)^2
    Δd=1.054x10^-6 m.

    Anyways I rounded it to 1.1*10^-6 m since it wanted the answer to two significant figures. This was wrong and I really don't know where I went wrong. Any help would be greatly appreciated
     
  2. jcsd
  3. Oct 29, 2012 #2

    haruspex

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    Please specify all units and say what the numbers refer to. 8.4 needles - one was broken? 8.4km long needles? 8.4 sq mm cross-section needles?
    Wrong units for an impulse. Also, please show working.
     
  4. Oct 30, 2012 #3
    sorry. 8.4 mg needles. It took 81 ms to stop and my impulse was 3.1x10^-3 kgm/s

    Here's my work through with numbers:

    ΔP=J
    m(v2-v1)=J
    v2-v1=J/m
    v1=-J/m + v2
    v1=-(3.1x10^-3)/(8.4x10^-3)+0
    v1=-0.369m/s

    v2=v1+aΔt
    a=(v2-v1)/Δt
    a=(0+0.369)/(81x10^-3)
    a=4.56 m/s^2

    Δd=v1Δt+(0.5)aΔt^2
    Δd=(-0.369)(81x10^-3)+(0.5)(4.56)(81x10^-3)^2
    Δd=-1.4943x10^-2 m
    Δd=-1.5x10^-2 m ; rounded to 2 sig figs.

    Anyways, that still isn't the right answer
     
  5. Oct 30, 2012 #4

    haruspex

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    I used a slightly simpler approach. I would normally show all units, but you still haven't told me the units for the '38' number, so I can't do that.
    accn = -38/84
    time = 81
    final velocity = 0
    s = vt - at2/2 = 1484
    So to two sig figures I confirm your result. The question then is the order of magnitude. If the 38 is in microNewtons, I get 1.5 10-3 m.
     
  6. Oct 30, 2012 #5
    So sorry, yeah the units on the 38 is mN. I'll try that. I got a 1.5x10^-2 so I know that I've got the right digits, just a problem with my powers. Thanks a lot for the confirmation/help!
    woah, just tried 1.5x10^-3 and it was wrong. And it was my last guess. I have no idea what was wrong
     
  7. Oct 30, 2012 #6

    haruspex

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    No, sorry - the power of 10 error was mine.
    Are you sure the answer is required to 2 sig digits only? And in what units?
    You said mN (milliNewtons). Did you mean microNewtons (μN)? mN gives me 15m!
    If the answer is required in mm, try 15 or 14.8.
     
  8. Oct 30, 2012 #7
    It's definitely milliNewtons (mN). Anyways I had three guesses, my first was an arithmetic error. And guess two and three were 1.5x10^-2 and 1.5x10^-3 respectively. It was also definitely asking for the answer to two sig figs, and when you get close to an answer but not exactly it tells you you're close and to check rounding etc. That message never came up. Also the program I'm doing these questions on is called "masteringphysics" and it doesn't care what units you use as long as they're equivalent. Ie I could use 10cm or 0.1m and it wouldn't care

    Anyways, i'm just really curious as to what the answer would be, incase a question like this came up on a test or something. Thanks for the help by the way.
     
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