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Spring Constant Question

  1. Dec 4, 2007 #1
    I have a problem in which a bullet is fired into a wooden block on a spring. At the time of the impact the block and the bullet inter into SHM. I am given amplitude, the mass of both the block and bullet, and the velocity of the bullet and am asked to give the spring constant.

    I was using Hooke's law where mg=-kx. I was calculating using the amplitude for x and the mass of the block (which is attached to the spring) for m. However, this did not seem to use all of the information given and I was wondering if I am missing something or if it is just a trick question.

    I then tried using m1v1=(m1+m2)v2 and (1/2)(m1+m2)v2^2=(1/2)kA^2, but when I plugged that value I found back into Hooke's law it did not give me any of my mass combinations.

    What am I supposed to do?
     
  2. jcsd
  3. Dec 4, 2007 #2

    Doc Al

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    That looks OK to me. You have an inelastic collision, after which energy is conserved. Looks to me that you've used all the information given: speed, masses, and amplitude.
    What do you mean when you say you plugged it back into Hooke's law? To find what?
    What do you mean by "did not give me any of my mass combinations"?
     
  4. Dec 4, 2007 #3
    Well my values were mass of bullet=5g mass of block=2.kg amp=11 cm and v of bullet=650m/s

    I used m1v1=(m1+m2)v2 to find that v2=1.3m/s (I'm not sure if that's right though).

    Then plugging that into the second formulat I cound K=350N/m

    I put that in kx=mg (350)(.11)=(m)(9.81) and found m=3.92kg. This did not seem right because my two masses combined were 2.505kg.
     
  5. Dec 4, 2007 #4

    Doc Al

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    The equation is correct, but recheck your arithmetic.

    Why would you use that equation? I presume that the block and spring are arranged horizontally (on a frictionless table, perhaps)? Why would the weight (mg) be relevant?

    You would use kx = mg if you had a mass hanging from a spring and wanted to find the equilibrium position.
     
  6. Dec 4, 2007 #5
    Alright. So my math in the first section is (.005kg)(65)=(2.kg+.005kg)(v2) Thus v2 does equal 1.3m/s. Which results in k=350N/m. I guess the spring constant sounds right but it just seems drastic that when the bullet hit the block it would drop from 650m/s to 1.3m/s. Where that bothered me was I guess I've watched too many movies where bullets have more effect then that. :) Does that seem right?
     
  7. Dec 4, 2007 #6

    Doc Al

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    Is the bullet speed 650 or 65 m/s?
     
  8. Dec 4, 2007 #7
    Sorry, 650m/s. I used that when finding finding 1.3 m/s as v2. I just typed it wrong here.
     
  9. Dec 4, 2007 #8

    Doc Al

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    Redo the calculation one more time.
     
  10. Dec 4, 2007 #9
    (.005)(650)=3.25=(2.505)(v2)

    v2=3.25/2.505=1.3m/s
     
  11. Dec 4, 2007 #10

    Doc Al

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    Is the mass of the block 2 or 2.5 kg?
     
  12. Dec 4, 2007 #11
    2.5 kg is the mass of the block and the mass of the bullet is 5 g or .005kg.
     
  13. Dec 4, 2007 #12

    Doc Al

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    No problem. In a previous post I thought you had written 2kg.
     
  14. Dec 4, 2007 #13
    Oops, sorry. Well I did think 2.5 kg. :) Thank you for your help.
     
  15. Dec 4, 2007 #14
    One more question. Since I used KE=PE to find the v2 would the total energy still be KE+PE or just one of them?
     
  16. Dec 4, 2007 #15

    Doc Al

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    Both! :wink:

    Total energy always equals KE + PE. But initially PE = 0, so Total Energy = initial KE + 0. And when at full amplitude, KE = 0, so Total Energy = 0 + final PE. So, initial KE = final PE.
     
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