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Period for SHM of a block on a spring shot by a bullet

  1. Apr 16, 2012 #1
    So I've been working on this problem for a while and can't seem to come up with anything and I don't really understand why I'm wrong in what I did. I could just get the answer online, but I actually want to understand what's happening.

    1. The problem statement, all variables and given/known data

    A rifle bullet with mass 8.12g and initial horizontal velocity 250m/s strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0cm. After the impact, the block moves in SHM. Calculate the period of this motion.

    2. Relevant equations

    (Period) T = 2pi*sqrt(m/k)
    (Kinetic Energy) K = .5*m*v^2
    (Potential Energy) U(spring) = .5*k*x^2
    (Momentum) P = mv

    3. The attempt at a solution
    I'll use m for the mass of the bullet and M for the mass of the block.

    So first I used conservation of momentum to assume that m*v(bullet) = (M+m)(v(final)).
    Plugging in I got (.00812kg)(250m/s) = (.922kg+.00812kg)*v(final)
    v(final) = 2.182514m/s

    So then I assumed that the energy that the system has right after the bullet makes contact is fully kinetic, since in SHM all of the energy is kinetic when at the origin. I think I'm right in assuming this, but maybe that's where I went wrong. But anyways, continuing with what I took as the assumption, I then used the formula for kinetic energy.
    K = .5*(M+m)(v(final))^2 So,
    K = .5*(.922kg+.00812kg)*(2.182514m/s)^2 = 2.214492J

    So then I went and assumed that all energy would be potential when the spring is fully compressed.

    U = .5*k*x^2
    2.214492J = .5*k*(.18m)^2
    k = 136.697066N/m

    So then I simply plugged the k into the formula for period and got:
    T = 2*pi*sqrt((M+m)/k)
    T = 2*pi*sqrt((.922kg+.00812kg)/136.697066N/m)
    T = .518286s

    Apparently this solution is wrong though, so I was wondering if anyone could help me out so that I could better understand this stuff. : /
     
    Last edited: Apr 16, 2012
  2. jcsd
  3. Apr 17, 2012 #2

    PeterO

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    I would certainly go about this problem the same way you have - so I would suggest re-checking your calculations in case there is a slip somewhere.
     
  4. Apr 17, 2012 #3

    ehild

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    Is the mass of the block 0.992 kg or 0.922 kg?

    ehild
     
  5. Apr 17, 2012 #4
    AAHHH!!!!! I can't believe that I didn't catch that, I reworked the entire problem for hours and couldn't find a way to get .557s, which was apparently the answer, but that was my mistake every single time. The correct mass is .992kg, I feel like such an idiot, thank you so much for the help!
     
  6. Apr 17, 2012 #5

    ehild

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    Gold Member

    Such things happen with everybody...:smile:


    ehild
     
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