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Simple harmonic motion: mass on a spring is hit by a bullet

  1. Nov 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A 4.0kg block is suspended from a spring with force constant of 500N/m.
    A 50g bullet is fired into the block from directly below with a speed of 150m/s and is imbedded in the block.
    Find the amplitude of the resulting simple harmonic motion.

    2. Relevant equations

    F=-kx
    E_p=mgh
    E_k=(mv^2)/2
    E_pspring=(kx^2)/2
    m1v1=m2v2

    3. The attempt at a solution

    Using conservation of momentum:
    0.05(150)=(4+0.05)v2
    v2=1.852 m/s

    MLYBAGY.png
    I am lost on how the decide where the equilibrium position is. Is it the resting position with the block considered the equilibrium position, or is it without the block, or is it with the block and the bullet?

    I said that the kinetic energy
    0.5*(4.05)(1.852)^2 is equal to 0.5k(x)^2 and completely ignored mgh(potential energy). Then i assumed x is the amplitude. This seems incorrect of course.

    Any help would be appreciated, thanks!
     
  2. jcsd
  3. Nov 24, 2015 #2

    haruspex

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    The new equilibrium position will, by definition, be wherever the block with embedded bullet would hang at rest. How does that change your answer?
     
  4. Nov 25, 2015 #3
    I'm under the assumption that at equlibium
    I am confused on whether i should consider there is already a spring potential on the block, before hit by the bullet, or if it has no spring potential at that point
     
  5. Nov 25, 2015 #4

    haruspex

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    The spring is not relaxed, so it certainly has potential energy then.
     
  6. Nov 25, 2015 #5
    I used the position at block+spring as the equilibrium point, and solved it using two methods to get an answer. Even though i didnt consider the block+spring potential energy, since that was my equilibrium point, would my answer be correct?

    0G6jxxP.jpg
    BnmbBHJ.jpg

    Sorry about the pictures instead of latex. I will learn it someday :D
     
  7. Nov 26, 2015 #6

    haruspex

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    I would take into account that the spring length on impact of bullet is different from the equilibrium length for block+bullet. You can determine the spring PE and GPE of each of those two positions. That will give you a little more KE when subsequently at the equilibrium length. (You can see that by considering what would happen if the bullet had no speed when it stuck to the block. The block+bullet would bob up and down a little.)
     
  8. Nov 26, 2015 #7
    I think i see what you mean. I will try to include the GPE as well and see what i end up with.
    Thanks!
     
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