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Homework question on Simple Harmonic Motion (the bullet question)

  1. Jan 12, 2009 #1
    1. The problem statement, all variables and given/known data

    A 1.08 kg block is attached to a horizontal spring with spring constant 3650 N/m. The block is at rest on a frictionless surface. A 10.6 g bullet is fired into the block, in the face opposite the spring, and sticks.

    What was the bullet's speed if the subsequent oscillations have an amplitude of 7.10 cm?

    2. Relevant equations

    This is the only formula I can think of but I do not have time?
    Vx=-VtsinO=-Awsinwt

    3. The attempt at a solution
    I do not know which formula I could use to start the problem...

    HELP!
     
  2. jcsd
  3. Jan 12, 2009 #2

    Dick

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    Use conservation of momentum to find the initial velocity of the mass. Find w in terms of the mass and spring constant. If the motion is described by x(t)=A*sin(wt) then what is v(t)? Differentiate x(t). Then the velocity you found using conservation of momentum should equal v(0). Solve for A. Try it.
     
  4. Jan 13, 2009 #3
    hmm...i didnt get it right..

    can some one show me how to do it step by step??
     
  5. Jan 13, 2009 #4

    Dick

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    Call t=0 the time the bullet hits the block. x(t)=A*sin(wt), v(t)=w*A*cos(wt). Put t=0 into that and conclude v(0)=w*A. Compute v(0) using conservation of momentum. Show us what you did.
     
  6. Jan 14, 2009 #5
    Okay so I solved for V(0)= w * A= 412.723

    The conservation of momentum is: V0= ( m1+ m2/ m1) * Vf
    Vf= 412.723/ 10.8
    = 38.166

    But that answer is wrong...so what do I do now?
     
  7. Jan 14, 2009 #6

    Dick

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    You really aren't showing enough of your work. I can't tell what you are actually doing. What did you get for w? From what you've got so far, it looks like Vf should be set equal to v(0)=w*A.
     
  8. Feb 24, 2009 #7
    I calculated the force required to push the block by the amplitude .071meters,
    F=259.15N
    then i calculated what the speed of the bullet would have to be to achieve this force, i got a strange answer of 24km/s which seems incredibly fast......
     
  9. Feb 24, 2009 #8

    Dick

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    That doesn't sound right. But then neither does you method of solving it.
     
  10. Feb 28, 2009 #9
    Hello,

    I believe the initial Amplitude is one of hellokitty's givens,

    "What was the bullet's speed if the subsequent oscillations have an amplitude of 7.10 cm?"

    Why is A being solved for?

    Edit: I just solved the problem. Perhaps just "solving" for amplitude left some confused, so I will leave some hints.
    1. Final steps don't include use of momentum equations, but equations for simple harmonic motion. Think on the reasoning behind this hint, is the ending momentum "constant" or in other words, is there a true "final" portion of there scenario, such as there is when two billiard balls hit each other?
    2. When differentiating the position equation for SHM, don't forget the chain rule! (There is an important constant inside the trig. function of x(t))
    3. After solving for the initial velocity of the mass and plugging that into v(t), look at the variable you wish to solve for and ask yourself how you may get rid of any other unknown constant.

    It may be helpful to think of this as a three-stage momentum problem involving SHM.
    You have:
    Stage 1: The bullet is moving, block is at rest.
    Stage 2. The instant when the block is hit by the bullet
    Stage 3: System is now modeled by SHM; momentum is constantly changing with respect to the block's distance from the equilibrium point.

    I hope this was helpful :)
     
    Last edited: Feb 28, 2009
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