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Homework Help: Bullet-in-Block with Spring

  1. Dec 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A pellet gun fires a bullet into a stationary block of wood that is attached to a spring on a frictionless surface. When the bullet enters the wood, it reamins inside, and the bullet and block enter into simple harmonic motion with amplitude = 11cm. The bullet ( m = 5g) was initially travelling at 650 m/s before hitting the block of wood ( m = 2.5kg).

    A: What is the spring constant of the spring?
    B: What is the total energy of the system after the collision?(I know how to solve this one, I just need A to solve it.)
    C: What is the maximum acceleration of the bullet/block system once it begins its oscillation?
    D:Where will the bullet and block reach 0 velocity?(I think this is at the equilibrium position, but I'm not sure how to solve for it)

    We know:
    the amplitude of the block-and-bullet is .11m
    the mass of the bullet is .005kg
    the mass of the block is 2.5kg
    the mass of the bullet-and-block is 2.505kg
    the velocity of the bullet before it hits the block is 650 m/s.

    2. Relevant equations
    I know for B that you use the equation for total energy, kinetic energy + potential energy.

    3. The attempt at a solution
    My attempt at A; it wasn't so good. I tried using F = -kx, but that of course didn't work since we don't know the acceleration to solve for F. Then I considered kx=mg, but that doesn't work either because that only works for a spring that is suspended. So I'm not really sure where to go with this :eek:
     
  2. jcsd
  3. Dec 13, 2008 #2

    Doc Al

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    Hint: What's the speed of "block+bullet" immediately after the collision?
     
  4. Dec 13, 2008 #3
    :yuck:I really don't know Al.
     
  5. Dec 13, 2008 #4

    Doc Al

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    Hint: What's conserved in any collision?
     
  6. Dec 13, 2008 #5
    Errrrr it's either energy or momentum. I think momentum.
     
  7. Dec 13, 2008 #6

    Doc Al

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    Good. Use a conservation law to figure out the speed of "block+bullet" after they collide.
     
  8. Dec 13, 2008 #7
    Alright. p=mv, so then .005*650=3.25; so 3.25/2.505= 1.297.
    So 1.297 would be the new velocity?
     
  9. Dec 13, 2008 #8
    Is that correct?
     
  10. Dec 13, 2008 #9

    Doc Al

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    Perfect!

    Now see if you can make use of that speed to solve part A (and part B, for that matter).
     
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