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Velocity of bullet

  1. Nov 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Agent Arlene devised the following method of measuring the muzzle velocity of a rifle (the figure below). She fires a bullet into a 4.064-kg wooden block resting on a smooth surface, and attached to a spring of spring constant k = 164.5N/m . The bullet, whose mass is 7.870 g, remains embedded in the wooden block. She measures the maximum distance that the block compresses the spring to be 9.460cm .
    (Figure 1)

    http://session.masteringphysics.com/problemAsset/1696765/3/Giancoli7.ch11.p22.jpg

    What is the speed v of the bullet?


    2. Relevant equations

    v=√(k/m)*x

    3. The attempt at a solution

    i tried using that equation and different variations but my mastering physics says i'm incorrect.
     
  2. jcsd
  3. Nov 6, 2013 #2

    phinds

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    Show your work
     
  4. Nov 6, 2013 #3
    i tried using √164.5/(4.064+.00787) *.0946


    i also tried using .5mv^2=.5kx^2
     
  5. Nov 6, 2013 #4
    Mech. energy is not conserved in inelastic collisions. Use momentum conservation and then use Energy conservation.
     
  6. Nov 6, 2013 #5
    i cant tell if the question want the velocity before the collision or after the collision.
     
  7. Nov 6, 2013 #6
    It says "muzzle velocity". So it is not the velocity after collision.
     
  8. Nov 6, 2013 #7
    so then should i not use the equation i did ?
     
  9. Nov 6, 2013 #8
    what equation should i use
     
  10. Nov 6, 2013 #9
    You should use energy conservation.
    The kinetic energy of the bullet before collision = the potential energy of spring after collision
     
  11. Nov 6, 2013 #10
    so 1/2mv2=mgh?
     
  12. Nov 6, 2013 #11
    no thats wrong
     
  13. Nov 6, 2013 #12
    The potential energy of the spring not the gravitational potencial energy(which would stay constant as the system is horizontal)
     
  14. Nov 6, 2013 #13
    .5mv^2=.5kx^2?
     
  15. Nov 6, 2013 #14
    that's right
     
  16. Nov 6, 2013 #15
    would x be the compression of the block after the bullets hit?
     
  17. Nov 6, 2013 #16
    Yes,right.x would be the compression after bullet hits it.
     
  18. Nov 6, 2013 #17
    mass should be in kg or gr?
     
  19. Nov 6, 2013 #18
    Wait i think we cant use the conservation of energy here as the mechanical energy is not conserved.Instead we will use conservation of momentum
     
  20. Nov 6, 2013 #19
    so then i would use m1v1right?
     
  21. Nov 6, 2013 #20
    My answer is coming to be 329m/s.
     
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