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Homework Help: 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)


    What is the speed v of the bullet?

    2. Relevant equations


    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


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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
  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.
  22. Nov 6, 2013 #21
    First ignore the spring and calculate the momentum as
    mv =(m+M)v1
    Now apply the enrgy coonservation
    find the value of v1 and then put it in first equation.You will get the answer.
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