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Conservation momentum/energy

  1. Mar 16, 2012 #1
    1. The problem statement, all variables and given/known data
    A bullet of mass 1.4×10−3 {\rm kg} embeds itself in a wooden block with mass 0.999 {\rm kg}, which then compresses a spring (k = 110 {\rm N/m}) by a distance 5.5×10−2 {\rm m} before coming to rest. The coefficient of kinetic friction between the block and table is 0.46.
    a)what is the initial speed of the bullet?
    b)What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block? (answer: ΔK/K


    3. The attempt at a solution
    i started by drawing a diagram labeling part A to be where the spring is compressed, part B to be where the block started pushing the spring, part C to be where the bullet impacts the block, and part D the firing of the bullet.
    I then used energy conservation, stating that the E@A=E@B. 1/2kx^2=1/2mv^2, solving for v and getting sq.rt of (kx^2/m). plugging in the numbers i get v=.5767 m/s. Then i attempt to use energy conservation from B to C, 1/2mv^2=1/2mv^2-(work by friction+work by spring) but i kind of just put that equation together myself because i know energy is lost by work done) and i dont really know what to do.
     
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  3. Mar 16, 2012 #2

    gneill

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    Staff: Mentor

    While the spring is being compressed, where is the KE of the block+bullet going? Is it only into the spring's PE?
     
  4. Mar 16, 2012 #3
    well there is also friction, so im guessing some of the KE from the block+bullet is being lost due to work done by friction.
     
  5. Mar 16, 2012 #4

    gneill

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    Staff: Mentor

    Correct. So when you want to find the velocity of the block+bullet when it first encounters the spring, both energies need to be accounted for. How much energy should be attributed to friction while the spring compresses?
     
  6. Mar 16, 2012 #5
    um...the energy attributed to friction is equal to the work done by friction? so im going to say FΔx, the force of friction is equal to mass of bullet+block x g x μ_k x distance spring compressed= .248 J
     
  7. Mar 16, 2012 #6

    gneill

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    Staff: Mentor

    Correct again :smile:

    So what's the total kinetic energy that the block+bullet had when it first hit the spring?
     
  8. Mar 17, 2012 #7
    ok so the total kinetic energy that the block+bullet system has is equal to 1/2mv^2+W(spring)+W(friction). if that is correct i understand the equation, however im not understanding exactly how that equation works. when i substitute the numbers in to solve for the velocity the system has before it hits the spring i get the sq. rt of a negative answer. i understand the energy part of this question but im not sure when it comes to actually applying the formula.

    what i did was: (EPE=elastic potential energy) EPE_A=KE_B+Ws+Wf, that becomes 1/2kx^2=1/2mv^2+(.248+.333). (the .333 is the work done by the spring, the force of the spring is F=kx=6.05N, W=FΔx=(6.05)(.055)=.333.) substituting the values i get 1/2(110)(.055)^2=1/2(1.0004)v^2+.581, doing that i get [sq.rt -.4146/.5002] which cant be.
     
  9. Mar 17, 2012 #8

    gneill

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    Staff: Mentor

    When the spring is fully compressed, the energy stored in it is what was left of the original KE minus what was lost due to friction:

    KE = PEspring + Wfriction

    Remember, the energy lost to friction while the spring was compressing never makes it into the spring's potential energy -- it was lost.
     
  10. Mar 17, 2012 #9
    wow. i got it, thank you. so i kind of understood it incorrectly, the KE at time of impact is = to the energy now in the spring plus the work done by friction. the work done by the spring isn't important here because that energy is going into the elastic potential energy not being lost by work.
     
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