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Speed of a bullet?

  1. Oct 26, 2008 #1
    A 12 g bullet is fired into a 8.0 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table. The coefficient of kinetic friction for wood sliding on wood is 0.20.

    What was the speed of the bullet?


    Relevant equations
    My professor told us that we would need these equations:
    (1/2)mv2 = mgx

    mvi = (m1+m2)v2


    The attempt at a solution
    v2 = 2(9.81)(0.20)(0.05)
    v = 0.443

    0.012vi = (8.0+0.012)(0.443)2
    vi = 131 m/s

    This seemed like a logical process and the answer seemed fairly reasonable (bullets move fast) however, this is not the right answer and I don't know what I am doing wrong.
     
  2. jcsd
  3. Oct 26, 2008 #2

    CompuChip

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    Do you understand what the equations you use are telling you?

    You are basically using conservation of energy. Before the bullet hits the block, the block is at rest and the bullet has a kinetic energy [itex]E = 1/2 m_\text{bullet} v_\text{bullet}^2[/itex].
    After the bullet hits the block, the bullet and the block start moving together. So they get some kinetic energy. How much (expressed in E)?
    There acts a frictional force on the block (with bullet) which will bring it to a stop. How much energy will the bullet+block have when it has come to rest? What happened to the energy that is missing? Can you write down a formula for the work done by the frictional force?
     
  4. Oct 26, 2008 #3
    Wow, I've never heard of a professor just giving out equations like that. I think you might have copied his/her second equation wrong. That equation looks like conservation of momentum for an inelastic collision, in which case you should not have a v squared.
     
  5. Oct 15, 2009 #4
    In other words: there is a kinetic energy imparted into the block when the bullet collides with it. That kinetic energy is the kinetic energy of the new block-bullet system. Next the block moves a distance d until it reaches a stop. This means that friction does work on it over that distance d. Work is equal to change in kinetic energy. The initial kinetic energy is just the kinetic energy imparted to the block by the bullet while the final is zero because it comes to a stop.
     
  6. Oct 16, 2009 #5

    CompuChip

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    That's a nice explanation NovaKing, but why on earth did you revive a thread that's been dead for almost a year?
     
  7. Oct 17, 2009 #6
    My friend needed help on the question and when he googled it he got this. I thought I would give an alternative explanation for anyone else in his class who also googled it. heh just helping I suppose...if I was right that is.
     
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