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Inelastic Collision of a bullet and block

  1. Oct 19, 2008 #1
    1. The problem statement, all variables and given/known data
    A bullet is fired vertically into a 1.20 kg block of wood at rest directly above it. If the bullet has a mass of 24.0g and a speed of 540 m/s , how high will the block rise after the bullet becomes embedded in it?


    2. Relevant equations

    1/2mv^2+mgh=1/2mvf^2+mgh
    mv=mvf
    3. The attempt at a solution

    I know that for this i must use momentum to find the final v which would be

    mV=(m+M)V'
    then i use that V' to find height

    1/2mV'^2=(m+M)gh

    I solved this and I got an answer of 413 meters this is wrong. I do not know what im doing wrong.
     
  2. jcsd
  3. Oct 19, 2008 #2

    Doc Al

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    Staff: Mentor

    Good!

    Make sure you use "m+M" on both sides. (They cancel, of course.)

    If that's not the problem (and I don't think it is), show your calculations, step by step.
     
  4. Oct 19, 2008 #3
    ok i did that and it is still wrong. i did .5(.240+1.20)(90)^2=(.240+1.20)(9.80)h

    i put the left side in the calc. and got 5832=(.240+1.20)(9.80)h then divided the right numbers by the left and my answer is 413.26m which is wrong according to mastering physics.
     
  5. Oct 19, 2008 #4

    Doc Al

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    Staff: Mentor

    Show how you calculated that speed.
     
  6. Oct 19, 2008 #5
    I multiplied .240kg(540m/s) then divided that by (.240+1.2).
     
  7. Oct 19, 2008 #6

    Doc Al

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    Staff: Mentor

    There's the problem. The bullet's mass is 24 grams = 0.024 Kg (not 0.24 Kg).
     
  8. Oct 19, 2008 #7
    wow thank you!
     
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