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Momentum conservation

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A bullet(mass 100 g) strikes a block of 4 kg kept on a table. The friction coefficient of table-block is 0.25. The bullet gets embedded and moves a distance of 20 m. Find speed of bullet.

    2. Relevant equations



    3. The attempt at a solution
    Using Conservation of momentum,
    speed of block + bullet= v/401 (v is the speed of the bullet)
    now equating Energy;
    Work one by friction = kinetic energy of system
    0.25*4.01*10*20=0.5*4.01*(v/401)^2
    v=4010
    am i right?
     
  2. jcsd
  3. Sep 22, 2008 #2
    Is applying conservation of momentum correct here???
     
  4. Sep 22, 2008 #3

    PhanthomJay

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    Your method is fine, but you missed a decimal point.
    100 grams is 0.1kg, not .01kg.
     
  5. Oct 6, 2008 #4
    solved
     
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