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Linear impulse and linear momentum

  1. Oct 1, 2007 #1
    1. The problem statement, all variables and given/known data
    given mass of bullet and box and given initial and final velocity of projectile, find kinetic friction coefficient

    2. Relevant equations

    G1 + integral of sum force respect to time = G2

    3. The attempt at a solution

    I setup like this...

    mv(projectile) + mv(block, which is 0) = mv(projectile) + mv(block)

    I was trying to use the equation under 2 but i don't know how that can help me anything.
    Last edited: Oct 1, 2007
  2. jcsd
  3. Oct 1, 2007 #2

    Chi Meson

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    drop this one

    I can't tell what it's supposed to mean, but the other equation will help. Use that to find the speed of the block after the collision.

    Then the block slides to a stop as friction does the (negative) work to take away the kinetic energy.
  4. Oct 1, 2007 #3
    yes i've found the velocity of the block.
    i don't know how to go from there...
    do i use work energy equation to finish it?
    1/2mv^2(for block) + 1/2mv^2(for projectile) - work done by friction = 1/2mv^2(for block) + 1/2mv^2(projectile)?
  5. Oct 1, 2007 #4

    Chi Meson

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    The work done by friction will just be on the block, forget the bullet after the collision, it's no longer your concern. So the KE' of the block is taken away by the work done by friction. KE = W
  6. Oct 1, 2007 #5
    i forgot to mention what cancells out from the equation above.
    well anyways, i think my brain is dead...
    -(coefficient of kinetic)mgd = 1/2mv^2(velocity that i got from the above).
    i got coefficent, which is 0.3 something but it's negative...
    do i take the absolute value of it since it depends on how i draw it?
    thank you very much.
  7. Oct 2, 2007 #6

    Chi Meson

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    the negative cancels out because the change in KE is negative (lost)
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