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Collision, Bullet fired into block of wood

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A 20.0 g bullet moving horizontally at 70.0 m/s strikes a 9.00 kg block of wood lying at rest on the pavement. The bullet embeds itself in the block of wood during the collision. Find
    a. the speed of the block of wood immediately after the collision,
    b. the frictional force between the pavement and the block of wood if the block of wood moves 2.50 m before stopping.

    2. Relevant equations
    conservation of momentum and kinetic energy
    3. The attempt at a solution
    Using conservation of momentun:
    A)MbVb+MwVw=(Mb+Mw)V
    MbVb+0=(Mb+Mw)V
    (.0200kg)70m/s=(9.00kg+.0200kg)V
    V=[(.0200kg)(70m/s)]/(9.02kg)
    V=.1552106....m/s
    Using conservation of Kinetic energy:
    .5MbVb^2+0=.5(Mb+Mw)V^2
    (.0100kg)(70m/s)^2=4.51kgV^2
    V^2=[.0100kg(4900m/s)]/4.51kg
    V^2=10.86474501....
    V=3.29617...m/s

    Should'nt these values be the same have I done something wrong? I used both to check that my answers were correct and then I got this result. the second part should be easy once we figure ths out.
     
  2. jcsd
  3. Dec 2, 2008 #2

    LowlyPion

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    Conservation of kinetic energy won't apply because there is an inelastic collision.

    Momentum is conserved, and the initial velocity is the V that you would use to solve the second part of the question.

    V2 = 2*a*x

    With you acceleration (deceleration in this case and the distance covered) then you should be able to figure the retarding force of the pavement and hence the coefficient of kinetic friction.
     
  4. Dec 2, 2008 #3
    Excellent! ok so I have one more quick question is this an inelastic collision because the bullet was embedded into the wood and so the wood absorbed all the energy of the bullet?
     
  5. Dec 2, 2008 #4
    Yes the wood was deformed and heated by the bullet
     
  6. Dec 2, 2008 #5

    LowlyPion

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    No. Some of the energy was converted to moving the block the 2.5 m. Friction brought it to rest. At that point the kinetic energy budget got drawn to 0.
     
  7. Dec 2, 2008 #6
    what I meant was all the kinetic energy of the bullet was directly transferred to the block. ignoring the production of heat and noise and what not else. I am not forgetting that the block was set in motion as a result. would the collision have been elastic if the pavement had been frictionless. or would it have been elastic if they bounced off eachother? I am just trying to get a clear picture/definition in my head of an inelastic collision using this example.
     
  8. Dec 2, 2008 #7

    LowlyPion

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    Basically you have it. If the coefficient of restitution is not 1, then it will have been an inelastic collision.

    Perhaps these would help?
    Inelastic collision:
    http://en.wikipedia.org/wiki/Inelastic_Collision

    Elastic collision:
    http://en.wikipedia.org/wiki/Elastic_Collision#Equations

    Coefficient of restitution:
    http://en.wikipedia.org/wiki/Coefficient_of_restitution#Further_details
     
  9. Dec 2, 2008 #8
    Ty! :)
     
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