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Conservation of Momentum and Mechanical Energy

  1. Feb 18, 2009 #1

    TG3

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    1. The problem statement, all variables and given/known data
    A projectile with a mass of 20 g has an initial horizontal velocity of 100 m/s when it hits and stops in a wood block of mass 0.402 kg. The block is sitting on a horizontal frictionless surface and is attached to a massless spring, initially relaxed, with spring constant 143 N/m. What is the maximum compression of the spring?

    2. Relevant equations
    P= MV
    K=1/2 MV^2
    F=-kx

    3. The attempt at a solution
    Initial Momentum of the Bullet = .02 x 100 = 2
    Initial momentum of the Bullet/Wood system = 2 due to conservation of momentum.
    2= .404 V
    V= 4.95 m/s is the velocity of the block/bullet system.
    K = 1/2 (.404) 4.95^2
    K= 4.95 Due to conservation of Energy, this is also the amount of energy the spring exerts.
    4.95 =143X
    Wrong. I suspect the error in my calculation is near the end, but I don't know that for certain.
     
  2. jcsd
  3. Feb 18, 2009 #2

    LowlyPion

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    Homework Helper

  4. Feb 19, 2009 #3

    TG3

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    Thanks- once I used energy not force it was easy to find the distance. (.26 meters.)
    In the second part though, there is friction (with a coefficient of friction of .25).

    How do I deal with that using the energy relationship?
     
  5. Feb 20, 2009 #4

    Delphi51

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    KE of block = PE gained by spring + work done against friction
     
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