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Conservation of energy

  1. Oct 27, 2008 #1
    1. The problem statement, all variables and given/known data
    An 8 g bullet is shot into a 4 kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it by 8.7cm. the force constant of the spring is 2400N/m. The initial velocity of the bullet is closest to
    A. 1190m/s B. 1020m/s C. 1110m/s D. 1150m/s E. 1070m/s

    2. Relevant equations
    U= 1/2Kx^2
    K = 1/2mv^2


    3. The attempt at a solution
    I calculated the potential energy from the spring compressing using U=1/2(2400)(0.087^2) and set that equal to the kinetic energy of the block moving with the bullet lodged into it K=1/2(4.008)V^2. solved for v then set that kinetic energy equation equal to the kinetic energy of the bullet K=1/2(0.008)v^2 but the value of v for the bullet I get is much much smaller than any of the choices
    1/2(2400)(0.087^2)=1/2(4.008)v^2=1/2(0.008)v^2
     
  2. jcsd
  3. Oct 27, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good!
    There's your error. The KE of the bullet does not equal the KE of block + bullet. KE is not conserved during the collision. But something else is.
     
  4. Oct 27, 2008 #3
    The momentum is conserved. Thanks Dr Al
     
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