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Springs, Simple Harmonic Motion

  1. Nov 18, 2007 #1
    1. A 25.0g bullet strikes a .600kg block attached to a fixed horizontal spring whose spring constant is 6.70x10^3 and sets it into vibration with an amplitude of 21.5cm. What is the speed of the bullet before impact if the two objects move together after impact?

    2. V=A[tex]\sqrt{k/m}[/tex]

    3. I used the equation above, adding the masses together (.625kg). I got the velocity to be 22.26m/s... but now i can't figure out how to get the speed of the bullet.

    The answer in the book says that the speed of the bullet is 557m/s.
    Last edited: Nov 18, 2007
  2. jcsd
  3. Nov 18, 2007 #2


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    Staff: Mentor

    One determines the maximum spring deflection from the amplitude, and from that the stored energy in the spring at max deflection, which by conservation of energy (and neglecting friction) should equal the initial kinetic energy of the bullet and block.

    From the KE of the bullet and block, one can determine the velocity of the combination.

    Since the bullet striking the block is an example of an inelastic collision, what about the momentum of the bullet and block?
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