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Conservation of Energy of a bullet

  1. Nov 7, 2003 #1
    hey ppl
    A 4.0 kg block is suspended from a spring with a force constant of 500 N/m. A 50g bullet is fired into the block from directly below with a speed of 150 m/s and is imbedded in the block.

    So basically... there are 3 sections here.

    A. bullet is travelling up and hasn't touched the motionless block yet.

    B. bullet impacts the box and is embedded in the box. energy is lost, but momentum is conserved

    C. box and bullet travels upward with the resulting momentum from the original bullet.

    So.. A to B, it is conservation of momentum. from B to C, it is conservation of energy.

    ok... I found the velocity of the block and the bullet when the block is at the equilibrium position. (just when it is impacted) by using the conservation of momentum.

    (m1)(vi) = (m1+m2)(vf)

    then i use that velocity in the conservation of energy equation. Therefore....

    (1/2)(m1+m2)V^2 + 1/2kx^2 = (m1+m2)gh + 1/2KA^2

    that may look like a blob to you but my question is simply this.. in this conservation of energy equation... do i include the change in graviational potential energy? I mean.. the block has moved up, right? so it should be included.... supposedly.

    A couple of my friends disagree with me. (At least according to one of them, he already had a peek at the answer key from the prof and he claims that gravitational potential energy isn't included.)

    So I'm confused why it shouldn't be included. Help. Plz. Thx.
  2. jcsd
  3. Nov 7, 2003 #2


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    Feel free to add a term for gravity, but remember that the block will then also be streching the spring a little. IIRC the two will cancel.
  4. Nov 7, 2003 #3
    Tell your friends they're wrong.

    Well, I guess it depends on the question. If it only asks for the speed at the instant after the collision, use your conservation of momentum equation & you're done.

    But beyond that point you are correct in that you must use conservation of energy, & you can't ignore gravity. Take your equation:

    (1/2)(m1+m2)V^2 + 1/2kx^2 = (m1+m2)gh + 1/2KA^2
    The only thing you left out is the FINAL kinetic energy + (1/2)(m1+m2)Vf^2.
    x is the distance the spring has been stretched (you can easily determine this distance from the given information, & substitute the actual value into the equation). Let h = 0 at the position of the block just before the bullet hits, so your A = x - h. (I'm not sure why you capitalized the K on the right side of the equation. I assume that was unintentional, & I'm switching it to lowercase to make it clear that the k on both sides is the same k.

    Now your equation becomes:
    (1/2)(m1+m2)V^2 + (1/2)kx^2 = (m1+m2)gh + (1/2)k(x-h)^2 + (1/2)(m1+m2)Vf^2
    (1/2)(m1+m2)V^2 + (1/2)kx^2 = (m1+m2)gh + (1/2)k(x^2-2xh+h^2) + (1/2)(m1+m2)Vf^2
    (1/2)(m1+m2)V^2 + (1/2)kx^2 = (m1+m2)gh + (1/2)kx^2 - xh + (1/2)kh^2 + (1/2)(m1+m2)Vf^2
    (1/2)(m1+m2)V^2 = (m1+m2)gh - xh + (1/2)kh^2 + (1/2)(m1+m2)Vf^2
    Last edited: Nov 7, 2003
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