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Finding the expression of the bullet's speed, vB

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data

    You have been asked to design a "ballistic spring system" to measure the speed of bullets. A spring whose spring constant is k is suspended from the ceiling. A block of mass M hangs from the spring. A bullet of mass m is fired vertically upward into the bottom of the block. The spring's maximum compression d is measured.

    Find an expression for the bullet's speed vB

    2. Relevant equations

    So basically i started with

    Ki + Ui = Kf + Uf

    and then manipulated adding in the equation for the spring constant, k

    and ended up with:

    rad[(2Mg +m)k/m)] * d but its totally wrong


    help please!
     
  2. jcsd
  3. Oct 23, 2008 #2

    mgb_phys

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    So the spring is extended holding the mass of the block (so the tension spring force is pulling the block up) - is the block going to move far enough that the spring is neutral or in compression?
     
  4. Oct 23, 2008 #3
    I would think that the block is going to far enough that the spring compresses right??
     
  5. Oct 23, 2008 #4

    mgb_phys

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    That would make it complicated - sorry i just noticed this is a homework question not a real experiment. So you cna probably assume that the spring is massless and is in tension all the way.
     
  6. Oct 23, 2008 #5
    so how would i go about finding the expression for the bullet's speed?
     
  7. Oct 23, 2008 #6

    mgb_phys

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    Conservation of energy.
    Start = gravitational potential energy of block (Mgh) and stored spring energy (1/2 kx^2) + ke of the bullet (1/2mv^2)
    End = more potential energy of block Mg(h+distance moved), less spring energy 1/2 k (x-distance moved)^2 + no ke of the bullet.

    Be careful of the signs, think about which energy is increased or decreased.
     
  8. Oct 27, 2008 #7
    I'm also working on this problem. I don't understand why energy is conserved, because isn't this an inelastic collision, in which some kinetic energy must be lost?
     
  9. Oct 27, 2008 #8
    You can use energy conservation for the last part of the problem - once the bullet has finished embedding itself in the block (though bear in mind that the bullet ends up embedded in the block, so the initial collision must be inelastic, as you say).

    Something else is conserved as well when the collision actually takes place, which you always need to take into account when doing a ballistic pendulum-style problem.
     
  10. Oct 27, 2008 #9
    Right, conservation of momentum sets up the equation: mVb=(m+M)Vf, where Vf is the final velocity of the block w/ embedded bullet.

    And for the energy conservation part, I have this:

    .5mVb^2 + Mgy + .5kx^2 = Mg(y+d) + .5kd^2

    I'm confused about the spring. Initially it has a potential elastic energy, but I don't understand how the distance (x) relates to the distance (d) that it moves.

    If I can get the energy conservation right, then I can just substitute in from the momentum equation.
     
  11. Oct 27, 2008 #10
    What do the letters 'd' and 'x' represent in each context?
     
  12. Oct 27, 2008 #11
    The problem states that d is the distance the the spring compresses when the bullet hits it, and x is what i'm unsure of, but I think it is the amount that the spring is already stretched because of mass M hanging from it.
     
  13. Oct 27, 2008 #12

    mgb_phys

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    The simplest way woul dbe to define the spring extention to be 0 at the top of it's motion when the bullet hits. Then spring extention = distance moved (except for signs)
     
  14. Oct 27, 2008 #13
    "The simplest way woul dbe to define the spring extention to be 0 at the top of it's motion when the bullet hits. Then spring extention = distance moved (except for signs)"

    And in that case the initial elastic spring force (.5ks^2) = 0?
     
  15. Oct 27, 2008 #14

    mgb_phys

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    No the initial spring force would be = the weight of the block.
    the initial spring energy would be 1/2 k x^2 and the final spring energy = 0.
     
  16. Oct 27, 2008 #15
    Ok, so the initial spring force doesnt matter right?

    And potential energy of the block, since we aren't given the height the block is above the ground, how does that get involved? I'm pretty lost . . .
     
  17. Oct 27, 2008 #16

    mgb_phys

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    Just using conservation of energy. You only care about changes in energy - you can pick any starting level, after all when you do the potential energy of a block you don't consider the altitude !

    So to make life easy we will set the potential energy of the block when it is hanging down to be zero, and the spring energy at the highest point to be zero. Then we only have one distance term to worry about.

    So initially we have:
    ke bullet + pe block(=0) + spring energy (=1/2 kx^2)

    Afterward we have:
    ke bullet(=0) + pe block (mgx) + spring energy(=0)

    That makes all the bullet ke + spring energy -> block pe.
    A bit of simple rearrangemnt (watch the signs) and you are set.
     
  18. Oct 27, 2008 #17
    Grateful for your help, and the problem is clearing up in my mind. One more thing about signs though. I thought that energy values were scalars and not vectors. So signs only come into play when there is a vector in the equation? Are the only variables affecting the signs the velocity of the bullet, g, and x?
     
  19. Oct 27, 2008 #18

    mgb_phys

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    Scalers have signs. If you slow down your ke change is negative.
    It is harder to get wrong than in a forces/vector calculation - thats why conservation of energy is often a good appraoch.
     
  20. Oct 27, 2008 #19
    i have question on the same problem.
    does this mean the answer is sqrt[(2Mgx - kx^2)/m]
    can someone please check my answer thanks so much
     
  21. Oct 27, 2008 #20

    Are the height for the gravitational potential energy and the change in position for the spring energy the same? How come?

    Thank you so much for your help.
     
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