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One-Dimensional Motion, Bullet-through-a-board-type question

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data

    36 g bullet, speed of 350 m/s strikes a 8 cm fence post. It is retarded by an average Force is 3.6 x 10[tex]^{6}[/tex] N while going through the post.

    a. Speed of bullet when it emerges?
    b. How many boards could the bullet penetrate?


    2. Relevant equations

    v[tex]^{2}[/tex]=v[tex]^{2}_{0}[/tex]+2a[tex]\Delta[/tex]x
    F=ma
    collision equations, maybe?

    3. The attempt at a solution

    I took the equation F=ma and plugged in (3.6x10[tex]^{6}[/tex])=(.03 kg)a and then got a=1.0x10[tex]^{8}[/tex] m/s[tex]^{2}[/tex].
    I then plugged that into this equation, v[tex]^{2}[/tex]=v[tex]^{2}_{0}[/tex]+2a[tex]\Delta[/tex]x and got
    v[tex]^{2}[/tex]= (350 m/s)[tex]^{2}[/tex] + 2(1.0x10[tex]^{8}[/tex] m/s[tex]^{2}[/tex])(.08 m)
    but the v I calculated is a bigger velocity than when the bullet started.
    so...what is going wrong here? thank you so much in advance :)
     
  2. jcsd
  3. Oct 30, 2007 #2
    Given an x-axis,

    -------------------------------->x
    Vector v, that is the speed of the bullet, goes along with x. The force F that retards(sp?) the bullet is in the opposite direction. In your equations, you're assuming F is positive.
     
  4. Oct 30, 2007 #3
    Ah, it's always the positives and negatives. Thanks.

    for part b, I am just going to use the same equation, with the delta x as the variable to be found, setting the final velocity to 0?
    so, setting that up, it would be

    v[tex]^{2}[/tex]=v[tex]^{2}_{0}[/tex]+2a[tex]\Delta[/tex]x

    0= (350 m/s)[tex]^{2}[/tex] - 2(1.0 x 10[tex]^{6}[/tex]) ([tex]\Delta[/tex]x)

    and then using that number, dividing it by the 8 cm known of the thickness of the post, ending up with the answer...
     
  5. Oct 30, 2007 #4
    Yeah, your logic is right. You can always use the other formulas for accelerated motion, but that's the easy way out.
     
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