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Homework Help: Stopping a bullet

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data

    A .22 rifle bullet, traveling at 357 m/s, strikes a block of soft wood, which it penetrates to a depth of 0.131 m. The block of wood is clamped in place and doesn't move. The mass of the bullet is 1.65 g. Assume a constant retarding force.

    What is the average speed of the bullet while it is being stopped by the wood?

    2. Relevant equations

    F=m(a)
    V=d/t
    d=x_o + V_o*(t) + 1/2at^2

    3. The attempt at a solution

    I thought the average speed while it is beings stopped would just be the 357 m/s but its not :(

    any help on how to find this? i dont really like the wording of this question. it's kind of confusing to me.

    Thanks
     
  2. jcsd
  3. Feb 6, 2010 #2
    Since the retarding force is constant, the deceleration is constant as well. What is the average speed for a constant acceleration (deceleration)? (357 m/s would definitely not be the average speed. That is the initial speed, while the final speed is 0 m/s)

    Also, note that:

    [tex]
    \vec{F}t=\Delta \vec{p}
    [/tex]

    where [tex]\vec{p}[/tex] is momentum.
     
  4. Feb 6, 2010 #3
    So i found the time using the equation:
    [itex]
    d = \frac{V_i + V_f}{2)} * t
    [/itex]
    0.131 = 357/2 * t

    t=0.00073389 seconds

    then i found the speed using the equation:

    speed = distance/t

    speed = 0.131/0.00073389

    speed = 178.5 m/s

    Does this look correct?

    Thank you
     
    Last edited: Feb 6, 2010
  5. Feb 6, 2010 #4
    Yep. Looks fine.

    You could've solved it a bit quicker though. For constant acceleration:

    [tex]
    d = v_{avg}t
    [/tex]
     
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