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Bullet Force

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A bullet of mass 0.0022 kg initially moving at 504 m/s embeds itself in a large fixed piece of wood and travels 0.72 m before coming to rest. Assume that the acceleration of the bullet is constant.

    What force is exerted by the wood on the bullet?

    F=N

    2. Relevant equations

    Ok, so I need to find acceleration to plug into the equation for force, which is F=m*a. The equation I am using is x=x0 + v0*t + .5*a*t^2.



    3. The attempt at a solution

    x=x0 + v0*t + .5*a*t^2 so this equation need to be set to solve for a. This is what I am plugging in:

    x0 = 0
    v0 = 504
    t = 0.0014 (from d/s = t)

    My answer is a = 0.706

    The equation vf^2 = v0^2 + 2a (delta x), and set to solve for a but I am a bit confused about delta x? In this equation, vf = 0.72 and v0 = 504

    I am either having trouble setting up the first equation for a, or I am using the wrong one in which case, I am not sure by what is meant by delta x in the second. Some direction is appreciated!
     
  2. jcsd
  3. Oct 11, 2008 #2
    It was looking good, but you said this:

    "x0 = 0
    v0 = 504
    t = 0.0014 (from d/s = t)"

    This assumes constant velocity, which it's not.

    You have to use this formula instead : [tex] v^2 = u^2 + 2as [/tex] where v = final velocity
    u = initial velocity
    a = acceleration
    s = displacement.

    Solve for acceleration, then multiply by mass.
     
  4. Oct 11, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    You could use this method if you found the right time. To find the time you need to use the average speed, not the initial speed.

    This equation is a better choice, but 0.72m is the distance (Δx) not the final speed. (The final speed is zero, of course.)
     
  5. Oct 12, 2008 #4
    Thanks for the help!

    I got the correct answer of 388.08 N by taking the equation v^2=u^2 + 2as and solving for a and then multiplying by mass. Both of your suggestions were very helpful in helping me to understand these concepts better!

    :smile:
     
  6. Oct 12, 2008 #5

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    It often is helpful in this kind of problem to use a graph. A graph of speed vs. time makes it very easy. The slope is the acceleration (constant -) and the distance is the area under the graph.

    AM
     
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