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Homework Help: How far does arrow penetrate into board

  1. Jan 20, 2008 #1
    [SOLVED] How far does arrow penetrate into board

    1. The problem statement, all variables and given/known data
    An arrow is fired with a speed of 21.0 m/s at a block of Styrofoam resting on a smooth surface. The arrow penetrates a certain distance into the block before coming to rest relative to it. During this process the arrow's deceleration has a magnitude of 1560 m/s^2 and the block's acceleration has a magnitude of 450 m/s^2.
    a)How long does it take for the arrow to stop moving with respect to the block?
    b)What is the common speed of the arrow and block when this happens?
    c)How far into the block does the arrow penetrate?

    2. Relevant equations
    x=Xi+Vi*t+ 1/2at^2
    x=0+21m/s (1.04x10^-2 s)+ 1/2 (-1560m/s^2)(1.04*10^-2 s)^2

    3. The attempt at a solution

    I got parts a and b. But part c is giving me problem. I tried using the equation above to get the final position of the arrow but the answer was wrong. Am I using the wrong equation. Or is it a math error?
  2. jcsd
  3. Jan 20, 2008 #2


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    Is the deceleration of the arrow (-1560 m/s^2) with respect to the block or some common reference point?
  4. Jan 20, 2008 #3
    Unfortunately, the problem doesn't specify that. I'm guessing since the problem gives the deceleration of the arrow while saying "during this process" it might be with respect to the block.
  5. Jan 20, 2008 #4
    if it is wid respect to the block...then use conservation of momentum ....
    go ahead..
  6. Jan 21, 2008 #5
    ok. But just out of curiosity, what if it was just with respect to a common point. What would I do then?
  7. Jan 21, 2008 #6
    looking over conservation of momentum, i don't think that's how i need to solve the problem seeing as how I haven't been taught that. Besides that, I have no mass included in the problem.

    I'm pretty sure I have to use one of the equations that relate velocity, time, position and acceleration. I used one of them( see above), but got the wrong answer.
  8. Jan 21, 2008 #7
    try using this formula
    x = x0 + 1/2 (v0 + v) t
  9. Jan 21, 2008 #8
    that one worked. thanks amarett0!
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