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A bullet pushing a knot out of a wood block

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  1. Sep 29, 2016 #1
    1. The problem statement, all variables and given/known data
    You shoot a 0.0050-kg bullet into a 2.0-kg wooden block at rest on a horizontal surface (Figure 1) . After hitting dead center on a hard knot that runs through the block horizontally, the bullet pushes out the knot. It takes the bullet 1.0 ms to travel through the block, and as it does so, it experiences an x component of acceleration of -4.9 × 105 m/s2. After the bullet pushes the knot out, the knot and bullet together have an x component of velocity of +10 m/s. The knot carries 10% of the original inertia of the block.
    What is the initial velocity of the bullet?
    2. Relevant equations
    Conservation of momentum equation
    pi=pf
    m1v1i+m2v2i=m1v1f+m2v2f

    3. The attempt at a solution
    v1f= (0.2)(10)+0.005(10)/ 0.005
    =410m/s
    the 0.2 came from the 10% of 2kg wooden block (knot)
    this answer is incorrect but i am unsure where my error is and what simple step i have missed
     
  2. jcsd
  3. Sep 29, 2016 #2

    Bystander

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    Does the momentum of the block change?
     
  4. Sep 29, 2016 #3
    no the momentum of the block would not change since its still at rest, only the knot moves
     
  5. Sep 29, 2016 #4

    PeroK

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    Do you know the required answer?

    In any case, I suspect the rest of the block does move, and as you don't know its speed you can't use conservation of momentum.
     
    Last edited: Sep 29, 2016
  6. Sep 29, 2016 #5
    no i don't know the final answer. if i do not use conservation of momentum, will i be using something with the acceleration
     
  7. Sep 29, 2016 #6

    PeroK

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    That's not a bad idea.
     
  8. Sep 29, 2016 #7
    how though? f=ma?
     
  9. Sep 29, 2016 #8

    I like Serena

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    Hey Emily! ;)

    There's quite some distracting information there.
    But let's see...
    Initially the bullet has some unknown speed ##v_0##.
    Then it was decelerated with ##-4.9 × 10^5\text{ m/s}^2## during ##1 \text{ ms}##.
    And afterwards it has speed ##10\text{ m/s}##.
    Is that sufficient information? :sorry:
     
  10. Sep 29, 2016 #9
    i am now confused as to what approach to take
     
  11. Sep 29, 2016 #10

    I like Serena

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    How about ##v=v_0 + at## (since we have constant deceleration)?
     
  12. Sep 29, 2016 #11
    so to rearrange for
    vi=vf/at
    =10/(−4.9×105 m/s2)(0.001s)
    =0.02
    i don't think i did this correctly because the bullet would need to travel much faster then that
     
  13. Sep 29, 2016 #12

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    I agree. The bullet must travel much faster than that - at least more than 10 m/s.
    Perhaps the rearrangement isn't quite correct?
     
  14. Sep 29, 2016 #13
    you are right my rearrangement was incorrect
     
  15. Sep 29, 2016 #14
    i divided when i should've subtracted, that means the vi=500m/s which is more realistic and correct. thanks
     
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