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A conservation of mechanical energy and momentum problem

  1. Jun 3, 2014 #1
    1. The problem statement, all variables and given/known data
    a 6-kg bullet is fired into a 2-kg block initially at rest at the edge of a table of height 1 m. The bullet remains in the block, and after the impact the block lands 2 m from the bottom of the table. Determine the initial speed of the bullet.


    2. Relevant equations
    m1v1i + m2v2i = (m1 +m2)bf

    1/2mvi2 + mgyi = 1/2mvf2 +mgyf

    3. The attempt at a solution
    I attempted to do this problem by first putting the information into the equation for the conservation of momentum. So (.006)v1i +0 = (2.006)(vf). Unfortunately this equation has two unknowns. I know that in order to substitute for the unknowns I must use the formula for the conservation of mechanical energy. I don't know how to do this based on the information provided. Any help would be greatly appreciated.
     
  2. jcsd
  3. Jun 3, 2014 #2

    Nathanael

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    I think that's a typo, or it's just a huge bullet haha

    Not a problem, because you can figure out what [itex]V_{f}[/itex] has to be.

    There's only one [itex]V_{f}[/itex] that satisfies the constraints of the problem (moving 2 meters horizontally in the time it takes to fall 1 meter)

    So the only unknown is the one you're solving for. (But first you need to write another equation and solve for [itex]V_{f}[/itex])
     
    Last edited: Jun 3, 2014
  4. Jun 4, 2014 #3
    How do I find [itex]V_{f}[/itex] based off the fact that it moves two meters horizontally in the time it takes to fall one meter? And its supposted to be grams my bad.
     
  5. Jun 4, 2014 #4

    CWatters

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    The time taken to do both is the same.
     
  6. Jun 4, 2014 #5
    How does that knowledge help me solve this problem?
     
  7. Jun 4, 2014 #6

    BiGyElLoWhAt

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    Until the block actually slides off the table, the vertical component of the velocity is 0. can you solve for the time it takes an object to hit the ground if it starts from rest at a height of 1m?
     
  8. Jun 4, 2014 #7
    Yes I do according to my calculations it would take .45 seconds to hit the ground. Now what do I do?
     
  9. Jun 4, 2014 #8

    Nathanael

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    Find the horizontal speed necessary to move 2 meters in 0.45 seconds.

    Does that speed play any special role in this problem?
     
    Last edited: Jun 4, 2014
  10. Jun 5, 2014 #9

    CWatters

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    See replies from others.

    Looks like you should get some practice with ballistic problems. You know the sort of thing... a projectile is thrown at an angle Θ to the ground with an initial velocity u. How far does it go? That sort of thing.
     
  11. Jun 5, 2014 #10
    OK what I did now was realize that if it moved 2 meters in .45 seconds it must be traveling at a rate of .225 m/s. Then I plugged that in to the original equation for the final velocity
    (.006)v1f = (m1+m2)(.225)
    I found the answer to be 75.225 which is clearly wrong. What did I do wrong?
     
  12. Jun 5, 2014 #11

    Nathanael

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    If it moves 0.225 meters in 1 second, (0.225 m/s) then how does it move 2 meters in 0.45 seconds?

    (You divided 0.45 by 2 instead of 2 by 0.45)
     
  13. Jun 5, 2014 #12
    I see my mistake. Whoops! Thanks I got it right now!
     
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