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Suspended Particle Questions

  1. Mar 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Particle A and B are attached by an inextensible string with masses 0.5 and 0.4 respectively

    10N
    ^
    |
    {a}
    |
    |
    {b}

    take a to be 9.8 downwards

    When the 10N is applied the particles move upwards. Ginf the tension in the lower string and the acceleration of the system.

    2. Relevant equations
    Kinematic Ones
    Ignore air resistance

    3. The attempt at a solution

    A:
    10 - T = 0.5a

    B:T + 10 = 0.4a


    BUT i think straight away this is wrong as what about the efffect of gravity???

    Thanks
     
    Last edited: Mar 29, 2008
  2. jcsd
  3. Mar 29, 2008 #2

    tiny-tim

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    Hi thomas49th! :smile:

    Hint: work out the acceleration of the whole system (the tensions won't matter, since they're internal).

    Then work out the tensions! :smile:
     
  4. Mar 30, 2008 #3
    yeh i got it in the end.

    but now i have a harder question :(

    Two particles P and Q of masses 3kg and 6kg respectively are attached to the ends of a light inextensible string. The string passes over a smooth pulley. The system is released from rest wit both masses a distance of 2m above the horizontal floor. Find how long it takes for particle Q to hit the floor. Assuming particle P does not reach the pulley find its greatest height above the floor.

    The time was easy to find (1.11s)
    but i dont know how to find the maximum height particle P reaches. I would of though 4 metres (2+2) but it must be travelling at such a speed that it continues after particle Q has hit the floor.

    Can someone please so me how to do the second part please.
    Thanks :)
     
  5. Mar 30, 2008 #4

    tiny-tim

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    Hi thomas49th! :smile:

    Hint: speed of P at 4m must be same as speed of Q at 0m.

    From then on, P is just decelerating under gravity.

    So … ? :smile:
     
  6. Mar 30, 2008 #5
    errr Q at 0m must be

    V= 0 + (1/3)(g)(1.1)
    = 3.629....
    and then use
    s= ut + 0.5at²

    where u = 3.629..

    a = (1/3)(9.8) and t = 1.11111..
    but that gives me 6. point somthing when the answer is 4.67 is the back off the book :(

    where is my error(s)?

    Cheerz :)
     
  7. Mar 30, 2008 #6

    tiny-tim

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    … conservation of energy …

    Hi thomas49th! :smile:

    Not sure what you've done …

    But shouldn't a be the full 9.8 for P on its own?

    And why not use conservation of energy? :smile:
     
  8. Mar 31, 2008 #7
    ill try this step by step:

    when Q hits the ground it's v becomes P intial velocity (u)
    the accleration of the system, is a 1/3 of g because i worked it out using f = ma OR does it change back to the orginal 9.8??? Can you explain.

    Thanks
     
  9. Mar 31, 2008 #8

    tiny-tim

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    … new ball-game … !


    Yes! :smile:
    Forget the original acceleration … this is a completely new ball-game!

    There's nothing moving except P … and the string is completely slack (no tension, so no force).

    So it's the same as if P was in free-fall (which it is!).

    So you would definitely use g.

    But if you use conservation of energy, the acceleration won't come into it, will it? :rolleyes:
     
  10. Mar 31, 2008 #9
    how can i apply the conservation of energy

    energy before = energy after

    in class i dont think we've used the conservation of energy apart from calculations involving elastic collsions between particles.

    Thanks :)
     
  11. Mar 31, 2008 #10

    tiny-tim

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    ah … so you haven't done PE + KE = constant?

    (potential energy + kinetic energy = constant.)
     
  12. Mar 31, 2008 #11
    we've dont avbout PE and KE being transfered, but that is in the m2 silbeius
    this is the m1 silibus, thus easier.

    I tried using a = 9.8 but my answer was wrong:

    a = 9.8
    u = 3.629
    t = 1.11
    the plug them into s = ut + 0.5at²

    and i get 10.069...
    so then i tried swapping the sign around for a and i got -2.005
    :S Im stuck
    Where next?

    Thanks
     
  13. Mar 31, 2008 #12

    tiny-tim

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    ok, we'll do it the way in the m1 syllabus! :smile:
    erm … where did that come from?? :rolleyes:
     
  14. Mar 31, 2008 #13
    well the first part of the question was to find the time for particle Q to hit the floor.

    Tht was
    Q:
    6g - T = 6a

    P: T + 3g = 3a

    so combine and rearrange giving a to be 1/3(9.8) - can you see why i was using that

    then stick it in s = ut + 0.5at² giving t = 1.11s
    :)
     
  15. Mar 31, 2008 #14

    tiny-tim

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    Yes I know! … but what's it doing here?

    1.11 is the time for Q to hit the ground. What has that to do with the time for P to reach the top?

    (There is a relationship, obviously … but are they the same?)
     
  16. Mar 31, 2008 #15
    well do i need to find the time using v = u+at
    t = u/a EDIT(v = 0)
    then put this into s = ut + 0.5at²
    giving me 2.6786.. then add this onto 2 metres giving me 4.69 which is close, but i have rounded when using u = 3.629 coudl this of made that 0.2 of a difference

    Thank you ;)
     
  17. Mar 31, 2008 #16
    I wouldn't get into the habit of using different decimal places for each part of any equation, that'll likely make your answer off, sometimes significantly. If the answer is to 2dp, I'd calculate everything to 3dp and then round up to give the answer to 2dp's. Try it for yourself, see if you can get the answer by being consistent with rounding.
     
  18. Mar 31, 2008 #17
    Noooooooooooooooooooooo i get 1.9999 when not rounding and that + 2 = 3.9999. WHAT? I dont know how to solve this question. Can you walk me through please?
    Thanks :)
     
  19. Mar 31, 2008 #18

    tiny-tim

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    Hi thomas49th! :smile:

    Looks bad. :frown:

    Show us your full working (starting from t = 1.106, u = 3.614), so we can see what went wrong. :smile:
     
  20. Mar 31, 2008 #19
    so t = [tex]\sqrt{\frac{12}{\frac{9.8}{}}}[/tex]

    i found v of Q when it hits the ground using
    v = u + at
    = [tex]0 +\frac{9.8}{3}[/tex]

    then v of Q would be u of P

    so u = 3.6147...
    then i need to find t using v = u + at
    t = v - u / a NOTE that v will be 0 at greaest height
    = u / a
    = -3.6147 / -9.8 (are the signs right?)
    = 0.36884...
    s = ut + 0.5at²
    i get
    s = 3.6147...(0.36884...) + 0.5(-9.8)(0.36884...)²

    and i get the wrong answer :( Where have i gone wrong!
     
  21. Mar 31, 2008 #20

    tiny-tim

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    oh thomas49th … why didn't you put the last line in?

    It's 1.3333 - .6666 = .6666.

    For some reason, you added, despite your own minus-sign! :cry:

    Incidentally, you could also have "run the film backwards", and just used u = 0, which would have given the same result! :smile:

    (Now why should that be? :rolleyes:)
     
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