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Kinetic energy and potential energy

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data
    IMG_3785.jpg
    the diagram shows two bodies X and Y connected by a light cord passing over a light, free-running pulley. X starts from rest and moves on a smooth plane inclined at 30 degrees to the horizontal.
    What will be the total kinetic energy of the system when X has travelled 2.0m along the plane?(g=9.8m/s2)


    2. Relevant equations
    E=mgh


    3. The attempt at a solution
    my teacher ask me to find the perpendicular distance of y and x, then use formula E=mgh to find the differences between potential energy between X and Y, and the differences will equal to the total kinetic energy since energy is conserved during the whole motion
    BUT, i cant even find the perpendicular distance of X and Y




    1. The problem statement, all variables and given/known data
    A stationary thoron nucleus (A=220, Z=90) emits an alpha particle with kinetic energy Ea. What is the kinetic energy of the recoiling nucleus?


    2. Relevant equations



    3. The attempt at a solution
    i totally have no idea about it.


    1. The problem statement, all variables and given/known data
    Particles X (of mass 4 units) and Y ( of mass 9 units) move directly towards each other, collide and then separate. If [tex]\lambda[/tex]Vx is the change of the velocity of X and [tex]\lambda[/tex]Vy is the change in velocity of Y, the magnitude of ratio [tex]\frac{\lambdaVx}{\lambdaVy}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    well the answer state is [tex]\frac{9}{4}[/tex]
    but i get [tex]\frac{2}{3}[/tex], just by rationing both magnitude right?
     
    Last edited: Oct 21, 2009
  2. jcsd
  3. Oct 21, 2009 #2

    tiny-tim

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    Hi look416! :smile:

    When Y moves down vertically 1 m, how much higher does X get? :wink:
     
  4. Oct 21, 2009 #3
    but in this case does we assume that Y is moving down 1m?
    ops, i think i have found the height of y,
    but i still couldnt find the perpendicular height of x
     
    Last edited: Oct 21, 2009
  5. Oct 21, 2009 #4
    just to push to thread up
     
  6. Oct 22, 2009 #5

    tiny-tim

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    (just got up … :zzz:)

    Do you mean "vertical"? (But height is vertical anyway)

    The height of x is the length along the slope, multiplied by the sine of the angle, isn't it?

    What's worrying you about that?​
     
  7. Oct 22, 2009 #6
    lolz you just reminded me of that
    btw it just afternoon for here
    ok i will try it ltr as im in cc now
     
  8. Oct 22, 2009 #7
    tiny-tim
    for your information i have just added another 2 questions,
    so could you teach me how to do it?
     
  9. Oct 22, 2009 #8
    push push
     
  10. Oct 22, 2009 #9

    tiny-tim

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    (It would better to start a new thread)

    An alpha particle has 4 nucleons (ie protons and neutrons, with the same mass), and the thoron nucleus has 220 nucleons, ie 55 times as massive.
    ?? :confused: Show us your calculations.
     
  11. Oct 22, 2009 #10
    Still cant get it very well,
    does that mean to calculate the kinetic energy of recoiling nucleus?
    Btw, could you explain what is recoiling nucleus?
    which is the one it refers to?
    the thoron nucleus or the alpha particle?


    For another question, here is how i did it
    by here i assume that after they collide particle X and particle Y move in the direction that oppose the direction before they collide (elastic collision) and considering the direction to the right is positive.
    therefore for paricle X
    mxu1-mxv1=0

    and for particle y
    my-u2+myv2=0

    after that i extracted it out
    mx(u1-v1)

    my(-u2+v2)

    the ratio between velocity of X and Y
    = [tex]\frac{{mx(u1-v1)}}{{my(-u2+v2}}[/tex]

    but they want the magnitude of ratio of change in speed so by putting the value of mass X and Y into the equation and pull it out
    [tex]\frac{4}{9}[/tex]
    then simplify it then i get [tex]\frac{2}{3}[/tex]
     
  12. Oct 23, 2009 #11

    tiny-tim

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    Hi look416! :smile:

    (in LaTeX, mx is m_x, and mx1 is m_{x1} :wink:)
    the thoron nucleus … it ejects part of itself, so it recoils, just like a gun recoils when it ejects a bullet :wink:
    You're only using conservation of momentum, and that applies to all collisions …

    you don't need it to be elastic

    (elastic is only necessary for conservation of energy)
    I don't understand … how exactly did you get 2/3 ? :confused:
     
  13. Oct 23, 2009 #12
    well i just take the [tex]\frac {m_{x}}{m_{y}}[/tex] out and simplify it
    thats all
     
    Last edited: Oct 23, 2009
  14. Oct 23, 2009 #13

    tiny-tim

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    take it out of what? simplify how? :confused:
     
  15. Oct 23, 2009 #14
    [tex]\frac {m_x}{m_y}[/tex] is the coefficient of the ratio change between Vxand vy, since mx values 4 units and my values 9 units then it turns out to be [tex]\frac {4}{9}[/tex], so i just simplify it
     
    Last edited: Oct 23, 2009
  16. Oct 23, 2009 #15
    about the second question
    i still cant figure it out sorry for im stupid:tongue2:
    which equation are we using
    kinetic equation which resembles
    [tex]\frac{1}{2}[/tex]mu2+[tex]\frac{1}{2}[/tex]mu12=[tex]\frac{1}{2}[/tex]mv2+[tex]\frac{1}{2}[/tex]mv12
     
  17. Oct 24, 2009 #16

    tiny-tim

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    oh I see … but 6/9 is 2/3.

    get some sleep! :zzz:​
    (btw, the element with Z = 90 is thorium … thoron was an old name for one of the isotopes of radon :wink:)

    Although energy is conserved, you don't know the exact mass of the decay product (A = 216, Z = 88), and a very small error in the m makes a big difference in mc2, which is what you need in the conservation of energy equation.

    So instead, use conservation of momentum (which of course applies to all collisions), to find the momentum, and then use m = 216 to find the kinetic energy (the small error in m won't make much difference here).
     
  18. Oct 24, 2009 #17
    lolz its pretty obvious im lack of sleep haha
    for the question regarding thoron, thx ya for your little info
    and i have got the answer hurray

    what abt the question regarding the two particles hoho?
    i am still circling inside it
     
  19. Oct 24, 2009 #18

    tiny-tim

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    I suspect you're keeping the equation in your head instead of bothering to write it out properly:

    if ∆x and ∆y are the changes in the velocities of x and y,

    then mxx = -myy,

    so |mx/my| = … ? :smile:
     
  20. Oct 24, 2009 #19
    oic now i got it
    yea
    so happy to find out the answer
     
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