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Two marbles colliding elastically on massless strings

  1. Jul 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Two marble spheres of masses 30 and 20 grams, respectively, are suspended from the ceiling by massless strings. The lighter sphere is pulled aside, as shown in the diagram, through an angle of 75° and let go. It swings and collides elastically with the other sphere at the bottom of the swing.
    a. To what maximum angle will the heavier sphere swing?
    b. To what maximum angle will the lighter sphere swing?

    -the length of the string is 95cm, see the diagram attached

    2. Relevant equations

    M=30g, V= speed of M, θ1 =angle of M
    m=20g, v= speed of m, θ2 =angle of m

    Law of conservation of kinetic energy:
    1/2mv^2 + 1/2MV^2 = 1/2mv'^2 + 1/2MV'^2

    Law of conservation of momentum:
    x component: mv(cos75) =mv'(cosθ2) + MV'(cosθ1)
    y component: mv(sin75) =mv'(sinθ2) + MV'(sinθ1)


    3. The attempt at a solution

    First I tried to find the velocity of m after it was released with the equation:
    v =sqrt(2gLsinθ)
    and got 4.24m/s

    Then I plugged that into the law of conservation of KE and got:
    1/2(20g)(4.24^2) = 1/2(20g)v^2 + 1/2(30)V^2
    179.776 = 10v^2 + 15V^2

    Then I plugged in the variables for the momentum equations:
    x component: 20(4.24)(cos75) =20v'(cosθ2) + 30V'(cosθ1)
    y component: 20(4.24)(sin75) =20v'(sinθ2) + 30V'(sinθ1)

    I have tried all sorts of algebra to solve for the unknowns but keep getting complex equations that seem useless. Any help would be greatly appreciated!
     

    Attached Files:

    Last edited: Jul 22, 2012
  2. jcsd
  3. Jul 22, 2012 #2
    For conservation of momentum, the values should be calculated just after the collision, θ=0.
     
  4. Jul 22, 2012 #3
    After release the potential energy in the initial position is being converted into kinetic energy. The two balls then collide and the kinetic energy is shared.

    How is the KE and PE related? Are there any points where the relationship between KE and PE is easy to solve?
     
  5. Jul 22, 2012 #4
    So if I was to use my momentum equations from before in this way I would get

    x component: 20(v)(cos75) =20v'(cos0) + 30V'(cos0)
    20(v)(cos75) =20v' + 30V'

    y component: 20(v)(sin75) =20v'(sin0) + 30V'(sin0)
    20(v)(sin75) = 0 ?

    I am a little confused. Do I use the initial velocity of the smaller mass (4.24m/s?) to figure out momentum before the collision? Or do I use the above equations to calculate the marbles momentum after the collision and ignore the initial velocity of the smaller marble?
     
  6. Jul 22, 2012 #5
    KE and PE are related through the equation PE = KE, mgh=1/2mv^2 (is that right in this case?). That is what I used to find the initial velocity of the 20g marble. But I am not sure if I needed to do that? I am also unsure of how else I could use the equation mgh=1/2mv^2.
     
  7. Jul 22, 2012 #6
    PE+KE = total energy = constant as there is no friction or other loss mechanism to consider.

    Where is PE highest? Where is PE lowest? Same questions for KE
     
  8. Jul 22, 2012 #7
    But how will I be able to find angles from this?
     
  9. Jul 22, 2012 #8

    Doc Al

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    Staff: Mentor

    Try this. How fast is the lighter marble moving just before it hits the heavier marble? Then use the conservation laws to find their speeds immediately after the collision.
     
  10. Jul 22, 2012 #9
    To Doc Al:
    I have figured out the velocity of the 20g marble is 3.715m/s just as it is about to hit the 30g marble using h= L- sqrt(L^2-x^2) and PE=KE.

    I know that I have to use the law of conservation of KE and momentum to figure out the rest, but, I do not know when I should be breaking things into x and y components of momentum.
    I have been trying to use these formulas:

    x: mvcos(75)= mv'cosθ + MV'cosθ
    y: mvsin(75) = mv'sinθ + MV'sinθ

    KE law: 10v^2 = 10v'^2 + 15V^2
    where v = 3.715m/s
    Should I have broken this question into x and y components like this?
     
  11. Jul 22, 2012 #10

    Doc Al

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    Staff: Mentor

    At the point of collision, the marbles are moving horizontally. No need for anything but a single direction.
    No. The 75° angle was the initial position of the marble. It has nothing to do with the collision.
     
  12. Jul 22, 2012 #11
    Okay, thanks!

    So this is what I came up with:

    v= 3.715m/s (From my previous post) m=20g
    V=0 M=30g

    momentum:
    mv=MV' + mv'
    20g(3.715m/s) = 30gV' + 20gv'
    74.3 = 30V' +20v'

    KE:
    1/2mv^2 = 1/2mv'^2 +1/2MV'^2
    10(3.715)^2 = 10v'^2 + 15V'^2 (sqrt root both sides)
    37.15 = 10v' + 15V'
    v' = (37.15 - 15V')/10

    But whenever I substitue v' into the momentum equation (those parts are in red). everything cancels out and I get v' = v'.....
     
  13. Jul 22, 2012 #12

    TSny

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    Homework Helper
    Gold Member

    Check that step.
     
  14. Jul 22, 2012 #13
    Ok, I have changed that to:

    10(3.715)^2 = 10v'^2 +15V'^2

    sqrt(138) = sqrt(10) v'^2 + sqrt(15) V'^2

    And when I solved for V' I got -0.000159? I think I messed up again,
    I apologize for my rusty algebra skills...
     
  15. Jul 23, 2012 #14
    Do I have the right steps though? Or are my steps and math wrong?

    (thank you for all your help so far!)
     
  16. Jul 23, 2012 #15

    TSny

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    Gold Member

    It looks like the algebra is giving you a little bit of difficulty. When you take the square root of the right hand side of the equation you should actually get
    [itex]\sqrt{10v'^{2}+15V'^{2}}[/itex]

    This cannot be simplified. In particular [itex]\sqrt{a+b}[/itex][itex]\neq[/itex][itex]\sqrt{a}[/itex]+[itex]\sqrt{b}[/itex]. So, taking the square root of both sides does not help.

    A good approach is to go back to your momentum equation 74.3 = 30V' +20v' and solve it for v' in terms of V'.
    Then substitute this expression for v' into the energy equation 10(3.715)2 = 10v'2 +15V'2. It's a bit messy, but you should then be able to solve for V'.
     
  17. Jul 24, 2012 #16
    Awesome!! I finally got values for the velocities. V'= 2.972m/s, v'=-0.743m/s.

    So this means the large marble moves forwards with ~3m/s and the small marble goes back the way it came with ~1m/s?

    Then I can find the heights of each marble after the collision:
    small: 1/2mv^2 =mgh +1/2mv'^2
    10(3.715)^2 = 20g(9.80)h + 10(0.743m/s)^2
    h= 0.676m

    large: 1/2mv^2 =mgh
    15(2.972)^2 = 30g(9.8)h
    h=0.451m

    Can I then use the formula h= sqrt(0.95^2 -x^2) to find the horizontal distances the marbles travel (see the diagram again). and then use the length of the string and x to find the angle?

    Ok, I attached a diagram showing how I used x and L to find theta for each marble. Is this logical?

    Thanks for takin the time :)
     

    Attached Files:

    Last edited: Jul 24, 2012
  18. Jul 24, 2012 #17
    redraw the diagram of the final positions with all the heights that you know and all the lengths that you know.

    Does it look simpler now?

    Hint try to make a 1/10th scale drawing , don't forget to use a compass.
     
  19. Jul 24, 2012 #18
    Pheobz, I can see you have added a triangle and want to calculate a horizontal displacement. But do you need that horizontal displacement, since its not part of the answer you have to give?


    What lengths do you know? What heights do you know? What do you know about the shape of the paths of the two marbles? where are your heights measured from?

    If you don't have a compass use a bit of string tied around a pencil. If the algebra/ trigonometry is unclear then often the best way to see a way forward is to make a drawing. Geometry really can help you here.
     
    Last edited: Jul 24, 2012
  20. Jul 24, 2012 #19

    TSny

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    That looks right!

    Yes.

    No. Here you want to think about a new problem where the small mass starts at the lowest point with a velocity of -0.743 m/s and swings up to it's max height h. In fact, you want to do the same calculation that you did below for the large mass.

    Yes, this is how you should have also done the small mass.

    [Note added: Strictly speaking, you should be converting grams to kg if you're working in the SI system of units. However, since the mass occurs in each term of your equations, you can get by without converting.]
     
  21. Jul 24, 2012 #20
    I found the horizontal displacement so that I could find the angle theta, I found 'x' using the formula h= L- sqrt(L^2-x^2). Then I went sin(theta) =x/L to find the angle for each marble, does this work?

    And thank you TSny, I think I have the right value for the small marble now, 0.028m!
     
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