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    (Original post by Electric-man7)
    Can someone explain the circle question? What was the final answer?
    8 and something else but it pretty much came downto your ability to make a good sketch to gain some insight
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    Hmmm 8 and something is suspiciously close to sqrt69, I'll assume it sqrt69 so I can have some sense of relief.
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    (Original post by Electric-man7)
    Hmmm 8 and something is suspiciously close to sqrt69, I'll assume it sqrt69 so I can have some sense of relief.
    Noo one solition was 8 and the other some other number smaller than 8
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    Wait redoing the problem I got sqrt61 and sqrt69
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    what were the radii and centres of the two circles?
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    (Original post by Quantum42)
    what were the radii and centres of the two circles?
    (-3,2) r=1,5 (5,1) r=2,5
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    Circle question: There are THREE tangents. One of them had the length 1 (sketch the circles and you will see that y axis is ALSO a tangent.

    Parametrics question - They, I think, are expecting a GENERAL solution. Hence, there are 2 SETS of values, not JUST 2 values. (That is, omega*t can take the values pi/6, 2pi pi/6, 4pi pi/6......and pi - pi/6, 3pi - pi/6, 5pi - pi/6.....)

    Yes. The spring was super easy. So was the trigonometry question. (The answers above are correct, imo)

    The answer to the stars question: v3 = sqrt(sqrt(3))*2*v2. (Yes, there is sqrt IN sqrt)
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    The parachute question: They asked to Calc terminal velocity, which is easy - F = mdv/dt = 0. That will yield v terminal. However, they had asked to calculate work done by resistive force throughout the journey.
    I did Net Work = Work by resistance + Work by gravity = Change in KE. Till there it's okay. But.... What do I put for change in KE?! I know the FINAL velocity but not the initial. I assumed it to be zero BUT that's just crazy. They said the parachute "JUMPED" off the plane!! They could have said "DROPPED"! Anyway, I used initial v = 0 and solved that equation to get the answer.
    What did u guys do?
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    (Original post by Electric-man7)
    Can someone explain the circle question? What was the final answer?
    8 and root 60 iirc
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    Wait. I think the centers of the circles were (-3,2) and (5,1) and the radii were 3 and 5 respectively. (The y Coordinates of the radii maybe interchanged. I don't rem exactly)
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    You are right, I made a stupid mistake about the circle question. how many marks will I lose?
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    Wait, who is right?! Me or the other person? Please quote!
    The circle question was worth 9 marks I think.
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    (Original post by cooltejaskd)
    Circle question: There are THREE tangents. One of them had the length 1 (sketch the circles and you will see that y axis is ALSO a tangent.

    Parametrics question - They, I think, are expecting a GENERAL solution. Hence, there are 2 SETS of values, not JUST 2 values. (That is, omega*t can take the values pi/6, 2pi pi/6, 4pi pi/6......and pi - pi/6, 3pi - pi/6, 5pi - pi/6.....)

    Yes. The spring was super easy. So was the trigonometry question. (The answers above are correct, imo)

    The answer to the stars question: v3 = sqrt(sqrt(3))*2*v2. (Yes, there is sqrt IN sqrt)
    Actually that was my solution, but this is actually wrong. You probably took the distance for the gravitational force as being from the center of the orbit instead of the distance between the actual stars. So for the 2 star system, you should have got v^2=Gm/4r and for the 3 star system you should have got v^2= Gm*sqrt3/3r
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    Your ratio should come out to something nice in the end.
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    It is 8 for the external tangent and something for the transverse tangent (which I totaly forgot about)
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    (Original post by cooltejaskd)
    The parachute question: They asked to Calc terminal velocity, which is easy - F = mdv/dt = 0. That will yield v terminal. However, they had asked to calculate work done by resistive force throughout the journey.
    I did Net Work = Work by resistance + Work by gravity = Change in KE. Till there it's okay. But.... What do I put for change in KE?! I know the FINAL velocity but not the initial. I assumed it to be zero BUT that's just crazy. They said the parachute "JUMPED" off the plane!! They could have said "DROPPED"! Anyway, I used initial v = 0 and solved that equation to get the answer.
    What did u guys do?
    Oh, this is a brilliant solution, I believe you are correct, and it is safe to assume that vi is zero, because we call it a parachute jump, when its realy parachutists dropping of a plane. How did you find work by gravity, what was the distance?
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    (Original post by Electric-man7)
    Actually that was my solution, but this is actually wrong. You probably took the distance for the gravitational force as being from the center of the orbit instead of the distance between the actual stars. So for the 2 star system, you should have got v^2=Gm/4r and for the 3 star system you should have got v^2= Gm*sqrt3/3r
    Yes but if you do it correctly you get 2 *3^1/4 so the solution with the mistake is a factor 2 off
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    (Original post by Electric-man7)
    Actually that was my solution, but this is actually wrong. You probably took the distance for the gravitational force as being from the center of the orbit instead of the distance between the actual stars. So for the 2 star system, you should have got v^2=Gm/4r and for the 3 star system you should have got v^2= Gm*sqrt3/3r
    I'd like to change my answer to:
    V3 = sqrt(1/sqrt(3))*2*v2
    Do you think that's correct?
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    That's V_2 and V_3 (Names of variables. NOT v SQUARED and v CUBED!)
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    (Original post by cooltejaskd)
    Circle question: There are THREE tangents. One of them had the length 1 (sketch the circles and you will see that y axis is ALSO a tangent.

    Parametrics question - They, I think, are expecting a GENERAL solution. Hence, there are 2 SETS of values, not JUST 2 values. (That is, omega*t can take the values pi/6, 2pi pi/6, 4pi pi/6......and pi - pi/6, 3pi - pi/6, 5pi - pi/6.....)

    Yes. The spring was super easy. So was the trigonometry question. (The answers above are correct, imo)

    The answer to the stars question: v3 = sqrt(sqrt(3))*2*v2. (Yes, there is sqrt IN sqrt)
    With paramatics you had to solve for x i believe. Were the values for t not +- 1/6pi +2kpi since you had to solve cos(x)=root3 /2?
 
 
 
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