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Oxford Physics: PAT test discussion

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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)


I tink due to symmetry there always 2 or 4 tangents never 3
Original post by Electric-man7
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?


The distance was GIVEN!!! As h! Lol. I legit took 1 full minute to check how to get the distance! But then found out that it's given in the question itself!!
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?


I think thats why it said estimate in stead of calcuate, the v of juming would be small relative to terminal v when jumping out of an airplane
Original post by Tomyil12345
I tink due to symmetry there always 2 or 4 tangents never 3


2 TOUCHING circles. Touching at one point. They have 3 tangents - one line above the circles' centers. One below. And one line which passes through the touching point and is perpendicular to the line joining the centers of the circles.
If you still don't get it, please tell me. I'll post a picture.
Original post by cooltejaskd
2 TOUCHING circles. Touching at one point. They have 3 tangents - one line above the circles' centers. One below. And one line which passes through the touching point and is perpendicular to the line joining the centers of the circles.
If you still don't get it, please tell me. I'll post a picture.


They were not touching though...
Original post by cooltejaskd
2 TOUCHING circles. Touching at one point. They have 3 tangents - one line above the circles' centers. One below. And one line which passes through the touching point and is perpendicular to the line joining the centers of the circles.
If you still don't get it, please tell me. I'll post a picture.


The circles didnt touch, the radia were root 6,25 and root 2,25 which is 2,5 and 1,5
Original post by cooltejaskd
I'd like to change my answer to:
V3 = sqrt(1/sqrt(3))*2*v2
Do you think that's correct?


Does the answer have 3^1/4 or (1/3)^1/4?

And someone said that my solution is incorrect. I took the distance between the starts itself, NOT the radius and got the above mentioned answer. I'm not sure if I had got 1/3^1/4 or 3^1/4 in my answer.
Original post by cooltejaskd
I'd like to change my answer to:
V3 = sqrt(1/sqrt(3))*2*v2
Do you think that's correct?

No
When I get back home, I'll write up a proper solution and post it online
Original post by Tomyil12345
The circles didnt touch, the radia were root 6,25 and root 2,25 which is 2,5 and 1,5


I'm pretty sure the radii were 3 and 5! And they were touching at y axis.
Anyone remembers the question completely? (Probably not) What I did was - add and subtract some constants to the equation, complete the square and get the equations to the circle in general form. And with that, I had the radii as 3 and 5.
Original post by cooltejaskd
I'm pretty sure the radii were 3 and 5! And they were touching at y axis.
Anyone remembers the question completely? (Probably not) What I did was - add and subtract some constants to the equation, complete the square and get the equations to the circle in general form. And with that, I had the radii as 3 and 5.


You can only devide and factorize not substract
Original post by cooltejaskd
I'm pretty sure the radii were 3 and 5! And they were touching at y axis.
Anyone remembers the question completely? (Probably not) What I did was - add and subtract some constants to the equation, complete the square and get the equations to the circle in general form. And with that, I had the radii as 3 and 5.


Radius was 1.5 and 2.5.
Original post by Tomyil12345
You can only devide and factorize not substract


AND and subtract the SAME constants.
Original post by cooltejaskd
AND and subtract the SAME constants.

You only do that after factorizing to compensate for the extra terms independent of x and y
Original post by Tomyil12345
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?


Yes. To solve for x, there was one term which was (omega*t) in the X equation. Hence, that omega*t can take any value that satisfies the general solution. Hence, there are 2 SETS of answers.
Original post by cooltejaskd
Yes. To solve for x, there was one term which was (omega*t) in the X equation. Hence, that omega*t can take any value that satisfies the general solution. Hence, there are 2 SETS of answers.

Yes i got the same, but i didnt see an omega ,wasnt x=a(t -sin(t))
Original post by Tomyil12345
You only do that after factorizing to compensate for the extra terms independent of x and y

Yes. Exactly what I did.... But I got the radii as 3 and 5... Idk if I made a mistake!
Original post by Tomyil12345
Yes i got the same, but i didnt see an omega ,wasnt x=a(t -sin(t))


Well, in that case, t can take many values in the X equation.
I THINK the X equation was x = a(omega*t - sin(omega*t))
Original post by cooltejaskd
Well, in that case, t can take many values in the X equation.


Yes but you had to solve for x so you also got a 1/2 and in the other general one -1/2 i think
Original post by cooltejaskd
I THINK the X equation was x = a(omega*t - sin(omega*t))


I dont recall omega but i didn’t have the time to check my anwsers so maybe is misread

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