Oxford Physics: PAT test discussion

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?

Reply 168

Tomyil12345

9

Original post by Electric-man7

Can someone explain the circle question? What was the final answer?

8 and root 60 iirc

Reply 169

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)

Reply 170

You are right, I made a stupid mistake about the circle question. how many marks will I lose?

Reply 171

Wait, who is right?! Me or the other person? Please quote!
The circle question was worth 9 marks I think.

Reply 172

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

Reply 173

Your ratio should come out to something nice in the end.

(edited 6 years ago)

Reply 174

It is 8 for the external tangent and something for the transverse tangent (which I totaly forgot about)

Reply 175

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?

(edited 6 years ago)

Reply 176

Tomyil12345

9

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

Reply 177

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?

(edited 6 years ago)

Reply 178