June 2011 G485-Fields, Particles and Frontiers of Physics

Reply 1040

susan23

3

I wish I revised stupid density of the universe and fission stuff....lol

Reply 1041

ViralRiver

16

Original post by Oh my Ms. Coffey

I did tan^-1(1.5x10^-4/4.5x10^-4) and got like 18 degrees.

exactly what I did - except I used my values instead of their 'approx' ones and got 18.3. Using their values gives 18.8 if I recall correctly.

Reply 1042

Oh my Ms. Coffey

I said nuclear power plant would come up :wink:

Reply 1043

ViralRiver

16

Original post by susan23

lol did you..I just did a straight line constant gradient from the origin...had a ****ing guess lol

Actually I have a feeling the Ek should be constant after 0.08 - there's no acceleration from the E-field - but then there is acceleration due to gravity etc.. not sure how involved the question was =\ .

Reply 1044

Franklin

What were the graphs like?

The KE graph?
and the voltage capacitance graph?

Reply 1045

teachercol

9

Well I'm glad I went through about half that paper in the last two days with my class. Transformers / PET / Rutherford were banker questions.

Overall - not too bad but as usual some tricky bits.

Q1 Straightforward - No of turns was 219 or 220.
Tricky bit baout the wires - you had to work out the PD acorss the wires as 12-11.8 = 0.2v and use that.

Q2. Most seem have got the PD worng. Series so Q is same so PD is inversely prop to C so PD across small cap is 4.5v. Energy ratio was 3:1 and stays constant.

Q3 Error was a bit obvious fortunately - you cant get plates 0.5 mm apart. First bit was striaghtforward. I wish I'd spent more time on these projectile style questions but easy if you spotted horiz motion is constant v and vert motin is v=u + at. Graph is tricky. Doesnt start at zero! Goes up in a stright line then flat after x=0.08m.

Q4Seen this one before. Routine if you did the past paers. Theta is 18.4.

Reply 1046

apo1324

11

I thought that its kinetic energy would increase due to it accelerating and then be constant after leaving the electric field.

Reply 1047

susan23

3

Original post by Franklin

What were the graphs like?

The KE graph?
and the voltage capacitance graph?

My voltage capcitance graph was exponentially decreasing....

Reply 1048

Ben Adams

Original post by Lengalicious

what exactly was impossible about this paper ¬.¬ its the simplest one theyve created since the newest spec

Alright buddy don't be big headed or anything will you? ****ing hell >.>

Im trying to see where i went wrong on the angle though. Most people i spoke to got 73 degrees too :s-smilie:

And does anyone know the correct answer to the ratio? Is it 1/27?
And how about the worded question immediately after?

Reply 1049

Oh my Ms. Coffey

Original post by Franklin

What were the graphs like?

The KE graph?
and the voltage capacitance graph?

I got the voltage one as an exponential graph working out the voltage values using v=v0e^-t/CR

Reply 1050

mattyatkinson

for the KE graph i put that it stated with some KE due to instant horizontal velocity and then increased to 0.08m then was constant

Reply 1051

Oh my Ms. Coffey

Original post by teachercol

Well I'm glad I went through about half that paper in the last two days with my class. Transformers / PET / Rutherford were banker questions.

Overall - not too bad but as usual some tricky bits.

Q1 Straightforward - No of turns was 219 or 220.
Tricky bit baout the wires - you had to work out the PD acorss the wires as 12-11.8 = 0.2v and use that.

Q2. Most seem have got the PD worng. Series so Q is same so PD is inversely prop to C so PD across small cap is 4.5v. Energy ratio was 3:1 and stays constant.

Q3 Error was a bit obvious fortunately - you cant get plates 0.5 mm apart. First bit was striaghtforward. I wish I'd spent more time on these projectile style questions but easy if you spotted horiz motion is constant v and vert motin is v=u + at. Graph is tricky. Doesnt start at zero! Goes up in a stright line then flat after x=0.08m.

Q4Seen this one before. Routine if you did the past paers. Theta is 18.4.

Are you sure? I got those values.

Reply 1052

toxp

Original post by Oh my Ms. Coffey

I got the voltage one as an exponential graph working out the voltage values using v=v0e^-t/CR

I did this, you can also use the values with the time constant, divide the initial voltage by e, and that is the value for 5s, divide initial by e^2 that is value for 10s.

Reply 1053

apo1324

11

The other graph was where we had to explain the terms open, closed and flat when describing the possible evolution of the universe.

Reply 1054

mattyatkinson

Original post by Ben Adams

Alright buddy don't be big headed or anything will you? ****ing hell >.>

Im trying to see where i went wrong on the angle though. Most people i spoke to got 73 degrees too :s-smilie:

And does anyone know the correct answer to the ratio? Is it 1/27?
And how about the worded question immediately after?

i got 1/27

Reply 1055

teachercol

9

Q5 The force on the rod is downwards
The force on the magnets is upwards so you subtract 0.016 from 2.4N
When dia is halved R is 4X so current is 1/4 and F is 1/4

Q6 Routine Rutherford. Nice and easy Q about forces. Easy calc. Assume spherical.

Q7. Easy crit density Q.

Q8 Easy PET Q - just book work
Q9 easy book ultrasound Q

Q10 Easy fission Q . Can use Water, heavy water or graphite as moderator.
Used to convert KE of neutrons to heat .

Reply 1056

Franklin

Original post by susan23

My voltage capcitance graph was exponentially decreasing....

Same, but in the previous part of the question you had to work out time constant, as 5.0s so did you use this to plot points on the graph and then draw it exponentially?

Reply 1057

theoiskl

Original post by Kalamari Dave

Yeah, there was another one. Where it said calulate the acceleration to be about 1x10^16 which came out to 1.2x10^16,
(then work out the time as 1x10^-9)
and then the next one where it says (use this to) now calculate the velocity to be 1.2x10^7 ms^-1
You cannot get the correct speed unless you get the acceleration correct.

Usually when they give you values, its to help candidates who don't calculate it, but it'll still allow them a chance to do the second part.
This one didn't.

(I found this out because first time run through i didn't work out the accn, used the value given, and didn't get the speed it wanted)

thats not a mistake they do that in all the past papers and other questions....

Reply 1058

susan23

3

I said for that energy stored on the capacitor question the ratio bit..that the ratio value increases when switch closes..because the 450 capacitor is connected to the resistor via negative terminal and the 150 is conneccted at positive terminal, so the electrons flow from 450 to 150 making the charge stored on 150 greater (neg) and 450 less (pos) so the energy stored on 150 greater than 450 so ratio increases. LOL sounds so wrong.

Reply 1059

mattyatkinson

Original post by teachercol

Well I'm glad I went through about half that paper in the last two days with my class. Transformers / PET / Rutherford were banker questions.

Overall - not too bad but as usual some tricky bits.

Q1 Straightforward - No of turns was 219 or 220.
Tricky bit baout the wires - you had to work out the PD acorss the wires as 12-11.8 = 0.2v and use that.

Q2. Most seem have got the PD worng. Series so Q is same so PD is inversely prop to C so PD across small cap is 4.5v. Energy ratio was 3:1 and stays constant.

Q3 Error was a bit obvious fortunately - you cant get plates 0.5 mm apart. First bit was striaghtforward. I wish I'd spent more time on these projectile style questions but easy if you spotted horiz motion is constant v and vert motin is v=u + at. Graph is tricky. Doesnt start at zero! Goes up in a stright line then flat after x=0.08m.

Q4Seen this one before. Routine if you did the past paers. Theta is 18.4.

how is the ratio 3:1

June 2011 G485-Fields, Particles and Frontiers of Physics

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