June 2011 G485-Fields, Particles and Frontiers of Physics Watch

susan23
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#1041
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#1041
(Original post by ViralRiver)
I completely had to take a guess (well educated guess). I drew a straight, increasing line up to 0.08m - each 0.04m increasing by by whatever my answer to the Ek question was). Then a similar, but decreasing line for Ek between 0.08m and 0.12m.
lol did you..I just did a straight line constant gradient from the origin...had a ****ing guess lol
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susan23
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#1042
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#1042
I wish I revised stupid density of the universe and fission stuff....lol
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ViralRiver
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#1043
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#1043
(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.
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Oh my Ms. Coffey
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#1044
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#1044
I said nuclear power plant would come up
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ViralRiver
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#1045
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#1045
(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 =\ .
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Franklin
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#1046
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#1046
What were the graphs like?

The KE graph?
and the voltage capacitance graph?
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teachercol
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#1047
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#1047
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.
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apo1324
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#1048
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#1048
I thought that its kinetic energy would increase due to it accelerating and then be constant after leaving the electric field.
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susan23
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#1049
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#1049
(Original post by Franklin)
What were the graphs like?

The KE graph?
and the voltage capacitance graph?
My voltage capcitance graph was exponentially decreasing....
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Ben Adams
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#1050
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#1050
(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

And does anyone know the correct answer to the ratio? Is it 1/27?
And how about the worded question immediately after?
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Oh my Ms. Coffey
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#1051
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#1051
(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
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mattyatkinson
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#1052
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#1052
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
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Oh my Ms. Coffey
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#1053
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#1053
(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.
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toxp
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#1054
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#1054
(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.
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apo1324
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#1055
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#1055
The other graph was where we had to explain the terms open, closed and flat when describing the possible evolution of the universe.
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mattyatkinson
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#1056
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#1056
(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

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
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teachercol
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#1057
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#1057
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 .
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Franklin
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#1058
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#1058
(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?
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theoiskl
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#1059
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#1059
(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....
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susan23
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#1060
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#1060
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.
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