ADVANCED HIGHER PHYSICS: Post Exam

Maybe I did but the whole question was rounding to 2sf max so I did too. They both gave the same answer and were in opposite directions so I cancelled them. I thought that was a good way to go as well as they had been discussing it in the question above.

I think my reasoning is sound. I take it you did it to 3/4 sf then subtracted. Maybe, but that answer is only as good as the figures (2sf) they gave you. Meh, either way it can't be that bad. It was just too nice that they had been discussing it above and it worked (for 2sf) that I could neglect it.

Yeah I only rounded to 2 sf and then when they cancelled I thought maybe they don't actually and I've rounded too soon so I went back to check. But yeah you could be right I suppose. I guess it'll depend on how the SQA intended us to interpret it again

Exactly. The discussion above made it too much of a coincidence though!

I suppose. That was discussing why it couldn't be though, albeit that was inbetween the charges. I just didn't want to write zero when I knew it wasn't actually zero lol :/.

True, but I think you are being slightly seduced by the fact that it is numbers before the decimal point. Imagine you got 3.241 one way and 3.237 the other. That would make me inclined to say its 0.

Yeah I see your point, but because the difference was 1100N/C I didn't feel it small enough to just ingnore it. Oh well, I guess we actually won't ever know unless we look at the past papers when they're published haha... by which time I most definately will not remember this question at all .

Remember 1C is a lot of charge. Anyway, you're right. I should stop worrying. I'm pretty sure I got high 80s low 90s which will do. It was a relatively hard paper so the boundaries will be lower than usual.

No one will care soon, you're right.

Yeah same, so should be fine . Do you think they'll be lower? I was worrying they may be raised a little.. didn't seem to difficult a paper. Although the last few papers I did before the exam were 2002 and 2003, so maybe my judgement is off as they were harder.

Haha sometimes I think it's really annoying that we don't get our papers back and see where we actually went wrong.. but then I think by august I won't remember or care.

Maybe I did but the whole question was rounding to 2sf max so I did too. They both gave the same answer and were in opposite directions so I cancelled them. I thought that was a good way to go as well as they had been discussing it in the question above.

I think my reasoning is sound. I take it you did it to 3/4 sf then subtracted. Maybe, but that answer is only as good as the figures (2sf) they gave you. Meh, either way it can't be that bad. It was just too nice that they had been discussing it above and it worked (for 2sf) that I could neglect it.

I think what you have done is fine, in that rounding mid-question is allowed. I personally just leave my anwers on the calculator and don't do any rounding until writing down the final answer; it must be said 0 is a nicer answer than 1100

Here are my answers, iv shortened the explanations and tried to make them as simple as possible to read:

1ai) 6.2t+4.1
1aii)11s
1aiii)6.2msE-1
1bi)Escape velocity greater than 3E8msE-1 => No light can escape planet, making it appear black
1bii) Minimum velovity required to escape a bodies gravitational field
1biii)Ek+Ep=0
0.5mvE2-GMm/r = 0
v=root(2GM/r)
1biv)Rearrange v=root(2GM/r) = 6.79E3M
1bv) V=4/3 pi rE3 = 1.28E12. And Density=Mass/Volume = 4.58E30/1.28E12 = 3.5E18kgmE-3


2ai) y=48(5.8)-12 = 266.4rpm
=>(266.4/60)x pi = 28radsE-1
2aii)Apply Same formula to find new angular velocity. Then use equations of angular motion to obtain -2.65radsE-2
2bi)I=1/3mlE2 = 2.12E-5
2bii) 6.47E-5
2c)-1.71E-4
2d) The ice causes cups to have a mass which has to be considered => a greater moment if inertia so when wind is recorded, it will not be accurate as the cups wont rotate at a angular velocity which is able to be compared to the wind speed.


3ai)y=2.1E-3 sin(207t)
3aii)0.43msE-1
3b)1.2E-3m
3ci)1.07E-8J
3cii) It Should have the kinetic energy at a maximum when displacement is zero and should be in the shape of a cos graph, cutting the x axis at points of maximum amplitude.

4ai) charge in centre, straight lines coming out of charge perpendicular to charge with arrows pointing away from charge
4aii) Charges are unequal and have different signs so E is in one direction at any point between the charges(to the right) and hence always has a non zero value
4aiii)0NCE-1
4b) The strong nuclear force is much greater than the electrostatic force, rendering the electrostatic force almost negligible in comparison.
4ci) Equate baryon numbers and charges to show that a proton consists of these quarks
4cii) 1 Down, 1 Anti up


5a) Use F-qvB and change the subject to find V using 3.2E-19 as the charge
5b) mg=QE use this and get 1.53E-7NC-1 (remember the mass is 5.02E-27)
5C) Answer = B simply use r=mv/qb and double radius since it is the diameter which is required.
5d) It will take a path with a smaller radius => it will strike the detector surface closer to S. And it will strike at the other side of S due to the fact that it is negatively charged.


6ai) current should be an increasing curve which levels off at 0.1A
6aii) Not enough voltage from battery alone.
6aiii) When the current rapidly decreases, the magnetic field through and around the inductor collapses, this induces a large emf. This emf is what causes the bulb to flash
6aiv) use P=I/T then use E=0.5LIE2 to get a value of 0.06H
6bi) To keep the voltage constant
6bii) Construct a current against frequency graph with the data
6biii) As frequency increases current decreases inverseley (im not sure about this one, i think my wording may be off)
6c) At high frequencies, the reactance of the capacitor is low and the reactance of the inductor is high, so at high frequencies, the capacitor allows current through, activating LS2. Opposite occurs for low frequencies.


7ai)1.06E-34
7aii)2E-24
7aiii)3.32E-10
7bi)Radius Increases
7Bii)Quantum


8a)0.21A
8b) when a force of one newton acts perpendicularly to one metre of wire carrying a current of one ampere.
8c)3.36E-7Nm-1


9ai) When a single beam is split into two beams, a reflected beam and a transmitted beam.
9aii)1.93E-4m ( im not sure what delta X is because i divided the fringe separations by 10, some people may have divided by 9, if so the answer will vary slightly, im not sure which is correct)
9aiii) 8.7%
9bi) The thickness of the soap film is a quarter wavelength of the light. so the light travels a total distance of a half wavelenth inside the soap film. It also undergoes a phase change of pi rads at x, when the reflected ray meets the incedent ray, they interfere destructively.
9bii) 1.16E-5
9biii)1.19E-5

10a) The incident waves reflect off of the plunger and undergo a phase change of pi rads, they interfere with the incident waves and form a stationary wave.
10b)a node is a point of minimum amplitude, an antonode is a point of maximum amplitude
10c) They collect at nodes since the oscillations of the antinodes will vibrate(almost push) the beads out of the way to fill the spaces which are the nodes.
10d)332ms-1


Now, these are my official answers, some of them may be wrong, please correct me if im wrong lol.

For the question about the graph of energy against displacement I drew a graph of the form y= -x^2 + c. I wasn't 100% sure but thats what I thought it would be from the equations. Just checked online (http://www.iop.org/activity/education/Projects/Teaching%20Advanced%20Physics/Vibrations%20and%20Waves/SHM/page_4479.html) and I think what I did was correct as I drew an quivalent of the blue line.