What do you think of STEP Physics this year? I have just had a look at them, and they seems to be OK although there are a few questions I didn't get the answer.

Kerwin

I'd agree with your summary on the whole. Looked straight-forward but still some hard (or really fiddly) parts.

Question 1a - real pain. Why not just call it Nuffield A-Level? Likewise b. c) easy. d) - How are we supposed to estimate the capacitance of the Earth? I can't believe we're supposed to do the integration for part of 4 marks. Part iii) - it would decrease it? e) Very fiddly. Didn't have time to finish off f).

2) I think I got this all right. b)ii) is the only challenging part. I did it using the principle of time reversibility.

Didn't do 3-4.

5 - part c)ii) (e-1)/e? as a recall. I didn't understand what c)iii) was on about - I decided they wanted to minimise the rate of of gain of KE for the burnt fuel/rate of gain of KE of rocket.

Then I did 8). Very straight-forward, but I simply can't draw graphs! I changed the scale about 6 times. Hopeless. Overall, I think probably a 1, possibly S if I'm very lucky (as with Maths III). Which parts couldn't you do?

Yours,

Michael

Q1:a) Indeed that is Nuffield Physics(did you do OCR Nuffield or London Nuffield?) It should come out as a grid but I can't be bothered to draw the whole thing. Calculation is easy.
d)i)

1) a) We do OCR Nuffield unfortunately. (I didn't even know there was a London Nuffield - ignorance is bliss I guess). The main reason I did STEP Physics was to counteract the Nuffield style. The calculation is quite ridiculously easy - so easy I was sure it couldn't be right and wasted time checking it.

d) i) I didn't know that formula for capacitance, and it's not on the sheet! How do you define capacitance for isolated objects? My guess is that it is the quantity of excess charge required by an object to carry, in order to raise its potential (that is the gain energy for a uninegative particle starting at from an infinite distance, and ending up at the sphere) to 1V. (Obviously I'm messing around with dimensions here, but that's how they always define things. Resistance is the p.d. required to drive 1A of current through a component. 1st ionisation energy is energy required to remove 1 mol of electrons from 1 mol of gaseous atoms at stp. These statements are both dimensionally incorrect, but standard.)

If indeed that is the correct definition I'll have a go at deriving the formula for the capacitance for a sphere.

2) Very, very fiddly last part. I just hope my time reversibility argument worked.

3) I'm doing 3 for the first time now. iii) 2 - perhaps you could post your reasoning. I make it 8h/l + sqrt(3). Penultimate part - crucial word is friction. I don't understand what the last part is on about. What is a drive shaft and for what purpose does it supply torque?

4) Again, looking at this for the first time. The graph in iii) - is it constant then zero? I'm really not sure about that. b)i)) is just superposition. They really should have used partial derivatives because t is also a factor. (But it may seem comfusing I suppose.)

ii) I worked out eventually when I remembered that cos(kx+wt) differentiates wrt x as -k sin(kx+wt) NOT k cos(kx+wt).(!!) A3/A1 = 2Kp/(Kp+Kq)

iv) The speed of the incident and reflected the wave is w/Kp. The speed of transmitted wave is w/Kq. Now power = k² A² xspeed. So what we want to prove is:

speed of incident wave xk² A1² = speed of reflected wave xk² A2² + speed of transmitted wave xk² A3² . And I believe it works. (My algebra is always a bit suspect though.)

That is by far the best question on the paper. I'll try the others a bit later...

Yours,

Michael

Kerwin - in my message I messed up the last part somewhat. I'll try again. We want to show why it is consistent with energy. Well in other words we want to show:

Power dissipated in reflected wave + Power in transmitted wave = Power gained from incidence wave.

Now power = speed of wave x energy per length = w/k x k² A² = wkA²

Power dissipated in reflected wave = wK_P A₂ ²
Power dissipated in transmitted wave = wK_Q A₃ ²
Power received from incoming wave = wK_P A₁ ²

Now I'll leave you to confirm that the first two add up to the last, using the ratios. I think it does...

Yours,

Michael

Michael -

I saw what I got wrong now - I forgot the wheel! So the answer does not come out right.

Cheers,

Kerwin.