Advanced Higher Physics 2011-2012 Discussion/Help Thread

Ive attached the missing marking instructions for AHP2005-6 papers
together with a terse set of cheatsheets - useful for revision

AHpastpaper2001.pdf
AHspecimen_solutions2003.pdf
AHspecimen_solutions2002.pdf
AHspecimen_solutions2001.pdf
AHpastpaper2006.pdf
AHpastpaper2005.pdf
AHpastpaper2004.pdf
AHpastpaper2003.pdf
AHpastpaper2002.pdf
AHspecimen_solutions2004.pdf

That's 2001-2006 i think, any solutions not with them you should be able to find on the SQA website. Hope this helps! :P

EDIT: Here's a better solution for 2001 I just found on Google: http://www.hswd.co.uk/trinity/downloads/003.pdf?PHPSESSID=687a76c6bfc763cd7ce0529cd04c1223

Just to let you guys know that the "specimen" answers are horrible. Littered with poor physics and mistakes.

Just to let you guys know that the "specimen" answers are horrible. Littered with poor physics and mistakes.

The answer links aren't the handwritten ones anymore. They look like they are official now

http://mrmackenzie.co.uk/advanced-higher/

The answer links aren't the handwritten ones anymore. They look like they are official now

http://mrmackenzie.co.uk/advanced-higher/

The answer links aren't the handwritten ones anymore. They look like they are official now

http://mrmackenzie.co.uk/advanced-higher/

This particular derivation is quite complex and the best way is by simply memorising the mehtod and understanding it.

$\alpha =\frac{\Delta v}{\Delta t}$

$\alpha =\frac{2vsin(\theta/2)}{r\theta/v}$

The "sin" can be taken out as for small values $\sin\theta = \theta$ therefore:

$\alpha =\frac{2v(\theta/2)}{r\theta/v}$

Then by cancellation:

$\alpha =\frac{v^2}{r}$

And by using the formula; $v = \omega r$ and finally you have derived the equation!:

$\alpha =\omega^2 r$

Hope this helped!

Hi guys,

Just noticed this equation on scholar for polarisation:

I = I₀cos²θ

Would I be right in saying we don't need to know about this?

I don't think they will ask you to calculate intensity (I) based on that equation. But you should be aware of the fact that when you change θ, I will follow a cyclic pattern (full I, then half I, no I, then half I again, full I).

Hi guys,

Just noticed this equation on scholar for polarisation:

I = I₀cos²θ

Would I be right in saying we don't need to know about this?

And yeah, there's I varies A² for waves etc. Thanks!

In 2004 question 8(c) the paper makes reference to a magnetic induction of 5.0 mT but then in the marking instructions it is subbed in as 5 * 10^(-5), and then the calculation is done as though it were 5 * 10^(-6). Am I missing something fairly big or has the infallible SQA made a mistake? :P

(To clarify what I did, I subbed it in as 0.0050 and got an answer in the form X * 10^(-9) whereas the SQA's answer is * 10^(-6), but their calculation gives * 10^(-7)...)

I think that must be a mistake.

In 2004 question 8(c) the paper makes reference to a magnetic induction of 5.0 mT but then in the marking instructions it is subbed in as 5 * 10^(-5), and then the calculation is done as though it were 5 * 10^(-6). Am I missing something fairly big or has the infallible SQA made a mistake? :P

(To clarify what I did, I subbed it in as 0.0050 and got an answer in the form X * 10^(-9) whereas the SQA's answer is * 10^(-6), but their calculation gives * 10^(-7)...)