Physics past paper help

Here's the past paper: http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202013%20-%20QP/6PH02_01_que_20130605.pdf

Here's the mark scheme:
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202013%20-%20MS/6PH02_01_msc_20130815.pdf

There are many things which don't make sense to me so please bear with me.

Q2. I put A because I thought that it's the kinetic energy of the motor that mattered. Also, D, the other option, (which is correct) that I was contemplating, didn't involve the steady speed v which was given in the question.

Q3. Y is clearly 90degrees ahead of X but apparently, it's the other way round. I've seen this in other past papers and my answer is normally correct.

Q4.

Q7. I got A: 18000. (3*60)/0.01

Q8: The angle of i is 57degrees and knowing the refracive index of glass is 1.5, I found r to be 34 hence I put C.

Q10: How is superposition involved? I thought when something reflects, it's partially polarised.

2. Efficiency is energy output over energy input. Energy input is the energy used by the motor over that time period. Energy output is the work required to lift the mass to that height. Hint: kinetic energy is completely irrelevant to this question.
3. Y is displaced to the right in the graph, but in terms of time, it's lagging behind. Remember the waves are moving towards the right. Have a think about this, it's a bit of a sneaky question - correct, but I'd probably also have mistakenly gone for D.
4. This isn't in my syllabus but according to a formula on this page, drift velocity is inversely proportional to the cross sectional area and not dependent on length.
7. Think about how you calculate power again.
8. The angle of i is not 57 degrees. Think about what the angle if incidence is.
10. The reflected light would be polarised but that wouldn't affect the production of the colours seen. The rays travelling back towards the eye (which are shown parallel on the diagram) would actually be occupying the same space and thus interfering. Superposition is, if anything, the most important phenomena here.

First of all, thanks for the help.
I understand q7 now. How do I know which power formula to use? As there's also a VIt one too.
Oh silly me, I forgot it's between the normal! There's an awful lot of things to do for one mark!

Despite my poor performance in this paper, I somehow still got a B. An A grade is 48/80. The grade boundaries are surprisingly low.

Power is the the rate of work being done. We know the total work and the time over which that work was done, so we do energy time = 3J/60s=0.05J/s=0.05W. 1W=1J/s

Power is the the rate of work being done. We know the total work and the time over which that work was done, so we do energy time = 3J/60s=0.05J/s=0.05W. 1W=1J/s

Hello, I need some help again.

Q10
Q16 I know how to work it out but isn't it: sini/sinr=n? In this case, to get the correct answer, you have to do sinr/sini?


http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-qp/6PH02_01_que_20100609.pdf

Q10: What's your problem?
Q16: It flips around when you go from a slower medium to air.

I just don't know where to start which is why I left it blank.
I read 0.45A at 9V. Also 9/20 is 0.45. However, this isn't an option.

For Q10: Remember that the pd (9V) is shared across the two components. The voltage across the resistor and the voltage across the filament lamp must add together to make 9V. Look on the graph and find the current at which the voltages add to make 9.

What do you mean add to make 9V? That could be any combination of voltages. such as 7+2, 8+1 5+4

The graph shows how the voltage varies with current. Given that the current is constant at all points in a series circuit, there is only one value for A at which the voltages add to make 9.

its ok i found the answer 0.3A. What I meant was how do I know what that one value is?

Do you understand why it's 0.3A? And what do you mean?