Correct me if I'm having a brain fart. If not, I wonder if they caught that and corrected it, or if teachers caught it themselves when marking, otherwise it's awful to think what effect on grades they'd have.
3biii) It asks, hence, for the overall percentage uncertainty in k, that is to combine the answers to 3bi and 3bii, which are the percentage uncertainties of T and d respectively (because k = Td).
However in the mark scheme it combines the answer for 3a (0.8) and 3b ii (0.1), but 3a was working out what k is, so they've combined the actual value of k with one of the percentage uncertainties... (That or they've got confused and used the value of d from the previous question for some reason...)
4b) The mark scheme references '3aii', a question that doesn't exist, the answer to which is supposedly 0.8%. 0.8 was the answer to 3a, which (as above) was just the value of k (or d in the working of 3bii), not even a percentage!
2d) The gradient. With all the points plotted correctly, and the correct calculation, I came to -0.82. The answer given is anywhere between -0.97 and -1.05. What did anyone else get? Because to get that I can't think of a line where the points are equally both sides.
Wow, the AQA P10 Unit 6 ISA mark scheme makes a lot of mistakes Watch
- Thread Starter
- 19-04-2016 21:20
- 23-04-2016 21:59
3biii) The 0.8 does not come from the value for "k" but for the %uncertainty for T. You see, to work out the %uncertainty for "k", you must rearrange the formula to get k=Td. %uncertainty k =%uncertainty T + %uncertainty d. aka 3bi) + 3bii).
4b) The mark scheme does incorrectly quote 3aii). It is actually referring to 3bi). This is where the 0.8% comes from, just as we saw in 3biii).
2d) I gave this question a go. The first time I fell short and got -0.953 (incorrect graph), As you can see, for this one I started to the right of the rightmost point and underneath the leftmost point. I tried again and got a gradient of exactly -1 (correct graph); for this one I started at the rightmost point and just above the leftmost one. I think I good idea is to make sure your line goes through the point closest to the origin so you can use it as an anchor. I attached my incorrect and correct graphs.
Hope this helps, it definitely helped my revision.
- 23-04-2016 22:00
ps sorry about the upside down graph
- 27-04-2016 21:23
Could anyone pm me the document/ms just need to check something, thanks Sam
- 28-04-2016 20:29