OCR A Physics AS Breadth 24/5/16

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I disagree with question 8, the force constant changed so you can't use that formula, you need to use E=1/2Fx - since the force is the same, E is proportional to x and so it's E/3

Agreed - edited.

Reply 21

teachercol

Original post by julesaquilina

for the very last question about number of photons could you have used e=pt and divided e by 1.6x10^-19? Cuz they gave you the power and they said it was one second

I don't think so - its photons not electrons

Reply 22

wuhuapple001

I have different opinion on one of the questions.

For question 27, the graph is current against voltage (If my memory still work). So the gradient of the graph should be 1/R. The first two bits of the answer is agreed, first the resistance is infinite and then it starts to decrease. However, after that the graph is a straight line going upwards for the last part. If it is a straight line, it suggests that the resistance is constant. Although the R value is less than before, but it is constant, not continue to decrease.

Reply 23

teachercol

Original post by kuashdfkus

also for question 11 did people not get current clockwise?

No - it has to go + to - for the bigger emf

Reply 24

Jamvicious

Original post by NotGambit

For Q1, would it not be correct to say "rate of work done" which i think was A ? :/

It asked for a definition of the watt not a definition of power

Reply 25

Parhomus

Original post by kuashdfkus

also for question 11 did people not get current clockwise?

It was anticlockwise because the EMF of the cell on the left was greater; and because they were facing opposite directions and conventional current goes from + to - you had to find the difference; which ended up being anticlockwise.

Reply 26

Parhomus

Original post by wuhuapple001

It's not constant because the scales for V and I were different I think; the fact that it was linear would have meant that it decreased at a constant rate.

Reply 27

teachercol

Original post by wuhuapple001

This is a common mistake.
R is not 1 / gradient; its V/I.
Its only = 1 / gradient for straight lines through the origin.

Reply 28

Parhomus

Super happy as it turns out I got them right; probably dropped a couple marks here and there; especially with the asteroid question because I crossed out what I initially wrote about the force being in the opposite direction; but I did draw a reflection in the x axis. Not bad indeed.

Reply 29

Parhomus

Original post by teachercol

This is a common mistake.
R is not 1 / gradient; its V/I.
Its only = 1 / gradient for straight lines through the origin.

Does the linear gradient mean it decreases at a constant rate?

Reply 30

teachercol

Original post by kuashdfkus

are you sure about 11 + 20

Yes - you have to use angles to the normal and work out c for glass.

Reply 31

voltz

Original post by teachercol

Hi,

Thanks for posting these :smile:

Based on the questions in this Breadth paper, do you have any predictions for topics (especially practical questions) which are likely to come up in the Depth one?

(edited 7 years ago)

Reply 32

teachercol

Original post by Parhomus

Does the linear gradient mean it decreases at a constant rate?

I don't think so. Draw lines from the origin.

Reply 33

teachercol

Original post by voltz

Hi,

Thanks for posting these :smile:

Based on the questions in this Breadth paper, do you have any predictions for topics (especially practical questions) which are likely to come up in the Depth one?

Moments torques and centre of gravity
Measuring Youngs Modulus.
LDRs thermistors and potential dividers
Resistivity and series / parallel

Reply 34

sanchit117

Original post by Parhomus

Yes, that's exactly what they wanted people to think, at least in my opinion; because clearly if the same force is applied and extension is different then the value of the force constant will be different.

Yeah, so if the force constant is different, you can't use the equation EPE=1/2kx^2 as k is also different.

Reply 35

kuashdfkus

where could we find the questions?

Reply 36

teachercol

Original post by kuashdfkus

where could we find the questions?

You cant - until the paper is released in a years time.

Reply 37

wuhuapple001

Original post by teachercol

This is a common mistake.
R is not 1 / gradient; its V/I.
Its only = 1 / gradient for straight lines through the origin.

Indeed R=V/I, but as the graph is plotted as current against voltage, gradient will be dI/dV, which means that it is equal to 1/R.

It does not matter whether the line is going through the origin or anywhere else, the gradient of the graph represents the change of current with repect to voltage. As dV/dI measures resistance, dI/dV represents 1/resistance.

Reply 38

JN17

Judging from this I can only see possibly 3 marks lost, mostly on multiple choice, but I could almost guarantee the actual mark scheme will be very specific for some things, not allowing answers which are correct because of a keyword missing, or not stating a fairly obvious point, kind of seems to be the trend with the new spec :/

Fairly happy though, all calculations questions were fine, didn't completely screw up anything.

Reply 39

wuhuapple001

Original post by Parhomus

It's not constant because the scales for V and I were different I think; the fact that it was linear would have meant that it decreased at a constant rate.