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# The Physics PHYA2 thread! 5th June 2013 watch

• View Poll Results: What mark do you think you got out of 70?
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1. (Original post by lebron_23)
Okay this is going to be a little difficult to put into context, but imagine a seesaw. Seesaws have a pivot at the centre, which is the point about which the seesaw rotates. When a child site at one end, he makes the seesaw move in a direction, either clockwise or anticlockwise. If another child were to sit on the other end then he would cause the seesaw to rotate in the other direction. Now, how much the seesaw rotates would depend on the masses of the children and where they say in relation to rate pivot, but the point is that a force applied to one side of a pivot will cause it to move in that direction.

Now imagine two pivots, one at the centre and one at one end of the seesaw. The pivots both exert reaction forces but when taking moments about the centre, the centre pivot's reaction force cancels out. So you now have one force acting down on one side of the pivot and a force acting up as well as a force acting down on the other side of the pivot. Now it's rather obvious that as the force that is on it's own moves down, the second force (which is the reaction force of the second pivot) moves upwards, hence they act in the same direction.

I appreciate that my explanation may have been terrible so I've attached two diagrams that might just help out. Sorry for the bad handwriting though.

Also, they may be upside down

Attachment 220486
Attachment 220487
Moments is probably the only thing I don't get, but I think I sort of get it now, thanks so it depends on where the pivot is to decide which one is clockwise and anticlockwise moments? On most last papers, they don't use examples like these, it more the arrow pointing one way is either AC and the other arrows are CW.

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2. This is a random question, but could someone please explain why a control wire used in Searles apparatus negates the effect of temperature change?

I think the answer is very obvious, but I just can't see it!

This isn't an actual question btw, I'm just trying to understand why this is so.

Thank you! :')

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3. Are there any formulas which aren't on the formula sheet?
4. Please could someone explain what to draw for q6a on June 2010???
5. (Original post by Claree)
Please could someone explain what to draw for q6a on June 2010???
Well lambda is the path difference and d is your slit separation.

Can you go from there?
6. (Original post by lebron_23)
Well lambda is the path difference and d is your slit separation.

Can you go from there?
I get it now . But why is lambda the path difference?
7. Can anyone explain 4B on page 96 in the AQA textbook please, just can't see it!!
8. (Original post by Claree)
I get it now . But why is lambda the path difference?
I wouldn't be able to tell you WHY it's called lambda but I can imagine that it's because it has something to do with distance and (not sure) but has something to do with the light arrive in phase/out of phase
9. Does anyone know why you divide the spring constant by 2 when calculating the extension in series? Help would be much appreciated

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10. (Original post by lebron_23)
I wouldn't be able to tell you WHY it's called lambda but I can imagine that it's because it has something to do with distance and (not sure) but has something to do with the light arrive in phase/out of phase
I thought that lambda meant the wavelength of the light? And therefore that the wavelength=path distance? Or is this a separate meaning for lambda?
11. (Original post by hannahclaire)
Can anyone explain 4B on page 96 in the AQA textbook please, just can't see it!!
The weight of the object is equal to two times the vertical component of the tension.

2*Tcos(73) = 4

T = 2/cos(73)

Hope that cleared it up for you
12. lol, this exam. I got full UMS on this piece of cake in January.
13. (Original post by Claree)
I thought that lambda meant the wavelength of the light? And therefore that the wavelength=path distance? Or is this a separate meaning for lambda?
Lord knows, physicists have a thousand names for one thing so it could be anything

Don't worry about it as I don't think that sort of in depth knowledge is required for the exam. It'll only drive you nuts hehe
14. (Original post by GeneralOJB)
lol, this exam. I got full UMS on this piece of cake in January.
Congratulations good sir, any helpful tips?
15. (Original post by lebron_23)
Congratulations good sir, any helpful tips?
Unfortunately not. I just did all the past papers from the years before and read over my notes and revision guide.
16. (Original post by lebron_23)
The weight of the object is equal to two times the vertical component of the tension.

2*Tcos(73) = 4

T = 2/cos(73)

Hope that cleared it up for you
oh yeah, seems so obvious now! Thanks for your help
17. (Original post by GeneralOJB)
Unfortunately not. I just did all the past papers from the years before and read over my notes and revision guide.
Yeah did this Jan 13 paper as my mock, it was pretty easy part from the last question.

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18. (Original post by GeneralOJB)
Unfortunately not. I just did all the past papers from the years before and read over my notes and revision guide.
Ah fair enough. I guess that's what I'm doing too, just going through all of the material I can find. I really want full UMS on this one as its got the lowest grade boundaries by far. 60/70 for an A pftt hehe

(Original post by hannahclaire)
oh yeah, seems so obvious now! Thanks for your help
No problem. I managed to confuse myself too, I was using 4g instead of 4 and ending up with ridiculous answers.
19. In the double slit experiment, how is Maxima and path difference increased or decreased?

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20. I'm a bit confused by the graph described by the mark scheme for q4c on the specimen paper. Is anyone able to explain what I'm meant to draw?

I think it's the graph from page 204 in the textbook, but is it meant to be the bold red line or the dotted red line?

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