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A2 Physics Unit 4 and 5 Revision Thread

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Original post by jgmuzic
Hi, was wondering if anyone could help me out on a small question - unit 4 aqa, multiple choice.

When the length of a simple pendulum is decreased by 600mm, the period of oscillation is halved. What is the original length of the pendulum?

A-800mm
B-1000mm
C-1200mm
D-1400mm

I don't do maths for a level, as you might have guessed. Is this something to do with proportions or rearranging the simple pendulum equation?
Any feedback is appreciated.


The answer is A.

We know that T=12\frac{1}{2}To To = original period of oscillation.
We also know that L = Lo -600x10^-9 Lo = original length.

So T = 12\frac{1}{2}To = 2π\piLo−600x10−3/g\sqrt {Lo-600x10^-3/g}

We also know that To =2π\piLo/g\sqrt {Lo/g} = 4π\piLo−600x10−3/g\sqrt {Lo-600x10^-3/g} ⇒\Rightarrow

12\frac{1}{2}Lo−600x10−3/g\sqrt {Lo-600x10^-3/g} = Lo/g\sqrt {Lo/g}
⇒\Rightarrow


14\frac{1}{4}Lo9\frac{Lo}{9} = Lo−600x10−39\frac{Lo-600x10^-3}{9} ⇒\Rightarrow

14\frac{1}{4}Lo = Lo - 600x10^-3 ⇒\Rightarrow 600x10^-3 = 34\frac{3}{4}Lo

Lo = 0.8m = 800mm
Does anyone have any good notes for relativity? I'm doing AQA Unit 5 with turning points in physics. And none of the revision guides seem to have detailed enough explanations.
Reply 82
Avatar for babuchang
babuchang
2
Hey, this should help for the relativity stuff...
Turning Points - Special Relativity.pdf601.0KB
Exam Style Questions.pdf857.4KB
Reply 83
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emmaar
2
does anyone has A2 physics hodder buk.....................plssss share!!!111
Reply 84
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jwest
aqa phys 5 turning points question if any of you can help
for special relativity questions involving time dilation, length contraction and the mass energy equivalence, is the t0 or l0 the view of the observer or the actual particle?
Original post by babuchang
Hey, this should help for the relativity stuff...


Do you have more notes?
Looking for as many questions as possible for Astrophysics.. anyone help with that??
Original post by Smithy-92

Original post by Smithy-92
Looking for as many questions as possible for Astrophysics.. anyone help with that??


http://cid-5432f573d3fc3f5a.office.live.com/view.aspx/.Public/Physics/Topics/Astrophysics%20Q.doc

http://cid-5432f573d3fc3f5a.office.live.com/view.aspx/.Public/Physics/Topics/Astrophysics%20A.doc

:h:
heyy
can anyone post some particle physics notes
i NEED em =)
Original post by Vampire-Love4ever


Thank you!

Where did you get these from?
Original post by soutioirsim

Original post by soutioirsim
Thank you!

Where did you get these from?


Welcome. I had them saved in my favs. last year and completely forgot about it and across it just today!
Original post by Vampire-Love4ever


Thanks very much! I assume since you just came across them again, you don't remember where they're from/what spec they belong to/etc?
Original post by Smithy-92

Original post by Smithy-92
Thanks very much! I assume since you just came across them again, you don't remember where they're from/what spec they belong to/etc?


Welcome. Edexcel- Old syllabus.
Questions? :smile:
Original post by Smithy-92

Original post by Smithy-92
Looking for as many questions as possible for Astrophysics.. anyone help with that??


Check these too.

http://www.khanacademy.org/#cosmology-and-astronomy
Reply 95
Avatar for RSM93
RSM93
You know the equation

Q = Qoe^-t/RC

and that time constant = RC

This is a stupid question maybe, but doesnt that mean that -t/RC in the first equation = 1?

Or is the t in the original equation the time that has passed by? Or something different
(edited 12 years ago)
Original post by RSM93
You know the equation

Q = Qoe^-t/RC

and that time constant = RC

This is a stupid question maybe, but doesnt that mean that mean -t/RC in the first equation = 1?

Or is the t in the original equation the time that has passed by? Or something different


RC is the time constant, as you said. The amount of charge left, in the capacitor, is 0.367 (e^-1) it's original value (at t= 0) So yes, -t/RC = 1
Once t = RC the amount of charge fall by 36.7% of the orginal value.
Reply 97
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RSM93
Original post by Summerdays
RC is the time constant, as you said. The amount of charge left, in the capacitor, is 0.367 (e^-1) it's original value (at t= 0) So yes, -t/RC = 1
Once t = RC the amount of charge fall by 36.7% of the orginal value.


Thanks :biggrin: seems simple when you understand it :p:
Reply 98
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babuchang
2
Original post by Summerdays
Do you have more notes?


Yeah which ones do you want? Astro, medical?
Reply 99
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RSM93
Original post by babuchang
Yeah which ones do you want? Astro, medical?


Do you have astro and the mandatory AQA unit 5 topic?

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