AS and A-Level physics resources thread

This thread is intended to be used for anyone to make requests for any resources for AS and A-Level Physics (and related subjects) - including past exam papers, mark schemes, etc. Please keep any requests in this thread; any requests posted elsewhere will be merged into this thread. Please note that in replying to any requests, materials should not be posted as attachments to posts, but instead as links to external hosts.

For requests for resources for other levels and qualifications, please use the following threads:


Past paper sites - external

http://www.freeexampapers.com/
http://www.thepaperbank.co.uk/
https://eiewebvip.edexcel.org.uk/pastpapers/
AQA A bank
AQA B bank
OCR bank
WJEC bank
Edexcel bank
CIE bank

Past Paper links - on TSR

http://www.thestudentroom.co.uk/showpost.php?p=20828230&postcount=488 (june 10, jan 10, june 09, jan 09)
http://www.thestudentroom.co.uk/showpost.php?p=8180200&postcount=63

Specifications

AQA A
AQA B
OCR
Edexcel
WJEC
CIE

Examiners reports

AQA A
AQA B
OCR
Edexcel

Speciman papers

AQA A
AQA B
OCR
Edexcel
CIE

Grade Boundries





Useful revision websites


Revision notes on our very own wiki

Would anyone be able to help me with these questions. I don’t understand the solution

This is the solution

Would anyone be able to help me with these questions. I don’t understand the solution

Hi, Can you verify the number that is circled in the picture - is it 4.0 m or 40 m? I would explain solution once the number is confirmed. Thank.

It is 4.0! Sorry for drawing a line through it

I assume that you are familiar with the principle of moment. If not, please state so.

(i) The distance AC is found by using trigonometry ratio and right angle triangle.

Triangle ADC is a right angled triangle. AD is equal to 4.0 m.

Using $\tan 30{}^\circ =\dfrac{AC}{AD}=\dfrac{AC}{4.0}$

You would get AC as 2.31 m.


(ii) The tension is found by using the principle of moment at point B. Why at point B? Because we don’t know the force at point B. So when we take the pivot at B, we can ignore the turning effect of the force at B.



To find the perpendicular distance from B to the wire, we use the right-angled triangle CBE as shown in the diagram above and sin ratio.

$\sin 60{}^\circ =\dfrac{CE}{CB}=\dfrac{CE}{BA+AC}=\dfrac{CE}{2.0+2.31}$

This would give CE as 3.73 m.

Apply principle of moment about B,
sum of anticlockwise moment of forces = sum of clockwise moment of forces
tension × CE = weight × 2.0

The tension can be calculated from the above equation as 268 N.

If you have any doubt about the explanation, let me know first before I explain the (iii) and (iv).

Oh no I get what you’ve just explained!!! Thank you so much!!! Yeah, we covered moments earlier in the year but I just didn’t really grasp harder questions like these!

Oh no I get what you’ve just explained!!! Thank you so much!!! Yeah, we covered moments earlier in the year but I just didn’t really grasp harder questions like these!

Try to attempt (iii) and (iv) to test your understanding. It is about static equilibrium.