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Solved physics Paper 5??

Soldier313

Soldier313

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ash369 said:
how do I add a picture here? I've taken a snapshot of a question that's confusing me but i'm not sure how to upload it here? !
Click to expand...

there's a button "upload a file", tap on it and you can upload your file.
 
A

Ash1994

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I have a few questions regarding paper 5 Physics:
1. Are the error bars in the graph all of the same length?
2. Can Error bars be really large or are they meant to be really small?
3. When calculating the gradient ,do we draw the triangle on the graph between the two points that LIE on the line of best fit?
4. Should the reading and absolute uncertainty have the same no of significant figures

Also can someone tell me how to calculate the uncertainity in l^2 question 2 in
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s07_qp_5.pdf
 
Soldier313

Soldier313

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raamish said:
i mean that i calculated percentage uncertainty like this: [error/value of g + 2*0.01(error of t)/value of t]* 100
my answer came : [24.1/987 + 0.01*2/1] * 100
=2%+2.44%=4.44%

is this method correct?
and i also want to know if a value is like 3 sig figures should the error be 3 sign figures as well or should it have same decimal places as the value. e.g 2.41+-0.19 or 2.41+-0.20.
Click to expand...

yes for the percentage error i used the same method, i believe our answers are different because the value if gravity i found was slightly smaller.

About the no of significant figures in the error, as far as i know, they have to be equal in number to the original error significant numbers,
eg if the original error was 0.1, then the calculated error should be to 1 sgf too

raamish said:
and one another question i always calculate my absolute uncertainty from the percentage uncertainty in questions such as qs 2b of n12/53.
so how do we calculate the absolute uncertainty directly when values both having absolute uncertainties are being divided, multiplied or added. Like values of t/s and l/cm
. In this question they are being divided so how will we calculate the absolute uncertainty of them. I know its v2 so the uncertainty will be multiplied by 2 afterwards.
Click to expand...


since the formula is v = A (square root of k/M)
First square both the sides ( i find it simple that way) , you'll get v^2 = A^2 (k/M)
so basically the error formula will be :
[2(error in v) / v ] = [ 2 (error in A)/A] + [ (error in k)/k ] + [ (error in M) /M ]
since mass is constant it doesn't have an error
So yu cancel out the portion of [ (error in M) /M ]

and then substitute your values, (make error in v) the subject and you'll get the answer
Remember whatever function is used, +, - , x or /
the errors will always be added (this is as far as i know)
Hope it helped!
 
Praveena

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Can anyone plz help me to find percentage uncertainity of mayjune 2012 p5 last part of r and m????:(
 
Wanderer

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Praveena said:
Can anyone plz help me to find percentage uncertainity of mayjune 2012 p5 last part of r and m????:(
Click to expand...

Since formula for r is r=underroot(u/p into pie)..therefore uncertainty in r will be half into the uncertainty in u or p..We have only been given the error in u so simply do 1/2 into percentage uncertainty in u....as you have already calculated %age uncertainty in 'u' in part d-ii.,,,,simply divide this answer by 2...that's your ans for e-ii :)
 
biba

biba

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Wanderer said:
Do we draw gradient traingle for the line of worst fit as well?
Click to expand...

yes we.. in order to find the uncertainty of the gradient we have to calculate the gradient of worst fit cuz:
uncertainty= gradient of best fit - gradient of worst fit
 
Wanderer

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biba said:
yes we.. in order to find the uncertainty of the gradient we have to calculate the gradient of worst fit cuz:
uncertainty= gradient of best fit - gradient of worst fit
Click to expand...

Yeah the gradient has to be calculated for the worst fit line as well but i meant that drawing the traingle on the graph is necessary?:)
 
biba

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Wanderer said:
Yeah the gradient has to be calculated for the worst fit line as well but i meant that drawing the traingle on the graph is necessary?:)
Click to expand...

yeah the cie examiner would want to know how did u calculate ur uncertainty :p
so working has to be shown :)
 
ash369

ash369

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Alright, um, how do you find the gradient and the y-intercept in this case? I mean, I know how they're concluded via the y=mx+c equation but how do you apply that here? :-/


Okay so it says 'there was a problem uploading my file!' whenever I try to upload ANYTHING! :S So can somebody please just help me with 2012/O/N/question 2 part b! My query remains as explained above!
JazakAllahu Khairan!
 
R

raamish

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ash369 said:
Finally! Anyone?
Click to expand...

use y=mx+c and then its saying v on the y axis and 1/lambda on x axis so rearrange the equation according to y=mx+c. like: hc/lambda/e- B/e=V. now for gradient mx is hc/lambda/e so 1/lamda is x so the rest of it in mx must be m(gradient)
 
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