Physics P5 uncertainty help

[Richard]

Yo, does any one know if uncertainties have to be written as 1s.f of can they be more? Cheers

 

[W-inx]

Example: Find lg R and its error and record in the table

R =231 +/- 10 (3s.f)
lg R should be to 3 sf

lg 231 = 7.85174904142
=> lg 231 = 7.852 (to 3 sf)

Notice in lg 7.852 is to 3 s.f and not 4. The numbers before the . do not count
E.g :
0.039 = 3 sf
3.430 = 3sf
3.4503 = 4sf
0.0002 = 4sf

Now the error in lg R = |lg Rmax - lg R|
Rmax = 231+10 = 241
Lg Rmax = Lg(241) = 7.91288933623 = 7.913 (to 3sf)

Therefore error in Lg R = |lg Rmax - lg R| = |7.913 - 7.852| = 0.061

So you record Lg R in the table as follows: 7.852 +/- 0.061

As you will see error above is to 3 s.f. Exceptionally in P5, the errors IN THE TABLE can be to more than 1s.f.

However when you will be calculating gradient and y-intercepts, you should bring the the error to 1s.f
Example for final answers of gradient and y-intercept:
gradient = 5.02 +/- 0.03 << Error to 1 s.f
y-intercept = 1.20 +/- 0.01 << Error to 1 s.f

So if it happens when u calculate gradient u get:
5.321 +/- 0.324 << This answer is not good and won't be accepted. You should work out the answer to the correct SF for the error. So you proceed as follows:

=>5.321 +/- 0.324 , error should be to 1 sf
=> 5.321 +/- 0.3 , now error in to 1 sf, but the gradient value is not good. Error is 1 SF but 1 D.P, so gradient should be to 1 D.P
=> 5.3 +/- 0.3 <<<<< That's the final answer for gradient


So to summarise:
> error in calculated values IN TABLE can exceptionally be to more than 1 sf.
> error in final answers for gradient or y-intercept should IMPERATIVELY be to 1 sf. Then you proceed as I explained above to get the final answer.


I suggest you go and work Nov 2009 P51. You will understand everything based on that paper. It's really the paper which has the most difficult data I've ever encountered.

If you have more problems reply here, and I will try to help.

Disclaimer: I did my best to help you. If you or someone who read this and found I made a mistake please correct me. I'm going for that paper too tomorrow

Good Luck.

 

[xyz!]

Example: Find lg R and its error and record in the table

R =231 +/- 10 (3s.f)
lg R should be to 3 sf

lg 231 = 7.85174904142
=> lg 231 = 7.852 (to 3 sf)

Notice in lg 7.852 is to 3 s.f and not 4. The numbers before the . do not count
E.g :
0.039 = 3 sf
3.430 = 3sf
3.4503 = 4sf
0.0002 = 4sf

Now the error in lg R = |lg Rmax - lg R|
Rmax = 231+10 = 241
Lg Rmax = Lg(241) = 7.91288933623 = 7.913 (to 3sf)

Therefore error in Lg R = |lg Rmax - lg R| = |7.913 - 7.852| = 0.061

So you record Lg R in the table as follows: 7.852 +/- 0.061

As you will see error above is to 3 s.f. Exceptionally in P5, the errors IN THE TABLE can be to more than 1s.f.

However when you will be calculating gradient and y-intercepts, you should bring the the error to 1s.f
Example for final answers of gradient and y-intercept:
gradient = 5.02 +/- 0.03 << Error to 1 s.f
y-intercept = 1.20 +/- 0.01 << Error to 1 s.f

So if it happens when u calculate gradient u get:
5.321 +/- 0.324 << This answer is not good and won't be accepted. You should work out the answer to the correct SF for the error. So you proceed as follows:

=>5.321 +/- 0.324 , error should be to 1 sf
=> 5.321 +/- 0.3 , now error in to 1 sf, but the gradient value is not good. Error is 1 SF but 1 D.P, so gradient should be to 1 D.P
=> 5.3 +/- 0.3 <<<<< That's the final answer for gradient


So to summarise:
> error in calculated values IN TABLE can exceptionally be to more than 1 sf.
> error in final answers for gradient or y-intercept should IMPERATIVELY be to 1 sf. Then you proceed as I explained above to get the final answer.


I suggest you go and work Nov 2009 P51. You will understand everything based on that paper. It's really the paper which has the most difficult data I've ever encountered.

If you have more problems reply here, and I will try to help.

Disclaimer: I did my best to help you. If you or someone who read this and found I made a mistake please correct me. I'm going for that paper too tomorrow

Good Luck.

hiee..ty fr al d efforts u tuk...
and can u evn help me out % uncertainty?
i m ataching 52/o/n/09 ....pls solve Q.2)e)ii) for me...or if u cud just explain me the method..
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_52.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_ms_52.pdf
thnx!

 

[Richard]

Thanks you so much, I understood all of that. Very well explained. 2009 paper p51 I kind of got stuck on because my points had a very poor correlation and my worst line didn't even end up crossing the line of best fit, all the other papers I have done have been fine so I'm not too worried about that one. I have two more questions if you are still online.
1. For paper 51 november part two. I understand now that the values of lg(r) have to be to "4 sf" and not two, however the values such as 8.111 is impossible to plot, given that it goes up in .02 increments does this mean for the values we plot to 2dp?
2. What are the necessary connections to a hall probe (voltmeter? and what else?) and should we use the symbol for a hall probe or just a box and label it?
Thanks for your help

 

[W-inx]

hiee..ty fr al d efforts u tuk...
and can u evn help me out % uncertainty?
i m ataching 52/o/n/09 ....pls solve Q.2)e)ii) for me...or if u cud just explain me the method..
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_52.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_ms_52.pdf
thnx!

That;s a nice question u got there, lemme go and solve it and I will come back with answers

 

[W-inx]

Thanks you so much, I understood all of that. Very well explained. 2009 paper p51 I kind of got stuck on because my points had a very poor correlation and my worst line didn't even end up crossing the line of best fit, all the other papers I have done have been fine so I'm not too worried about that one. I have two more questions if you are still online.
1. For paper 51 november part two. I understand now that the values of lg(r) have to be to "4 sf" and not two, however the values such as 8.111 is impossible to plot, given that it goes up in .02 increments does this mean for the values we plot to 2dp?
2. What are the necessary connections to a hall probe (voltmeter? and what else?) and should we use the symbol for a hall probe or just a box and label it?
Thanks for your help

1. You do not do to 2 dp. If it's 8.111 try your best to put it near to 8.111 that is a very little bit higher than 8.11. You should not neglect any of the dps, it's really important.

2. I will give u that in a while I don't think there will be hall probe last yr they gave it

 

[W-inx]

hiee..ty fr al d efforts u tuk...
and can u evn help me out % uncertainty?
i m ataching 52/o/n/09 ....pls solve Q.2)e)ii) for me...or if u cud just explain me the method..
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_52.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_ms_52.pdf
thnx!

Okay I have checked the paper. Well here is the expln:

TO find b it's simple, you take 1 point on your Line of best fit(which you used to calculated gradient) and then you take g best and you replace in euqation to find t

Now for the undertainty, use the equation in the Mark scheme.

When u see t^2 know that error in t = 2Δt/t
if % uncertainty in t the 2Δt/t x 100

E.g Find error in X^9

=> error in X^9 = 9ΔX/X
% error = 9ΔX/X x 100

That's 1 thing. Now in an equation:
b=m/(R^2), find %uncertainty in R

Given the following Data:
b = 3 +/- 1
m=5 +/- 2


Whether its multiplication of division, you ALWAYS add errors.

Let's solve it:

Δb/b x 100 = Δm/m x 100 + 2ΔR/R x 100

You always divide error by true value.

=> Δb/b x 100 = Δm/m x 100 + 2ΔR/R x 100
=> [(1/3) x100] =[(2/5) x 100] + [2(ΔR/R) x 100]

You need to know that % uncertainty in R = ΔR/R x 100 = Let's call it a

=> [(1/3) x100] =[(2/5) x 100] + 2a

Make a S.O.F you get the % uncertainty in R. Now apply this to your question and Voila

 

[xyz!]

Okay I have checked the paper. Well here is the expln:

TO find b it's simple, you take 1 point on your Line of best fit(which you used to calculated gradient) and then you take g best and you replace in euqation to find t

Now for the undertainty, use the equation in the Mark scheme.

When u see t^2 know that error in t = 2Δt/t
if % uncertainty in t the 2Δt/t x 100

E.g Find error in X^9

=> error in X^9 = 9ΔX/X
% error = 9ΔX/X x 100

That's 1 thing. Now in an equation:
b=m/(R^2), find %uncertainty in R

Given the following Data:
b = 3 +/- 1
m=5 +/- 2


Whether its multiplication of division, you ALWAYS add errors.

Let's solve it:

Δb/b x 100 = Δm/m x 100 + 2ΔR/R x 100

You always divide error by true value.

=> Δb/b x 100 = Δm/m x 100 + 2ΔR/R x 100
=> [(1/3) x100] =[(2/5) x 100] + [2(ΔR/R) x 100]

You need to know that % uncertainty in R = ΔR/R x 100 = Let's call it a

=> [(1/3) x100] =[(2/5) x 100] + 2a

Make a S.O.F you get the % uncertainty in R. Now apply this to your question and Voila

hey thnx sooooo much! dat was really very helpful!

 

[jarjar]

for paper oct nov 2011/51, can someone tell me how do we draw the error bar? since the uncertainty for the last question is o.oo5 but in the graph,the increment is o.o2 only. correct me if im wrong

 

[xyz!]

for paper oct nov 2011/51, can someone tell me how do we draw the error bar? since the uncertainty for the last question is o.oo5 but in the graph,the increment is o.o2 only. correct me if im wrong

yeah i faced d same problm..wt my teachr said is dat v shud sumhow show the error bar wid a sharp pencil...it wnt b distinguishable from the plotted point..bt den its ok...v do draw..cos only then wil v get the mark...

 

[xyz!]

W-inx pls help me wid 53/O/N/11 Q.2)b) ...the table...its very confusing..pls help asap!
here are the lnks for the QP and MS
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_53.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_53.pdf

 

[jarjar]

yeah i faced d same problm..wt my teachr said is dat v shud sumhow show the error bar wid a sharp pencil...it wnt b distinguishable from the plotted point..bt den its ok...v do draw..cos only then wil v get the mark...

oh i see. thank you so much

 

[xyz!]

oh i see. thank you so much

no problem!

 

[jarjar]

another question! how if the increment is o.25 and my uncertainty is 0.4? shall i just make error bar o.5? bcos its pretty hard to estimate o.4

 

[W-inx]

another question! how if the increment is o.25 and my uncertainty is 0.4? shall i just make error bar o.5? bcos its pretty hard to estimate o.4

Hehehe I bet that Nov 2009 p51
u just use calculator and find the increment of 1 little square using proportion u can find 0.4 is how many squares

 

[jarjar]

lol. so much trouble to get one mark huh. now i love p4.

 

[W-inx]

lol. so much trouble to get one mark huh. now i love p4.

I had a hard time with that 2009 paper

1 point was not even on the line

 

[jarjar]

yeap. the very 1st point. its quite difficult to do a best fit that fits all the error bar too. how u did it? im worried if something like this will happen tomorrow!

 

[W-inx]

W-inx pls help me wid 53/O/N/11 Q.2)b) ...the table...its very confusing..pls help asap!
here are the lnks for the QP and MS
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_53.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_53.pdf

Oh that's hard I dunno myself :S Im going to do this :O

 

[W-inx]

yeap. the very 1st point. its quite difficult to do a best fit that fits all the error bar too. how u did it? im worried if something like this will happen tomorrow!

I think it may happen, they made easy papers i think they will rip us with tomorrows paper. Prepare for worst. I wasnt going to revise for that, but I guess i will go grab my notes