Physics Paper 5 tips

[kiara15]

1. calculate the gradient of best fit line
2. calculate the gradient of worst fit line
3. error=worst gradient-best gradient (BECAUSE OTHERWISE THE CHANGE IS NEGATIVE)

The third step is only done if the answer of error in (gradient= gradient of best fit line - gradient of worst acceptable line) is negative so u can decide which one to use on the bases of seeing which value is bigger because u need a positive error value


Hope this helps

oh we onlly need uncertainity to be positive wether gradient of best fit is less than or greater than gradient to worst fit? ok i got that point and we wil subtract likewise

 

[kiara15]

U getting negative answer from best gradient-worst gradient in this question

worst gradient= 1.33 and best line gradient=1.28 so should i subtract best lline gradient by worst line gradient??

 

[kiara15]

Read my comment #9
https://www.xtremepapers.com/community/threads/physics-paper-5-notes.27194/#post-585614

And to join them u need to draw a straight line from the the bottom of the first error bar to the top of the last error bar and make sure your line passes through all the error bars.

can you show the pic of worst acceptable line joining error bars? plz

 

[fb.junks]

1.33-1.28 is this case specifically

 

[fb.junks]

can anyone plz help me on how to draw the worst fit line in the graph ?

need to draw a straight line from the the bottom of the first error bar to the top of the last error bar and make sure your line passes through all the error bars.
Tell me if u dont get it and I might be able to clear it up

 

[fb.junks]

can you show the pic of worst acceptable line joining error bars? plz

Its does look great but gives a rough idea that the worst acceptable line should start for the lowest point of the first error bar to the top of the highest error bar.
See the image!!!

 

[fb.junks]

can anyone plz help me on how to draw the worst fit line in the graph ?

See my above comment!!!

 

[Silent Hunter]

can anybody please fill out the errors section in this paper's question 2 so that i can match ...... MS doesnt give the error values :\ :|

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_52.pdf

thank you

fb.junks
FRENZYAMU
biba

or any one?

thank you

 

[FRENZYAMU]

can anybody please fill out the errors section in this paper's question 2 so that i can match ...... MS doesnt give the error values :\ :|

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_52.pdf

thank you

fb.junks
FRENZYAMU
biba

or any one?

thank you

0.07,0.04,0.02,0.01,0.01,0.005

 

[Silent Hunter]

0.07,0.04,0.02,0.01,0.01,0.005

am getting 0.0069 instead of the blue one ..... and the error bars .....

the error bar for the last three errors are coming so small .... how to make the bars then?

thanks again

 

[fb.junks]

am getting 0.0069 instead of the blue one ..... and the error bars .....

the error bar for the last three errors are coming so small .... how to make the bars then?

thanks again

To get answer correct .5(lg((1420+15)x10^3)-lg((1420-15)x10^3))=.005

 

[Silent Hunter]

To get answer correct .5(lg((1420+15)x10^3)-lg((1420-15)x10^3))=.005

thank you and about the error bars?> how to draw them as they are coming very small :\

 

[fb.junks]

U have to manage it.Nothing can be done I suppose.

 

[afoo.666]

how did you DRAW these error bars on graph ?

Its does look great but gives a rough idea that the worst acceptable line should start for the lowest point of the first error bar to the top of the highest error bar.
See the image!!!
View attachment 28044

 

[sagar65265]

about the error bars?> how to draw them as they are coming very small :\

The scale has already been given on the paper, so each small millimeter box on the y axis is equivalent to 0.2/10 = 0.02, and since the smallest uncertainty is about 0.005, that's about a fourth of a millimeter on either side of the point. So the uncertainty is really, really small, and it's best to make the point really fine (just a dot) and make a thin line just above and below the point, about a fourth of a millimeter on both sides of the point; if you were allowed to choose the axis scalings (the values and the range of the axis) then you could choose it to make the scale value of a millimeter smaller and thus make the error bar more noticeable, but since the scale is given here, there's not much more that can be done aside from approximating the position of the error bars.

Good Luck for your exams!

 

[Silent Hunter]

The scale has already been given on the paper, so each small millimeter box on the y axis is equivalent to 0.2/10 = 0.02, and since the smallest uncertainty is about 0.005, that's about a fourth of a millimeter on either side of the point. So the uncertainty is really, really small, and it's best to make the point really fine (just a dot) and make a thin line just above and below the point, about a fourth of a millimeter on both sides of the point; if you were allowed to choose the axis scalings (the values and the range of the axis) then you could choose it to make the scale value of a millimeter smaller and thus make the error bar more noticeable, but since the scale is given here, there's not much more that can be done aside from approximating the position of the error bars.

Good Luck for your exams!

yeah the scales so small ..... i was afraid i might be doing something wrong thats why asked and thanks for the clearance o the doubt .......

 

[sagar65265]

how did you DRAW these error bars on graph ?

Imagine you're drawing a graph of terminal velocity versus mass of an object.

Suppose you have a value of the terminal velocity of an object as 50 +- 10 ms^-1. This basically means that the mass is recorded as 50 ms^-1, but due to any kinds of errors creeping into the reading, the mass might actually be any value between 40 ms^-1 and 60 ms^-1. These are, respectively, the minimum and the maximum values that the terminal velocity of the object can be. Since the terminal velocity is actually recorded as 50 ms^-1, that is the point that you draw on the graph. To include the possibilities due to any errors, right above the point representing 50 ms^-1, for the same value of the mass, draw another point at 60 ms^-1 and another at 40 ms^-1. These are the extremes, but since the terminal velocity can be anywhere between these two values, join the two points with a vertical line between them. Drawing a horizontal line at 60 ms^-1 and 40 ms^-1 is just to maintain a boundary of the uncertainty, to ensure that the line you need to draw doesn't go outside the confines of the two extremes, in this case 40 ms^-1 and 60 ms^-1.

Hope this helped!

Good Luck for all your exams!

 

[afoo.666]

Imagine you're drawing a graph of terminal velocity versus mass of an object.

Suppose you have a value of the terminal velocity of an object as 50 +- 10 ms^-1. This basically means that the mass is recorded as 50 ms^-1, but due to any kinds of errors creeping into the reading, the mass might actually be any value between 40 ms^-1 and 60 ms^-1. These are, respectively, the minimum and the maximum values that the terminal velocity of the object can be. Since the terminal velocity is actually recorded as 50 ms^-1, that is the point that you draw on the graph. To include the possibilities due to any errors, right above the point representing 50 ms^-1, for the same value of the mass, draw another point at 60 ms^-1 and another at 40 ms^-1. These are the extremes, but since the terminal velocity can be anywhere between these two values, join the two points with a vertical line between them. Drawing a horizontal line at 60 ms^-1 and 40 ms^-1 is just to maintain a boundary of the uncertainty, to ensure that the line you need to draw doesn't go outside the confines of the two extremes, in this case 40 ms^-1 and 60 ms^-1.

Hope this helped!

Good Luck for all your exams!

uhm, im asking how am I suppose to know that the error is either horizintal or vertical and how long should we extend our error bars?

 

[Soldier313]

uhm, im asking how am I suppose to know that the error is either horizintal or vertical and how long should we extend our error bars?

If for example the two variables we're dealing with are VOltage and Current, and the errors provided are for say current, then on plotting, if we place current on the y-axis, the error bars will be vertical, and if we place current on the x-axis, then the error bars will be horizontal
The length of the error bars depends on the scale of the graph as well as the magnitude of the error itself

 

[FRENZYAMU]

am getting 0.0069 instead of the blue one ..... and the error bars .....

the error bar for the last three errors are coming so small .... how to make the bars then?

thanks again

i know its CRAP....