# CIE AS Level Physics Paper 3 PracticalWatch

Announcements
#1
In the Physics practical paper there is always a question in the 2nd experiment about calculating percentage uncertainty of a certain measurement made (angle/length etc.). How do we calculate this?

This is also followed by a question asking you to justify the number of sig. figs. used.
0
9 months ago
#2 where is the absolute uncertainty (use a reasonable estimated value, or relate it to the apparatus used), x is the measured value
e.g rebound height can be taken as x +- 0.5 cm

Justifying number of sf: Relate the sf in the calculated value to the sf used in the values in the calculation. e.g v = x/t. x = 10.00 cm, t = 3.9 s. t and x are measured to 4 and 2 sf respectively, so v can be given to 2 or 3 sf. (same sf as least precise value or 1 more)
0
#3
Thanks that's really helpful. The absolute uncertainty of a measurement is 1/2 of the most precise measuring of the instrument used right? For example, if a ruler is used, the absolute uncertainty would be +-0.5 mm because the ruler can measure to the nearest mm. Is this method accepted by CIE?
(Original post by BobbJo) where is the absolute uncertainty (use a reasonable estimated value, or relate it to the apparatus used), x is the measured value
e.g rebound height can be taken as x +- 0.5 cm

Justifying number of sf: Relate the sf in the calculated value to the sf used in the values in the calculation. e.g v = x/t. x = 10.00 cm, t = 3.9 s. t and x are measured to 4 and 2 sf respectively, so v can be given to 2 or 3 sf. (same sf as least precise value or 1 more)
0
9 months ago
#4
(Original post by Varun2111)
Thanks that's really helpful. The absolute uncertainty of a measurement is 1/2 of the most precise measuring of the instrument used right? For example, if a ruler is used, the absolute uncertainty would be +-0.5 mm because the ruler can measure to the nearest mm. Is this method accepted by CIE?
a metre rule measures to the nearest 1 mm
the uncertainty is +-1mm because 0.5mm at one end of the object and 0.5 mm at the other end => 1 mm

generally the quantities measured in question 2 of P3 are (relatively) hard to measure; such as rebound height, short time of oscillation, oscillations which are very damped, small forces, small resistances, small temperature rises. care should be taken
in general it is not as simple as 1/2 the smallest division
e.g time is +-(reaction time)
it should be according to the quantity you measured
Last edited by BobbJo; 9 months ago
0
#5
So how would, for example, the reaction time be measured? It's very subjective isn't it? Will the examiner have a certain range within which the student's answer must lie?
(Original post by BobbJo)
a metre rule measures to the nearest 1 mm
the uncertainty is +-1mm because 0.5mm at one end of the object and 0.5 mm at the other end => 1 mm

generally the quantities measured in question 2 of P3 are (relatively) hard to measure; such as rebound height, short time of oscillation, oscillations which are very damped, small forces, small resistances, small temperature rises. care should be taken
in general it is not as simple as 1/2 the smallest division
e.g time is +-(reaction time)
it should be according to the quantity you measured
0
9 months ago
#6
(Original post by Varun2111)
So how would, for example, the reaction time be measured? It's very subjective isn't it? Will the examiner have a certain range within which the student's answer must lie?
Yes it is subjective. You should estimate the reaction time (around 0.2-0.3s is reasonable). The examiner will have a range in which the absolute uncertainty must lie
0
#7
There is also another question which asks you to explain whether results support a given relationship. How are we expected to answer that?
(Original post by BobbJo)
Yes it is subjective. You should estimate the reaction time (around 0.2-0.3s is reasonable). The examiner will have a range in which the absolute uncertainty must lie
1
9 months ago
#8
(Original post by Varun2111)
There is also another question which asks you to explain whether results support a given relationship. How are we expected to answer that?
You will be asked to calculate two values of a constant.
first value of k = ...
second value of k = ....
You have to calculate the % difference in k. If it is less than 10% or 20% (or whatever%; you decide an appropriate criterion to test against), then the results support the relation.

Let's say k = 2d/t^2
e.g first value of k = 1.1 x 10^-3
second value of k = 1.3 x 10^-3
3 ways to calculate % difference in k
% difference in k =
(i) difference as a percentage of the smaller value: (1.3 - 1.1) x 10^-3/ 1.1 x 10^-3 x 100% = 18%
(ii) difference as a percentage of the larger value: 0.2/1.3 x 100% = 15%
(iii) difference as a percentage of the mean value: 0.2/1.2 x 100% = 17%
All are accepted.

Decide on an appropriate limit for experimental errors.
You could even calculate the % error in k for one value
e.g %k = %d + 2%t = 1 + 6 x 2 = 13%. The experimental errors (18% above) is larger than this so the experiment does not support the relation
The question is only worth 1 mark and calculating the error in the constant is much hard work for 1 mark. So usually, you take the acceptable limit as 10% or 20%.

You could also take the % uncertainty calculated in the earlier part.

It is easiest to take 10% or 20%.
0
#9
How do we calculate the two different values of K?

Sorry if I'm asking too many questions, but I am actually homeschooled and don't have much practical experience. Thanks for being so prompt and helpful.
(Original post by BobbJo)
You will be asked to calculate two values of a constant.
first value of k = ...
second value of k = ....
You have to calculate the % difference in k. If it is less than 10% or 20% (or whatever%; you decide an appropriate criterion to test against), then the results support the relation.

Let's say k = 2d/t^2
e.g first value of k = 1.1 x 10^-3
second value of k = 1.3 x 10^-3
3 ways to calculate % difference in k
% difference in k =
(i) difference as a percentage of the smaller value: (1.3 - 1.1) x 10^-3/ 1.1 x 10^-3 x 100% = 18%
(ii) difference as a percentage of the larger value: 0.2/1.3 x 100% = 15%
(iii) difference as a percentage of the mean value: 0.2/1.2 x 100% = 17%
All are accepted.

Decide on an appropriate limit for experimental errors.
You could even calculate the % error in k for one value
e.g %k = %d + 2%t = 1 + 6 x 2 = 13%. The experimental errors (18% above) is larger than this so the experiment does not support the relation
The question is only worth 1 mark and calculating the error in the constant is much hard work for 1 mark. So usually, you take the acceptable limit as 10% or 20%.

You could also take the % uncertainty calculated in the earlier part.

It is easiest to take 10% or 20%.
0
9 months ago
#10
(Original post by Varun2111)
How do we calculate the two different values of K?

Sorry if I'm asking too many questions, but I am actually homeschooled and don't have much practical experience. Thanks for being so prompt and helpful.
The symbol for the constant does not have to be k. It can be anything.

You will be given the relation between quantities you have measured.

Let's say T = Cr^2
You will be asked to calculate C.

1st experiment: r = 1.6 cm, T = 0.85 s
C = T/r^2 = 0.85/1.6^2 = 0.33
2nd experiment: r = 3.5 cm, T = 1.32 s
C = 1.12 / 3.5^2 = 0.11

You don't need to give units at this stage.
0
#11
Ok. So each experiment in question 2 will ask you to make 2 and only 2 sets of measurement right? So there is no ambiguity in choosing which measurments..
(Original post by BobbJo)
The symbol for the constant does not have to be k. It can be anything.

You will be given the relation between quantities you have measured.

Let's say T = Cr^2
You will be asked to calculate C.

1st experiment: r = 1.6 cm, T = 0.85 s
C = T/r^2 = 0.85/1.6^2 = 0.33
2nd experiment: r = 3.5 cm, T = 1.32 s
C = 1.12 / 3.5^2 = 0.11

You don't need to give units at this stage.
0
9 months ago
#12
(Original post by Varun2111)
Ok. So each experiment in question 2 will ask you to make 2 and only 2 sets of measurement right?
Yes
Last edited by BobbJo; 9 months ago
0
#13
It doesn't matter whether we use 10 or 20% right? Either one is accepted for all experiments? And we dont have to justify the usage of 10 or 20, do we?
(Original post by BobbJo)
You will be asked to calculate two values of a constant.
first value of k = ...
second value of k = ....
You have to calculate the % difference in k. If it is less than 10% or 20% (or whatever%; you decide an appropriate criterion to test against), then the results support the relation.

Let's say k = 2d/t^2
e.g first value of k = 1.1 x 10^-3
second value of k = 1.3 x 10^-3
3 ways to calculate % difference in k
% difference in k =
(i) difference as a percentage of the smaller value: (1.3 - 1.1) x 10^-3/ 1.1 x 10^-3 x 100% = 18%
(ii) difference as a percentage of the larger value: 0.2/1.3 x 100% = 15%
(iii) difference as a percentage of the mean value: 0.2/1.2 x 100% = 17%
All are accepted.

Decide on an appropriate limit for experimental errors.
You could even calculate the % error in k for one value
e.g %k = %d + 2%t = 1 + 6 x 2 = 13%. The experimental errors (18% above) is larger than this so the experiment does not support the relation
The question is only worth 1 mark and calculating the error in the constant is much hard work for 1 mark. So usually, you take the acceptable limit as 10% or 20%.

You could also take the % uncertainty calculated in the earlier part.

It is easiest to take 10% or 20%.
Last edited by Varun2111; 9 months ago
0
9 months ago
#14
(Original post by Varun2111)
It doesn't matter whether we use 10 or 20% right? Either one is accepted for all experiments?
Any specific criterion used should be specified by the candidate.
0
#15
Alright. But just specified, not justified?
(Original post by BobbJo)
Any specific criterion used should be specified by the candidate.
0
9 months ago
#16
(Original post by Varun2111)
Alright. But just specified, not justified?
No justification required
It is only 1 mark
0
#17
Ok. Thank you so much for your help!
(Original post by BobbJo)
No justification required
It is only 1 mark
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Bournemouth University
Wed, 19 Feb '20
• Buckinghamshire New University
Wed, 19 Feb '20
• University of Warwick
Thu, 20 Feb '20

### Poll

Join the discussion

Yes (46)
31.08%
No (76)
51.35%
Don't know (26)
17.57%