Practical Exam

1.

(a) (i) check that the metre rule labelled r is of uniform width and thickness by taking suitable measurements. your method and all your measurements should be shown below. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. (4)

(ii)

the arrangement shown below has been set up ready for you to use. X

y

Y

Z

x

M

adjust the height of the pulley and/or the position of the stands so that the string yz is horizontal. measure the distances x and y. x .................................................................................................................................. y .................................................................................................................................. the principle of moments gives the equation

y w = 2mg x where w is the weight of the suspended rule and mg is the weight of the slotted masses. you may assume m = 0.200 kg. use this equation to find the weight w. …………………………………………………………………………………….. …………………………………………………………………………………….. (5) ealing, hammersmith and west london college

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(iii)

explain how you measured x and y, adding to the diagram above if you wish. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. (3)

(b)

(i) set up the test circuit as shown in the diagram. leads x and y are labelled. before you switch on the power supply, have your circuit checked by the supervisor. you will be allowed a short time to correct any faults, but if you are unable to set up the circuit, the supervisor will set it up for you. you will only lose two marks for this. X

m A

A V B T e s t c irc u it

Y

U n k n o w n c irc u it (2)

(ii)

connect x to a and y to b to measure the current in and the voltage across the unknown circuit. repeat this with x connected to b and y connected to a. summarise your results in the table below. complete the table by inserting appropriate units and determining the resistance in each case.

x connected to

y connected to

a

b

b

a

current/

voltage/

resistance/

(4)

ealing, hammersmith and west london college

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(iii)

a technician makes up the following 3 circuits for an examination such as this. 100 Ω

100 Ω A

100 Ω

100 Ω

A

B

B

100 Ω J

100 Ω K

100 Ω A

B

100 Ω

100 Ω L

explain carefully which of the circuits is the one you tested. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. (3)

(iv)

deduce the resistance of the diode when it is conducting. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. (3) (total 24 marks)

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2.

(a) (i) w± 0.3 mm of supervisor’s value, with unit (1) t ± 0.3 mm of supervisor’s value, with unit (1) [w and t both to 0.1 mm or better] both from ≥ 3 readings (1) at different points along the rule (1) [may score mark from diagram] [use unit penalty once only] (ii)

(iii)

(b)

x and y to mm precision, with unit seen at least once (1) y by difference method (1) w correct, 2/3 significant figures, + unit (1) ± 0.1 n of supervisor’s value (1) (1) [± 0.2 n gets 1 mark only]

(iii)

(iv)

5

check heights vertical by using set square on bench [may be on diagram] (1) for x place rule on bench and note positions vertically below x and y (1) statement of how it was ensured that positions were vertically below x and y (1) 3 [last two marks can by scored by using pythagoras with xy = 80 cm] (i) circuit set up correctly without help (1) (1) [ignore meter polarity errors. polarity of cell –1]

(ii)

4

all units correct (1) v ≈ 1.5 v both to 0.01 v or better and iab ≈ 16 ma and iba ≈ 8 ma, both to 0.1 ma or better (1) correct calculation of r 2/3 significant figures [ecf] (1) rba = 200 ω ± 10 ω [no ecf] (1) rba ≈ 200 ω (1) hence circuit l (1) either diode arm does not conduct when b positive (1) or for circuit j, rba = 50 ω and circuit k rba = 100 ω (1) correct formula (1) diode arm resistance = rd + 100 ω (1) [allow ecf from wrong circuit] correct calculation of diode resistance [ecf] (1)

2

4

3

3 [24]

sample results (a)

(i) w = 28.4, 28.1, 28.1, 28.2 mm W = 28.2 mm t = 6.3, 6.3, 6.0, 6.2 mm t = 6.2 mm (ii)

x = 797 – 30 = 767 mm y = 400 – 196 = 204 mm 204 767 w= 2 × 0.2 × 9.81 × = 1.04 n

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(iii)

(b)

check heights vertical by using set square on bench. for x place rule on bench and note positions vertically below x and y. statement of how it was ensured that positions were vertically below x and y. (i)

circuit set up correctly without help.

(ii) x connected to a b (iii)

(iv)

3.

y connected to b a

current/ma

voltage/v

resistance/ω

16.5 7.9

1.55 1.57

93.9 199

rba ≈ 200 ω hence circuit l either diode arm does not conduct when b positive or for circuit j, rba = 50 ω and circuit k rba = 100 ω

1 1 1 + = X 199 93.9 [candidates may write 1/200 for second term] x = 178 ω resistance of diode = x – 100 = 178 – 100 = 78 ω

the first part of this question tested similar skills as question 1a, with corresponding responses from candidates. the majority could measure the width and thickness of the rule correctly and did a number of repeat readings whilst the best candidates drew a sketch of the rule to indicate the positions where the repeat readings had been taken. when determining the distances x and y candidates lost marks for a number of reasons: • distances were recorded only to the nearest cm • the scale readings that lead to the value of y were not shown • units were omitted from the measured values. again, as in question 1a, the weight of the rule was given in kg by a large number of candidates; this was treated as a unit error and candidates could obtain credit for a value which was close to the supervisor’s value. a number of candidates omitted “g” from the given formula and so obtained an answer that was in kg; this lost the mark for the calculation of the weight but could still gain the marks for accuracy.

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the explanation of how x and y were determined was poorly attempted. good candidates obtained the allocated mark by drawing the set square and vertical rule on the diagram. when determining x the majority of candidates had a horizontal metre rule above the bench, which would have been difficult to maintain in this position. only a small number of candidates correctly used the horizontal metre rule on the bench and recorded the readings on this rule vertically below x and y by use of a vertical rule and set square. in order to find x, a number of candidates measured xy and then used pythagoras with their previously determined value of y in order to calculate x; this technique was given credit. the second part of the question was very similar to question 1a and so the same comments apply as there. as in question 1a, examiners allowed error carried forward from incorrect values in the table in order to give the candidates as much benefit as possible in the analysis of their circuits and their attempts to calculate the diode resistance.

4.

(a) (i) check that the rod part of the stand labelled s is of uniform crosssection by taking suitable measurements. your method and all your measurements should be shown below. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. (4)

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(ii)

balance the stand on the knife edge so that the rod is horizontal. mark on the piece of tape the position c of the balance point. now set up the arrangement shown below.

x

y

T

C M a s k in g ta p e

S ta n d S

B

A

W support the rod with the newton-meter at a point b on the tape near the end. mark this point. adjust the height of the newton-meter so that the rod is horizontal. explain how you ensured that the rod was horizontal. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. (1)

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(iii)

the principle of moments shows that

W=

T ( x + y) x

where w is the weight of the stand and t is the newton-meter reading. measure and record the distances x and y, and record the value of t. x .................................................................................................................................. y .................................................................................................................................. t .................................................................................................................................. use these data to calculate a value for the weight w of the stand. ..................................................................................................................................... ..................................................................................................................................... (4)

(iv)

estimate the percentage uncertainty in your value for y. discuss the difficulty of estimating the percentage uncertainty for x. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (3)

(b)

(i) set up the test circuit as shown in the diagram. leads x and y are labelled. before you switch on the power supply, have your circuit checked by the supervisor. you will be allowed a short time to correct any faults, but if you are unable to set up the circuit, the supervisor will set it up for you. you will only lose two marks for this. X

m A

A V B T e s t c irc u it

Y

U n k n o w n c irc u it (2)

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(ii)

connect x to a and y to b to measure the current in and the voltage across the unknown circuit. repeat this with x connected to b and y connected to a. summarise your results in the table below. complete the table by inserting appropriate units and determining the resistance in each case.

x connected to

y connected to

a

b

b

a

current/

voltage/

resistance/

(4)

(iii)

a technician makes up the following 3 circuits for an examination such as this. 100 Ω

100 Ω A

100 Ω

100 Ω

A

B

B

100 Ω J

100 Ω K

100 Ω A

B

100 Ω

100 Ω L

explain carefully which of the circuits is the one you tested. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. (3)

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(iv)

deduce the resistance of the diode when it is conducting. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. (3) (total 24 marks)

5.

(a) (i) diameter to 0.01 mm precision and ± 0.4 mm of supervisor’s value (1) three values with unit (1) checked at points along length (1) and in two perpendicular (different) directions (1) [can get both the last two marks from diagram]

4

(ii)

height of rod above bench near each end measured (1) [allow “different points”]

(iii)

x and y to mm precision and unit and t to 0.1 n or better, with unit (1) w correctly calculated to 2/3 significant figures plus unit (1) w ± 2 n of supervisor’s value [w± 3 n gets 1 mark] (1) (1)

4

δy ± 1 mm or ± 2mm (1) [or > 2 mm if from range] correct calculation of percentage [ecf] (1) difficult to know the exact point about which the base pivots or difficulty in making measurement due to difference in levels of a and c (1)

3

(i) –1 if polarity of cell was changed for candidate [no penalty if “polarity of meter(s)”] circuit set up correctly without help (1) (1)

2

(iv)

(b)

(ii)

all units correct (1) v ≈ 1.5 v to 0.01 v or better and iab ≈ 20 ma and iba ≈ 15 ma, both to 0.1 ma or better (1) correct calculation of r to 2/3 significant figures [ecf] (1) rba = 100 ω ± 5 ω [no ecf] (1) as rba ≈ 100 ω (1) it is circuit k [dependent mark] (1) no current through diode in reverse or why other circuits are eliminated (1) 3

ealing, hammersmith and west london college

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4

10

(iv)

correct numerical parallel formula (1) indication somewhere that diode arm = rd + 200 (1) correct calculation (1) or i in diode arm = iab – iba (1) calculation of resistance of diode arm (using either v) (1) rd = r – 200 (1) 3 [in both cases, ecf from part (iii), even if answer negative or silly]

3

[24]

sample results (a)

(i) diameter/mm = 12.65, 12.67; 12.67,12.67; 12.67, 12.66 checked at points along length and in two perpendicular directions (ii)

height of rod above bench near each end measured

(iii)

x = 16.6 cm y = 59.5 cm t = 3.7 n 3.7(16.6 + 59.5) 16.6 w= = 17.0 n

(iv)

δy = 0.2 cm

0.2 % uncertainty = 59.5 ×100 = 0.3% (b)

(i)

circuit set up correctly

(ii) x connected to a b (iii)

(iv)

y connected to b a

current/ma

voltage/v

resistance/ω

19.6 15.3

1.51 1.52

77.0 99.3

as rba ≈ 100 ω it is circuit k no current through diode in reverse

1 1 1 + = x 99.3 77 ⇒ x = 343 ω rd = 343 – 200 ω = 143 ω

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6.

many candidates were unable to read a micrometer correctly, but were awarded a mark for recording three readings. the better candidates said that these readings were taken at different points along the rod; very few realised the importance of taking readings at different orientations to check for uniformity. several candidates wasted time by working out actual areas of crosssection or taking an average of their readings - the latter missing the whole point of the question. a significant number of candidates appeared not to have removed the prism for part (ii). the instructions were quite clear: “now set up the arrangement shown below”, in which no prism was shown. many candidates tried to determine whether the rod was horizontal by using a set square in various ways instead of measuring the height above the bench at each end. in determining the weight of the stand, errors in units and significant figures were legion: • x and y were recorded to the nearest centimetre and t in whole newtons • yet the weight was given to 4 or more sf • with units ranging from g and kg to nm (maybe because they were using a newton-meter) [indeed, some supervisors quoted the weight of the stand on the front cover in g or kg.] credit was given for the accuracy of an answer even if it was stated in incorrect units, although candidates were often well outside the prescribed limits. although many candidates stated an uncertainty in y of 0.5 mm, it was considered that this was not appropriate in the context of this experiment and that 1 mm or 2 mm was more realistic. most candidates were then able to calculate the percentage uncertainty correctly, for which credit was given. few candidates described the difficulty in estimating the percentage uncertainty in x in terms of the difficulty of making the measurement - most simply said that the reason was because x was small. whilst this makes the percentage uncertainty larger, it is not what makes it difficult to determine. in part (b) most candidates were able to set up the circuit unaided and obtained correct values for i, v and r. however, common errors were • recording i in a instead of ma • not recording i to 0.1 ma precision • quoting resistance values to an inappropriate number of significant figures, for example as 0.1 kω (only 1 sf) or 100.7 ω (4 sf) when the data was recorded to 3 sf. those candidates making a mistake in the table invariably had difficulty answering the remaining parts of the question. for example, if they had incorrectly quoted i in a, they then obtained a value of r from b to a of, say 0.099 ω instead of 99 ω; this made it difficult to reconcile with any of the three circuits. in some cases candidates chose to ignore the decimal point and argued logically for circuit k. others were awarded marks for analysing the three circuits correctly. for example, for saying that the resistance from b to a was 50 ω, 100 ω and 200 ω respectively when the diode was in reverse mode and not conducting, even if they could not relate their very low value for r to any of the circuits. the mathematically more able candidates successfully calculated a value for the diode resistance, but in general the ability of candidates to handle parallel circuits was weak, many leaving this section blank.

7.

(a) (i) using the voltmeter provided measure the potential difference between the terminals of the cell. because the voltmeter has a very high resistance, you may assume that this reading is the e.m.f. e of the cell. e = ..................................................................................................................... (1)

ealing, hammersmith and west london college

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(ii)

set up the circuit shown in the diagram below. before you connect your circuit to the cell, have your circuit checked by the supervisor. you will be allowed a short time to correct any faults. if you are unable to set up the circuit, the supervisor will set it up for you. you will lose only two marks for this. A 68 Ω

V

(2)

(iii)

connect the cell to your circuit and adjust the variable resistor until the potential 1 difference v1 across the variable resistor is approximately 2 e. record v1 and the corresponding current i1. v1 = ................................................................................................................... i1 = .................................................................................................................... hence calculate the resistance r1 of the variable resistor at this setting. ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (3)

(iv)

using your knowledge of potential dividers, what is the expected value of r1 when 1 v1 is 2 e? ............................................................................................................................ ............................................................................................................................ calculate the percentage difference between your value of r1 and the expected value. ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (2)

(v)

repeat the experiment with the potential difference v2 across the variable resistor set

ealing, hammersmith and west london college

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1 to approximately 4 e. record v2 and the corresponding current i2. v2 = ................................................................................................................... i2 = .................................................................................................................... calculate the resistance r2 of the variable resistor at this setting. ............................................................................................................................ ............................................................................................................................

1 what value would you expect to get for r2 when v2 = 4 e? ............................................................................................................................ ............................................................................................................................ (3)

(vi)

suggest three reasons why your experimental values of r1 and r2 differ from the expected values. ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (3)

(b)

(i) determine accurate values for the internal diameter di and the external diameter de of the washer provided. state any special precautions which you took to ensure that the diameters were as accurate as possible. ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (5)

(ii)

determine an accurate value for the thickness t of the washer.

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............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (2)

(iii)

calculate the volume v of the washer given that

1 2 2 v = 4 π(d e – d i )t ............................................................................................................................ ............................................................................................................................ using the top pan balance measure the mass m of the washer. m = ..................................................................................................................... hence determine the density of the material from which the washer is made. ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (3) (total 24 marks)

8.

(a)

(i)

approximately 1.5 v measured to 0.01 v or better + unit (1)

(ii)

circuit set up correctly without help

(iii)

≈ 0.8 v to 0.01 v or better + unit (1) 7.0 ma → 12.0 ma to 0.1 ma or better + unit (1) correct calculation to 2 to 4 significant figures + unit (1) [allow ecf]

1 2

3

[apply unit penalty once only for each quantity (i, v and r) in part (a)] [r values to 2 to 4 significant figures else –1 once only] (iv) (v)

value + unit (1) correct percentage difference with expected value as denominator (1)

2

sensible values (v2 < v1, i2 > i1) with v to 0.01 v or better (1) and i to 0.1 ma or better + units correct calculation to 2 to 4 significant figures + unit (1) [allow ecf from wrong i and v or r1] value + unit (1) 3

ealing, hammersmith and west london college

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(vi)

(b)

any 3 of: resistor values will have a tolerance; potential difference across connecting wires / wires in the circuit will have resistance; ammeter will have resistance / potential difference across ammeter; cell has / potential difference across internal resistance of cell; error in meters including not accurate; voltmeter has finite resistance / current in voltmeter (1) (1) (1) [do not accept resistors heating up]

max 3

(i) di = ± 0.03 cm from supervisor’s value from repeat measurements (1) (1) [± 0.05 cm from repeats (1) or ± 0.03 cm from single reading (1)] de = ± 0.03 cm from supervisor’s value from repeat measurements (1) (1) [± 0.05 cm from repeats (1) or ± 0.03 cm from single reading (1)] checked in perpendicular directions (1) [allow different directions] zero error check here or in (ii) (1)

) ) )

max 1

5

[allow centre average value if no supervisor’s data] (ii)

(iii)

± 0.03 mm from supervisor’s value from repeat measurements (1) (1) [± 0.05 mm from repeats (1) or ± 0.03 mm from single reading (1)]

2

correct calculation ≥ 2 significant figures + unit (1) correct substitution into mass / volume, (1) value ≥ 2 significant figures + unit and in range –3 –3 6.0 → 10.0 g cm or ± 2.0 g cm of supervisor’s density (1)

3 [24]

sample results (a)

(i) (iii)

e = 1.52 v

v1 = 0.76 v i1 = 10.62 ma r1 = 0.76 / 10.62 × 10 = 71.6 Ω

–3

(iv)

expected value = 68 Ω (71.6 − 68 / 68) × 100% = 5.3%

(v)

v2 = 0.38 v i2 = 15.93 ma r2 = 0.38 / 15.93 × 10 = 23.9 Ω expected value = 1/3 × 68 = 22.7 Ω

–3

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(b)

9.

(i) di = 1.66 cm, 1.66 cm average di = 1.66 cm de = 3.47 cm, 3.47 cm average de = 3.47 cm (ii)

thickness 1.81 mm, 1.79 mm, 1.85 mm average t = 1.81 mm

(iii)

v = ¼ π (3.47 − 1.66 ) × 0.181 3 = 1.32 cm m = 10.32 g density = 10.32 / 1.32 –3 = 7.8 g cm

2

2

most candidates successfully set up the circuit. there were three reasons for the loss of marks in (a) (i), (iii) & (v); omitted units for any of the quantities v, i or r. in fact, some candidates went through the whole question without quoting a single unit. on most gcse papers units are often given on the question paper and a skill that must be developed at as level is to always quote a unit with an answer. either or both of the quantities v and i not quoted to the precision of the instrument that was being used to measure them. values of r not quoted to an appropriate number of significant figures. in (a) (iv) examiners expected candidates to realise that in an ideal potential divider circuit the expected value of r1 should be equal to the other resistor (68 Ω) in the circuit if the voltage across r1 was to be equal to half the e.m.f. of the power supply. only the best candidates realised this. in (a) (v) the value of r2 should have been 1/3 of 68 Ω to give a potential difference of ¼ of the e.m.f. of the cell. those candidates who had achieved success in (a) (iv) were often caught out here and stated a value which was ¼ of 68 Ω. for questions such as (a) (vi) candidates should be encouraged to give a location for any additional resistance. answers such as “the meters have resistance” were popular but gained no marks. much better answers stated that the resistance of the ammeter was not negligible or that the resistance of the voltmeter was not infinite. equivalent answers such as there was a small potential difference across the ammeter or a small current in the voltmeter were obviously perfectly acceptable. good candidates correctly determined the volume and the density of the washer, obtaining a value in the required range. weaker candidates made the following errors; mixed cm and mm were used in the calculation of the volume. units were omitted from the value of the volume. a significant number of candidates thought that the value of the fraction in the formula was ½ and not ¼ . incorrect density values were obtained because candidates had tried to make a conversion from 3 3 g/cm to kg/m , centres should note that such conversions are not required for density in the practical paper.

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10.

(a) (i) set up the circuit shown in the diagram below. use the red led (light emitting diode). before you connect your circuit to the power supply, have it checked by the supervisor. you will be allowed a short time to correct any faults. if you are unable to set up the circuit, the supervisor will set it up for you. you will lose no more than two marks for this. A

+ V

470 Ω (2)

(ii)

connect the circuit to the power supply. record the potential difference vr across the red led and the corresponding current ir. vr = .................................................................................................................... ir = ..................................................................................................................... hence calculate the resistance rr of the red led in this circuit. ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (3)

(iii)

replace the red led with the green led. ensure that the positive end of the led is connected as in the previous circuit. record the potential difference vg across the green led and the corresponding current ig. vg = .................................................................................................................... ig = ..................................................................................................................... hence calculate the resistance rg of the green led in this circuit. ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (3)

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(iv)

repeat your measurements and calculations with the 220 Ω resistor connected into the circuit in place of the 470 Ω resistor. 1

red led with 220 Ω resistor

vr´ = ................................................................................................................... ir´ = .................................................................................................................... ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ rr´ = ................................................................................................................... 2

green led with 220 Ω resistor

vg´ = ................................................................................................................... ig´ = .................................................................................................................... ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ rg´ = .................................................................................................................... (5)

(v)

write three conclusions based on your experimental observations. ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (3)

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(b)

(i) determine an accurate value for the diameter d of the marble. explain, with the aid of a diagram, how you tried to ensure that an accurate value for the diameter was found. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (3)

(ii)

the metre rule has a small piece of blu-tack attached to it at the 5 cm mark. balance the rule on the knife edge to determine the position of the centre of mass of the rule and blu-tack combination. scale reading at centre of mass = ......................................................................

(iii)

secure the marble to the metre rule at the position of the blu-tack. set up the arrangement shown in the diagram below. C e n tre o f m a s s M a rb le

S lo tte d m a s s

B ench

M e tre ru le P iv o t

determine the mass m of the marble using the principle of moments. using the above diagram, show carefully the measurements which you took in order to determine the mass m. record all your measurements and calculations in the space below. the slotted mass has a mass of 10 g ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ ............................................................................................................................ (5) (total 24 marks)

ealing, hammersmith and west london college

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11.

(a) (ii)

(i)

circuit set up correctly without help (1) (1)

≈ 1.8 v to 0.01 v or better + unit (1) sensible values to 0.1 ma or better + unit (1) correct calculation to 2 to 4 significant figures + unit (1) [allow ecf]

(iii)

vg › vr and ≈ 2 v to 0.01 v or better + unit (1) ig ‹ ir and sensible value to 0.1 ma or better + unit (1) correct calculation to 2 to 4 significant figures + unit and rg › rr (1) 3 [apply v unit penalty, i unit penalty, r unit penalty and significant figure penalty once only in part(a)]

(iv)

vr ′ › vr and ≈ 2 v to 0.01 v + unit (1) ir ′ ≈ 2ir and to 0.1 ma + unit (1) vg ′ › vg and ≈ 2 v to 0.01 v + unit (1) ig′ ‹ ir ′ and ≈ 2 ig to 0.1 ma or better + unit (1) correct trend in r values rr ′ ‹ rr, rg′ ‹ rg and rr ′ ‹rg (1)

(v)

2

3

5

any 3 of: resistance of led changes; it decreases as current in it increases (or vice versa); resistance of a green led is greater than resistance of a red led ; – at a given current ; as current doubles: resistance of led approximately halves / voltage across led remains ≈ constant; brightness of led increases as current through it / voltage across it increases (1) (1) (1) max 3 (i) ± 0.03 cm of supervisor’s value to 0.1 mm from repeat (1) (1) measurements + unit [± 0.05 cm from supervisor’s value (1)] at least two perpendicular directions or zero error check (1)

(b)

(ii)

3

all measurements to nearest mm or better (allow 5 cm) (1) C e n tre o f m a s s M a rb le

S lo tte d m a s s M e tre ru le

B ench 5 .0 c m (iii)

4 9 .9 c m

P iv o t 7 2 .1 c m

centres of mass clearly shown on diagram (1) scale readings shown (1) correct calculation giving m ± 0.3 g of supervisor’s value (1) (1) [± 0.05 g from supervisor’s value (1)] [no ecf, –1 if no unit]

5 [24]

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sample results (a)

(i) (ii)

circuit set up correctly without help

vr = 1.87 v ir = 8.48 ma rr = 1.87 / 8.48 × 10 = 221 Ω

(iii)

–3

vg = 1.96 v ig = 8.28 ma rg = 1.96 / 8.28 × 10 = 237 Ω

(iv)

–3

vr ′ = 2.00 v ir ′ = 17.0 ma rr ′ = 2.0 / 17 × 10 = 118 Ω vg′ = 2.06 v ig′ = 16.77 ma

–3

rg′ = 2.06 / 16.77 × 10 = 123 Ω (b)

–3

(i) d = 1.55 cm, 1.54 cm, 1.545 cm average d = 1.545 cm measured in at least 2 perpendicular directions d d (ii)

1

2

scale reading at centre of mass = 49.9 cm

(iii) C e n tre o f m a s s M a rb le

S lo tte d m a s s M e tre ru le

B ench 5 .0 c m

4 9 .9 c m

P iv o t 7 2 .1 c m

10 (72.1 − 49.9) = x (49.9 − 5.0) x = 10 × 22.2 / 44.9 = 4.94 g

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12.

(a) almost all candidates set up the circuit successfully without help. candidates of all abilities also found it relatively easy to take four sets of voltage and current readings and calculate the corresponding resistances correctly. the majority gained maximum marks on sections (a) (i) to (iv). as with 6733/2a, the most common ways of losing marks were the omission of units, wrong units for current (ma for a), lack of precision on voltage readings and quoting resistances to 5 significant figures. some centres, using diodes other than those specified in the practical instructions, found that their characteristics were rather different. the mark scheme was modified to try to accommodate these variations, but in some cases the red and green diodes appeared to give virtually identical readings. this made things difficult for both candidates and examiners. in section (a) (v) most candidates scored 1 or 2 of the 3 marks available by noting that the resistance of the green diode was greater (in each case) than that of the red diode, and that the resistance of the diodes was not constant. (b)

(i) most candidates gained the precaution mark for either checking for zero error or taking repeat readings in different planes. a significant number either did not show any repeats or failed to calculate an average value.

(b)

(iii) this section discriminated well, with better candidates showing scale readings taken to millimetre precision. this enabled them to calculate the mass of the marble to within 0.3g of the supervisor’s value. candidates who attempted to use ‘g’ in their calculations often made arithmetic errors or confused ‘g’ with the 10g mass given. very few candidates used the diagram well, most failing to show (measurements to) the centres of mass.

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13.

(a) the supervisor has set up the apparatus shown in the diagram below. the newton meter is clamped vertically but the supervisor has not made the section bc of the string horizontal. A

n e w to n m e te r l h1

p u lle y w heel

0 .5 0 0 k g m ass

h2 B

la rg e m a s s to p re v e n t s ta n d tiltin g

C

G c la m p

do not move the stand that is clamped to the bench. adjust the separation of the stands and the height of the nail at a until the section bc of the string is horizontal and the angle θ is between 20° and 40°. you have been provided with a 30° set square so that you can easily estimate an angle in this range. explain how you checked that bc was horizontal. you may add to the above diagram if you wish. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2)

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(b)

measure the vertical height h1 of the point a above the bench and the vertical height h2 of the point b above the bench. also record the distance, l, between the points a and b. hence blank calculate a value for the angle θ using: sin θ = (h1 – h2) / l ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (4)

(c)

if the string is horizontal the principle of moments may be used to show that w = 2t tan θ – mg where kg, and

w = the weight of the metre rule, t = the reading on the newton meter, m = the mass which is suspended from the rule, which may be taken as 0.500 g = the gravitational field strength.

record t. t = …………………….. hence, using your result from part (b), calculate the weight of the metre rule. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... weight = ..................................... (4)

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(d)

set up the apparatus as shown in the diagram below using the second metre rule, which is identical to the suspended rule. use one of the 100 g masses from the first experiment. adjust the position of the pivot so that the system rests in equilibrium with the metre rule horizontal. c e n tre o f m a s s 100 g m ass b e n c h su rfa c e p iv o t

W

the centre of mass of the rule may be taken to lie at the 50.0 cm mark. take such measurements as are necessary to find the weight w of the rule. show these measurements on the diagram. now use the principle of moments to calculate w. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (5)

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(e)

the equation in part (c) may be rewritten in the form

1 1 mg + W 2 t tan θ = 2 a student suggests that the value of w may be found by a graphical method if the value of m is varied. plan an experiment to investigate this suggestion. your plan should include (i)

a description of how the experiment would be performed;

(ii)

a sketch of the graph to be plotted;

(iii)

how you would use the graph to find a value for w.

..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (9) (total 24 marks)

14.

(a)

measure the height of the string above the bench in two places (1)

use right angle of set square against bench to ensure vertical height measured or correct use of set square alone (1)

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2

27

(b)

h1 and h2 recorded to nearest mm or better with unit seen once (1) l = 98.0 cm ± 0.5 cm and height difference between 33 and 63 cm (1) correct calculation of sin θ (1) θ found with unit (1)

(c)

4

t recorded to 0.1 n and in range 3.5 n to 8.0 n (1) correct substitution [allow g = 10 n/kg] (1) correct calculation 2/3 s.f. + unit supervisor’s value ± 0.5 n

(d)

4

100 g mass > 20 cm from pivot (1) 2 distances or 3 scale readings shown on diagram with at least one reading to nearest mm (1) distances or scale readings shown to the centres of the mass and the rule (1) correct calculation of w or m [ignore units] (1) value ± 0.10 n of supervisor’s value with unit and > 2s.f. (1)

(e)

(i)

5

vary m

increase separation of the clamps/change height of the newtonmeter/change the height of the nail (1) string (bc) horizontal (1) record the reading on the newtonmeter/t (1) record the heights h1 and h2 (1) find/measure θ or sin θ (1) (ii)

max 5

using several/many/ 5 or more/ range readings (1) [5 or more can be scored from table or graph] plot suitable graph e.g. t tan θ against mg (1) straight line with positive gradient and with positive intercept [or intercept consistent with expected graph] (1)

(iii)

intercept = 1/2 w or consistent with graph (1)  If experiment in part(d) used. (1)  Vary m ( mgx = Wy) (1)  Plot suitable graph e.g. m against y/x (1)  Straight line through origin (1)  Suitable gradient e.g. W/g (1) 

9   Max 4     [24]

sample results (b)

h1 = 68.5 cm h2 = 10.5 cm l = 98.0 cm sin θ = (68.5 – 10.5)/98.0 = 0.592 ∴ θ = 36.3°

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(c)

(d)

t = 4.3 n –2 w = 2 × 4.3 n × tan (36.3) – 0.5 kg × 9.81 m s = 1.41 n 3 .0 c m

2 7 .6 c m

5 0 .0 c m

c e n tre o f m a s s

1 00 g m ass b e n c h su rfac e p iv o t W 100 (27.6 – 3.0) = m × (50 – 27.6) ∴ m = 109.8 g ∴ w = 0.1098 × 9.81 = 1.08 n 15.

good candidates showed, on the diagram, the vertical metre rule with the set square between it and the bench. many candidates showed this in two positions and gained the two marks without the need for a description. if it is suggested that candidates may add to the diagram if they wish, then often full marks can be obtained simply by adding full details to the diagram. most candidates obtained a correct value for the angle. typical errors from weaker candidates included: •

quoting the heights to the nearest cm,

•

obtaining a height difference that was outside the expected range because candidates had not followed the instruction to set the angle at approximately 30°.

most candidates obtained a correct value for w. typical errors from weaker candidates included: •

only recording the newton meter reading to the nearest newton,

•

quoting the calculated value of w to an inappropriate number of significant figures.

calculating the weight of the metre rule by balancing the rule on the knife edge produced a wide range of marks. good candidates obtained all 5 marks, but weaker candidates made the following errors: •

not showing the measurements on the diagram with sufficient precision,

•

calculating the mass of the rule rather than the weight,

•

obtaining a value outside the expected range, in most cases because the distances on the rule had not been measured to the centre of mass of the 100 g mass,

•

unit problems such as quoting the weight of the rule as 1.2 kg rather than 1.2 n.

many candidates omitted the required detail in the planning part of the experiment.

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candidates should not be afraid to describe the techniques that they have used in the first part of the second experiment in order to answer the planning part. the major weakness of all candidates is not explaining how procedures are carried out, for example most candidates stated that the string needed to be made horizontal but few went on to say that this could be achieved either by increasing the separation of the stands, or by changing the height of the nail, or by changing the height of the newton meter. despite hints in the stem of the question, weaker candidates had difficulty identifying the variables in the given equation. many plotted t against m without stating that θ would need to be held constant. such a graph could not score marks for an intercept of ½ w because the intercept is now ½ w / cos θ , hence candidates who plotted such a graph rarely gained any marks for the processing of the results.

16.

(a) (i) you have been provided with a 4.0 m length of constantan wire which has the same diameter as the constantan wire which is attached to the metre rule. using the top pan balance provided, measure the mass m of the 4.0 m length of wire. m = .................................................................................................................... carefully separate the turns of the 4.0 m length and measure the diameter d of the wire. state any precautions that you took to ensure that an accurate value of d was obtained. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (3)

(ii)

estimate the percentage uncertainty in your value for d. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2)

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(iii)

determine a value for the density of constantan given that

π d 2l volume of wire = v = 4 where l = length of wire = 4.0 m. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (5)

(i) set up the circuit as shown in the diagram below using the 100 ω resistor. before you connect your circuit to the power supply, have your circuit checked by the supervisor. you will be allowed a short time to correct any faults. if you are unable to set up the circuit, the supervisor will set it up for you. you will lose only two marks for this.

(b)

p o w e r s u p p ly 100 Ω

A

V c ro c o d ile c lip X 4 m m p lu g (2)

(ii)

connect the crocodile clip to the wire at the zero end of the rule. connect the power supply and use the 4-mm plug labelled x to make a connection to the wire at the 20.0 cm mark. to make good electrical contact the 4-mm plug should be pressed firmly against the wire. measure the current i in the circuit and the potential difference v across the 20.0 cm length of wire. i = ...................................................................................................................... v = ..................................................................................................................... (3)

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(iii)

hence calculate the resistance r1 of a 20.0 cm length of wire. ........................................................................................................................... ........................................................................................................................... (2)

(iv)

repeat parts (ii) and (iii) to find the resistance r2 of an 80.0 cm length of wire. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (3)

(v)

use your results from parts (iii) and (iv) to determine the resistance r of a length x = 60.0 cm of wire, where r = r2 – r1. hence determine a value for the resistivity ρ Rπ d 2 of the wire given that ρ = 4 x .

........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (4) (total 24 marks)

17.

(a) (i) d to 0.01 mm or better and ± 0.03 mm of supervisor’s [unit seen somewhere] repeat shown (2)

(ii)

other precaution e.g. measured in 2 perpendicular directions/ measured in different places/ used ratchet to avoid overtightening micrometer/ avoided kinks in wire (1)

3

sensible δd consistent with d (1) correct calculation of percentage (1)

2

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(iii)

sensible mass and correct formula for density (1) correct substitution into volume formula with consistent length units (1) density to 2/3 s.f. + unit (1) x

value 8.4 → 9.8 × 10 (1) correct x value (1) 5  e.g. − 3 for g mm -3    0 for g cm - 3    + 3 for kg m -3   (d)

(i) (ii)

circuit set up correctly without help

2

i to nearest ma or better and approximately 50 ma with unit (1) v to nearest mv or better and in range 50 mv to 90 mv with unit (1) v repeated (1)

(iii)

(iv)

3

correct calculation of r (1) V r from I to 2/3 s.f. + unit (1)

2

i ≤ i1, to nearest ma or better + unit (1) v = 3 × to 4 × v1 to nearest mv or better + unit (1) correct calculation of r ≥ 2 s.f. + unit (1) [only penalise a particular unit error once in (ii), (iii) and (iv)]

(v)

correct calculation of r (1) correct substitution for ρ with consistent length units (1) unit of ρ (1) –7 value 4.2 → 5.2 × 10 Ω m and ≥ 2 s.f. (1)

3

4 [24]

sample results (a)

(i) m = 2.74 g d = 0.31, 0.31, 0.31 mm d = 0.31 mm (ii)

δd = 0.01(mm)

0.01 mm percentage uncertainty = 0.31 mm × 100% = 3.2(%) (iii)

V=

π (0.31 × 10 - 3 m) 2 × 4.0 m 4 –7

3

= 3.02 × 10 m

2.74 × 10 -3 kg −7 3 ρ = 3.02 × 10 m = 9080 kg m

–3

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(b)

(ii) i = 49 ma v = 89, 90, 88 mv V = 89 mv (iii) (iv)

89 mV r1 = 49 mA = 1.82 Ω m i = 48 ma v = 266, 265, 267 mv V = 266 mv

V 266 mV I r2 = = 48 mA =5.54 Ω (v)

r = 5.54 Ω – 1.82 Ω = 3.72 Ω 3.72Ω × π × (0.31 × 10 −3 m) 2 4 × 0.6 m ρ= –7

= 4.7 × 10 Ω m 18.

this question in group 2 proved to be more straightforward the a question in group 1. however this was compensated for by a more difficult question b. the mean marks and standard deviations for the two groups were approximately the same. the measurement of the diameter of the wire proved to be more straightforward than the measurement of the thickness of the foil. sufficient precautions were allowed to enable most candidates to score full marks in the section. in the same way the measurement of the percentage uncertainty was straightforward and the majority of candidates scored the full 2 marks for this section. the volume calculation did not have the problem of the confusion between 16t and t, however weaker candidates still had expressions that contained a mixture of units, normally l in metres and d in mm. this lead to a density value which had an incorrect power of 10 but the majority of candidates could obtain the first of the two value marks. contact resistance was less critical in this group but examiners still expected candidates to repeat their measurement of v, few did this. the calculation of resistance produced similar problems to those described in group 1. the calculation of resistivity lead to the same problems as described in group 1. most candidates determined a correct value for r but weaker candidates mixed units of mm and cm in the formula and did not give the correct units for resistivity. because of the problems of the foil in group 1, any correctly calculated value of resistivity was allowed, here candidates had to obtain an answer within a range of about + 10% of the true value, good candidates had no difficulty with this but weaker candidates often obtained values outside the range.

19.

(a) determine the position of the centre of mass of the metre rule labelled x by balancing it on the pivot so that it is approximately horizontal. position of centre of mass = ....................................................................................... (1)

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(b)

set up the apparatus as shown in the diagram below using the metre rule labelled x. the system should rest in equilibrium with the metre rule approximately horizontal. c e n tre o f m a s s o f ru le 100 g m ass bench

p iv o t W

take such measurements as are necessary to find the weight w of the rule. show these measurements on the diagram. now use the principle of moments to calculate w. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (6)

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(c)

set up the apparatus as shown in the diagram below using the metre rule labelled x with the nail passing through the hole at the 1.0 cm mark. the loop of thread from which the 1.00 kg mass is suspended should be placed in the position of the centre of mass of the rule.

n e w to n m e te r

n a il s u p p o rtin g ru le x

y 1 kg m ass bench adjust the height of the newton meter until the metre rule is horizontal. explain how you ensured that the rule was horizontal. you may add to the above diagram if you wish. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2)

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(d)

by applying the principle of moments to the horizontal rule it can be shown that: (w + mg)x = ty where and

w = the weight of the metre rule, mg = weight of the 1.00 kg mass = 9.81 n, t = the reading on the newton meter, x and y are the lengths shown in the diagram.

record the reading on the newton meter and determine values for x and y. hence calculate a second value for w. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (4)

(e)

calculate the percentage difference between your two values for w. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... which value of w do you consider to be more accurate? give a reason for your answer. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (3)

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(f)

a student wishes to investigate how the force supporting the rule varies with the distance of the newton meter from the nail. the equation in part (d) may be rewritten in the form

(W + mg ) x t=

y

you are to plan this investigation. your plan should include (i)

an indication of the values in the equation which would need to be kept constant;

(ii)

a description of how the experiment would be performed;

(iii)

a sketch of the graph to be plotted;

(iv)

an indication of the expected results.

..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (8) (total 24 marks)

20.

(a)

recorded to the nearest mm with unit (1)

ealing, hammersmith and west london college

1

38

(b)

100 g mass > 20 cm from pivot (1) 3 scale readings or 2 clear distances shown on diagram [at least one to nearest mm] (1) distances or scale readings shown to centres (1) correct calculations of w with unit (1) value ± 0.05 n of supervisor’s value and > 2 s.f. (1) (1) [if ± 0.10 n (1)]

6

[if mass unit used for w, penalise calculation of w but give range marks appropriate to value] –2

[allow g = 10 m s in (b)] (c)

(d)

measure the height of the rule above the bench in 2 places (1) use the right angle of the set square against the bench to ensure the vertical height is measured (1)

2

t recorded to 0.1 n or better with unit and 2/3 s.f. (1) [and approximately 6 n] x and y recorded to nearest mm or better + unit (1) correct calculation of w with unit and 2/3 s.f. (1) value ± 0.5 n of supervisor’s value (1)

(e)

4

correct calculation of percentage difference (1) first value is more accurate (1) because newtonmeter uncertainty is likely to be large (1) or because the effect of the suspended 1 kg mass on the (1) newtonmeter reading > >effect of mass of rule

(f)

(i) (ii)

3

keep m and x constant (1)

move position of newtonmeter (1) obtain different values of y (1) ensure newtonmeter is vertical (1) adjust height of newtonmeter/clamp to make rule horizontal (1) record reading on newtonmeter (1) (obtain) several/many/5 or more/ range of reading (1) [5 or more can be scored from table or graph] adjust newtonmeter to set zero correctly (1)

(iii)

plot t against 1/y (1)

(iv)

straight line through origin expected (1) slope = (w + mg) x (1) [wrong experiment max 5]

8 [24]

sample results (a)

position of centre of mass = 50.0 cm

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(b)

3 .0 c m

2 7 .6 c m

5 0 .0 c m

c e n tre o f m a s s o f ru le

1 00 g m ass bench

p iv o t

W –2

0.1 kg × 9.81 m s × (27.6 cm – 3.0 cm) = w (50.0 cm – 27.6 cm) w = 1.08 n (d)

(e)

t = 5.9 n x = 50.0 cm – 1.0 cm = 49.0 cm y = 95.0 cm – 1.0 cm = 94.0 cm (w + 9.81 n) 49.0 cm = 5.9 n × 94.0 cm w+ 9.81 n = 11.32 n ∴ w = 11.32n – 9.81n = 1.51 n

1.51 N – 1.08 N percentage difference = 2 (1.51 N + 1.08 N) × 100 % 1

= 33 (%) 21.

sections (a) to (d) in this group have comparable sections in group 1 and the comments made there are equally valid in this group. in the calculation of the percentage difference examiners insisted that the average value of the weight of the rule was used in the denominator of the percentage difference calculation, this meant that only a small proportion of candidates gained the mark. the second part of this section did not discriminate between candidates. most thought, incorrectly, that the horizontal metre rule with the newton meter attached would give the most accurate value because it was more stable, they ignored points such as •

the mass of the rule had a very small effect on the newton meter reading because the 1 kg mass was so much greater than the mass of the rule.

•

the newton meter could only be read to a precision of 0.1 n in about 6.0 n.

•

the newton meter is far less accurate than the metre rule.

candidates were clearly worried about the fact that it was difficult to balance the metre rule on the knife edge. there was no appreciation of the fact that, say, a 1 mm movement of the pivot could produce a change from a resultant clockwise moment to a resultant anticlockwise moment so that the distance measurements were very precise, say, a 1 mm error in 200 mm. candidates do not need to waste time achieving perfect balance, in fact a 1 mm movement of the pivot will change from resultant clockwise to resultant anticlockwise moment. also the calibration of the metre rule is likely to be far more accurate than that of the newton meter.

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