Mark Scheme and Instructions for A-level Physics Exam Questions, Study notes of Physics

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AS

Physics

7407/2 - Paper 2

Mark scheme

June 2018

Version/Stage: 1.0 Final

Physics - Mark scheme instructions to examiners

1. General

The mark scheme for each question shows:

 the marks available for each part of the question  the total marks available for the question  the typical answer or answers which are expected  extra information to help the Examiner make his or her judgement and help to delineate what is acceptable or not worthy of credit or, in discursive answers, to give an overview of the area in which a mark or marks may be awarded.

The extra information is aligned to the appropriate answer in the left-hand part of the mark scheme and should only be applied to that item in the mark scheme.

At the beginning of a part of a question a reminder may be given, for example: where consequential marking needs to be considered in a calculation; or the answer may be on the diagram or at a different place on the script.

In general the right-hand side of the mark scheme is there to provide those extra details which confuse the main part of the mark scheme yet may be helpful in ensuring that marking is straightforward and consistent.

2. Emboldening

2.1 In a list of acceptable answers where more than one mark is available ‘any two from’ is used, with the number of marks emboldened. Each of the following bullet points is a potential mark. 2.2 A bold and is used to indicate that both parts of the answer are required to award the mark. 2.3 Alternative answers acceptable for a mark are indicated by the use of or. Different terms in the mark scheme are shown by a / ; eg allow smooth / free movement.

3. Marking points

3.1 Marking of lists

This applies to questions requiring a set number of responses, but for which candidates have provided extra responses. The general principle to be followed in such a situation is that ‘right + wrong = wrong’.

Each error / contradiction negates each correct response. So, if the number of errors / contradictions equals or exceeds the number of marks available for the question, no marks can be awarded.

However, responses considered to be neutral (often prefaced by ‘Ignore’ in the mark scheme) are not penalised.

3.2 Marking procedure for calculations

Full marks can usually be given for a correct numerical answer without working shown unless the question states ‘Show your working’. However, if a correct numerical answer can be evaluated from

An answer in surd form cannot gain the sf mark. An incorrect calculation following some working can gain the sf mark. For a question beginning with the command word ‘Show that…’, the answer should be quoted to one more sf than the sf quoted in the question eg ‘Show that X is equal to about 2.1 cm’ – answer should be quoted to 3 sf. An answer to 1 sf will not normally be acceptable, unless the answer is an integer eg a number of objects. In non-practical sections, the need for a consideration will be indicated in the question by the use of ‘Give your answer to an appropriate number of significant figures’.

3.9 Unit penalties

An A-level paper may contain up to 2 marks (1 mark for AS) that are contingent on the candidate quoting the correct unit for the answer to a calculation. The need for a unit to be quoted will be indicated in the question by the use of ‘State an appropriate SI unit for your answer ’. Unit answers will be expected to appear in the most commonly agreed form for the calculation concerned; strings of fundamental (base) units would not. For example, 1 tesla and 1 Wb m–^2 would both be acceptable units for magnetic flux density but 1 kg m^2 s–^2 A–^1 would not.

3.10 Level of response marking instructions

Level of response mark schemes are broken down into three levels, each of which has a descriptor. The descriptor for the level shows the average performance for the level. There are two marks in each level.

Before you apply the mark scheme to a student’s answer read through the answer and annotate it (as instructed) to show the qualities that are being looked for. You can then apply the mark scheme.

Determining a level

Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the descriptor for that level. The descriptor for the level indicates the different qualities that might be seen in the student’s answer for that level. If it meets the lowest level then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer. With practice and familiarity you will find that for better answers you will be able to quickly skip through the lower levels of the mark scheme.

When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level and then use the variability of the response to help decide the mark within the level. i.e. if the response is predominantly level 2 with a small amount of level 3 material it would be placed in level 2.

The exemplar materials used during standardisation will help you to determine the appropriate level. There will be an answer in the standardising materials which will correspond with each level of the mark scheme. This answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer with the example to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner’s mark on the example.

You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate.

Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other valid points. Students do not have to cover all of the points mentioned in the indicative content to reach the highest level of the mark scheme

An answer which contains nothing of relevance to the question must be awarded no marks.

OR

(and ) set A ’s average velocity is smaller / set A ’s velocity at gate 1 is smaller/ Set A ’s velocity is smaller at both gates 2 

Alternative Method

values of u and v are

calculated ( condone POT

error ) and corresponding

values for each s

determined; 1  a comparison of their distances leading to conclusion that set B

produced when s is

largest OR

ratio ( 1 2

1 2

3 t t

t t

t

 ) is

proportional to distance s and B’s ratio is greater 2 

3

( ) t

u v

s 

 OR

a

v u

s

 ; OR

( s = ) 1 2

1 2

3 t t

t t

t

u /ms-^1 v /ms-^1 s /m

𝑣^2 − 𝑢^2 /

𝑚^2 𝑠−^2

set A 0.16 4 0.2 38 0.3 56 0.

set B 0.18 1 0.2 70 0.4 76 0.

1 2

1 2

3 t t

t t

t

set A 7. 12

set B 9.

Allow ecf for acceleration where used to find s Using a =0.042: sA = 0.354 and sB= 0. Treat a larger change in velocity as neutral

Question Answers Additional Comments/Guidelines Mark

Continuous, ruled straight best fit line through 1st^ and last points 1 

Gradient from

step

step

x

y

seen

and G = 0.045 range (0.042 to

n =4 point below and n =7 above, other points cut by line of best fit. Line must not be thicker than half a square grid Line must have no variation in thickness Do not accept more than one line drawn, do not accept discontinuities

steps at least half the height and half the width of the grid; (at least 3 squares horizontally and at least 5 squares vertically)

allow

change in y

change in x

where points are on line and are at

least half drawn line apart (∆𝑥 ≥ 3 and ∆𝑦 ≥ 0. 175 )

Ignore any units given for G Allow 1 sf answers of 0.04 or 0.05 where correct working is shown.

their

G



( h = 9.2 × 10 −^3 m)

Ecf from part 1. Expect 2 sf normally. Penalise 3 or more sf.

Condone 1 sf answers where correct working is shown in part 4.1 and where their G is quoted to 1sf. In this case, allow use of their rounded G or full carry value.

idea that the intercept can be found by calculating a  Gn where a and n are values read-off (from a point on the line) and G is the gradient ; intercept compared to 0, 0 (OWTTE in a general y=mx +c description)

or

Read-off points (of line of best fit for) x 1 and x 2 compare with corresponding y 1 and y 2 , compares the ratio of the x terms to the ratio of the y terms; if equal then directly proportional or Determine the constant of proportionality for at least two points (on line of best fit) and compare, where constant exists then directly proportional. 

simply explaining how to find the intercept does not fully answer the question and gets no credit must describe the comparison aspect; do not accept idea of extrapolation off the grid or re-plotting on axes that include (0, 0)

Idea that a and n will share a common factorial increase. (^1)

Total 9

substitution of D = 59.90, d = 19.32 and t = 12. into

t

D d V (^)  

 

 

   4 4

^2 ^2

OR

𝑉 = their ∆𝐴 × 12. 09

OR Correctly finds difference in their volumes 2 

3.1 x 10-^5 / 3.05 x 10-^5 / 3.05 3 x 10-^5 (m^3 ) 3 

Or equivalent Correct substitution into

t

D d V (^)  

 

 

   4 4

^2 ^2

receives the

first two marks (allow POT) Expect values: VD = 3.41 x 10 -^5 (m^3 ) Vd = 3.54 x 10 -^6 (m^3 )

no limit on maximum sf

Correct answer scores 3. Allow 3rd^ sf round error where answer rounds to 3.1 x 10-^5 when correct method seen

Question Answers

Additional Comments/Guidelines Mark

Procedure: MAX 2 Take more measurement(s) of h for additional / different masses (of clay)  Convert (total) mass into weight (and equal to the repulsive force of magnet A on magnet B)  Describe method to measure h using ruler or set square 

Analysis : Plot a graph of F against 1/ h^3 

Should be a straight line of best fit 

Determination of k : MAX 1 Measure gradient and set equal to k 

More than one added mass, allow varies amount of clay

condone 1/ h^3 against F or equivalent (in this case determination of k must be consistent with graph)

This mark can be awarded if seen by drawing of straight line with positive gradient on sketch of graph.

Allow one mark for plot of F against h^3 and statement that area under graph is k. Mark Procedure as scheme.

5

Substitute (total) weight into formula and rearrange to find k 

Must be consistent with graph

Total 11

Quest ion

Answers Additional Comments/Guidance Mark

Mass of alpha particle =

19 7

 ^ 

=6.6(53) ×

10 −27^ (kg)

OR

Correctly re-arranged k.e. equation (with v^2 or v as subject) with 8.1 x 10 -^13 (J) substituted correctly 1 

1.56 × 107 seen 2 

Allow mass = 2 × m p + 2 × m n = 6. 696

× 10 -^27 kg

Allow mass = 4 x 1.66 x10-^27 kg =

6. 64 × 10 -^27 kg

Allow mass = 4 x 1.67 x10-^27 kg =

6. 68 × 10 -^27 kg

Allow slight rounding on mass (must be correct to 2 sf)

Condone incorrect mass in

otherwise correct substitution with v

or v^2 recognisable as subject.

Alternative approaches are:

v E k^ specific charge

e

v^2 E k

m 

Must see answer to at least 2 sf

Must see attempt to use one of the alternative approaches to support correct answer.

2

8.1 x 10-^13 ÷ 1.785 × 10^5

Or

5.06 x 10^6 ÷ 1.785 × 10^5 seen 2 

28 (.4) (eV) 3 

8.1 x 10-^13 ÷ (5.1 × 10^4 x 3.5) is

worth 1st^ and 2nd^ marks

Condone POT errors in second mark

Correct answer obtains 3 marks

99(.3) (eV) scores 1 mark

03.4 ( Q^ =)^ 0.^85 × 10

−3 × 1.2 × 10−9 = 1.02 × 10−1 2

OR n = (their Q ) ÷ 1.6 × 10−19^1 

n = 6.4 × 10^6 (c.a.o.) 2 ^

Condone one POT error for one mark_._

2

At 3.5 cm the pd drops / the current begins OR

When the source is 10 cm away no ionisation

occurs in the air gap (because the alpha

particles have insufficient range to reach the air gap)

OR 

When the radioactive source is close enough

(approx. 5 cm) ionisation occurs 

OR

When beyond 3.5 cm no change in pd / current equals zero  When ionisation occurs / charge carriers are liberated in the air gap:

resistance has decreased OR current increases (from zero) OR the potential difference decreases (with a maximum current) (to its minimum value) (across the air gap)

Must be sense of abrupt change

MAX 3

Allow more ionisation for second mark

3

From 10 cm separation until 5 cm (approx) separation nothing changes / appreciates that pd is 4500 V / pd across gap = 4500 V until ionisation occurs   

Current is produced: the pd across 5 MΩ

resistor is 4250 V / most pd is across the 5 MΩ

resistor / small pd across air gap 

Current is produced and the pd across the air gap is 250 V  

Current is produced and the pd across the air gap is

250 V

Ques tion

Answers Additional Comments/Guidance

Mark

Use of n A =

c

c A

to make c A the subject of the equation

or

speed in glass A = 2.05( 2 )× 10^8 ms−^1 1 

Speed in glass B = 1.98 5 ( 3 ) × 108

or

their speed in glass A × 0.967 48 ( or equivalent )

or

Alternative 1st^ and 2nd^ marks

Use of n A/ n B = c B/ c A by substitution for 𝑛𝐴 1 

Use of n A/ n B = c B/ c A by substitution for 𝑛𝐴

and c B = c A × 0.96748 2 

Or

n B = 1.461 / 0.96748 1  2 

Condone truncation without appropriate rounding mid- calculation.

Condone use of c= 3x10^8 But must see answer to 4 sf answer

Values obtained using c= 3x10^8 :

 speed in glass A=

2.05(3)× 10^8 ms−^1

 speed in glass B =

1.98(7) × 108

 n= 1.

watch for maths errors: dividing by 1.03252 ≠ multiplying by 0.

Keys to Objective Test Questions (each correct answer is worth 1 mark)

Q 5 6 7 8 9 10 11 12 13 14 15 16 17

A B C B C C D D C B D D A D

Q 18 19 20 21 22 23 24 25 26 27 28 29 30

A C A C C B B D A D A A D C

Q 31 32 33 34

A A A B B