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Cambridge International AS & A Level

MATHEMATICS 9709/43
Paper 4 Mechanics October/November 2024
MARK SCHEME
Maximum Mark: 50

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge International will not enter into discussions about these mark schemes.

Cambridge International is publishing the mark schemes for the October/November 2024 series for most
Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level
components.

This document consists of 19 printed pages.

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Generic Marking Principles

These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the
mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles.

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:
• the specific content of the mark scheme or the generic level descriptors for the question
• the specific skills defined in the mark scheme or in the generic level descriptors for the question
• the standard of response required by a candidate as exemplified by the standardisation scripts.

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:
• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the
syllabus and mark scheme, referring to your Team Leader as appropriate
• marks are awarded when candidates clearly demonstrate what they know and can do
• marks are not deducted for errors
• marks are not deducted for omissions
• answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as
indicated by the mark scheme. The meaning, however, should be unambiguous.

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

GENERIC MARKING PRINCIPLE 5:

Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited
according to the quality of the candidate responses seen).

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in
mind.

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Mathematics-Specific Marking Principles

1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required
then no marks will be awarded for a scale drawing.

2 Unless specified in the question, non-integer answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the
degree of accuracy is not affected.

3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points.

4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).

5 Where a candidate has misread a number or sign in the question and used that value consistently throughout, provided that number does not alter the
difficulty or the method required, award all marks earned and deduct just 1 A or B mark for the misread.

6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.

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Mark Scheme Notes

The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons
outside the scope of these notes.

Types of mark

M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units.
However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea
must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula
without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method
mark is earned (or implied).

B Mark for a correct result or statement independent of method marks.

DM or DB When a part of a question has two or more ‘method’ steps, the M marks are generally independent unless the scheme specifically says otherwise;
and similarly, when there are several B marks allocated. The notation DM or DB is used to indicate that a particular M or B mark is dependent on
an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full
credit is given.

FT Implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are
given for correct work only.

• A or B marks are given for correct work only (not for results obtained from incorrect working) unless follow through is allowed (see abbreviation FT above).
• For a numerical answer, allow the A or B mark if the answer is correct to 3 significant figures or would be correct to 3 significant figures if rounded (1
decimal place for angles in degrees).
• The total number of marks available for each question is shown at the bottom of the Marks column.
• Wrong or missing units in an answer should not result in loss of marks unless the guidance indicates otherwise.
• Square brackets [ ] around text or numbers show extra information not needed for the mark to be awarded.

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Abbreviations

AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent

AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)

CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)

CWO Correct Working Only

ISW Ignore Subsequent Working

SOI Seen Or Implied

SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)

WWW Without Wrong Working

AWRT Answer Which Rounds To

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Question Answer Marks Guidance

1 1 B1 For either correct.

KEbefore =  m  52
1
2 Do not allow  m  ( 6 − 5) .
2

1 2
KE after =  m  62
2 1
Note: Difference =  m  11 .
2

WD against resistance B1 Do not allow if errors such as e.g. ( m − 24 )  50 .

= 50  24  = 1200 

( )
1 M1 Attempt at work energy equation with 4 relevant terms;
 m  62 − 52 + 50  24 = 1541 dimensionally correct. Allow sign errors.
2
5.5m = 1541 −1200 1
M0 for  m  ( 6 − 5) .
2

341 A1
m= = 62
5.5

SC for constant acceleration method (question only gives total work done, and does not suggest constant force)

a = 0.11 SCB1 From 62 = 52 + 2  a  50 SOI.

m = 62 SCB1 1541
From − 24 = m  0.11 or 30.82 − 24 = m  0.11.
50

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Question Answer Marks Guidance

2 Resolving either direction M1 With correct number of relevant terms. Allow sin/cos mix. Allow
sign errors. Do not allow ‘forces to the left = forces to the right’
e.g. 12cos30 − 8sin30 = 16sin30 unless subsequently ‘corrected’.

 (12sin30 + 24 + 8cos30 − 16cos30 )  = Fx or F cos or F sin   A1

(
Fx =  30 − 4 3 )  = 23.07 .

 (12cos30 − 8sin30 − 16sin30) = Fy or F sin  or F cos   A1

(
Fy =  6 3 − 12 ) = −1.607 .

(6 ) + (30 − 4 3 )
2 2 M1 Attempt to find F.
F= 3 − 12 Must have correct number of relevant terms. (Forces must have or
not have components as required). All forces resolved/not resolved
30 − 4 3
F= as appropriate, but allow consistent sin/cos muddle.
cos(their  )
6 3 − 12 Allow use of their  provided correctly derived from equations
F=
sin ( their  ) with the correct number of relevant terms.

 6 3 − 12  M1 Attempt to find  .
 = tan −1   Must have correct number of relevant terms. (Forces must have or
 30 − 4 3  not have components as required). All forces resolved/not resolved
 30 − 4 3  as appropriate, but allow consistent sin/cos muddle. Allow upside
 = cos −1   Note: this will not give the correct answer
their F  30 − 4 3 
  down so tan −1 
 6 3 − 12 
.
unless F given to several significant figures  
 6 3 − 12  Allow use of their F provided correctly derived from equations
 = sin −1   with the correct number of relevant terms.
 their F 
 6 3 − 12   1.607 
Note: watch for use of sin −1  or sin −1 
 30 − 4 3   which
 23.07 
 
leads to correct answer of angle 4.0° scores M0A0.

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Question Answer Marks Guidance

2 F = 23.1 N A1 [23.1277…] Both correct Allow 4.0° but not simply 4° .

 = 3.99 above the negative x-axis oe 3.986... Allow answers about the direction such as ‘Above the
west’, ‘north of west’ etc, or clockwise 183.99 from x axis, or
resultant sketch with angle indicated. If not specified in working
please check original diagram to see if direction specified there
instead. Allow a bearing of 274 .0 °. Allow explanation of direction
that could be drawn uniquely.
Or e.g. 86.0° to left of the y-axis or 176.0° from the positive x-
axis.

Question Answer Marks Guidance

3(a) Resolving up slope. M1 Must have correct number of relevant terms (weight component
If correct should see and 240 N resistance). Allow sign errors. Allow cos 4.58 or cos
D F = 240 + 1600 g  0.08  = 240 + 1280 = 1520 4.6. Do not allow g missing. Using sin −1 0.08 or sin 0.08 scores
M0B0A0.
Must have either 0.08 or sin 4.58 or sin 4.6, not just sin .

Power = their (1520 )  32 B1 Power

OE. E.g. = their 1520 .
32
Allow any driving force provided it has a resistance and a weight
component.

Power = 48640 W A1 Allow 48 600 W or 48.64 kW or 48.6 kW. Must state units if given
in kW.

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Question Answer Marks Guidance

3(b) 0.95  their 48640 46208 5776 B1FT Power

DF= or = or 1925.3 DF = oe e.g. 0.95 their 48640 = DF × 24
24 24 3 v
FT their power from part (a) Do not allow if not using power from
part (a)
Note: candidates who use sin 0.08 in part (a) should get a DF of
332.3 N, which can score B1FT and use of sin −1 0.08 should get a
DF of 93299 N, can score B1FT. If candidate uses 48600 DF =
46170
= 1923.75
24
Candidates who omit the weight component in part (a) should get
a DF of 304 N, and can score B1FT

their DF − 240 − 1600 g  0.08 = 1600a M1 N2L Must have correct number of relevant terms (weight
component and 240 N resistance). Allow sign errors. Must be
dimensionally correct. Allow without using 95% or with using
5%. Must have either 0.08 or sin 4.58 or sin 4.6, not just sin
or sin −1 0.08 or sin 0.08.

a = 0.253 ms −2 A1
Allow
19
Note: 0.25 scores A0.
75
If candidate uses 48600 they must get 0.252(34…) rather than
0.253.

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Question Answer Marks Guidance

4 For attempt at use of conservation of momentum. *M1 Must have three non-zero terms. Allow sign errors. Must have
correct masses with relevant velocities. Their v may be in the
opposite direction.
If g included with the masses:
Allow M1A0A1 for first three marks. Can then score M1M0A0
for final three marks.

3  8 = 3  2 + 6v A1
Or 3  8 = −3  2 + 6v

v=3  v = 5 A1 Allow finding only v = 3, but if both speeds found they must both
be correct

KE after DM1 Allow use of any vB even if  0

= 0.5  3  22 + 0.5  6  their vB 2  = 33 if correct Using speed of5, KEafter = 0.5  3  22 + 0.5  6  their 52  = 81
Candidates may work out the loss for each particle separately
which only scores DM1 when losses subsequently added together.

( (
KEloss = ± 0.5  3  82 − 0.5  3  22 + 0.5  6  their 32 )) DM1 If two speeds found then FT their lower speed, even if later choose
the wrong loss.
 = 96 − 33 if correct If only one speed found then FT their speed.
If the candidate thinks that the particles coalesce then score M0
here

Loss = 63 [J] A1 If both losses found, must state which is the greater.
Only award this mark if no errors – e.g. the KE loss for v = 5
should be 15 J. If wrong then A0.
Allow −63 [J].
Can score full marks even if only v = 3 is found.

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Question Answer Marks Guidance

5(a) Attempt at resolving in at least one direction *M1 Correct number of relevant terms with T resolved; allow sign
errors; allow sin/cos mix.
Can score M1 for any F = Tcos30 .
Do not allow g missing in the equation for R. Must have 12, not
just m.
Could see R as part of an equation for F. E.g.
F = 0.5 (12 g − T sin 30 ) .

R + Tsin30 = 12g A1 Both correct.

F = Tcos30

Use of F = 0.5R to form an equation in T or R only *DM1 Allow sign errors in R; allow consistent sin/cos mix in R but no
other errors. Must be two term R as a linear combination of weight
and a component of T, and F must be a single term which is a
component of T. Do not allow g missing.
3
If correct T cos30 = 0.5 (120 − T sin 30 ) or T = 60 − 0.25T .
2
If no working shown to eliminate T or R, then DM2 for getting T
value correct for their equations and A1 if fully correct. Could use
0.5R = Tcos30 and solve simultaneously.

Attempt to solve for T DM1 Allow consistent sin/cos mix and allow sign errors. Must get to
'T = ' .
Dependent on both previous M1s.

T = 53.8 N A1 53.7622
Note: For sign errors:
R − Tsin30 = 12g answer should be 97.3985…
R − Tsin30 = −12 g answer should be -97.3985…
R + Tsin30 = −12 g answer should be -53.7622……
Each of the above would usually get M1A0M1M1A0.

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Question Answer Marks Guidance

5(b) Tcos30 − F = 12  0.2 *M1 Attempt at N2L; correct number of relevant terms with T resolved;
allow sign errors; allow sin/cos mix, but can be F or any
reasonable attempt at friction.

Use of F = 0.5R to form an equation in T and solve DM1 Must be a two term R as a linear combination of weight and a
component of T. Allow sign errors and consistent sin/cos mix.
Must get to 'T = ' .
The equations if correct should be
T cos30 − 0.5 (120 − T sin 30 ) = 12  0.2
3  3 
or T − 60 + 0.25T = 2.4 or T  + 0.25  = 62.4 and these
2  2
 
must be solved.
Any use of T or R from part (a) scores DM0 here.

T = 55.9 N A1 T = 55.9127
Note: For sign errors:
R − Tsin30 = 12g answer should be 101.294…
R − Tsin30 = −12 g answer should be –93.5026…
R + Tsin30 = −12 g answer should be –51.6117…
Each of the above would usually get M1M1A0.

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Question Answer Marks Guidance

6(a) 0.6 2 B1 0.6 2

v =  0.6t dt = t  = 0.3t 2   +0 AG Must see t or 0.3t 2 and 4 must be actually shown
2  2
t = 4 V = 0.3  42 or 0.3 16 = 4.8 substituted. Merely stating t = 4 is not enough to score this mark.
1
Do not allow use of s = ut + at 2 which leads to 4.8 if a = 0.6 is
2
used.

1
5 v

6(b) Quadratic with correct curvature starting from (0, 0) to (4, 4.8). B1 4

t
2 4 6 8 10 12 14 16 18 20

Note: the grid is for reference – not shown in QP.

Their graph does not need to be to scale.

Horizontal line at from (4, 4.8) to (15, 4.8) B1 The points should be specified somehow, but for an accurate
sketch allow a line just below 5 without specifying 4.8.

Line from (15, 4.8) to (20, 0) B1 The points should be specified somehow, but for an accurate
sketch allow a line just below 5 without specifying 4.8. Allow all 3
marks if using V instead of 4.8. ISW any extra out of the range for
t of 0 to 20.
If no marks scored then SC B1 for a correct shaped graph with no
numbers.
If using a value of v  4.8 , allow SC B1 for the first section
correct and SC B1 for the second and third both correct.

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Question Answer Marks Guidance

6(c) Attempt to find acceleration M1 0 − 4.8

For calculation oe  = −0.96 Allow +0.96 .
20 − 15

v =19.2 − 0.96t A1 Oe e.g. Allow  v = 4.8 − 0.96 ( t − 15 ) .

6(d)  4 2  0.3 3 *M1 Attempt to integrate their v from part (a) provided this came from
 
 00.3t dt  3 t  = 0.1t 

3
integration, but allow a restart here. The power of t must increase
by 1 with a change of coefficient. Use of s = vt scores M0. No
need for limits.
If no integration seen allow SCM1 for answer of 6.4 in place of
M1M1.

0.1 43  −0.1 03  DM1 Correct use of correct limit(s) (expect 6.4).

4.8 11 + 0.5  4.8  5 B1 Both correct and added. May be done in one go using a trapezium
 = 52.8 + 12 = 64.8 (11 + 16 )  4.8
.
2
Could do the last stage by integration.
Maximum B1 for final answer 74.4 from thinking the first section
is also straight.

Distance = 71.2 m A1

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Question Answer Marks Guidance

7(a) T − 3g = 3a *B1 For one correct equation.

5 g − T = 5a
*B1 For any two correct consistent equations. If tensions TA and TB
5 g − 3g =  ( 3 + 5) a
both stated must at some point state or imply that they are equal to
score the second B1.

Attempt to solve for T DM1 May find a first  a = 2.5 Must get to 'T = ' . Dep on both B
marks.

75 A1 Allow without working.

T = 37.5 N or N
2

Correct use of suvat with their a and solve for t DM1 E.g. 2 = 0.5  2.5  t 2 Must get to ' t = ' . Dep on both B marks or
the B1 for the equation 5 g − 3 g =  ( 3 + 5 ) a if this used to find a,
but not on first M1 mark.
0 + their 10
Could find v = 10 then use 2 = t.
2

40 2 10 A1 t = 1.2649 If candidates do not try to find T but do attempt to

t =1.26 or or find the time, they can score B1B1M0A0M1A1. Do not allow if
5 5
also give negative answer and do not discard. Note:
t = 1.6 only, scores A0 .

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Question Answer Marks Guidance

7(a) Alternative for final 2 marks, even if nothing scored earlier

Use of energy to find velocity at plane and then suvat to find t M1 1 1

PE loss = KE gain: 5g  2 − 3g  2 =  5v 2 +  3v 2 ,
2 2
Or PE loss – WD by tension = KE gain: 5g  2 − their 37.5  2 =
1
 5v 2 ,
2
Or WD by tension – PE gain = KE gain:
1
their 37.5  2 − 3g  2 =  3v 2 ,
2
0+v
v = 10 or v = 3.16 then 2 = t,
2
Dependent on both B marks only if candidate uses tension, but
otherwise not dependent on either B mark.

40 2 10 A1 t = 1.2649
t  = 1.6  =1.26 or or
5 5

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Question Answer Marks Guidance

7(b) Correct use of suvat before B hits the plane or when A has risen 2 *M1 v 2 = 0 + 2  2.5  2 . Must use s = 2 and u = 0 ,
m, to attempt to find velocity of A when string becomes slack
Using their a and/or their t. OR v = 0 + 2.5  1.6 Must use u = 0 ,
1
OR 2 = ( 0 + v ) 1.6 Must use s = 2 and u = 0 .
2
Must be complete method to find v or v 2 v = 10  .

Could have found v = 10 in part (a) and give M1 if used in part

(b).

Correct use of suvat for motion of A (between height of 3m and

3.25 m), to form an equation in t, using a = −g
*DM1
( )
3.25 − 3 = their 10 t + 0.5  ( −10 )  t 2 .

Solving a 3 term quadratic for t to get at least one (unsimplified) DM1 10 − 5 10 + 5

value using their 2.5 (or using any other correct method) If correct should get t = 0.093, 0.540 or , ,
10 10
0.09262… or 0.53978…
Could use formula and realise that the time for at least 3.25 m,
b 2 − 4ac 10 − 4  5  0.25
=2 =2 which gets DM1.
2a 25

5 1 A1 Time = 0.53978− 0.09262 = 0.44721…

Time = 0.447 s or s or
5 5  10 10 − 5 
2  −  = 0.44721…
 10 10 

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Question Answer Marks Guidance

7(b) Alternative for last 3 marks of Q7(b) finding max height

For attempt to find max height using correct suvat with a = −g *DM1
( )
2
02 = their 10 + 2  ( −10 ) s Where s is distance above 3
metres.
(which leads to a maximum height of 3.5 m).

For attempt to find time from .25 m below top to top DM1 5
t = 0.224 or [0.22360….] probably from
10
3.5 − 3.25 = 0t + 0.5 10t 2  . Dep on both previous M1s.
 

5 A1 For doubling.
Time = 0.447 s or s
5

Alternative for last 3 marks of Q7(b) by finding velocity at height of 3.25 m

w2 = their 10 + 2  ( − g )  0.25 .
Correct use of suvat for motion of A (between height of 3m and *DM1 2
 w2 = 5 or  w = 5  .
3.25 m), to form an equation to find speed at height of 3.25 m    

For correct use of suvat to find time to max height DM1  5

0 = 5 + ( − g )  t t =  .
 10 

5 A1 For doubling.
Time = 0.447 s or s
5

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Question Answer Marks Guidance

7(b) Alternative using energy

Correct use of energy before B hits the plane or when A has risen 2 *M1 1 1
m, to attempt to find velocity of A when string becomes slack PE loss = KE gain: 5g  2 − 3g  2 =  5v 2 +  3v 2 ,
2 2
using their T if necessary. Or PE loss – WD by tension = KE gain: 5g  2 − their 37.5  2 =
1
 5v 2 ,
2
Or WD by tension – PE gain = KE gain:
1
their 37.5  2 − 3g  2 =  3v 2 .
2
Must be complete method to find v or v 2 v = 10 or 3.162 .
Do not allow sign errors.

Correct use of energy to find velocity of A at height of 3.25 m *DM1 1 2 1

PE gain = KE loss: 3g  0.25 =  3  10 −  3w2 .
2 2
Must be complete method to find w or w w = 5 or 2.236
2

Do not allow sign errors.

For attempt to find time from .25 m below top to top DM1 5
t = 0.224 or [0.22360….] probably from 0 = 5 − 10t .
10
Dep on both previous M1s.
Do not allow sign errors.

5 A1 For doubling.
Time = 0.447 s or s
5

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