Cambridge International AS & A Level

MATHEMATICS 9709/42
Paper 4 Mechanics February/March 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 f or
Teachers.

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

Cambridge International is publishing the mark schemes f or the February/March 2024 series f or most
Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level
components.

This document consists of 22 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 point s.

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 appli cation 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 sign ificant 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(a) Total distance travelled = 50 + 2  10  = 70 B1 Any explicit expression for the distance,
x − 50
e.g. = 2 → x = 2 ( 20 − 10 ) + 50.
20 − 10

70 B1FT Oe.
Velocity = − = −3.5 m s–1 Ft their distance  50.
20
Must be negative.
Do not ISW if (e.g.) velocity = 3.5.
Do allow “3.5 m s–1 , directed towards O.”

1(b) Velocity 5 m s–1 for 0 t 10 B1 May be seen on diagram.

B1 Stepped diagram with four horizontal lines segments.

Ignore vertical line segments.

B1FT All correct.

10, 20, 40 and 60 indicated on the t-axis.
Their 5, 2 and their –3.5 (allow –3 and – 4 is
acceptable too with line segment halfway between
them) indicated on the v-axis, corresponding to the
position of the horizontal line segments.
FT their –3.5 m s–1 and/or FT their 5 m s–1 .
If their answer to (a) is positive, allow use of
negative their answer for this mark.

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Question Answer Marks Guidance

2(a) −5 = u − g  2 M1 For use of constant acceleration to get an equation in

or 5 = u + g  2 u only.
or 5 = 0 + gt  t = 0.5, so time to greatest height is T = 2 − 0.5 = 1.5. Using v = −5 , t = 2 and a =  g.
Hence 0 = u + ( − g )  1.5

Speed = 15 m s–1 A1 Must be positive.

2(b) 02 = 102 − 2gs  s = 5 M1 For use of constant acceleration formula(e) to find

distance travelled from height with speed 10 m s–1 to
or 102 = 02 +  2 gs  s = 5 maximum height with speed 0 m s–1 , a =  g.
1 Must be a complete method, e.g.
or 10 = 0 + gt  t = 1 , so s = 10 1 − g 12  s = 5
02 = ( their 15) − 2gs1  s1 = 11.25
2
2
1
or ( m ) 102 = ( m ) gh  h = 5 and 102 = ( their 15)2 − 2 gs2  s2 = 6.25
2
or
10 = ( their 15 ) − gt  t = 0.5 ,
1
s2 = ( their 15 + 10 )  ( their 0.5)
2
and with an attempt at s1 − s2  = 5.
Energy method with 2 terms, dimensionally correct.

Total distance = 10 m A1FT FT their 15 m s–1 if used.

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Question Answer Marks Guidance

3 8000 B1 P
Driving force = or 4000 Use of F = , oe. E.g. DF  2 = 8000.
2 v

R = 600 g cos30  = 3000 3 = 5196.152423 B1

their Driving force − 600 g sin30 − F = 0 *M1 For attempt to resolve parallel to the plane; 3 terms;
allow sign errors; allow sin/cos mix; allow g missing;
 F = 4000 − 3000 = 1000 F not resolved.
Allow with just DF.

Use of F =  R DM1 To form an equation in  only where R is a

component of weight or mass.

3 A1 Awrt 0.192 [0.1924500897].

 = 0.192 or 1
9 Oe, e.g. .
3 3
Allow 2sf.

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Question Answer Marks Guidance

4 For resolving in either direction *M1 Correct number of terms allow sign errors; allow
sin/cos mix on θ.
Forces that need resolving should be resolved.

F + 2F cos45 = 30sin A1

2F sin 45 + 3F = 30cos A1

 1 + 2cos 45  DM1 For attempt to find θ.

 = tan −1   Using their F which can be solved for θ.
 2sin 45 + 3 
From equations with correct number of relevant
 F + 2 F cos 45 
or  = sin −1   terms, forces that need resolving should be resolved.
 30  cos −1  2sin 45 + 3 
+
If tan  = , so have  = tan  ,
 2 F sin 45 3 F  sin   1 + 2cos 45 
or  = cos −1  
 30  then allow M1.

 F 2 ( 3 + 2sin 45 )2 + F 2 (1 + 2cos 45 )2 = 30 2   DM1 For attempt to find F.

  From equations with the correct number of relevant
302 terms, forces that need resolving should be resolved.
F= Using their θ.
( 3 + 2sin 45 )2 + (1 + 2cos 45 )2
30sin 
or F =
1 + 2cos 45
30cos
or F =
3 + 2sin 45

F = 5.96 [5.96270…] and  = 28.7 [28.6750…] A1 Awrt to 5.96 and 28.7.

Allow 5.97.

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Question Answer Marks Guidance

5(a) For attempt at integration *M1 Increase power by 1 and a change in coefficient in at
least one term (which must be the same term).
s = vt is M0.

1 3+1 9 1 3 A1 May be unsimplified.

( s =) t − t 2+1 + t  + c  = t 4 − t 3 + t  + c 
( 3 + 1) 2  ( 2 + 1) 4 2

 1  1  4 3  1 3  1   DM1 1 1
correctly, or substitute t = .
Distance =     −   +    −
 4  2  2  2   2 
( 0 ) Use limits 0 and
2 2
  M0 if including other regions.

21 A1 Oe, e.g. 0.328125. Condone 0.328.

= m
64

Special Case if no integration seen. Maximum 1/4

21 B1 Oe, e.g. 0.328125. Condone 0.328.

m
64

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Question Answer Marks Guidance

5(b) Attempt to differentiate ( a = ) 3t 2 − 9t  *M1 Decrease power by 1 and a change in coefficient in at
least one term (which must be the same term).
v
a = is M0.
t

Solve a = 0 to get t = 3 A1 Ignore t = 0 if not rejected.

1 DM1 Must be using an expression for s from integration.

Distance from t = to t = 3 =
2 1
Allow missing minus sign at start; use limits and
 1 3   1  1  3  1   1 
4 3 2
    34 −  33 + 3  −     −   +     = 1
 4

2   4  2  2  2   2    their 3 correctly, where  their 3  4.
2
 1 3  21   37 1125  21
    34 −  33 + 3  −  = 17 = = 17.578125 Or use limit 3 and find difference from their
 4 2  64   64 64  64
from part (a).

May see
 1  1  4 3  1 3  1    1 3 
2      −   +    −   34 −  33 + 3 
 4  2  2  2   2   4 2 
 

 21 37  29 A1FT 573
So total  = + 17  = 17 m Oe, e.g. , 17.90625. Condone 17.9.
 64 64  32 32
FT their positive integration value in part (a),
37 69
e.g. 17 + their (a) or + 2  their (a).
64 4

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Question Answer Marks Guidance

5(b) Special Case if no integration seen. Maximum M1A1B1 for 3 marks

Attempt to differentiate ( a = ) 3t 2 − 9t  M1 Decrease power by 1 and a change in coefficient in at

least one term (which must be the same term).
v
a = is M0.
t

Solve a = 0 to get t = 3 A1

29 B1 573
17 m Oe, e.g. , 17.90625. Condone 17.9.
32 32

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Question Answer Marks Guidance

6(a) Attempt at Newton’s second law on car or trailer or system *M1 Correct number of terms; allow sign errors; allow
sin/cos mix; sin  or  = 2.865 needs to be
substituted; allow with  = 2.87 or 2.9 or better;
allow g missing.

Car: 3000 − 1800g  0.05 − 800 − T = 1800a A1 Any two equations correct.
3000 − 900 − 800 − T = 1800a → 1300 − T = 1800 a 
Trailer: T − 300g  0.05 − 100 = 300a
T − 150 − 100 = 300a → T − 250 = 300a 
System: 3000 − 1800 g  0.05 − 300 g  0.05 − 800 − 100 = (1800 + 300 ) a
3000 − 900 − 150 − 800 − 100 = 2100a → 1050 = 2100a 

Solving for a or T DM1 From equations with correct number of relevant

terms, allow g missing.
If no working seen, must be correct for their
equations.

Acceleration = 0.5 m s–2 A1 Oe. Allow 0.499 from  = 2.87.

Tension = 400 N A1 Allow awrt 400 to 3sf, www.

Condone using car equation with 0.499 to get
T = 400.5.
Using exact a from an inexact angle gives T = 400.

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Question Answer Marks Guidance

6(b) Work done against resistance on car = 800  50  = 40 000 B1

or Work done by driving force = 3000  50  = 150 000 

PE change = (1800 + 300 ) g  50  0.05 = 52 500 B1 Allow 2100 g  50  sin 2.9.

Or 2100 g  50  sin 2.87.

1 B1
Initial KE =  (1800 + 300 )  202  420 000
2

1 1 M1 Attempt at work-energy equation; correct number of

 (1800 + 300 ) v 2 = 3000  50 − 800  50 − 6000 − (1800 + 300 ) g  50  0.05 +  (1800 + 300 )  202 relevant terms; dimensionally correct; allow sign
2 2
errors; allow sin/cos mix in relevant resolved terms.
1050v = 150 000 − 40 000 − 6000 − 52 500 + 420 000 
2
  Only PE must be from a component.
1050v 2 = 471500  Allow 2100 g  50  sin 2.9
 
or 2100 g  50  sin 2.87.

Speed = 21.2 m s−1 [21.1907…] A1 Awrt 21.2 to 3sf.

 = 2.87 gives 21.18909187.
 = 2.9 gives 21.17674855.

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Question Answer Marks Guidance

6(b) Alternative Method for Question 6(b): Using energy on Car or Trailer only (Must be finding Tension for this method)

Car: 3000 − 1800g  0.05 − 800 − T = 1800a B1

6000
Trailer: T − 300 g  0.05 − = 300a
50
2920
Solve to get T = = 417.1428571
7

Work done against resistance on car = 800  50  = 40 000 B1

or Work done by driving force = 3000  50  = 150 000 
2920  146000 
or Work done against tension =  50  =
7  7 

PE change = 1800 g  50  0.05  = 45 000 B1 Or 300 g  50  0.05 = 7500.

or Allow 1800 g  50  sin 2.9 or 1800 g  50  sin 2.87.
1
Initial KE = 1800  202 360 000 1
Or  300  202 60 000.
2 2

1 2920 1 M1 Attempt at work-energy equation; correct number of

1800v 2 = 3000  50 − 800  50 −  50 − 1800 g  50  0.05 + 1800  202 relevant terms; dimensionally correct; allow sign
2 7 2
errors; allow sin/cos mix in relevant resolved terms.
Only PE must be from a component.
M0 if using T from part (a).

Speed = 21.2 m s−1 [21.1907…] A1 Awrt 21.2 to 3sf.

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Question Answer Marks Guidance

6(b) Special case: use of constant acceleration. Maximum 2 marks

6000 6000  103  B1 Oe.

F= = 120 , 2100a = 3000 − − 800 − 2100 g  0.05  a = 210  6000
50 50   Allow 2100a = 3000 − − 800 − 2100 g  sin 2.9.
50
6000
Allow 2100a = 3000 − − 800 − 2100 g  sin 2.87.
50

 2 103  B1
v = 20 + 2  50  210 →  v = 21.2 m s
2 −1
 

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Question Answer Marks Guidance

7(a) ( m ) a = ( m ) g sin  = ( m ) g  0.6  a = 6 M1 For attempt at Newton’s second law; 2 terms; allow
sign errors; allow sin/cos mix, allow  = 36.9 or
better.
Allow  = 37. Allow for a = 6 seen/used.

v2 = 02  + 2  6  0.75  v = 3 A1 AG.

3 =  0 + 6t  t = 0.5 A1 AG.
Allow A2 for use of a = 6 with any 2 of v = 3 ,
t = 0.5 and s = 0.75 to get the third value.

Alternative Method 1 for Question 7(a)

( m ) a = ( m ) g sin  = ( m ) g  0.6  a = 6 M1 For attempt at Newton’s second law; 2 terms; allow

sign errors; allow sin/cos mix, allow  = 36.9 or
better.
Allow  = 37. Allow for a = 6 seen/used.

0 + 6  0.5 = 3 A1 AG.

1 A1 AG.
0  0.5 +  6  0.52 = 0.75
2 1
Allow A2 for 3  0.5 −  6  0.52 = 0.75.
1 2
or 3  0.5 −  6  0.52 = 0.75
2

Alternative Method 2 for Question 7(a)

1 M1 Attempt at conservation of energy, 2 terms,

( m ) v2 = ( m ) g  0.75  0.6 dimensionally correct, allow sin/cos mix, allow
2
 = 36.9 or better. Allow  = 37.

v =3 A1 AG.

Use constant acceleration to get t = 0.5 A1 AG. Must show a = 6.

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Question Answer Marks Guidance

7(a) Alternative Method 3 for Question 7(a)

1 M1 1
( s = ) ( 0 + 3)  0.5 Use s = ( u + v ) t with u = 0 with any 2 of v = 3,
2 2
1 t = 0.5 and s = 0.75.
or 0.75 = ( 0 + v )  0.5
2
1
or 0.75 = ( 0 + 3) t
2

Correctly identifies one value. A1

( s = ) 0.75 A1 AG.
or v = 3
or t = 0.5

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Question Answer Marks Guidance

7(a) Alternative Method 4 for Question 7(a)

( )
32 = 02 + 2a  0.75 → a = 6 B1 Use constant acceleration to get an equation in a
only, using u = 0 and any 2 of v = 3 , t = 0.5 and
or 3 = ( 0 + ) 0.5a → a = 6 s = 0.75.
1 Or using v = 3 , t = 0.5 and s = 0.75.
or 0.75 = ( 0  0.5 + ) a  0.52 → a = 6
2
1
or 0.75 = 3  0.5 − a  0.52 → a = 6
2

1 1 M1 Use constant acceleration formula in an attempt to

3 = ( 0 + ) 6t , 0.75 = ( 0  t + )  6  t 2 , 0.75 = 3t −  6  t 2 find the third unused value using a = 6.
2 2
1 1 Or use constant acceleration formula in an attempt to
or s = ( 0  0.5 + )  6  0.52 , s = 3  0.5 −  6  0.52 find u using a = 6.
2 2
( ) 1
or v = ( 0 + ) 0.5  6; v 2 = 02 + 2  6  0.75; 0.75 = 0.5v −  6  0.52
2
1
or 3 = u + 6  0.5 , 32 = u 2 + 2  6  0.75 , 0.75 = 0.5u +  6  0.52
2

Get correct unused value A1 Or u = 0 if using all 3 values to find a = 6.

AG.

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Question Answer Marks Guidance

7(b) For BC ( m ) a = ( m ) g sin  − ( m ) g  0.8  0.25 = ( m ) g  0.6 − ( m ) g  0.8  0.25  a = 4 *M1 For attempt at Newton’s second law; 3 terms; allow
sign errors; allow g missing; allow sin/cos mix, allow
 = 36.9 or better.

1 1 A1 For either s P or sQ .
For P , sP = 3t +  4t 2 For Q , sQ =  4 ( t + 0.5)
2

2 2 t is the time after P arrives at B.

1 1
Or for P , sP = 3(T − 0.5) +  4 (T − 0.5) For Q , sQ =  4T 2 T is the time from when both particles released.
2

2 2

1 1 DM1 For use of sP = sQ ; s P and sQ of correct form.

3t +  4t 2 =  4 ( t + 0.5)  3t + 2t 2 = 2t 2 + 2t + 0.5
2

2 2 Using their a  their 6 from part (a), a   g.

1 1
Or 3(T − 0.5) +  4 (T − 0.5) =  4T 2
2
Using same a in both s P and sQ .
2 2
( )
 3T − 1.5 + 2 T 2 − T + 0.25 = 2T 2 
 

t = 0.5, so total time = 1 s or time T = 1 s A1

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Question Answer Marks Guidance

7(b) Alternative Method for Question 7(b)

For BC ( m ) a = ( m ) g sin  − ( m ) g  0.8  0.25 = ( m ) g  0.6 − ( m ) g  0.8  0.25  a = 4 *M1 For attempt at Newton’s second law; 3 terms; allow
sign errors; allow g missing; allow sin/cos mix, allow
 = 36.9 or better.

1 A1 For either. t is the time after P arrives at B.

For P sP = 3t +  4t 2
2
1
Or Q has speed 2 m s–1 after 0.5 s, so sQ = 2t +  4t 2
2

1 1 1 DM1 1
3t +  4t 2 = 2t +  4t 2 +  4  0.52 For use of sP = sQ  their  4  0.52 ; s P or sQ of
2 2 2 2
correct form.
Using their a  their 6 from part (a), a   g.
Using same a in both s P or sQ .

t = 0.5 so total time = 1 s A1

Alternative Method 2 for Question 7(b): Using relative velocity

For BC ( m ) a = ( m ) g sin  − ( m ) g  0.8  0.25 = ( m ) g  0.6 − ( m ) g  0.8  0.25  a = 4 *M1 For attempt at Newton’s second law; 3 terms; allow
g missing; allow sign errors; allow sin/cos mix.

[In 0.5 s] Q has speed 2 m s–1 and has moved 0.5 m A1 For both.

0.5 DM1 Attempt at time from relative velocity using their 0.5
t= m and their 2 m s–1 .
3− 2

t = 0.5 so total time = 1 s A1

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 9709/42 Cambridge International AS & A Level – Mark Scheme February/March 2024
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Question Answer Marks Guidance

7(c)  1 B1 For either.

Immediately before the collision speed of P  = 3 + 4   = 5 ms −1
 2
Speed of Q  = 0 + 4  1 = 4 m s–1

5( m) + 4 ( m) = ( m) + ( m)  v  v = 4.5 *M1 Use of conservation of momentum; 4 non-zero

terms; allow sign errors; allow their 4m s–1 (  0 or 2 )
and 5m s–1 (  3 ) .
Use of mg then withhold final A mark.

1 B1 May be implied by 1.25 m.

Distance from B at collision =  4  12 = 2 m
2
1
OR = 3  0.5 +  4  0.52 = 2 m
2

1 DM1 Use of constant acceleration with u = their 4.5,

1.25 = 4.5t +  4t 2
2 a = their 4 (   g ) and s = 3.25 − their 2,
s  3.25, s  0.75, s  4, s  2.75, s  0.5.
If using two formulae, must be a complete method to
get an equation in t only.

t = 0.25s only A1 A0 if from use of mg in conservation of momentum.

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