Cambridge International AS & A Level
FURTHER MATHEMATICS
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Paper 3 Further Mechanics 3
May/June 2020
MARK SCHEME
Maximum Mark: 50
Published
Students did not sit exam papers in the June 2020 series due to the Covid-19 global pandemic.
This mark scheme is published to support teachers and students and should be read together with the
question paper. It shows the requirements of the exam. The answer column of the mark scheme shows the
proposed basis on which Examiners would award marks for this exam. Where appropriate, this column also
provides the most likely acceptable alternative responses expected from students. Examiners usually review
the mark scheme after they have seen student responses and update the mark scheme if appropriate. In the
June series, Examiners were unable to consider the acceptability of alternative responses, as there were no
student responses to consider.
Mark schemes should usually be read together with the Principal Examiner Report for Teachers. However,
because students did not sit exam papers, there is no Principal Examiner Report for Teachers for the June
2020 series.
Cambridge International will not enter into discussions about these mark schemes.
Cambridge International is publishing the mark schemes for the June 2020 series for most Cambridge
IGCSE™ and Cambridge International A & AS Level components, and some Cambridge O Level
components.
This document consists of 14 printed pages.
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Cambridge International AS & A Level – Mark Scheme
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Generic Marking Principles
May/June 2020
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 descriptors 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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Cambridge International AS & A Level – Mark Scheme
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May/June 2020
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, 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 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 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.
© UCLES 2020
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Mark Scheme Notes
May/June 2020
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
FT
© UCLES 2020
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.
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.
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May/June 2020
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
1
Answer
For greatest height, T =
At t =
=
M1
A1
u 3
2
Speed =
vv 2 + vh 2 =
Marks
B1
u
2g
2T
u 2Tg u
, ↑ vv = −
=
3
2
3
6
→ vh =
May/June 2020
M1
u 2 3u 2
+
36
4
A1
7
u
3
5
© UCLES 2020
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PUBLISHED
Question
2
Answer
For A: T = 3mg
May/June 2020
Marks
M1
For B: ↑ T cosθ = mg
Equate: 3mg cos θ = mg
1
cos θ =
3
A1
→ T sin θ = mrω 2 with r = ( a − x ) sin θ
M1
Equate: 3mg = m ( a − x ) ω 2
A1
x=
A1
a
4
5
Question
3(a)
Answer
Marks
T − mg = m.a
M1
1
5
T = 5mg . a / a = mg
2
2
M1
a=
A1
3
g (upwards) AG
2
3
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Question
3(b)
Answer
Gain in KE =
Marks
1 2
1
mv Gain in GPE = mga
2
2
B1
2
B1
1 
5mg .  a 
1
2 
Loss in EPE =
2
a
1 
5mg .  a 
1 2 1
1
2 
mv + mga =
2
2
2
a
v=
May/June 2020
M1
2
[
1 2 1
5
mv + mga = mga ]
2
2
8
A1
1
ga
2
4
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Question
Answer
4(a)
Area
Square
Triangle
Shape ABEFD
100
Centre of mass
from BC
5
Centre of mass
from DC
5
1
x
3
5
2
x
y
½ x. 15/2
100 −
15
x
4
May/June 2020
Marks
M1
Take moments about BC:
15 
15
1

100 − x σ . x = 500σ − xσ . x
4 
4
3

(M1 for all terms present)
x=
A1
400 − x 2
AG
80 − 3 x
Take moments about DC:
15 
15 5

100 − x  . y = 100 × 5 − x.
4 
4 2

y =
M1
A1
800 − 15 x
160 − 6 x
4
4(b)
Use condition: x  x
B1
2 x2 − 80 x + 400  0
M1
x = 20 − 10 2
A1
3
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
5(a)
Answer
May/June 2020
Marks
dv
= kdt
3u − v
M1
− ln ( 3u − v ) = kt + d
t = 0, v = u : d = − ln 2u
M1
v = 2u : t =
A1
1
ln 2
k
3
5(b)
v
B1
vdv
dv
= 3ku − kv [ 
= kdx ]
dx
3u − v
( − ( 3u − v ) + 3u ) dv = kdx so −v − 3u ln
3u − v
M1A1
( 3u − v ) = kx + c
x = 0, v = u : c = −u − 3u ln 2u
v = 2u : x =
M1
A1
u
( 3ln 2 − 1)
k
5
© UCLES 2020
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Cambridge International AS & A Level – Mark Scheme
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Question
6(a)
Answer
May/June 2020
Marks
Let components of velocity (parallel to plane and perpendicular) after impact be (x, y)
y = v cos α = eu sin α
B1
x = v sin α = u cos α
B1
Divide: tan α =
B1
1
1
: tan2 α = .
e
e tan α
3
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Question
6(b)
Answer
May/June 2020
Marks
B1
1
v2 = u 2
3
1 2
 u cos α 

 = u
3
 sin α 
M1
( tan α )2 = 3
M1
α = 60°
A1
1
3
A1
2
e=
Alternative method for 6(b)
KE after impact =
From (a) sin α = 1 /
KE =
(
1
1
2
2
m x 2 + y 2 = m ( u cos α ) + e 2 ( u sin α )
2
2
(
(1 + e )
)
and cos α = e /
)
B1
(1 + e )
A1
1 2 e
e2  1
2
+
mu 
 = mu e
2
1
1
+
+
e
e
2


This is equal to
M1
M1
1 1 2
1
× mu so e =
3 2
3
A1
tan α = 3, α = 60°
5
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Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Question
7(a)
Answer
( N + ) mg cosθ =
Marks
B1
mv 2
a
M1A1
1 2 1 7 ag
mv − m
= − mg ( a + a cosθ )
2
2
2
Loses contact when N = 0, so combine and simplify
cos θ =
May/June 2020
M1
A1
1
: θ = 60° AG
2
5
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Question
7(b)
Answer
When P is vertically below O, its horizontal displacement is a sin 60 , so time T =
Marks
a sin 60
6a
= a 3/v =
v cos 60
g
M1
A1
1
From (a), v 2 = ag
2
Vert: h =
May/June 2020
M1
v 3
1
T − g .T 2
2
2
A1
3
3a
a − 3a = −
2
2
This corresponds to the point A
A1
Alternative method for question 7(b)
y=x 3−
M1A1
4 x2
a
B1
1
3
Coordinates of A: x = a 3, y = − a
2
2
Substitute coordinates into y = x 3 −
4 x2
and show that these satisfy this equation
a
M1A1
5
© UCLES 2020
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