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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
9709/43
MATHEMATICS
Paper 4 Mechanics 1 (M1)
October/November 2014
1 hour 15 minutes
*9816445705*
Additional Materials:
Answer Booklet/Paper
Graph Paper
List of Formulae (MF9)
READ THESE INSTRUCTIONS FIRST
If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet.
Write your Centre number, candidate number and name on all the work you hand in.
Write in dark blue or black pen.
You may use an HB pencil for any diagrams or graphs.
Do not use staples, paper clips, glue or correction fluid.
DO NOT WRITE IN ANY BARCODES.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in
degrees, unless a different level of accuracy is specified in the question.
Where a numerical value for the acceleration due to gravity is needed, use 10 m s−2.
The use of an electronic calculator is expected, where appropriate.
You are reminded of the need for clear presentation in your answers.
At the end of the examination, fasten all your work securely together.
The number of marks is given in brackets [ ] at the end of each question or part question.
The total number of marks for this paper is 50.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying larger
numbers of marks later in the paper.
This document consists of 4 printed pages.
JC14 11_9709_43/RP
© UCLES 2014
[Turn over
2
1
A car of mass 1400 kg moves on a horizontal straight road. The resistance to the car’s motion is
constant and equal to 800 N and the power of the car’s engine is constant and equal to P W. At an
instant when the car’s speed is 18 m s−1 its acceleration is 0.5 m s−2 .
(i) Find the value of P.
[3]
The car continues and passes through another point with speed 25 m s−1 .
(ii) Find the car’s acceleration at this point.
[2]
2
Q
P
A
B
The tops of each of two smooth inclined planes A and B meet at a right angle. Plane A is inclined
at angle to the horizontal and plane B is inclined at angle to the horizontal, where sin = 63
65 and
16
sin = 65 . A small smooth pulley is fixed at the top of the planes and a light inextensible string passes
over the pulley. Two particles P and Q, each of mass 0.65 kg, are attached to the string, one at each
end. Particle Q is held at rest at a point of the same line of greatest slope of the plane B as the pulley.
Particle P rests freely below the pulley in contact with plane A (see diagram). Particle Q is released
and the particles start to move with the string taut. Find the tension in the string.
[5]
3
O
WN
7N
8N
Each of three light inextensible strings has a particle attached to one of its ends. The other ends of the
strings are tied together at a point O. Two of the strings pass over fixed smooth pegs and the particles
hang freely in equilibrium. The weights of the particles and the angles between the sloping parts of
the strings and the vertical are as shown in the diagram. It is given that sin = 0.8 and cos = 0.6.
(i) Show that W cos = 3.8 and find the value of W sin .
[3]
(ii) Hence find the values of W and .
[3]
© UCLES 2014
9709/43/O/N/14
3
4
A particle P starts from rest and moves in a straight line for 18 seconds. For the first 8 seconds of the
motion P has constant acceleration 0.25 m s−2 . Subsequently P’s velocity, v m s−1 at time t seconds
after the motion started, is given by
v = −0.1t2 + 2.4t − k,
where 8 ≤ t ≤ 18 and k is a constant.
5
(i) Find the value of v when t = 8 and hence find the value of k.
[2]
(ii) Find the maximum velocity of P.
[2]
(iii) Find the displacement of P from its initial position when t = 18.
[3]
A box of mass 8 kg is on a rough plane inclined at 5 to the horizontal. A force of magnitude P N
acts on the box in a direction upwards and parallel to a line of greatest slope of the plane. When
P = 7X the box moves up the line of greatest slope with acceleration 0.15 m s−2 and when P = 8X
the box moves up the line of greatest slope with acceleration 1.15 m s−2 . Find the value of X and the
coefficient of friction between the box and the plane.
[8]
6
v m s−1 4
Particle P
O
Q
P
hm
1.0
1.4
1.8
t s
Particle Q
−4
Fig. 1
Fig. 2
Particles P and Q have a total mass of 1 kg. The particles are attached to opposite ends of a light
inextensible string which passes over a smooth fixed pulley. P is held at rest and Q hangs freely, with
both straight parts of the string vertical. Both particles are at a height of h m above the floor (see
Fig. 1). P is released from rest and the particles start to move with the string taut. Fig. 2 shows the
velocity-time graphs for P’s motion and for Q’s motion, where the positive direction for velocity is
vertically upwards. Find
(i) the magnitude of the acceleration with which the particles start to move and the mass of each of
the particles,
[5]
(ii) the value of h,
[1]
(iii) the greatest height above the floor reached by particle P.
[2]
[Question 7 is printed on the next page.]
© UCLES 2014
9709/43/O/N/14
[Turn over
4
7
35 N
−1
4m
s
A
12
.5
m
O
A small block of mass 3 kg is initially at rest at the bottom O of a rough plane inclined at an angle to the horizontal, where sin = 0.6 and cos = 0.8. A force of magnitude 35 N acts on the block at
an angle above the plane, where sin = 0.28 and cos = 0.96. The block starts to move up a line
of greatest slope of the plane and passes through a point A with speed 4 m s−1 . The distance OA is
12.5 m (see diagram).
(i) For the motion of the block from O to A, find the work done against the frictional force acting
on the block.
[4]
(ii) Find the coefficient of friction between the block and the plane.
[3]
At the instant that the block passes through A the force of magnitude 35 N ceases to act.
(iii) Find the distance the block travels up the plane after passing through A.
[4]
Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable
effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will
be pleased to make amends at the earliest possible opportunity.
Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local
Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.
© UCLES 2014
9709/43/O/N/14
Cambridge International AS & A Level
MATHEMATICS
9709/42
Paper 4 Mechanics
October/November 2022
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 2022 series for most
Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 22 printed pages.
© UCLES 2022
[Turn over
9709/42
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Generic Marking Principles
October/November 2022
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.
© UCLES 2022
Page 2 of 22
9709/42
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
October/November 2022
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 2022
Page 3 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
Mark Scheme Notes
9709/42
October/November 2022
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
•
•
•
•
•
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.
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.
© UCLES 2022
Page 4 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
October/November 2022
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
© UCLES 2022
Page 5 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
1
Answer
Marks
Work done by cyclist = 50 × 100 (= 5000 J)
Their 5000 – 3560 =
October/November 2022
Guidance
B1
M1 Work energy equation. Three terms. Allow sign errors.
Dimensionally correct.
1
m  62
2
mass = 80 kg
A1
3
SC: Acceleration considered as a constant
© UCLES 2022
62 = 0 + 2a  50 [⇒ a = 0.36 ]
3560
100 −
= m  their 0.36
50
M1 From use of v = 6 , u = 0 , s = 50. Must be using
correct suvat formulae.
For equation involving mass using N2L with three
terms.
Allow sign errors in N2L.
mass = 80 kg
A1
Page 6 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
2(a)
Answer
Marks
Guidance
R = 0.4g cos 30  = 2 3  or F or  R = 0.4 g sin 30  = 2
B1 Use of m instead of 0.4 condoned.
0.4g sin 30 – µ 0.4g cos 30 = 0
M1 For using F = µR. Allow sin/cos mix.
Both must be different components of their weight
only, not a 2 term R. Allow sign errors. Allow g
omitted.
3
 4sin 30  1
=
.
3 or

3
 4cos30  3
A1 AG (exact answer only)
If zero scored then SC B1 for [Angle of friction = 30°
1
3.
so] µ = tan 30 =
3
Allow full marks if using m in place of 0.4 or W in
place of mg or 0.4g
1
3 , but A1(ISW) for
A0 for  = 0.577=
3
1
 = 3 = 0.577
3
 =
3
© UCLES 2022
October/November 2022
Page 7 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
2(b)
Answer
Marks
Guidance
7.2 – 0.4 g sin 30 – F = 0.4a
M1 Newton’s second law. Four terms. Second term must
be a component of their weight. F ≠ 0 and F   .
Allow sin/cos mix. Allow sign errors.
F must be a numerical expression
May use their F from part (a).
a=8
A1
1
1 = 0 +  ( their positive 8)  t 2
2
M1 For use of constant acceleration formula(e) and solving
for t. a  10, a   g .
Allow if a is negative in part (a) and use |a | here.
Time = 0.5 s
A1
4
© UCLES 2022
October/November 2022
Page 8 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
3
Answer
Marks
October/November 2022
Guidance
Attempt to resolve either direction
M1 Correct number of terms. Allow sin/cos mix. Allow
sign errors. Allow g missing.
0.3g + T cos  – 4 sin 60= 0 (T cos α° = 0.464…)
A1 OE
T sin  –4 cos 60= 0
A1 OE If the two Ts are different, award maximum
A1A0 unless subsequently stated that the two Ts are
the same.
(T sin α° = 2)
M1 Attempt to solve for α. No missing/extra terms. Allow
g missing. Must get to ‘α =’.

4 cos 60 
2 
−1 
 = tan 

 0.464 
 4 sin 60 − 0.3g 
 = tan −1 
T=
4cos 60
=
sin ( their )
( 4 cos 60 )2 + ( 4 sin 60 − 0.3 g )2
= 2 2 + ( 0.464)
2
Tension = 2.05 N α = 76.9
M1 OE Attempt to solve for T. No missing/extra terms.
Allow g missing. Must get to ‘T =’.
A1 For both AWRT 2.05, 76.9
(Tension = 2.05314… N α = 76.9356…)
Alternative method for Q3 using triangle of forces
Attempt at cosine rule from triangle of forces
M1 Must use lengths 4 and 0.3g with a suitable angle.
Allow g missing.
T 2 = 42 + ( 0.3g ) − 2  4  ( 0.3g )  cos30
A1
Tension = 2.05
A1 Tension = 2.05314… AWRT 2.05
Attempt at sin rule
M1 Must have angle 30° and another angle in terms of α
with correct numerators, but allow g missing.
Their T
4
Their T
0.3g
=
or
=
sin 30 sin (180 −  )
sin 30 sin ( − 30 )
A1 Correct. Allow sin  instead of sin (180 −  ) .
α = 76.9
A1 α = 76.9356… AWRT 76.9
2
© UCLES 2022
Page 9 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
3
Answer
Marks
October/November 2022
Guidance
Alternative method for Q3 using Lami’s theorem
Attempt at Lami’s theorem
M1 Must have numerators correct and at least one angle
correct. Allow g missing.
A1 A1 A1 for two parts second A1 for all three.
4
0.3g
T
=
=
sin  sin ( 210 −  ) sin (150 )
M1 For solving for α using compound angle formula. Must
be correct for their angles. Allow g missing.


4sin 210

 0.3g + 4cos 210 
 = tan −1 
T=
4sin (150 )
sin 
or T =
0.3g sin (150 )
M1 For solving for T using their α. Allow g missing.
sin ( 210 −  )
Tension = 2.05 N α = 76.9
A1 For both AWRT 2.05, 76.9
6
SC: Tension and the 4 N force considered in the wrong directions
Attempt to resolve either direction
M1 Correct number of terms. Allow sin/cos mix. Allow
sign errors. Allow g missing.
T cos 60 –4 sin = 0
And: T sin 60 – 4 cos   − 0.3g = 0
A1 For both
OE If the two Ts are different, they get SC A0 unless
they subsequently state that the two Ts are the same.
2
2
1 2 3 2
 T cos 60   T sin 60−0.3g 

 +
 = 1  T + T − 3 3T + 9 = 16
4
4
4
4

 

2
 T − 3 3T − 7 = 0  T = 6.31( or − 1.11)
3
OR: 4 3 sin  − 4cos  = 3 ⇒ 8sin ( − 30 ) = 3 ⇒  = sin −1 + 30
8
© UCLES 2022
Page 10 of 22
M1 OE Attempt to solve for T or α. No missing/extra
terms. Allow g missing. Must get to ‘T =’ or ‘α=’.
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
3
Answer
Marks
T = 6.31N  = 52.0
Guidance
A1 (T = 6.30617…, α=52.0243…)
6
© UCLES 2022
October/November 2022
Page 11 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
4(a)
Answer
October/November 2022
Marks
Guidance
P = D × 15
B1
D – 500 = 1200 × 0.8 (⇒ D = 1460)
M1 Attempt at Newton’s second law with three terms.
Allow sign errors.
Power = 21900 W
A1 Allow 21900 without units or 21.9 kW, but not simply
21.9 without units or with wrong units.
For any D. OE including
P
.
15
3
4(b)
B1 Sight of both KEs.
1
1
 1200  322 −  1200  152
2
2
 = 614400 − 135000 = 479400
[Change in KE =]
Work done by engine = 21900 × 53 ( = 1160700)
B1ft
OE e.g. 21900 =
WD
53
FT their 21900.
B1 AG Must come from 1160700 – 500d = 479400 OE
e.g. 500d = 681300 .
Distance AB = 1362.6 m
3
© UCLES 2022
Page 12 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
5(a)
Answer
Marks
October/November 2022
Guidance
T − 136.8 − 50 = 40a 
500 cos15 –80g sin 20 –T = 80a  482.96− 273.61− T = 80a 
500 cos15 –80 g sin 20 – 40 g sin 20 − 50 = ( 80 + 40 ) a
 482.96− 273.61− 136.8− 50 = 120a 
M1 Attempt at Newton’s second law for at least one case.
Allow sign errors. Do not allow g missing. Correct
number of terms. Allow sin/cos mix.
For attempt to solve for T or a
M1 From equation(s) with no missing/extra terms. Allow g
missing. Must get to ‘T =’ or ‘α =’.
Acceleration = 0.188 ms−2
A1 Allow AWRT 0.19.
Tension = 194 N
A1
T – 40g sin 20 –50 = 40a
A1 Any 2 equations.
5
5(b)
[1.2 = 0 + 0.188t]
M1 For use of constant acceleration formula(e) and solving
for t with their positive a, leading to a positive value of
t
a  10, a   g
Allow if a is negative in part (a) and use |a | here.
Time = 6.39 s
A1 Allow 6.38
Allow 6.32 from a = 0.19
2
© UCLES 2022
Page 13 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
6(a)
Answer
Marks
October/November 2022
Guidance
0.3  2  +0 = 0.3  0.6 + 0.4  v
M1 For use of conservation of momentum. Must be 3
terms. Allow sign errors.
Speed of B = 1.05 ms−1
A1 AG Allow M1 A0 if g included with the masses.
2
6(b)
0.4  1.05 +0 = ( 0.4 + m )  0.5
m = 0.44 or
M1 For use of conservation of momentum. Must be 3
terms. Allow sign errors.
A1 Allow M1 A0 if g included with the masses.
11
25
2
© UCLES 2022
Page 14 of 22
Cambridge International AS & A Level – Mark Scheme
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9709/42
Question
6(c)
© UCLES 2022
Answer
Marks
October/November 2022
Guidance
1.2 [m] or 0.9 [m]
B1 Must be a distance as some candidates get 1.2 from
0.6
.
0.5
0.5t
B1
1
Seen but not + at 2 unless later state that a = 0 .
2
B0 if only 0.5t = 2.1 (may see solving to find t = 3.5) .
0.6t
B1
1
Seen but not + at 2 unless later state that a = 0 .
2
Allow B2 in place of second and third B1 marks for
‘difference in speeds is 0.1 [ ms−1]’.
Distances equal so 0.6t − 0.9 = 0.5t and solve for t
0.9
Or t =
0.6 − 0.5
M1 OE Must get to ‘ t = ’. Allow ± their 0.9 but not ±1.2
or ± 2.1 or 1.5 . Do not allow 0.6t + 0.5t = 0.9 .
Do not allow M1 if either or both terms include
1
+ at 2 unless they state a = 0.
2
Time = 9 s
A1 CWO
Page 15 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
6(c)
© UCLES 2022
Answer
Marks
October/November 2022
Guidance
Alternative method for question 6(c) using time from start of motion
1 [m] or 1.1[m]
B1
0.6T
B1
1
Seen but not + aT 2 unless later state that a = 0 .
2
0.5T
B1
1
Seen but not + aT 2 unless later state that a = 0 .
2
Allow B2 in place of second and third B1 marks for
‘difference in speeds is 0.1 [ ms−1]’.
Distances equal so 0.6T −1.1 = 0.5T and solve for T
1.1
Or T =
0.6 − 0.5
M1 OE Must get to ‘ T = ’. Allow ± their 1.1 but not ±1.0
or ± 2.1. Do not allow 0.6T + 0.5T = 1.1 .
Do not allow M1 if either or both terms include
1
+ aT 2 unless they state a = 0
2
⇒ Time from BC collision = 11 − 2 = 9 s
A1 CWO
Page 16 of 22
Cambridge International AS & A Level – Mark Scheme
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9709/42
Question
6(c)
Answer
Marks
Guidance
Alternative method for question 6(c) using distance travelled from time when B and C collide
1.2 [m] or 0.9 [m]
Time taken for A is
Or d + 0.9 = 0.6t
B1
Or d = 0.5t
B1 FT Allow ± their 0.9 but not ±1.2 or ± 2.1 or 1.5 .
d + 0.9
0.6
Time taken for BC is
B1
d
0.5
d + 0.9 d
4.5
=
 d = 4.5  Time =
0.6
0.5
0.5
M1 Must get to ‘ t = ’.
Allow ± their 0.9 but not ±1.2 or ± 2.1.
Time = 9 s
A1 CWO
5
© UCLES 2022
October/November 2022
Page 17 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
7(a)
Answer
v=
Marks
3
3


0.3 2
t ( +c )  = 0.2t 2 ( +c ) 
1.5


Guidance
M1 For integration (do not penalise missing c)
The power of t must increase by 1 with a change of
coefficient. Use of v = at scores M0.
 8
Velocity = 1.6  =  ms−1
 5
A1 ISW any extra work using the second equation for a.
2
© UCLES 2022
October/November 2022
Page 18 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
7(b)
Answer
v=
Marks
1
1


−
−k − 2
t  + d   = 2kt 2  + d 
−0.5


1
1


−
−
−k
 4 2 + d  = 2k  4 2 + d  Their 1.6 = k + d 
−0.5


1
1

 
−
−
−k
k

0.3 =
 16 2 + d  = 2k  16 2 + d  0.3 = + d 
2
−0.5


 
Attempt to solve for k or d
−
1
2
− 1 or  v = 
Guidance
*M1 For integration. No need for constant. Allow use of
given value of k = 2.6 .
The power of t must increase by 1 with a change of
coefficient. Use of v = at scores M0.
Their 1.6 =
k = 2.6 v = 5.2t
October/November 2022
A1 FT For both equations in k and d ( Allow unsimplified ) .
DM1 Or substitute k = 2.6 into both equations and solve both
for d (with d  0 ).
Must get to ‘ k = ’ or ‘ d = ’.
A1 AG (AG for k, not for the expression).
Allow unsimplified expression for v and/or in terms of
k.
If k is substituted then both equations must be shown
−2.6 − 2
t −1
−0.5
1
−
1
to give a value of d = −1 and getting v = 5.2t 2 − 1
SC B1 for solving the correct equations simultaneously
with no working seen and getting correct expression
for v .
SC A1 for correct expression for v if only the first M1
is scored.
© UCLES 2022
Page 19 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
7(b)
Answer
Marks
October/November 2022
Guidance
Alternative method for question 7(b) from using limits
v=
1
1


−
−k − 2
t  + d   = 2kt 2  + d 
−0.5


*M1 For integration No need for constant. Allow use of
given value of k = 2.6 .
The power of t must increase by 1 with a change of
coefficient. Use of v = at scores M0.
−
−
−
−
−k
−k
4 2 −
 16 2 = 1.6 − 0.3 or 2k  4 2 − 2k  16 2 = 1.6 − 0.3
−0.5
−0.5
1
1
1
1
Attempt to solve for k
k = 2.6, v = 5.2t
−
1
2
DM1 Must be an equation using 0.3 and their k, and 16 and
4, Must be subtracting the limits to form the equation
in k , but may have sign errors in their 1.6 − 0.3 .
A1 AG (AG for k, not for the expression).
Allow unsimplified expression for v and/or in terms of
k.
−2.6 − 2
t −1
−0.5
1
− 1 or v =
A1 FT OE For correct unsimplified equation in k from using
limits for v as their 1.6 and 0.3 and limits for t as 4 and
16.
4
7(c)
5.2T
T=
−
1
2
M1 For solving for T. Must get to ‘ T = ’. Must come from
integration, with their 1 from part (b) or found here not
equal to zero. Do not allow made up value of d.
− their 1 = 0
A1 OE Must be exact. Allow both marks as long as
expression for v is correct, however obtained in
Q7(b).
676
or 27.04
25
2
© UCLES 2022
Page 20 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
7(d)
Answer
4

0
3
5
0.2t 2 dt =
0.2 2
t or
2.5
4
27.04

4
Marks
1
1


−
5.2 2
t − their 1 t
 5.2t 2 − their 1 dt =
0.5


27.04
 0.2 5   5.2 1 
=
t 2  +  t 2 − t
 2.5  0  0.5
 4
October/November 2022
M1
27.04
5 4
1
 

 

 = 0.08t 2  + 10.4t 2 − t 

 



0 
 4 

Guidance
3
−
1
For integration of 0.2t 2 oe or 5.2t 2 − their 1 . May
be in terms of k. Not from any other expression. Their
1 may be zero (or replaced by zero).
Their 1 may came from either part (b) or part (c)
The power of t must increase by 1 with a change of
coefficient in at least one term.
A1ft For both integrals (unsimplified) No need for limits
FT non-zero value of d. May be in terms of k.
Their 1 may came from either part (b) or part (c).
= 0.08  32 + (10.4  5.2 − 27.04 ) − (10.4  2 − 4 )  = 2.56 + 10.24
M1 For correct use of limits (0 and 4 then 4 and their
27.04) in both of their integrals, which have come from
3
−
1
integration of 0.2t 2 and 5.2t 2 − their 1 . Not from any
other expression. Their 1 may be zero (or replaced by
zero).
Their 1 may came from either part (b) or part (c).
Allow M1 for d = 0 (the final answer is 35.8 ).
=
© UCLES 2022
A1 oe Awrt 12.8 Allow if using 27(.0) rather than 27.04
Allow all 4 marks as long as expression for v is
correct, however obtained in Q7(b).
64
or 12.8
5
Page 21 of 22
Cambridge International AS & A Level – Mark Scheme
PUBLISHED
9709/42
Question
7(d)
Answer
Marks
Guidance
SC for using a calculator to integrate.
4

B1 AWRT 2.56
Allow 10.2
Must use 27.04 or 27(.0) if latter integral.
3
Either 0.2t 2 dt = 2.56
0
27.04
Or

4
1


−
2 − 1 dt = 10.24
5.2
t




Total distance = 12.8 m
B1 AWRT 12.8. Allow if using 27(.0) rather than 27.04
Allow both B marks as long as expression for v is
correct, however obtained in Q7(b).
4
© UCLES 2022
October/November 2022
Page 22 of 22
Oxford Cambridge and RSA
A Level Mathematics A
H240/03 Pure Mathematics and Mechanics
Friday 15 June 2018 – Afternoon
Time allowed: 2 hours
* 7 0 0 9 8 7 2 8 3 4 *
You must have:
• Printed Answer Booklet
You may use:
• a scientific or graphical calculator
INSTRUCTIONS
• Use black ink. HB pencil may be used for graphs and diagrams only.
• Complete the boxes provided on the Printed Answer Booklet with your name, centre
number and candidate number.
• Answer all the questions.
• Write your answer to each question in the space provided in the Printed Answer
Booklet. If additional space is required, you should use the lined page(s) at the end of
the Printed Answer Booklet. The question number(s) must be clearly shown.
• Do not write in the barcodes.
• You are permitted to use a scientific or graphical calculator in this paper.
• Final answers should be given to a degree of accuracy appropriate to the context.
• The acceleration due to gravity is denoted by g m s–2. Unless otherwise instructed, when
a numerical value is needed, use g = 9.8.
INFORMATION
• The total mark for this paper is 100.
• The marks for each question are shown in brackets [ ].
• You are reminded of the need for clear presentation in your answers.
• The Printed Answer Booklet consists of 16 pages. The Question Paper consists of
8 pages.
© OCR 2018 [603/1038/8]
DC (SC/TP) 164254/2
OCR is an exempt Charity
Turn over
2
Formulae
A Level Mathematics A (H240)
Arithmetic series
S n = 12 n ^a + lh = 12 n "2a + ^n - 1h d ,
Geometric series
a ^1 - r nh
1-r
a
S3 =
for r 1 1
1-r
Sn =
Binomial series
^a + bhn = a n + n C 1 a n - 1 b + n C 2 a n - 2 b 2 + f + n Cr a n - r b r + f + b n
n
n!
where n C r = n C r = c m =
r
r! ^n - rh !
^1 + xhn = 1 + nx +
n ^n - 1h f ^n - r + 1h r
n ^n - 1h 2
x +f+
x +f
2!
r!
Differentiation
f ^xh
f l^xh
tan kx
sec x
k sec 2 kx
sec x tan x
cot x
- cosec 2 x
- cosec x cot x
cosec x
du
dv
v -u
u dy
dx
d
x
Quotient rule y = ,
=
2
v dx
v
Differentiation from first principles
f l^xh = lim
h"0
f ^x + hh - f ^xh
h
Integration
c f l^xh
dd
dx = ln f ^xh + c
e f ^xh
n
n+1
1
; f l^xh`f ^xhj dx = n + 1 `f ^xhj + c
dv
du
Integration by parts ; u dx = uv - ; v dx
dx
dx
Small angle approximations
sin i . i , cos i . 1 - 12 i 2 , tan i . i where i is measured in radians
© OCR 2018
H240/03 Jun18
^n ! Nh ,
^ x 1 1, n ! Rh
3
Trigonometric identities
sin ^A ! Bh = sin A cos B ! cos A sin B
cos ^A ! Bh = cos A cos B " sin A sin B
tan _A ! Bi =
aA ! B ! ^k + 12h rk
tan A ! tan B
1 " tan A tan B
Numerical methods
y dx . 12 h "^y0 + y nh + 2 ^y1 + y 2 + f + y n - 1h, , where h =
f ^x nh
The Newton-Raphson iteration for solving f ^xh = 0 : x n + 1 = x n f l^x nh
Trapezium rule:
b
ya
b-a
n
Probability
P ^A , Bh = P ^Ah + P ^Bh - P ^A + Bh
P ^A + Bh = P ^Ah P ^B Ah = P ^Bh P ^A Bh or P ^A Bh =
Standard deviation
/^x - -xh
=
n
2
/ f ^x - -xh
=
/f
2
/ x2 -2
- x or
n
P ^A + Bh
P ^Bh
/ fx 2 - 2
/ f -x
The binomial distribution
n
n-x
If X + B ^n, ph then P ^X = xh = c m p x ^1 - ph , Mean of X is np, Variance of X is np ^1 - ph
x
Hypothesis test for the mean of a normal distribution
If X + N ^n, v 2h then X + N cn,
v 2m
n
and
X-n
+ N ^0, 1h
v n
Percentage points of the normal distribution
If Z has a normal distribution with mean 0 and variance 1 then, for each value of p, the table gives the value of z
such that P ^Z G zh = p .
p
0.75
0.90
0.95
0.975
0.99
0.995
0.9975
0.999
0.9995
z
0.674
1.282
1.645
1.960
2.326
2.576
2.807
3.090
3.291
Kinematics
Motion in a straight line
Motion in two dimensions
v = u + at
v = u + at
s = ut + 12 at 2
s = ut + 12 at 2
s = 12 ^u + vh t
s = 12 ^u + vh t
s = vt - 12 at 2
s = vt - 12 at 2
© OCR 2018
H240/03 Jun18
v 2 = u 2 + 2as
Turn over
4
Section A: Pure Mathematics
Answer all the questions.
1
A circle with centre C has equation x 2 + y 2 + 8x - 2y - 7 = 0 .
Find
(i)
the coordinates of C,
[2]
(ii)
the radius of the circle.
[1]
[3]
2
Solve the equation 2x - 1 = x + 3 .
3
In this question you must show detailed reasoning.
A gardener is planning the design for a rectangular flower bed. The requirements are:
•
•
•
the length of the flower bed is to be 3 m longer than the width,
the length of the flower bed must be at least 14.5 m,
the area of the flower bed must be less than 180 m2.
The width of the flower bed is x m.
By writing down and solving appropriate inequalities in x, determine the set of possible values for the width
of the flower bed.
[6]
4
In this question you must show detailed reasoning.
The functions f and g are defined for all real values of x by
f ^xh = x 3 and g ^xh = x 2 + 2 .
(i)
Write down expressions for
(a) fg ^xh ,
[1]
(b) gf ^xh .
(ii)
© OCR 2018
[1]
Hence find the values of x for which fg ^xh - gf ^xh = 24 .
H240/03 Jun18
[6]
5
5
(i)
Use the trapezium rule, with two strips of equal width, to show that
=
(ii)
4
1
11
dx .
- 2.
4
0 2+ x
Use the substitution x = u2 to find the exact value of
=
(iii)
4
1
dx .
0 2+ x
2
,
4
where k is a rational number to be determined.
7
[6]
Using your answers to parts (i) and (ii), show that
ln 2 . k +
6
[5]
[2]
It is given that the angle i satisfies the equation sin a2i + 14 rk = 3 cos a2i + 14 rk .
(i)
Show that tan 2i = 12 .
[3]
(ii)
Hence find, in surd form, the exact value of tan i, given that i is an obtuse angle.
[5]
The gradient of the curve y = f ^xh is given by the differential equation
^2x - 1h3
dy
+ 4y 2 = 0
dx
and the curve passes through the point ^1, 1h . By solving this differential equation show that
f ^xh =
ax 2 - ax + 1
,
bx 2 - bx + 1
[9]
where a and b are integers to be determined.
© OCR 2018
H240/03 Jun18
Turn over
6
Section B: Mechanics
Answer all the questions.
8
1
0
In this question c m and c m denote unit vectors which are horizontal and vertically upwards respectively.
0
1
2
A particle of mass 5 kg, initially at rest at the point with position vector c m m, is acted on by gravity and
45
15
-7
also by two forces c m N and c m N.
-8
-2
9
(i)
Find the acceleration vector of the particle.
[3]
(ii)
Find the position vector of the particle after 10 seconds.
[3]
A uniform plank AB has weight 100 N and length 4 m. The plank rests horizontally in equilibrium on two
smooth supports C and D, where AC = x m and CD = 0.5 m (see diagram).
4m
A
xm
C 0.5 m D
B
The magnitude of the reaction of the support on the plank at C is 75 N. Modelling the plank as a rigid rod,
find
(i)
the magnitude of the reaction of the support on the plank at D,
[1]
(ii)
the value of x.
[3]
A stone block, which is modelled as a particle, is now placed at the end of the plank at B and the plank is on
the point of tilting about D.
(iii)
Find the weight of the stone block.
(iv)
Explain the limitation of modelling
© OCR 2018
[3]
(a) the stone block as a particle,
[1]
(b) the plank as a rigid rod.
[1]
H240/03 Jun18
7
10
Three forces, of magnitudes 4 N, 6 N and P N, act at a point in the directions shown in the diagram.
6N
4N
2i
i
PN
The forces are in equilibrium.
(i)
Show that i = 41.4°, correct to 3 significant figures.
[4]
(ii)
Hence find the value of P.
[2]
The force of magnitude 4 N is now removed and the force of magnitude 6 N is replaced by a force of
magnitude 3 N acting in the same direction.
(iii)
11
Find
(a) the magnitude of the resultant of the two remaining forces,
[3]
(b) the direction of the resultant of the two remaining forces.
[2]
The velocity v m s-1 of a car at time t s, during the first 20 s of its journey, is given by v = kt + 0.03t 2, where
k is a constant. When t = 20 the acceleration of the car is 1.3 m s-2. For t 2 20 the car continues its journey
with constant acceleration 1.3 m s-2 until its speed reaches 25 m s-1.
(i)
Find the value of k.
[3]
(ii)
Find the total distance the car has travelled when its speed reaches 25 m s-1.
[7]
© OCR 2018
H240/03 Jun18
Turn over
8
12
One end of a light inextensible string is attached to a particle A of mass m kg. The other end of the string is
attached to a second particle B of mass mm kg, where m is a constant. Particle A is in contact with a rough
plane inclined at 30° to the horizontal. The string is taut and passes over a small smooth pulley P at the top
of the plane. The part of the string from A to P is parallel to a line of greatest slope of the plane. The particle
B hangs freely below P (see diagram).
P
A
B
30°
The coefficient of friction between A and the plane is n.
(i)
(ii)
It is given that A is on the point of moving down the plane.
(a) Find the exact value of n when m = 14 .
[7]
(b) Show that the value of m must be less than 12 .
[2]
Given instead that m = 2 and that the acceleration of A is 14 g m s-2, find the exact value of n.
[5]
END OF QUESTION PAPER
Oxford Cambridge and RSA
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OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a
department of the University of Cambridge.
© OCR 2018
H240/03 Jun18
Oxford Cambridge and RSA
Friday 14 June 2019 – Afternoon
A Level Mathematics A
H240/03 Pure Mathematics and Mechanics
Time allowed: 2 hours
* 7 6 6 3 7 9 6 7 6 1 *
You must have:
• Printed Answer Booklet
You may use:
• a scientific or graphical calculator
INSTRUCTIONS
• Use black ink. HB pencil may be used for graphs and diagrams only.
• Answer all the questions.
• Write your answer to each question in the space provided in the Printed Answer
Booklet. Additional paper may be used if necessary but you must clearly show your
candidate number, centre number and question number(s).
• You are permitted to use a scientific or graphical calculator in this paper.
• Give non-exact numerical answers correct to 3 significant figures unless a different
degree of accuracy is specified in the question.
• The acceleration due to gravity is denoted by g m s–2. Unless otherwise instructed, when
a numerical value is needed, use g = 9.8.
INFORMATION
• The total mark for this paper is 100.
• The marks for each question are shown in brackets [ ].
• You are reminded of the need for clear presentation in your answers.
• The Printed Answer Booklet consists of 16 pages. The Question Paper consists of
12 pages.
© OCR 2019 [603/1038/8]
DC (SC/TP) 173992/2
OCR is an exempt Charity
Turn over
2
Formulae
A Level Mathematics A (H240)
Arithmetic series
S n = 12 n ^a + lh = 12 n "2a + ^n - 1h d ,
Geometric series
a ^1 - r nh
1-r
a
for r 1 1
S3 =
1-r
Sn =
Binomial series
^a + bhn = a n + n C1 a n - 1 b + n C2 a n - 2 b 2 + f + n Cr a n - r b r + f + b n
n
n!
where n C r = n C r = c m =
r
r! ^n - rh !
^1 + xhn = 1 + nx +
n ^n - 1h f ^n - r + 1h r
n ^n - 1h 2
x +f+
x +f
2!
r!
Differentiation
f ^xh
tan kx
sec x
cot x
cosec x
f l^xh
k sec 2 kx
sec x tan x
- cosec 2 x
- cosec x cot x
du
dv
v -u
u dy
d
dx
x
Quotient rule y = ,
=
2
v dx
v
Differentiation from first principles
f ^x + hh - f ^xh
h"0
h
f l^xh = lim
Integration
c f l^xh
dd
dx = ln f ^xh + c
h
^
f
x
e
n
n+1
1
; f l^xhaf ^xhk dx = n + 1 af ^xhk + c
dv
du
Integration by parts ; u dx = uv - ; v dx
dx
dx
Small angle approximations
sin i . i , cos i . 1 - 12 i 2 , tan i . i where i is measured in radians
© OCR 2019
H240/03 Jun19
^n ! Nh,
^ x 1 1, n ! Rh
3
Trigonometric identities
sin ^A ! Bh = sin A cos B ! cos A sin B
cos ^A ! Bh = cos A cos B " sin A sin B
tan ^A ! Bh =
aA ! B ! ^k + 12h rk
tan A ! tan B
1 " tan A tan B
Numerical methods
y dx . 12 h "^y 0 + y nh + 2 ^y 1 + y2 + f + y n - 1h, , where h =
f ^x nh
The Newton-Raphson iteration for solving f ^xh = 0 : xn + 1 = xn f l^xnh
Trapezium rule:
b
ya
Probability
P ^A , Bh = P ^Ah + P ^Bh - P ^A + Bh
P ^A + Bh = P ^Ah P ^B Ah = P ^Bh P ^A Bh or P ^A Bh =
Standard deviation
/^x - -xh
=
n
2
/ f ^x - -xh
=
/f
2
/ x2 -2
- x or
n
b-a
n
P ^A + Bh
P ^Bh
/ fx 2 - 2
/ f -x
The binomial distribution
n
n-x
If X + B ^n, ph then P ^X = xh = c m p x ^1 - ph , mean of X is np, variance of X is np ^1 - ph
x
Hypothesis test for the mean of a normal distribution
If X + N ^n, v 2h then X + N cn,
v 2m
n
and
X-n
+ N ^0, 1h
v n
Percentage points of the normal distribution
If Z has a normal distribution with mean 0 and variance 1 then, for each value of p, the table gives the
value of z such that P ^Z G zh = p .
p
z
0.75
0.674
0.90
1.282
0.95
1.645
0.975
1.960
0.99
2.326
0.995
2.576
0.9975
2.807
Kinematics
Motion in a straight line
Motion in two dimensions
v = u + at
v = u + at
s = ut + 12 at 2
s = ut + 12 at 2
s = 12 ^u + vh t
v 2 = u 2 + 2as
0.999
3.090
0.9995
3.291
s = 12 ^u + vh t
s = vt - 12 at 2
s = vt - 12 at 2
© OCR 2019
H240/03 Jun19
Turn over
4
Section A: Pure Mathematics
Answer all the questions.
1
C
13.5 cm
8.3 cm
32°
A
B
The diagram shows triangle ABC, with AC = 13.5cm , BC = 8.3cm and angle ABC = 32° .
[2]
Find angle CAB.
2
A circle with centre C has equation x 2 + y 2 - 6x + 4y + 4 = 0.
(a) Find
(i) the coordinates of C,
[2]
(ii) the radius of the circle.
[1]
(b) Determine the set of values of k for which the line y = kx - 3 does not intersect or touch the
circle.
[5]
3
(a) In this question you must show detailed reasoning.
[4]
Solve the inequality x - 2 G 2x - 6 .
(b) Give full details of a sequence of two transformations needed to transform the graph of
y = x - 2 to the graph of y = 2x - 6 .
[3]
© OCR 2019
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5
4
y
P
O
1
2r
x
The diagram shows the part of the curve y = 3x sin 2x for which 0 G x G 12 r.
The maximum point on the curve is denoted by P.
(a) Show that the x-coordinate of P satisfies the equation tan 2x + 2x = 0.
[3]
(b) Use the Newton-Raphson method, with a suitable initial value, to find the x-coordinate of P,
giving your answer correct to 4 decimal places. Show the result of each iteration.
[4]
(c) The trapezium rule, with four strips of equal width, is used to find an approximation to
y0
1
2r
3x sin 2x dx.
Show that the result can be expressed as kr 2 ^ 2 + 1h, where k is a rational number to be
determined.
[4]
(d)
(i) Evaluate
y0
1
2r
[1]
3x sin 2x dx.
(ii) Hence determine whether using the trapezium rule with four strips of equal width gives
an under- or over-estimate for the area of the region enclosed by the curve y = 3x sin 2x
[1]
and the x-axis for 0 G x G 12 r.
(iii) Explain briefly why it is not easy to tell from the diagram alone whether the trapezium
rule with four strips of equal width gives an under- or over-estimate for the area of the
region in this case.
[1]
5
In this question you must show detailed reasoning.
(a) Prove that (cot i + cosec i) 2 =
1 + cos i
.
1 - cos i
(b) Hence solve, for 0 1 i 1 2r, 3 (cot i + cosec i) 2 = 2 sec i.
© OCR 2019
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[4]
[5]
Turn over
6
6
y
O
x
The diagram shows part of the curve y =
2x - 1
.
(2x + 3) (x + 1) 2
Find the exact area of the shaded region, giving your answer in the form p + q ln r, where p and q
are positive integers and r is a positive rational number.
[10]
© OCR 2019
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7
Section B: Mechanics
Answer all the questions.
7
8
A cyclist starting from rest accelerates uniformly at 1.5m s - 2 for 4 s and then travels at constant
speed.
(a) Sketch a velocity-time graph to represent the first 10 seconds of the cyclist’s motion.
[2]
(b) Calculate the distance travelled by the cyclist in the first 10 seconds.
[2]
A particle P projected from a point O on horizontal ground hits the ground after 2.4 seconds.
The horizontal component of the initial velocity of P is 53 d m s -1 .
(a) Find, in terms of d, the horizontal distance of P from O when it hits the ground.
[1]
(b) Find the vertical component of the initial velocity of P.
[2]
P just clears a vertical wall which is situated at a horizontal distance d m from O.
(c) Find the height of the wall.
[3]
The speed of P as it passes over the wall is 16 m s -1 .
(d) Find the value of d correct to 3 significant figures.
© OCR 2019
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[4]
Turn over
8
9
B
0.2 kg
0.5 kg
P
i
The diagram shows a small block B, of mass 0.2 kg, and a particle P, of mass 0.5 kg, which are
attached to the ends of a light inextensible string. The string is taut and passes over a small smooth
pulley fixed at the intersection of a horizontal surface and an inclined plane.
The block can move on the horizontal surface, which is rough. The particle can move on the inclined
plane, which is smooth and which makes an angle of i with the horizontal where tan i = 34 .
The system is released from rest. In the first 0.4 seconds of the motion P moves 0.3 m down the
plane and B does not reach the pulley.
(a) Find the tension in the string during the first 0.4 seconds of the motion.
[4]
(b) Calculate the coefficient of friction between B and the horizontal surface.
[5]
10 In this question the unit vectors i and j are in the directions east and north respectively.
A particle R of mass 2 kg is moving on a smooth horizontal surface under the action of a single
horizontal force F N. At time t seconds, the velocity v m s -1 of R, relative to a fixed origin O, is
given by v = ( pt 2 - 3t) i + (8t + q) j, where p and q are constants and p 1 0.
(a) Given that when t = 0.5 the magnitude of F is 20, find the value of p.
[6]
When t = 0 , R is at the point with position vector (2i - 3j) m .
(b) Find, in terms of q, an expression for the displacement vector of R at time t.
[4]
When t = 1, R is at a point on the line L, where L passes through O and the point with position
vector 2i - 8j.
(c) Find the value of q.
© OCR 2019
[3]
H240/03 Jun19
9
11
B
2a
d
C
h
30°
A
The diagram shows a ladder AB, of length 2a and mass m, resting in equilibrium on a vertical wall
of height h. The ladder is inclined at an angle of 30° to the horizontal. The end A is in contact with
horizontal ground. An object of mass 2m is placed on the ladder at a point C where AC = d.
The ladder is modelled as uniform, the ground is modelled as being rough, and the vertical wall is
modelled as being smooth.
(a) Show that the normal contact force between the ladder and the wall is
mg (a + 2d) 3
.
4h
[4]
It is given that the equilibrium is limiting and the coefficient of friction between the ladder and the
ground is 18 3 .
(b) Show that h = k (a + 2d ) , where k is a constant to be determined.
[7]
(c) Hence find, in terms of a, the greatest possible value of d.
[2]
(d) State one improvement that could be made to the model.
[1]
END OF QUESTION PAPER
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