M3 January 2012 Mark Scheme

7/31/2019 M3 January 2012 Mark Scheme

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Mark Scheme (Results)

January 2012

GCE Mechanics M3 (6679) Paper 1


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January 2012

Publications Code UA030776

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Pearson Education Ltd 2012


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General Marking Guidance

All candidates must receive the same treatment. Examiners must mark thefirst candidate in exactly the same way as they mark the last.

Mark schemes should be applied positively. Candidates must be rewarded forwhat they have shown they can do rather than penalised for omissions.

Examiners should mark according to the mark scheme not according to theirperception of where the grade boundaries may lie.

There is no ceiling on achievement. All marks on the mark scheme should beused appropriately.

All the marks on the mark scheme are designed to be awarded. Examinersshould always award full marks if deserved, i.e. if the answer matches the

mark scheme. Examiners should also be prepared to award zero marks if thecandidates response is not worthy of credit according to the mark scheme.

Where some judgement is required, mark schemes will provide the principlesby which marks will be awarded and exemplification may be limited.

When examiners are in doubt regarding the application of the mark schemeto a candidates response, the team leader must be consulted.

Crossed out work should be marked UNLESS the candidate has replaced itwith an alternative response.


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EDEXCEL GCE MATHEMATICS

General Instructions for Marking

1. The total number of marks for the paper is 75.

2. The Edexcel Mathematics mark schemes use the following types of marks:

M marks: method marks are awarded for knowing a method and attempting to apply it,unless otherwise indicated.

A marks: Accuracy marks can only be awarded if the relevant method (M) marks havebeen earned.

B marks are unconditional accuracy marks (independent of M marks)

Marks should not be subdivided.

3. Abbreviations

These are some of the traditional marking abbreviations that will appear in the mark schemesand can be used if you are using the annotation facility on ePEN.

bod benefit of doubt ft follow through the symbol will be used for correct ft

cao correct answer only cso - correct solution only. There must be no errors in this part of the questionto obtain this mark

isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark

4. All A marks are correct answer only (cao.), unless shown, for example, as A1 ft toindicate that previous wrong working is to be followed through. After a misread however,the subsequent A marks affected are treated as A ft, but manifestly absurd answersshould never be awarded A marks.


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General Principals for Core Mathematics Marking

(But note that specific mark schemes may sometimes override these general principles).

Method mark for solving 3 term quadratic:1. Factorisation

cpqqxpxcbxx =++=++ where),)(()( 2, leading to x= ....

amncpqqnxpmxcbxax ==++=++ andwhere),)(()( 2, leading to x=

2. FormulaAttempt to use correct formula (with values for a, b and c), leading to x=

3. Completing the square

Solving 02

=++ cbxx : ( )22 , 0bx q c q , leading to x=

Method marks for differentiation and integration:1. Differentiation

Power of at least one term decreased by 1. ( 1 nn xx )

2. IntegrationPower of at least one term increased by 1. ( 1+ nn xx )

Use of a formulaWhere a method involves using a formula that has been learnt, the advice given in recentexaminers reports is that the formula should be quoted first.Normal marking procedure is as follows:Method mark for quoting a correct formula and attempting to use it, even if there are mistakesin the substitution of values.Where the formula is not quoted, the method mark can be gained by implication from correctworking with values, but may be lost if there is any mistake in the working.


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January 20126679 Mechanics M3

Mark Scheme

Question

Number

Scheme Marks

1. EPE =2.1

5.0 2 B1

GPE lost = EPE gained M1 (used)

2.1

5.01.18.98.0

2

=

A1ft

4.41= N or 41 N A14


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Question

Number

Scheme Marks

2.

(a) 3,3

22===

T B1

8.12.092 === xa ms-2 towards C M1 A1

(3)

(b) 89.1)04.025.0(92222 === xav M1

v = 1.37 ms-1

A1

(2)

(c) 2.03sin5.0 == tx M1 A1ft

137.04.0sin3

1 1=

t s A1(3)

8


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Question

Number

Scheme Marks

3.

(a)( ) ( )

2 3d 10 10 100,d 6 6 6va v

x x x x = = =

+ + +

M1M1, A1

=( )

3

100 18014 6

=

+

ms-2

A1

(4)

(b) =++

= txxxt

xd10d6

6

10

d

d

M1

M1

[ ]14

2

12

6 102

Txx t

+ =

M1 A1

196 6 14 2 12 10 102

T+ = M1

178 17.8(s)T T= = A1

(6)

10


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Question

NumberScheme Marks

4.

(a)

gT 5.060cos = , T = g (1) M1, A1

Extension in the string =x,8.0

6.19 x

a

xT ==

B1

Using (1) , xxg ,5.24= = 0.4 m * M1, A1(5)

(b) 2

5.060sin = rT (2) M1 A1

Using (2) 260sin)4.08.0(5.060sin +=g M1 A1

3

5,2.1

22 gg == (4.04 or 4.0) A1(5)

10

0.5g

T

r

60

B

A


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Question

NumberScheme Marks

5.

(a) Distance of P from the centre of the Earth =R + x

( )2xR

kF

+=

x = 0, F =mg, ( )2Rmgk= M1 A1

( )2

2

xR

mgRF

+= * A1

(3)

(b) maF = ,( ) x

vv

xR

gR

d

d2

2

=+

M1 A1

( )

gR

22 2

2

V

d d

R

R

gRv v x

R x=

+ M1 A1

R

R

gR

V xR

gRv

22

22

2

1

+=

M1

A1

6232

1

22

122

2 gR

R

gR

R

gRV

gR== M1

225 5

2 4 6 12 6

V gR gR gR gRV= + = = , 6

5gRV = A1, A1

(9)

12


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Question

Number

Scheme

Marks

6.

(a)

GPE gained = ( )cos1 mgl

Conservation of energy: ( ) 22

1cos1

4

11

2

1mvmgl

glm += M1A1 A1

+=

+= cos2

4

3cos22

4

112glglv

Resolving towards the centre of the circle: M1

l

mvmgT

2

cos = A1 A1

+= cos2

4

3cos mgmgT M1

+=

+=

4

1cos3cos3

4

3 mgmgT * A1

(8)

(b)4

1cos0 == T M1

4cos2

4

32 glglv =

+= ,

4

glv = M1, A1

(3)

(c) Horizontal component of velocity atB

( )44

1180cos

4

glgl== B1ft

Extra height h42

1

642

1 glmglmmgh =+ M1 A1

128

15

128

1

8

1 llh =

= (0.117l) A1

OR: Using128

15

2

16

15

4

2

sin 22 l

g

gl

g

vh =

==

(4)

OR: Using128

15,2

64

150,2

22 lhgh

glasuv ==+=

15

l

11gl


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Question

NumberScheme Marks

7.

(a) ( ) +== xxxxxxxxy d12364

d6

4

d432222 M1 A1

=5

1024

45312

4

6

2

543

=

+

xxx M1

(160.8.)

( )22 3 3 4 5d 6 d 36 12 d

4 4y x x x x x x x x x

= = + M1 A1

=15

10496

46

1

5

129

4

6

2

654=

+

xxx M1

(549.5)

......416.31024

5

15

10496== x M1 A1

Required distance 3.42 2 = 1.42(cm) * A1(9)

(b) Base has radius 4422

1= cm B1

About to topple42.1

4tan = M1 A1

5.70 A1(4)

(c) Parallel to slope: sinmgF = Perpendicular to the slope: cosmgR = M1 A1

About to slip: RF =

== 7.16,3.0tan A1

(3)

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Documents

M3 January 2012 Mark Scheme