Mark Scheme (Results) January 2011

GCE

GCE Mechanics M3 (6679) Paper 1

Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH

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January 2011 Publications Code UA026583 All the material in this publication is copyright ÂŠ Edexcel Ltd 2011

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 have been 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 schemes. •

bod – benefit of doubt

•

ft – follow through

•

the symbol

•

cao – correct answer only

•

cso - correct solution only. There must be no errors in this part of the question to 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

•

will be used for correct ft

The second mark is dependent on gaining the first mark

January 2011 Mechanics M3 6679 Mark Scheme Question Number 1.

Scheme

v

dv = 7 − 2x dx

1 2

v 2 = 7 x − x 2 ( +c )

Marks M1 M1A1 A1

x = 0 v = 6 ⇒ c = 18

v = 0 x 2 − 7 x − 18 = 0 ( x + 2 )( x − 9 ) = 0

M1 A1

∴x = 9

[6]

GCE Mechanics M3 (6679) January 2011

1

Question Number

Scheme

Marks

2. (a)

Mass ratio

4m 3 r 8

Dist from O

B1 B1

(4 + k)m

km 1 − r 2

0

Moments about O: 3 1 r × 4m = r × km 8 2

M1

k =3

A1 (4)

(b)

30o O

G

λ mg

7mg

tan 30 = OG =

B1

OG r

λ

(7 + λ )

M1

× 2r

1 λ 1 = × 2r × √ 3 (7 + λ ) r

7 + λ = 2 √ 3λ 7 λ= 2 √ 3 −1

(

)

A1

A1

(o.e.) or 2.84

(4) [8]

GCE Mechanics M3 (6679) January 2011

2

Question Number

Scheme

Marks

3. (a)

2

M1

2

Vol = π ∫ y 2 dx = π ∫ e 2 x dx 1

1

M1 A1

2 1 = π ⎡⎣e 2x ⎤⎦ 1 2 1 = π ⎡⎣e 4 − e 2 ⎤⎦ 2

(b)

C of M

∫

2

1

∫ =

2

1

A1 (4)

π y 2 x dx vol

M1 A1

2

21 ⎡1 ⎤ e x dx = ⎢ xe 2 x ⎥ − ∫ e 2 x dx ⎣2 ⎦1 1 2 2x

2

M1

2

⎡1 ⎤ ⎡1 ⎤ = ⎢ xe 2 x ⎥ − ⎢ e 2 x ⎥ ⎣2 ⎦1 ⎣ 4 ⎦ 1 1 1 1 ⎞ ⎛1 = × 2e 4 − × 1e 2 − ⎜ e 4 − e 2 ⎟ 2 2 4 ⎠ ⎝4 1 ⎞ ⎛3 = ⎜ e4 − e2 ⎟ 4 ⎠ ⎝4 1 ⎞ ⎛3 π ⎜ e4 − e2 ⎟ 4 4 ⎠ = 1.656... C of M = ⎝ 1 4 2 π (e − e ) 2

A1 M1 A1

= 1.66

(3 sf)

GCE Mechanics M3 (6679) January 2011

(6) [10]

3

Question Number

Scheme

Marks

4. (a)

⎛ πt ⎞ x = 5sin ⎜ ⎟ ⎝ 3⎠

x = 5 ×

π

⎛ πt ⎞ cos ⎜ ⎟ 3 ⎝ 3⎠ M1A1

⎛π ⎞ ⎛ πt ⎞ x = −5 × ⎜ ⎟ sin ⎜ ⎟ ⎝3⎠ ⎝ 3⎠ 2

x=−

(b)

π2 9

period =

A1

(∴ S.H.M.)

x

(3)

2π π

=6

B1

3

B1

amplitude = 5 (c)

(d)

(2)

⎛ πt ⎞ cos ⎜ ⎟ 3 ⎝ 3⎠ 5π max . v = 3

... = 5 ×

π

At A x = 2

sin

π 3

tA =

M1

or vmax = aω

A1 (2) M1

⎛ πt ⎞ 2 = 5sin ⎜ ⎟ ⎝ 3⎠

t = 0.4 3

π

A1

× sin −1 0.4

At B x = 3

tB =

time A → B =

3

π

3

π

× sin −1 0.6

× sin −1 0.6 −

3

π

A1

× sin −1 0.4

= 0.2215... = 0.22 s accept awrt 0.22

A1 (4) [11]

GCE Mechanics M3 (6679) January 2011

4

Question Number

Scheme

Marks

5. A

Ta

Tb

B mg

(a)

B1

l √2 R ( ↑ ) Ta cos 45 = Tb cos 45 + mg

r=

M1A1

Ta − Tb = mg √ 2 R ( →)

(1)

Ta cos 45 + Tb cos 45 = mrω

M1A1

2

1 1 l + Tb × =m ω2 √2 √2 √2 Ta + Tb = mlω 2

(2)

Ta − Tb = mg √ 2

(1)

Ta ×

( 1 = m ( lω 2

) + g √ 2)

M1

2Ta = m lω 2 + g √ 2 Ta

2

A1

Tb = mlω 2 − Ta =

(b)

(

1 m lω 2 − g √ 2 2

Tb > 0

ω2 >

g√2 l

A1

)

(

(8) M1

)

1 m lω 2 − g √ 2 > 0 2

A1

*

GCE Mechanics M3 (6679) January 2011

(2) [10]

5

Question Number

Scheme

6.

A

Marks B

C

(a) 3 4

l

θ

Ta

P

Tb

3mg

5 length AP = length BP = l 4 1 kmg ( 4 l ) 1 Ta = Tb = = kmg 4 l R (↑)

Ta cos θ + Tb cos θ = 3mg

1 3 1 3 kmg × + kmg × = 3mg 4 5 4 5 3 kmg = 3mg 10 k = 10 *

B1

( or T = ...)

( or 2T cos θ = 3mg ) 3 ⎛ 1 ⎞ ⎜ or kmg × = 3mg ⎟ 5 ⎝ 2 ⎠

M1A1 M1A1 A1

A1 (7)

(b) 12 5

initial extn =

l

B1

13 8 l −l = l 5 5

λ x2

2

10mg ⎛ 8l ⎞ 128mgl E.P.E. lost = 2 × = 2× ⎜ ⎟ = 2l 2l ⎝ 5 ⎠ 5 12l 36mgl P.E. gained = 3mg × = 5 5 1 36mgl 128mgl × 3mv 2 + = 2 5 5 256 gl 72 gl v2 = − 15 15 ⎛ 184 ⎞ v = √⎜ gl ⎟ ⎝ 15 ⎠

M1A1

M1A1

A1 (6) [13]

GCE Mechanics M3 (6679) January 2011

6

Question Number

Scheme

7.

Marks

u v

mg (a)

mgl ( cos α − cos θ ) =

M1A1=A1

1 2 1 2 mv − mu 2 2

v 2 = u 2 + 2 gl ( cos α − cos θ )

A1

*

(4) (b)

⎛3 ⎞ v 2 = 2 gl ⎜ − cos θ ⎟ + u 2 ⎝5 ⎠ ⎛3 ⎞ At top θ = 360° v 2 = 2 gl ⎜ − 1⎟ + u 2 ⎝5 ⎠ 2 v 2 > 0 − 2 gl × + u 2 > 0 5 4 gl u2 > 5 gl * u>2 5 cos α =

3 5

M1A1 M1

A1 (4)

GCE Mechanics M3 (6679) January 2011

7

Question Number (c)

Scheme

Marks

Equation of motion along radius at lowest point: T1 − mg =

θ = 180

M1A1

mv 2 l

M1

⎛3 ⎞ v 2 = 2 gl ⎜ + 1⎟ + u 2 ⎝5 ⎠

16 gl + u2 5 m ⎛ 16 gl ⎞ T1 = ⎜ + u 2 ⎟ + mg l ⎝ 5 ⎠ 2 21mg mu = + 5 l v2 =

A1

At highest point: T2 + mg =

M1

mv 2 l

2 ⎛ 2 ⎞ mu T2 = 2mg ⎜ − ⎟ + − mg l ⎝ 5⎠ mu 2 9mg T2 = − l 5 T1 = 5T2

M1

⎛ mu 2 9mg ⎞ 21mg mu 2 + = 5⎜ − ⎟ 5 l 5 ⎠ ⎝ l 66mg 4mu 2 = 5 l 33 gl u2 = * 10

M1

θ = 360

GCE Mechanics M3 (6679) January 2011

A1

A1 (9) [17]

8

Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN Telephone 01623 467467 Fax 01623 450481 Email publications@linneydirect.com Order Code UA026583 January 2011 For more information on Edexcel qualifications, please visit www.edexcel.com/quals Edexcel Limited. Registered in England and Wales no.4496750 Registered Office: One90 High Holborn, London, WC1V 7BH

M3 MS Jan 2011

Published on Jun 11, 2011

GCE Mechanics M3 (6679) Paper 1 GCE Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, Lond...

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