M3 MS Jun 2008

GCE

Edexcel GCE Mathematics Mechanics 3 M3 (6679

June 2008

Mathematics

Edexcel GCE

Mark Scheme (Final)

General Marking Guidance •

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

•

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

•

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

•

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

•

All the marks on the mark scheme are designed to be awarded. Examiners should 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 the candidate’s response is not worthy of credit according to the mark scheme.

•

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

•

When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, the team leader must be consulted.

•

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

June 2008 6679 Mechanics M3 Mark Scheme Scheme

Question Number

Q1(a)

1 λ⎛1 ⎞ ⎜ L⎟ 2 L⎝2 ⎠

EPE stored =

1 m 2gL (= mgL) 2 λL = mg L EPE = KE ⇒ 8

Marks

2

B1

⎛ λL ⎞ ⎜= ⎟ ⎝ 8 ⎠

B1

KE gained =

(b)

*

i.e. λ = 8mg

M1A1cso (4)

M1

EPE

= GPE + KE 2

8mgL 1 8mg ⎛ 1 ⎞ L 1 = mg + mu 2 ⎜ L⎟ = 2 L ⎝2 ⎠ 8 2 2

mgL 2

=

m 2 u 2

∴u=

gL

A1A1 M1A1 (5) 9 Marks

Scheme

Question Number Q2 (a)

Marks

O A

T=3=

2π

B 2π 3 ) ; a = 0.12 , u 2 = a 2 ω 2 , u = 0.12 × ω = 0.251 ms −1 (0.25 m s-1)

∴ ω=

ω u = ω 2 (a 2 - x 2 2

M1A1 M1 A1

(b)

Time from O → A → O

= 1.5s

∴ t = 0.5

⎛π ⎞ ⇒ OP = 0.12 sin ⎜ ⎟ ⎝3⎠

x = a sin ωt

(

v 2 = ω2 a2 - x 2 v =

2π 3

)

0.12 2 − 0.104.... 2 =

B1 M1A1

Distance from B is 0.12 – OP = 0.12 – 0.104… = 0.016m

(c)

(4)

M1A1 (5)

M1 2π × 0.0598 3

= 0.13 ms −1

A1

(2)

11 Marks

Scheme

Question Number

Marks

Q3 (a)

M1A1

↑ →

θ l

h r

sin θ =

T cos θ + N = Mg

(1)

T sin θ = mrω 2 r l

sub into (1)

from (2)

M1A1

(2)

T = mlω 2 M1

ml cos θ ω 2 + N = mg

A1

N = mg − mhω 2 Since in contact with table N …0

∴ω 2 „

g h

*

M1A1 cso (8)

(b)

r : h :l = 3: 4:5

∴ extension =

mg 2mg h × = h 4 2 5mh 2 2g T = mlω 2 = ω ω= 4 5h T=

h 4

B1 M1A1 M1A1 (5) 13 marks

Question Number

Scheme

Q4 (a)

= Mass a 3

2 π ×: 3

C of M from O:

216 ×

Moment :

Marks

+

216

8

208

27

1

26

3 × 6a 8

3 × 2a 8

M1A1

x

Use of

3 r 8

2a × 3 6a × 3 = 8× + 208 x 8 8 480a 30a = 208 13

x =

M1

M1

*

A1 cso (5)

(b)

+

Mass

πa 3 × :

C of M:

Moments :

416 3 30 a 13

+

320a

y = (c)

+

=

S

24 =

488 3

9a =

+ 216a =

201 a 61

B1 B1

y

M1

488 y 3

*

A1 cso (4)

201 a 61 2a tan θ = ....

12a − tan θ =

2a 12a − 201 61 a

θ = 12.93..... so critical angle = 12.93…

∴ if θ = 12° it will NOT topple.

M1 M1 A1 A1√ (4) 13 marks

Scheme

Question Number

Marks

Q5(a)

T – mg cos θ =

F = ma

Sub for

v2 : a

M1A1

1 mv 2 = mga cos θ 2 v 2 = 2ga cos θ

Energy

T = mg cos θ + 2mg cos θ : θ = 60 ∴ T =

mv 2 a

M1A1

3 mg 2

M1A1 (6)

(b)

Speed of P before impact = → 2 ga PCLM :

2 ga

→u

→0

•

•

m

3m

B1

→

•

∴u =

2 ga = 4

ga 8

*

M1A1cso

4m

(3)

(c) (i)

At A v = 0 so conservation of energy gives: 1 4mu 2 = 4m ga (1- cos θ ) 2 ga = ga (1- cos θ ) 16 cos θ =

15 , θ = 20o 16

M1A1 M1 A1

(ii)

At A

T = 4mg cos θ

=

15mg 4

(accept 3.75mg)

M1A1 (6) 15 Marks

Scheme

Question Number Q6 (a)

3

F = ma (→)

(x + 1)

3

∫ ( x + 1) −

3

3

∀x

(c)

dx v= dt

∫

dv dx

M1A1

dx = 0.5 ∫ v dv

3

2(x + 1)

=

2

x = 0, v = 0 ⇒ c´ = –

(b)

= 0.5a = 0.5 v

Marks

1 2 v ( + c) 4 3 2

∴

⎛ 1 v 2 = 6⎜⎜1 − 2 ⎝ ( x + 1)

⎞ ⎟⎟ ⎠

*

2

6 ( x + 1) 2 − 1 x +1

=

x +1

dx =

( x + 1) 2 − 1

( x + 1) 2 − 1

M1A1 cso (6)

B1

(1)

M1

6 ∫ dt

6 t + c′

=

M1 A1

(Q ( x + 1) always > 0 )

v< 6

∴

v2 < 6

Separate and ∫

M1

M1 A1

t = 0, x = 0 ⇒ c′ = 0

M1

t = 2 ⇒ ( x + 1) − 1 = (2 6 ) 2 2

(x + 1)

2

= 25

⇒ x = 4 ( c′ need not have been found)

M1 A1 cao (7) 14 Marks