1306 m3 june 2013 - withdrawn paper mark scheme

Mark Scheme (Results) Summer 2013 GCE Mechanics M3 6679/01 Original Paper

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

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June 2013 6679 M3

Mark Scheme

Question Number Scheme Marks

1.

A

32

l

v

E P E gained 21 5 1 5

2 2 8mg mglll

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

M1A1

21 5 32 8 2

mgl mglmv + = M1A1ft

2 74glv =

1 72

v gl= oe A1

1 5


2

R α mg

( )R cosR mgα↑ = M1A1

( )2 215R sin ,

200mvR m

rα→ = = × M1,A1

Divide: 215tan

200gα = M1

6.548...α = 6.5, 6.55= (2 or 3 s.f. only) A1

1 6


3

(a) 2 2r r x′ = − B1

( )2 2

0d

rr x x xπ −∫ M1

2 2 4 4

02 4 4

rx r x rππ

⎡ ⎤= − =⎢ ⎥

⎣ ⎦ A1

4

32 34 3 8rx r rπ π= ÷ = * M1A1

(b) Mass 313

krπ ρ 32 33

rπ ρ× ( )31 63

r kπ ρ + B1

k 6 6k +

x 14

kr 38

− r x B1

( ) ( )21 3M 6 64 8

O k r r x k− × = + M1A1ft

( )( )

2 94 6k r

xk−

=+

Distance = ( )( )

2 9

4 6

k r

k

−

+ A1

(c) C of M must be at O 0x⇒ = M1

2 9 0k∴ − = ( )3 0k k= > A1

1 7


4

A B

(a) ( )2 0.41 2.5

0.8A

xT x

+= = + * M1A1

(b) ( ) ( )2 0.41 2.5

0.8A

xT x

−= = − B1

0.2B AT T x− = M1

5 51 1 0.22 2

x x x⎛ ⎞− − + =⎜ ⎟⎝ ⎠

A1

5 0.2x x− = 25x x= − M1

∴ SHM period 25π

= * A1

(c) cos 0.3cos5x a t tω= =

0.3a = and 0.2x = − B1

2cos53

t = − M1

11 2cos5 3

t − ⎛ ⎞= −⎜ ⎟⎝ ⎠

Time to D: 0.460st = A1

(d) ( )2 2 2 2v a xω= − ( )( )22 2 25 0.3 0.2Dv = − ± M1

New oscillation: ( )2

2 2 25 0.22Dv a⎛ ⎞= −⎜ ⎟

⎝ ⎠ M1

( ) ( )2

2 2 2 2 2 25 0.2 5 0.3 0.22Dv a⎛ ⎞= − = −⎜ ⎟

⎝ ⎠ M1dep

0.374... 0.37a = = m A1

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5 (a)

Energy A to C: ( )2 21 1 1 cos2 2

mv mu mga θ− = − M1A1

NL2 along radius at C: ( )2

cos mvmg Ra

θ − = M1

2cosmga mvθ = A1

( )21 1cos 1 cos2 2

mg mu mgaθ θ− = − M1dep

2cos 2 2 cosag u ga gaθ θ− = −

( )21cos 23

u gaga

θ = + A1

(b) 4cos5

θ = 2 24 2or 5 5

v ga u ga⇒ = = B1

Vert. motion: Initial speed 6sin5 5

agv θ= = Energy Solution:

(can be worked from top)

2

2 6 925 5 5

ag aV g⎛ ⎞

= + ×⎜ ⎟⎜ ⎟⎝ ⎠

21 4 9 12 5 5 2

am ga mg mV⎛ ⎞ + × =⎜ ⎟⎝ ⎠

M1A1

2 36 18 486125 5 125

ag agV ag= + = 2 225

V ag= A1

Horiz. speed 4 8cos 25 5 5 5

ag agv θ= = × = M1A1

Dirn: 486 8 486tan125 645 5

φ⎛ ⎞

= ÷ =⎜ ⎟⎜ ⎟⎝ ⎠

8 1cos5 22

φ = × M1

70.05... 70φ = = ° 70φ = °

A1

1 9


6

(a) 3

14

kxx

= − M1

3

1 d4 d

v kvx x= − M1dep

22

18 2

kv cx

= + M1depA1

11, 2 48 2

kx v c= = × = + M1A1

1 1 14,2 8 4 32

kx v c= = × = + A1

1 1 1 148 4 2 32

k⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1k = M1A1

(b) 1 1 02 2

c c= + ⇒ =

22

4vx

= M1

2v (or v) is never zero ⇒ direction of motion never reversed A1

(c) 2

1 416 x

= M1

2 64 8x x= = A1

d 2dxvt x

= =

8

1 0d 2d

Tx x t=∫ ∫

[ ]82

01

22

Tx t⎡ ⎤

=⎢ ⎥⎣ ⎦

M1A1

63 22

T= 3154

T = s or 15.75 s A1

1 10


7

60o

T

mg

(a) ( )R cos 60T mg↑ = 2T mg= B1

H.L. 62 mgxmgl

= M1

13

x l= A1

43

AB l= * A1

(b)

2sin 60T mrω= M1

4 sin 603

r l= B1

243

T mlω=

2423

mg mlω= M1

2 32gl

ω =

32gl

ω = A1

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1306 m3 june 2013 - withdrawn paper mark scheme

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