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PRE-LEAVING CERTIFICATE EXAMINATION, 2014

APPLIED MATHEMATICS

MARKING SCHEME

HIGHER AND ORDINARY LEVEL

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General Guidelines

1. Penalties of three types are applied to candidates’ work as follows:

Slips – numerical slips S(–1) Blunders – mathematical errors B(–3) Misreading – if not serious M(–1) Serious blunder or omission or misreading which over simplifies: – award the attempt mark only.

Attempt marks are awarded as follows: Higher Level 5 (att 2). Ordinary Level 5 (att 2), 10 (att 3).

2. The marking scheme shows one correct solution to each question In many cases there are other equally valid methods.

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Higher Level Solutions

1. (a) 1 2: 3 : 9t t = (5)

1 23t t⇒ =

1 2 14t t t+ = (5)

( )( )1 11 9 4 72002

t t⇒ = (5)

21 400t = (5)

1 220 , 60t s t s= =

80s⇒ (5)

(b) 2362PS t t= + (5)

( ) ( )( )2130 3 2 32QS t t= − + − (5)

( ) ( ) ( )2 2330 3 3 62Q PS S t t t t d− = − + − − − = (5)

2118 812

d t t⇒ = − −

' 18 0d t⇒ = − = (5)

18t⇒ =

81d m⇒ = (5)

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2. (a) (i) Shortest time 84 16.85

d sv

= = = (5)

Shortest path 2 2 2 2

84 215 3

d sv p

= = =− −

(5)

∴ Difference is 4.2s (5)

(ii) B lands downstream a distance of ( )( )3 16.8 50.4m= (5)

(b) (i) 8 0PV i j= + (5)

6 8QV i j= +

PQ P QV V V= −

2 8i j= − (5)

Magnit 68 / ;m s= Dir 1tan 4E S−

(ii) Time 150 1510

s= =

P is ( )150 15 8 30m− × = from junct (5)

30 sin 76 29.1md = × ° = (5)

(iii) Time 30 cos 76 0.8868

s× °= =

Time to new event ( )1 0.88 0.12s= − = (5)

For 0.12 10 1.2Q ⇒ × =

( )For 0.12 8 30 30.96P ⇒ × + =

( ) ( ) ( )( ) ( )2 22 1.2 30.96 2 1.2 30.96 0.6 915.3792d = + − × =

30.26l m⇒ = (5)

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3. (a) (i) ( )2 22sin 2

5u uRg g

α= = (5)

11.8 , 78.2α = ° ° (5)

(ii) 4Range 9gt ht⋅ = (5)

( ) ( )2 2

29 4sin sin 22u ug g

α α=

( )28 sin cosu

gα α= (5)

9sin 16cosα α=

16tan9

α =

1 16tan9

α − ⎛ ⎞= ⎜ ⎟⎝ ⎠

(5)

(b) (i) tan 1 tan11 2 tan tan 4 1 2 tan4

α αα β α

= =− ⎛ ⎞− ⎜ ⎟

⎝ ⎠

(5)

tan 11 41 tan2

α

α⇒ =

−

2tan9

α⇒ = (5)

(ii) 11 2 tan 04

α ⎛ ⎞− =⎜ ⎟⎝ ⎠

tan 2α⇒ = (5)

(iii) tan 11 31 2 tan4

α

α=

⎛ ⎞− ⎜ ⎟⎝ ⎠

(5)

2tan7

α⇒ = (5)

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4. (a) (i) (5)

4 4g T a− = (5)

3 35gT S a− − = (5)

2 25gS a− = (5)

(ii) 8 16; ;3 3 15g g ga T S= = = (5)

(iii) Force ( )2 2S T T= + +

8 7415

g= (5)

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(b) (i) (5) (ii)

4a T g= −

5 5b T g= − (5)

10 10 22

a b g T+⎛ ⎞ = −⎜ ⎟⎝ ⎠

(5)

225

a b g T+ = −

484 4 85134 95

a T g

a b g T

b g T

= −

+ = −

= −

5 5

5 5 10 2

5 15 3

b T ga b g T

a g T

= −+ = −

= −

3 1 13 23 9 25 4 5 5

g T g T g T⎛ ⎞− + − = −⎜ ⎟⎝ ⎠

12 4 65; ;17 17 17

g g ga b T−= = = (5)

}

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5. (a) Impact Velocity: ( ) ( )8cos30 8sin 30i j° + ° (5)

Rebound Velocity: ( ) ( )cos 45 sin 45V i V j− ° + ° (5)

8sin 30 sin 45V° = °

14 2V ms−= (5)

cos 45 18cos30 3V e− °⎡ ⎤⇒ − = =⎢ ⎥°⎣ ⎦

(5)

(b) (i) Before Mass After

A: u 3 p 3 6 3p q u+ =

B: 0 6 q 3up q −− =

4;9 9u up q= = (5)

A: 9u 3 g 3 243 6

9 9u fug h+ = −

B: 49fu− 6 h 1 4

3 9 9u fug h ⎛ ⎞− = − +⎜ ⎟

⎝ ⎠

0h = 1 3 24 43 9 9 27 27

u fu u fug −⎛ ⎞⇒ = − = −⎜ ⎟⎝ ⎠

3 24 1 4f f⇒ − = − −

15

f⇒ = (5)

(ii) 15ug = − (5)

(iii) ( ) 2 21 3. 32 2BK E u u= = (5)

( )2 21. 3

2 15 150Au uK E ⎛ ⎞= =⎜ ⎟

⎝ ⎠ (5)

Total loss: 2 2 23 112

2 150 75u u u= − = (5)

}

}

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6. (a) 2 cos30T Mg° = (5)

3MgT⇒ =

2sin 30 4ll

° = ⇒ = (5)

4 1.2 2.8s = − =

( )303 2.8

T Mgks

= ⇒ =×

(5)

14.8M kg= (5)

(b) (i) 1 63

R T g+ = (5)

22 23

T Mw r=

2 2

2

v vM r Mr r

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

26

2 2v=

294vT⇒ = (5)

(ii) 0 18R T g= ⇒ = (5)

29 18

4v g=

2 2v g⇒ =

2 2v g⇒ ≤ (5)

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(iii) 29cos 6

3rT g Tθ −= = (5)

2

189

gTr

⇒ =−

2 2

2

18sin 6 v vT Tr r

θ = × ⇒ =

22

18 18 3.29

g grr

×=−

2 5.76r⇒ =

1.8h m⇒ =

0.8m⇒ below (5)

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7. (a) (i) 1 2:53

a S W⎛ ⎞ =⎜ ⎟⎝ ⎠

(5)

2 3 40 35

WS = = (5)

(ii) ( ) ( ) 3sin 602

R S Sμ↔ = ° = (5)

( ) 1 1002

R S+ = (5)

11002

R S= −

200 32 2

S Sμ −⎛ ⎞⇒ =⎜ ⎟⎝ ⎠

15 3 322

μ +⇒ = (5)

(b) Diagram: (5) (i) A be the point where the string meets the walls. (ii) B be the point of contact on one wall. (iii) C be the point on the line of intersection of the walls at the same level as B. (iv) α , the inclination of the string to the wall.

(v) β , the inclination of the string to the vertical.

1sin 302 2

OB rOA r

α α= = = ⇒ = ° (5)

2 1sin 452 2

OC rOA r

β β= = = ⇒ = ° (5)

( ) cos 45 2T W T W° = ⇒ = (5)

( ) sin 302

WS T S R↔ = ° ⇒ = = (5)

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8. (a) Let M = mass per unit area

Mass of element 2M xdxπ=

Moment of inertia of the element 22M xdx xπ= (5)

Moment of inertia of the disc 3

02

rM x dxπ= ∫ (5)

4

0

24

rxMπ ⎡ ⎤

= ⎢ ⎥⎣ ⎦

(5)

4

24rMπ=

212

mr= (5)

(b) (i) 2 21 12 2

I w mv mgh+ = (5)

2

22

1 12 2

vI mv mghr

⎛ ⎞+ =⎜ ⎟

⎝ ⎠

2 2 2v u ah= +

2 2v ah= (5)

( )2

1 2 1 22 2

ahI m ah mghr

⎛ ⎞ + =⎜ ⎟⎝ ⎠

2

Ia m mgr⎡ ⎤+ =⎢ ⎥⎣ ⎦

2

2

mgraI mr

⇒ =+

(5)

mg T ma− = (5)

2

21 mrT mg ma mgI mr

⎡ ⎤= − = −⎢ ⎥+⎣ ⎦

2

mgITI mr

=+

(5)

(ii) 2

2 7mgr ga

I mr= =

+

2 27mr I mr= +

2 2162

mr Mr= 12M m= (5)

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9. (a) (i) 1.5 2.1 12.6 WB gs

⎛ ⎞= − = =⎜ ⎟⎝ ⎠

1.5 712.6 12.6 6W gs⇒ = = = (5)

(ii) 1.5 3.15 11.55L L WB s B g= = − = (5)

11.55 1112.6 12

LL

W

BsB

⇒ = = = (5)

(iii) ( )11100 0.85 10012

V V+ = +

1200 10.2 1100 11V V+ = + 3125V cm⇒ = (5)

(b) (i) x = depth of water Diagram: (5)

y = depth of oil

0.12x y+ = (5)

B oil + B water ( )( ) ( )( ) ( )3900 0.15 0.15 1000 0.15 0.15 600 0.15yg xg g= + = (5)

100 90 9x y+ =

100 100 12x y+ =

0.3y m⇒ = (5)

(ii) Pressure ForceArea Area

W= = (5)

( )( )

3

2

600 0.150.15

g=

90g= (5)

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10. (a) ( )6 4 1ye dy x dx= +∫ ∫ (5)

26 2ye x x c= + +

6c⇒ =

22 1

6 6y x xe⇒ = + +

22 6ln

6x xy

⎛ ⎞+ +⇒ = ⎜ ⎟

⎝ ⎠ (5)

(b) 2 2F cv Ma cv= − ⇒ = −

2a kv⇒ = − (5)

2dv kvdt

= −

10

230 0

tdv k dtv

= −∫ ∫ (5)

10

30

1 115

kt kv t

⇒ − = − ⇒ = (5)

22

vdv dvkv v kdsds v

= − ⇒ = −

10 150

30 0

1 dv k dsv

= −∫ ∫

10

30ln 150v k= −

1 ln 3150

k⇒ =

10 9.1ln 3

t s⇒ = = (5)

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(c) (i) Retardation 30 kx= + (5) 40 30 60k= +

16

k⇒ =

3 306

dv svds

= − −

1018

dv svds

= − −

1018s⎛ ⎞= − +⎜ ⎟

⎝ ⎠ (5)

(ii) 0

4 010

18S svdv ds⎛ ⎞= − +⎜ ⎟⎝ ⎠∫ ∫ (5)

02 2

4 0

102 36

Sv s s= − +

2 360 288 0S S⇒ + − =

80S cm⇒ = (5)

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Ordinary Level Solutions 1.

(i) 45 10 5v u at a= + ⇒ = + (5)

5 35a =

17a ms−⇒ = (5)

(ii) 0 45 9 5a a= + ⇒ = − (5)

15d ms−⇒ = (5)

(iii) 1 15 10 5 35 45 11 9 452 2

d = × + × × + × + × × (10)

50 87.5 495 202.5= + + +

835 m= (10)

(iv) 1835A.Sp 33.425

ms−= = (10)

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2. (i) 26 cos sinAV i jα α⎡ ⎤= +⎣ ⎦ (5)

24 10i j= + (5)

18 0BV i j= + (10)

(ii) AB A BV V V= − (5)

( ) ( )24 10 18 0ABV i j i j= + − + (5)

6 10i j= + (5)

(iii) 80cosBX θ= (5)

6802 34

= ×

24034

= (5)

120 3417

= (5)

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3. (a) (i) 212

S ut at= +

2140 3 sin 02

t gtθ − = (10)

80 3 sin 1210

t θ= =

3sin2

θ =

60θ⇒ = ° (10)

(ii) XS cosu tθ= (5)

( )140 3 122

⎛ ⎞= ⎜ ⎟⎝ ⎠

240 3=

416 m= (5)

(b) 21 452YS gt= − = − (5)

2 9t =

3t s= (5)

( )3 150 3XS x= = (5)

50 3x =

86.6 m= (5)

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4. (i) (5) (10) (10)

(ii) 4 4g T a− = (5)

32gT S a− − = (5)

22gS a− = (5)

(iii) 9 3a g=

2

3ga ms−⇒ = (5)

8 7;3 6g gT S= = (5)

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5. (i) ( ) ( ) ( )P.C.M : 3 10 5 7 3 5 9p+ = +

1263

p ms−= (10)

(ii) ( )1 2 1 2N.E.L : V V e u u− = − −

( )26 9 10 73

e⎛ ⎞− = − −⎜ ⎟⎝ ⎠

79

e⇒ = (10)

(iii) ( )( ) ( )( )2 21 1. 3 10 5 72 2BK E = +

272.5= (5)

( ) ( )( )2

21 2 1. 3 6 5 92 3 2AK E ⎛ ⎞= +⎜ ⎟

⎝ ⎠

16156

= (5)

1. 33LossK E = (5)

Fraction loss 4327

= (5)

(iv) Impulse ( ) 23 10 3 63

⎛ ⎞= − ⎜ ⎟⎝ ⎠

10 SN= (10)

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6. (a) ( ) ( ) ( ) ( )3 3 4 1 2 2 50

14x

x− + + − +

= = (10)

1.8x⇒ = (5)

( ) ( ) ( ) ( )3 1 4 1 2 2 50

14y

y+ + − +

= = (10)

0.6y⇒ = − (5)

(b) Area c.g.

Lamina: 144π ( )0,0 (5)

:abcΔ 36 ( )7, 2 (5)

Remainder: ( )144 36π − ( ),x y

( ) ( ) ( )144 36 144 0 36 7xπ π− = −

0.61x = − (5)

( ) ( ) ( )144 36 144 0 36 2yπ π− = −

0.17y = − (5)

Coordinates of c.g. ( )0.61, 0.17− −

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7. (a) (i) Diagram: (5)

(ii) ( ) sin 5y w wθ+ =

( ) cosx w θ↔ = − (5)

:p ( ) ( )5 sin 2w a w aθ = (5)

2sin5

θ =

23.6θ⇒ = ° (5)

(iii) 215

x w= −

235

y w= (5)

(b) (i) Diagram: (5)

(ii) :A ( ) ( )0.9 0.2 cos 45 0.75R g= °

5 5 263 2

R = = (5)

0.2 cos 45 sin 45g R N° = + °

13

N = (5)

0.2 sin 45 cos 45g R Nμ° = + ° (5)

1μ = (5)

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8. (i) 2 2 20.13 0.12 r= +

0.05r m⇒ = (10)

(ii) 12sin13

α = (5)

5cos13

α = (5)

(iii) Diagram: (10)

(iv) sin 3R gα =

12 313

R g⎛ ⎞ =⎜ ⎟⎝ ⎠

32.5R⇒ = (10)

(v) 2cosR mrwα =

( ) 2532.5 3 0.0513

w⎛ ⎞ =⎜ ⎟⎝ ⎠

5 303

w = (10)

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9. (a) (i) Diagram: (10)

(ii) 1000B Vg Vgρ= = (10)

1000 1300R Vg Vg+ =

300R Vg=

3000V= (10) (b) B W= (5)

( )1 1

3W

WS

= (5)

13

S =

1000

3ρ = (10)