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PRE-LEAVING CERTIFICATE EXAMINATION, 2017

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), 10 (att 3). Ordinary Level 5 (att 2), 10 (att 3), 15 (att 5).

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) (i) 15029001500 !"!=f

270900 !"=f

42029002400 !=f (5)

3031=f (5)

(ii) 0150sin15002900: =!"!! #gCar

0270sin900: =!!" #gCaravan (5)

2480sin2400 =#g

29431sin =#

°= 6# (5)

(b) (i) ;2.018 2ttSp += 205.012 ttSQ += (5, 5)

(ii) 18925.030 2 =+= ttd (5)

07561202 =!+ tt

( )( ) 06126 =!+ tt

st 6= (5)

(iii) 7.5605.012 2 =+ tt (5)

011342402 =!+ tt

st 6.4= (5)

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2. (a) (i) jlV A 399 !!=

jlV B 015 += (5)

jlVAB 3924 !!= (5)

smVAB /62.28= ; sWDiR !

833tan: 1 (5)

(ii) (a) 62.28

27cos530 °==ABV

AXtime

s5.16=

s5.6105.16 =! (5)

(b) tt 1515018350 +=!

20033 =t

st 06.6= (5)

(b) (i) CWCW VVV +=

iujxix +!!= 23

( ) jxixu 23 !!= (5)

also ( ) jujyiyVW +!= 23

( ) jyuiy 23 !+= (5)

yxu 33 =! and yux 22 !=!

12ux != ,

125uy =

juiuVW64

5 += (5)

(ii) WCWC VVV !=

iujuiu !+=64

5

juiu64

+= (5)

CV : magnitude : 0.3u

Direction : °7.33 North of East (5)

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3. (a) 15cos5 == tgSx # (5)

#Cosg

t 3= (5)

2

1592

35 2 =!=##

#gCos

gCosg

SingSy (5)

( )2

15tan129tan15 2 =+! ##

( ) 151930 2 =+! tt

08103 2 =+! tt (5)

( )( )0243 =!! tt

3/4tan 1! , S2tan 1! (5)

(b) (i) 0=Sy

02/14518 2 =! tgCostSin o # (5)

02/29 2 =! tCosgt #

#gCos

t 218= (5)

+=#

#cos

218sin45cos18g

giV o

#tan21829 +

29cos

218cos45sin18 !=!=#

#g

gjV o

51

tan2182929tan =

+=!=

#$

iVjV (5) o4.63=# (5)

(ii) 245=iV , 29ejV =

( ) 2114162245 22=+ e

41161624050 2 =+ e

933=e (5)

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4. (a) (i) aggA 633: ="!! (5)

aggB 16348: =!+" (5)

agg 223511 =!

2/04.1 sma = (5)

(ii) N2.6=" (5)

(b) (i) (5)

(ii) agSA 66: =! (5)

aSgB 222: =!"+ (5)

( )abgC !=!" 33: (5)

( )abgD +="! 44: (5)

bag 7+= ( )DC +

bag += 153 ( )DCBA +!+

2/265 smga = , 2/

263 smgb =

acc of 4kg mass 2/134 smg (5)

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5. (a) ( )15/8tan 1!=$

( ) ( ) juiuu oo $$ !!!!= 90sin90cos (5)

juiu $$ cossin !!

jvivV $$ sincos +!= (5)

1 $$ cossin vu = (5)

vu1715

178 = uv

158=

2 $

$cos

sinuve

OldNew

!=!= (5)

$$ cossin euv =

Dividing 2 by 1 $$

$$

coscos

sinsin

veu

uv =

22 euv =

euu =2

2

22564

loss%72 (5)

(b) ( ) mqmpomumPCM +=+ )(cos: # (5)

#cos: euqpNEL !=! (5)

( ) #cos12

eup != , ( ) #cos12

euq += (5)

( )( ) #

##$#cos1

2

sinsintaneu

up

u

!==+ (5)

( )e!=

1tan22tan $$ (B moves along i -axis after impact $# = )

e!

=! 1

tan2tan1tan2

2

$$

$

$2tan=e (5)

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6. (a) (i) gTo 3=

gke 3=

13 TgF !=

( )xekg +!3

kx!= (5)

xkmFa

3!==

3kW =

52

=%W (5)

%103

=k

88.2960=k (5)

(ii) 1keF =

013.088.2960 &=

N5.38= (5)

(b) (i) 222

211 2/12/1 mVmghmVmgh +=+

( ) ( ) 22/1coscos mVrmgormg +=+ $# (5)

( )$# coscos22 != grV

rmvRmg

2

cos =!$

( )Rmgrmv != $cos2

( ) rRmgrmgr !=! $$# coscoscos2

( )#$ cos2cos3 != mgR (ergo) (5)

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(ii) 0=R o6021cos

43cos === $$# (5)

rvt23

21 = (5)

vrt 3=

(iii) + 2

232

323

vrg

vrv (5)

&+=gr

rgr 21

322

3 2

rrr2143

23 =+

4r below centre (5)

7. (a) (i) 480:A ( )32,20

312:B ( )13,20 (5)

792:T ( )yx,

( ) ( )2031220480792 +=x

cmx 20= (5)

( ) ( )1331232480792 +=y

cmy 52.24= (5)

(ii) y

x!

=38

tan$

4837.152.2438

20 =!

o02.56=$ (5)

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(b) (i) ( ) ( ) oWTA 60sin5.34sin69:�ˆ =$ (5)

WT23sin2 =$ (5) (5)

( ) ( )°+= 30sin $TW

( )oTT 30sin23sin2 += $$

+= $$$ cos21sin

23

23sin2 (5)

$$ cos43sin

43 +=

$$ cos43sin

45 =

53tan =$

WT27= (5)

(ii) ( )3

25/113

153

30tantan130tantan30tan =

!

+=

°!°+=°+

$$$ (5)

693

260sin69

== xxo

( ) ( )22 606969 oSinL +=

cm2

769= (5)

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

mass of element = xdxM %2

moment of inertia of the element 22 xxdxM %= (5)

moment of intertia of the disc =rdxxM

0

32% (5)

r

o

xM4

24

%= (5)

4

24rM%=

2

21mr (5)

(b) (i) =a

a

dxxMI3

2

32% (5)

a

a

xM3

2

4

42%=

( ) ( )[ ]( ) ( )

423

2342

44

22

aaaa

m !&!

=%

%

226ma= (5)

(ii) (a) ( ) ( )3

6443/43/4:2

22 maammlRodIp == (5)

( )22 11426:)( ammaAnnulusIp + (5)

2510ma=

22

5103

64:)( mamasystemIp +=

3

1594 2ma=

To find h ( ) ( ) mghamgamg 51144: =+

mgamgh 485 =

5

48ah =

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==

5485

3/1594222

amg

mamghIT %%

sga

2797

3% (5)

(b) ga

gl

glT

727972 == %

al 07.11= (5)

9. (a) (i) Weight Ngmg 14715 ===

Bouyancy in water WtAppWt .!=

105147 !=

N42=

SS

WB 14742 ==

27=S (5)

(ii) Let x = mass of the gold x!'15 = mass of the aluminium

Wt of gold xNmg 8.9== (5)

Wt of aluminium ( )Nx!= 158.9

Bouyancy suffered by the gold NxxSW

97

6.128.9 === (5)

Bouyancy suffered by the alum. ( ) ( )Nxx !=!= 15928

15.3158.9

( ) 42159289/7 =!+' xx

( ) 54154 =!+ xx

2=x mass of gold = 2 kg; mass of aluminium = 13 kg (5)

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(iii) Total volume ( ) kgdensityTotalmassTotal

7003

10002/715 ===

Volume of gold ( )( ) kg6300

110006.12

2 ==

th271 of the total volume (5)

(b) Volume of shell 33 3/43/4 rR %% !=

!=3

3

253

34 %

3

691 m% (5)

Vol. of liq disp ( )[ ] 33 1833/421 m%% == (5)

Wt. of object NpgVpg 691%== (5)

Bouyancy ( ) Ngg %% 000,27150018 == (5)

Ngpg %% 000,27691 ='

3/21978.1780 mNp =

Relative density 7802.1= (5)

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10. (a) (i) 36

3VdsdvV != (5)

!= dsvdv

361

2

!=! dsdvV3612

104

09 36

11 SV

V

!=!

smV /31= (5)

(ii) 36

3Vdtdv !=

!= dtVdv

361

3 (5)

tV 36

1121 6

82 !=!

t361

641

361

21 =!

st327= (5)

(b) ( ) ( )200 !!=!!= TkTTkdtdT (5)

!=!

330

0

50

100 20dtk

TdT (5)

( ) 330

0

50

10020ln tkT !=! (5)

( ) 33020ln 50

100kT !=!

k33030ln80ln =!

330

3/8ln=k (5)

( ) tT330

3/8ln20ln 30

100

!=! (5)

t330

3/8ln8ln =

st 700= (5)

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Ordinary Level Solutions 1. (a) (i) atuV += Diag. (5) a62810 += (5)

2/3 smd = (5)

(ii) a41026 += (5)

2/4 sma = (5)

(iii) 20810448610318 +&+&+&+&=PQ (5)

20840326054 ++++=

m394= (5)

(iv) 18394=ASp

21= sm /98 (5)

(b) New time sRQ 5.6= (5)

New

2116

394=ASp

sm /9.23 (5)

2. (i) jicV °+°!= 45sin2845cos28 (15)

ji 88 +! (5)

(ii) If they are to meet 8!=x (10)

( )222 134=+ yx

20864 2 =+ y

1442 =y

12=y (10)

(iii) ( ) ( )tty 818 +=

tt 81812 += 184 =t

st 5.4= (10)

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3. (i) [ ]jiV $$ sincos130 +=

smji /12632 += (10)

(ii) 0126 =!= gtyV

st 6.12= (10)

(iii) 2

101265

10126126 !=Sy

m8.793= (10)

(iv) =5

12632Sx

m4.806= (10)

(v) 1445126 2 =!= ttSy (5)

01441265 2 =+! tt

( )( ) 02465 =!! tt

sst 24,56= (5)

4. (a) (i) (5) (5)

(ii) agT 164 =! (5)

aTg 2424 =! (5)

ag 4020 =

2/5 sma = (5)

(b) (i) 2

21 atuts +=

( )82110 a=

2/5.2 sma = (5)

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(ii) aTg 33 =! (5)

NT 5.22! (5)

(iii) xaxgT =! (5)

5.225.12 =x 8.1=x (5)

5. (i) qpMCP 424325:... +=&+& (5)

eqpLEN 2:... !=! (5)

;3

411 ep != 3

211 eq += (5)

( ) 44321131 =&!+ e

21=e (5)

(ii) smp /3= (5)

smq /4= (5)

(iii) ( )( ) ( )( )22 342152

21. +=BEK

J43= (5)

( )( ) ( )( )22 442132

21. +=AEK

J41= (5)

Loss in JEK 2. = (5)

(iv) N4352 =! (5)

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6. (a) (i) ( ) ( ) ( ) ( ) 516

72436125 =+!++= px (10)

3=p (5)

(ii) ( ) ( ) ( ) ( ) qpqy =+++=16

233675 (10)

5=q (5)

(b) Area c.g

ADC( 18 ( )7,1 (5)

ACB( 67.5 ( )3,6 (5)

Lamina ABCD 85.5 ( )yx,

( ) ( )65.671185.85 +=x

19184=x (5)

( ) ( )35.677185.85 +=y

19163=y (5)

7. (a) ( )PRgPRg != 5.02.05.0 (15)

mPR71= (10)

(b) Diag. (5)

( ) ( ) ( )CosACosASinAp 818016401680: =+ (5)

ASinA cos5.24 = (5)

85=TanA (5)

oA 32= (5)

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8. (a) (i) (10)

(ii) rmvT

2

= (5)

( )5.0

45 2

=

NT 160= (5)

(b) (i) 5

45 rTan o =

cmr 5= (5)

(ii) (5)

(iii) gRSin o 345 =

302

1 =&R

NR 230= (10)

(iv) 245 mrwRCos o =

10053

21230

2w&&=

210=w srad / (10)

9. (i) (10) (10)

(ii) pVgB = (5)

( ) ( )102.01000 3= (5)

N80=

gBT 40=+ (5)

80400 !=T

N320= (5)

(iii) 320400 =µ (5)

54=µ (5)

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