WMS23 - LC Applied Mathsmselkirk.weebly.com › uploads › 9 › 4 › ...lc-appliedmaths...1....
Transcript of WMS23 - LC Applied Mathsmselkirk.weebly.com › uploads › 9 › 4 › ...lc-appliedmaths...1....
Page 1 of 19
PRE-LEAVING CERTIFICATE EXAMINATION, 2017
APPLIED MATHEMATICS
MARKING SCHEME
HIGHER AND ORDINARY LEVEL
Page 2 of 19
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.
Page 3 of 19
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)
Page 4 of 19
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)
Page 5 of 19
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)
Page 6 of 19
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)
Page 7 of 19
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
$$
$
$2tan=e (5)
Page 8 of 19
6. (a) (i) gTo 3=
gke 3=
13 TgF !=
( )xekg +!3
kx!= (5)
xkmFa
3!==
3kW =
52
=%W (5)
%103
=k
88.2960=k (5)
(ii) 1keF =
013.088.2960 &=
N5.38= (5)
(b) (i) 222
211 2/12/1 mVmghmVmgh +=+
( ) ( ) 22/1coscos mVrmgormg +=+ $# (5)
( )$# coscos22 != grV
rmvRmg
2
cos =!$
( )Rmgrmv != $cos2
( ) rRmgrmgr !=! $$# coscoscos2
( )#$ cos2cos3 != mgR (ergo) (5)
Page 9 of 19
(ii) 0=R o6021cos
43cos === $$# (5)
rvt23
21 = (5)
vrt 3=
(iii) + 2
232
323
vrg
vrv (5)
&+=gr
rgr 21
322
3 2
rrr2143
23 =+
4r below centre (5)
7. (a) (i) 480:A ( )32,20
312:B ( )13,20 (5)
792:T ( )yx,
( ) ( )2031220480792 +=x
cmx 20= (5)
( ) ( )1331232480792 +=y
cmy 52.24= (5)
(ii) y
x!
=38
tan$
4837.152.2438
20 =!
o02.56=$ (5)
Page 10 of 19
(b) (i) ( ) ( ) oWTA 60sin5.34sin69:�ˆ =$ (5)
WT23sin2 =$ (5) (5)
( ) ( )°+= 30sin $TW
( )oTT 30sin23sin2 += $$
+= $$$ cos21sin
23
23sin2 (5)
$$ cos43sin
43 +=
$$ cos43sin
45 =
53tan =$
WT27= (5)
(ii) ( )3
25/113
153
30tantan130tantan30tan =
!
+=
°!°+=°+
$$$ (5)
693
260sin69
== xxo
( ) ( )22 606969 oSinL +=
cm2
769= (5)
Page 11 of 19
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 =
Page 12 of 19
==
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)
Page 13 of 19
(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)
Page 14 of 19
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)
Page 15 of 19
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)
Page 16 of 19
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)
Page 17 of 19
(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)
Page 18 of 19
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)
Page 19 of 19
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)
Page 20 of 19