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Scéimeanna Marcála Scrúduithe Ardteistiméireachta, 2007
Matamaitic Fheidhmeach Ardleibhéal
Marking Scheme Leaving Certificate Examination, 2007
Applied Mathematics Higher Level
Applied Maths Cvr 15/8/07 19:31 Page 1
Coimisiún na Scrúduithe StáitState Examinations Commission
LEAVING CERTIFICATE APPLIED MATHEMATICS
HIGHER LEVEL
MARKING SCHEME
App Maths HL Title 15/8/07 19:30 Page 1
M32
Coimisiún na Scrúduithe Stáit State Examinations Commission
Scéim Marcála Scrúduithe Ardteistiméireachta, 2007 Matamaitic Fheidhmeach Ardleibhéal Marking Scheme Leaving Certificate Examination, 2007 Applied Mathematics Higher Level
Page 2 of 27
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 oversimplifies: - award the attempt mark only. Attempt marks are awarded as follows: 5 (att 2). 2 The marking scheme shows one or more approaches to solving each question.
In many cases there are other equally valid methods of solution.
1. (a) A particle is projected vertically downwards from t = 0
t = 2
t = 3 29.9 m
the top of a tower with speed u m/s. It takes the particle 4 seconds to reach the bottom of the tower. During the third second of its motion the particle travels 29.9 metres.
Find (i) the value of u (ii) the height of the tower.
5 5 5 5 m 100
)16(9.4)4(4.5
(ii)
m/s4.5
5.249.29
)9(9.4)3(9.29
)4(9.4)2(
(i)
221
221
=
+=
+=
=
+=
+=+
+=
+=
ftuts
u
u
u h
uh
ftuts
20
Page 3 of 27
m/s 4.5
)2(8.925
:seconds First two25
)1(9.4)1(9.29
:second Third (i)2
21
u
uatuv
uu
ftuts
=
+=+=
=+=
+=
OR
5 5 5
1 (b) A train accelerates uniformly from rest to a speed v m/s.
It continues at this speed for a period of time and then decelerates uniformly to rest.
In travelling a total distance d metres the train accelerates through a distance pd metres and decelerates through a distance qd metres, where p < 1 and q < 1.
(i) Draw a speed-time graph for the motion of the train.
(ii) If the average speed of the train for the whole journey is bqp
v++
,
find the value of b.
(i)
t1 t2 t3
speed
time
v
5
5
5
5
5 5
1
1
22
speed Average
)(
321
321
2
121
=⇒
++=
+−−
+=
++=
=
−−=
=
b
qpv
vqd
vqdpdd
vpd
d
tttd
qdvtqdpddvt
pdvtii
30
Page 4 of 27
2. (a) Ship B is travelling west at 24 km/h. Ship A is travelling north at 32 km/h.
At a certain instant ship B is 8 km north-east of ship A.
(i) Find the velocity of ship A relative to ship B. (ii) Calculate the length of time, to the nearest minute, for which the ships
are less than or equal to 8 km apart.
North 53.13East :direction
km/h 04 :magnitude
32 24
0 24
32 0 )(
°
+=
−=
+−=
+=
ji
VVV
jiV
jiVi
BAAB
B
A
rr
rrr
rrr
rrr
5
5 5 5 5
5 30
A
8.13° 8
VAB
B X
minutes 24
hours 396.040
138cos16
2 time )(
==
°=
=
.VAX
iiAB
r
Page 5 of 27
2 (b) A man can swim at 3 m/s in still water. He swims across a river of width 30 metres. d
330°
5 30 m
He sets out at an angle of 30° to the bank. The river flows with a constant speed of 5 m/s parallel to the straight banks. In crossing the river he is carried downstream a distance d metres.
Find the value of d correct to two places of decimals.
( )
( )
metres 04.48
20402.2
2030cos35
seconds 20
30sin3
30 cross toTime
=
×=
×−=
=
=
d
5
5 5
5 20
Page 6 of 27
3. (a) A particle is projected with a speed of 57 m/s at an angle α to the horizontal.
Find the two values of α that will give a range of 12.5 m.
5 5 5
5 5 25
°°=⇒
°°=⇒
=
==
==
⎟⎟⎠
⎞⎜⎜⎝
⎛=
=
=⇒
=−
=
75or 15
150or 302212sin
5.122sin2512.5 Range
2sin25cossin50
sin514.cos57
.cos57 Range
sin514
0.sin57
02
21
α
α
α
α
α
αα
α
α
α
αα
g
t
gt
gtt
rjr
Page 7 of 27
3 (b) A plane is inclined at an angle 45° to the horizontal. A particle is projected up the plane with initial speed u at an angle θ to the horizontal. The plane of projection is vertical and contains the line of greatest slope.
The particle is moving horizontally when it strikes the inclined plane. Show that 2tan =θ .
( )( )
( )
( ) ( )
( ) ( )( )
( ) ( )
( )
( )( ) ( )
( ) ( ) ( )
( )
2tantan13tan3
31
tan11tan
3145tan
45sin245cos45sin45sin245cos
45sin1
tan4545 angle Landing
45sin45cos
45sin245cos45sin
45cos45sin 45sin245cos
45cos45sin245sin45cos
45sin45cos
45cos45sin2
45cos45sin 0
0 2
21
=⇒+=−
=+
−
=−
−−−=−−−−
−=
−=⇒°=
−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−−=
−−=−−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−−=
−−=
−=⇒
−−=
=
θθθ
θθ
θ
θθθθθ
θ
θ
θθ
θθθ
θθ
θ
θθ
u uuu u
uvv
u g
u. g u
.t g uvu u
gu. g u
.t g uv
gu t
.tg.t u
r
i
j
j
i
j
5 5 5 5 5
25
Page 8 of 27
4. (a) A particle slides down a rough plane inclined at to the horizontal. The °45
coefficient of friction between the particle and the plane is 43 .
Find the time of descending a distance 4 metres from rest.
5 5 5 5
20
( )
s.15.2
232
242
104
m/s 24
45cos4345sin
45sin 45cos
2
221
2
=
=
⎟⎠
⎞⎜⎝
⎛+=
+=
=
=−
=−=
gt
tg
ftuts
gf
mf mgmg
mfR mg mgR
μ
Page 9 of 27
4 (b) A light inextensible string passes over a small fixed pulley A, under a small moveable pulley B, of mass m kg, and then over a second small fixed pulley C. A particle of mass 4 kg is attached to one end of the string and a particle of mass 6 kg is attached to the other end. The system is released from rest.
(i) On separate diagrams show the forces acting on each particle and on
the moveable pulley B. (ii) Find, in terms of m, the tension in the string. (iii) If m = 9.6 kg prove that pulley B will remain at rest while the two
particles are in motion.
(i)
( )
( )
motion.in are particles two the rest whileat remain willBpulley
021 Bpulley ofon accelerati
0206
mass kg 6 ofon accelerati
0204
mass kg 4 ofon accelerati
8.4or 04.47T 9.6 )(
485
48642
212
6644 )(
⇒
=+=
≠−=−==
≠=−==
=⇒=
+=⇒
⎭⎬⎫
⎩⎨⎧
⎟⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ −=
⎭⎬⎫
⎩⎨⎧ +=−
=−=−
qp
g.gT q
g.gTp
gmiii
m
mgT
gTgTm
qp mT mg
q g T p g Tii
mg 4g
T T T T
6g
4 kg 6 kg
A
B
C
5,5
5
5 5
5
30
Page 10 of 27
5. (a) A smooth sphere P, of mass 2 kg, moving with speed 9 m/s collides directly with a smooth sphere Q, of mass 3 kg, moving in the same direction with speed 4 m/s.
The coefficient of restitution between the spheres is e.
(i) Find, in terms of e, the speed of each sphere after the collision.
(ii) Show that the magnitude of the momentum transferred from one sphere to the other is ( )e+16 .
5 5 5
5
( )
( )
( ) ( )
)1(666
)26(33(4) Impulse
OR
)1(6 66
36292 Impulse )(
26or 51030
36or 51530
49 NEL
324392 PCM )(
2
1
21
21
ee
e
ee
eii
ee v
ee v
e v v
vv ) ( i
+−=−−=
+−=
+=+=
−−=
++
=
−−
=
−−=−
+=+
20
Page 11 of 27
5 (b) A smooth sphere A, of mass 4 kg, moving
with speed u, collides with a stationary smooth sphere B of mass 8 kg. The direction of motion of A, before impact, makes an angle α with the line of centres at impact.
The coefficient of restitution between the spheres is 21 .
α ˚
u
A B
Find in terms of u and α
(i) the speed of each sphere after the collision (ii) the angle through which the 4 kg sphere is deflected as a result
of the collision (iii) the loss in kinetic energy due to the collision.
5 5 5 5 5
5
30
( ) ( )
( )
( )
( ){ } ( )
cos
cos)sin1(2 cossin22 KEin Loss
cossin2
cos218
2 1sin4
2 1 after KE
242 1 before KE )(
90 Angle )(
cos21 B of Speed
sinA of Speed
cos21 and 0
0cos21 NEL
8408cos4 PCM )(
22
2222
22222
2222
22
22
21
21
21
α
αα
αα
αα
αα
α
α
α
α
α
α
uuu
uuu
uu
uu
uuiii
ii
u
u
uv v
u v v
v v ui
=
−−=
−−=
+=⎭⎬⎫
⎩⎨⎧+=
==
−=
=
=
==⇒
−−=−
+=+
Page 12 of 27
6. (a) A particle of mass m kg is suspended from a fixed point p by a light elastic string. The extension of the string is d when the particle is in equilibrium. The particle is then displaced vertically from the equilibrium position a distance not greater than d and is then released from rest. (i) Show that the motion of the particle is simple harmonic. (ii) Find, in terms of d, the period of the motion.
5 5 5 5
5 gd
km
ii
mkωx
mkx
kxkxkd mg
xdk mg x
kd m kd T i
π
π
ωπ
2
2
2 Period )(
with 0about S.H.M.
onAccelerati
)(inc. of dirn.in Force
:position Displaced
g :positionmEquilibriu )(
0
=
=
=
==⇒
−=
−=−−=+−=
=⇒=
25
Page 13 of 27
d
c 6 (b) A bead slides on a smooth fixed circular hoop,
of radius r, in a vertical plane. The bead is projected with speed gr10 from the highest point c. It impinges upon and coalesces with another bead of equal mass at d. cd is the vertical diameter of the hoop. Show that the combined mass will not reach the point c in the subsequent motion.
5 5 5 5 5
( )
( )
( )
. reach not willmass combined thepossiblenot is This21
44
14
44
)2()2())(2(21)0()2())(2(
21
speed with reach to: mass combined For the
212)0(
at mass combined the: of speed thebe Let
14
)0(21)2()10(
21
at energy Total at energy Total reachesit when : of speed thebe Let
22
22
22
2
22
21
2
1
1
1
2
2
c
grv
grvgr
mgrvmvm
rgmvmgmvm
vc
vv
mvmmvdv
grv
mgmvrmggrm
dcdcv
⇒
−=⇒
+=
+=⎟⎟⎠
⎞⎜⎜⎝
⎛
+=+
=
=+
=
+=+
=
25
Page 14 of 27
7. (a) aob and cod are two uniform rods, each of weight W, freely hinged at o.
b
a c
d
o
l2== coao and l5== odob . The rods are in equilibrium in a vertical plane. The ends b and d rest on a smooth horizontal plane and are connected by a light inextensible string of length 5 . l
Find the tension in the string.
R1
T b
a c
o
R2
T
W W θ
d
5
( )
( ) ( )
W.WT
WTW
RTosW
odo
WRWR
RWosW
d
WRR
400or 310
7
25
235
43
60cos560sin560c23
: for about moments Take
and
560cos23560c
27
: systemfor about moments Take
60
2
1
12
2
21
=
⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟
⎠⎞
⎜⎝⎛
=+⎟⎠⎞
⎜⎝⎛
==
=⎟⎠⎞
⎜⎝⎛ ++⎟
⎠⎞
⎜⎝⎛
°=
=+
lll
lll
ll
ll
θ
5 5 5 5
25
Page 15 of 27
7 (b) A uniform disc of radius 25 cm and mass 100 kg rests in a vertical plane F
θ
10 cm
perpendicular to a kerb stone 10 cm high. A force F is applied to the disc at an
angle θ to the horizontal, where 34tan =θ .
(i) Draw a diagram showing all the forces acting on the disc. (ii) Find the least value of F required to raise the disc over the kerb stone.
R F
θ
10 cm
N
100g
α
5 5 5 5
5
N 784or 80
10054
53
43
0100sins vert
43
coscos horiz
53
2515sin
gF
gFF
NgNFinR
FR
FR
=
=⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛
==++
=
=
==
θα
θα
α
25
Page 16 of 27
8. (a) Prove that the moment of inertia of a uniform square lamina, of mass m and
side 2r, about an axis through its centre parallel to one of the sides is 2
31 mr .
{ }
{ }
231
434
3
-
2
2
m
M
3x 2 M
dx x 2 M lamina theof inertia ofmoment
xdx 2 M element theof inertia ofmoment
dx 2 M element of mass
areaunit per mass MLet
r
r
r
r
r
r
r
r
r
r
=
=
⎥⎦
⎤⎢⎣
⎡=
=
=
=
=
−
∫
5 5 5 5
20
Page 17 of 27
8 (b) (i) A uniform square lamina abcd of side 2r oscillates in its own plane
about a horizontal axis through a , perpendicular to its plane.
If the period of small oscillations is g382π , find the value of r.
(ii) If the lamina is released from rest
when ab is vertical, find the maximum velocity of corner c in the subsequent motion.
a
b c
d
( ) ( )
( )
( )
( )
m/s 87.54467.1
)2( 38
21
PEin Loss KEin Gain (ii)
2
2382
238
2
2
238
34
34 I (i)
22
221
2
2
22
==⇒=⇒
−=⎟⎠⎞
⎜⎝⎛
=
=
=⇒
=
=
=
=
=
+=
ωω
ω
ω
π
π
π
v
rrmgrm
mghI
r
gr
mgr
rm
MghIT
mgrMgh
rm
rmrm
5 5 5 5
5 5
30
Page 18 of 27
9. (a) A U-tube whose limbs are vertical and of
26.2 cm
equal length has mercury poured in until the level is 26.2 cm from the top in each limb. Water is then poured into one limb until that limb is full. The relative density of mercury is 13.6. Find the length of the column of water added to the limb.
26.2 + x
2x A B
5 5
5 5
( )( )
cm. 7.22 water ofcolumn ofLength
cm 1
10213600 102261000
Bat Pressure A at Pressure
6.22 water ofcolumn oflength Let the
2
2
=
=
=+
=
+=
x
x g x . g
x
-
-
20
( )( )
cm. 7.22 water ofcolumn ofLength
cm 1or m 01.0
213600 262.01000
Bat Pressure A at Pressure 6220. water ofcolumn oflength Let the
OR
=
=
=+=
+=
x
x g x g
x
5
5
5 5
Page 19 of 27
9 (b) A uniform solid sphere is held completely
immersed in 500 cm3 of water by means of a string tied to a point on the base of the container. The tension in the string is 0.0784 N. When 300 cm3 of another liquid, of relative density 1.2 is added and thoroughly mixed with the water, the volume of the mixture is 800 cm3 and the tension in the string is 0.1078 N. Find (i) the relative density of the mixture
(ii) the mass of the sphere.
5
5 5
( ) ( )
( )
( )
kg. 032.0
1078.040430784.0
1078.04043
or 1075 mixture
0784.01 or 1000 water
)(
075.1or 4043
108001000 103001200105001000
mixture of mass liquid of mass water of mass )(
6
66
=
+=⎟⎠⎞
⎜⎝⎛+
+=⎟⎠⎞
⎜⎝⎛
+=
+=
=
×××
=××+××
=+
−
−−
m
mgmg
mgs
mgVg
mgs
mgVg
mgTBii
s
s
i
m
m
5 5
30 5
Page 20 of 27
10. (a) Solve the differential equation
xydxdy sin2=
given that y = 1 when x = 2π .
x
x y
Cxy
Cx y
dxxydy
xydxdy
cos11 y
1cos 1
1 2
,1
cos 1
sin
sin
2
2
+=
+=
−=⇒==
+−=−
=
=
∫∫
π
5 5 5
20 5
Page 21 of 27
10 (b) The acceleration of a racing car at a speed of v m/s is
⎟⎟⎠
⎞⎜⎜⎝
⎛−
32001
2v m/s2
The car starts from rest. Calculate correct to two decimal places
(i) the speed of the car when it has travelled 1500 m from rest (ii) the maximum speed of the car.
( )[ ] [ ]
( )
m/s 57.56
03200
1
0 on accelerati )(
m/s 1244
3200
3200
1500 3200ln16003200ln1600
3200ln1600
3200
3200
32001 )(
2
1615
2
2
1500 0
v0
2
5001
0 2
0
2
=⇒
=−
=
=⇒
=−
=−−
=−−
=−
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
∫∫
v
v
ii
.v
e v
v
x v
dx dv vv
v dxdvv i
v
5
5 5 5 5 30 5
Page 22 of 27
Some Alternative Solutions
2 (b)
x 3
5 30 θ
d θ
30
( )( )
m 04.48
30985.31tan
985.31832.2
30sin3
sin
832.230cos53253 222
=
=°
°=
=
=−+=
dd
xx
θ
θ
5 5 5
20 5
3 (b)
( )( )
( )
2tancos2sin2sin
21cos
21sin22sin
45cos45sin2sin
sin
sin 0
0 plane horizontalConsider 45cos
45sin2
45cos45sin 0
plane inclinedon 0 2
21
=⇒−=
⎭⎬⎫
⎩⎨⎧
⎟⎠
⎞⎜⎝
⎛−⎟
⎠
⎞⎜⎝
⎛=
−=
=
=⇒
−=
=
−=⇒
−−=
=
θθθθ
θθθ
θθ
θθ
θθ
gu
gu
tt
gu t
gt u
v
gu t
.tg.t u
r
j
j
5 5
5 5
25 5
Page 23 of 27
3 (b)
( )( )
{ }
{ }
( ) ( )( ) ( )
{ }
{ }
2tansincos2
cos2sin2
cossin2sin
sin
..2sin.2
..245sinsin45coscos45sincos45cossin
..245cos45sin
.45cos45sin.45sin45cos
45tan
cossin2
45cos45sincos45cossin2
45cos45sin2
045cos45sin
plane inclinedon 0 2
21
=⇒=
−=
⎥⎦
⎤⎢⎣
⎡−=
=
=
=++−
=−+−
+−−=−−
−=⇒
−=
−=
−=
−=⇒
=−−
=
θθθ
θθ
θθθ
θ
θ
θθθθ
θθ
θθ
θθ
gu
ugugt
tgu
tgu
tg θuθu
tgθu tg θu
vv
vv
gu
gu
gu t
.tg.t u
r
ji
i
j
j
5
5 5 5 25 5
Page 24 of 27
5 (b)
( )
( ){ } ( )
cos
coscos2 KEin Loss
cos
cos218
2 104
2 1 after KE
cos2cos42 1 before KE
direction in theenergy kinetic of loss Take )(
22
2222
22
22
2222
α
αα
α
α
αα
u
uu
u
u
uu
iiii
=
−=
=⎭⎬⎫
⎩⎨⎧+=
==
r
5
10 5
6 (b)
. reach not willmass combined the 2 As4
74
142
))(2(21)()2(
KEin Loss PEin Gain : height maximumAt
212)0(
at mass combined the: of speed thebe Let
14
)0(21)2()10(
21
at energy Total at energy Total reachesit when : of speed thebe Let
21
1
1
1
2
2
crh
rh
grgh
vmhgm
vv
mvmmvdv
grv
mgmvrmggrm
dcdcv
⇒<
=
=
=
=
=
=+
=
+=+
=
5 5 5
5 5
25
Page 25 of 27
7 (a)
R1
7 b
a c
d
o
R2
7 W W
θ 6
( ) ( ) (
WR
xWxWxRb
xx, x,
=
+=
1
1
71320: systemfor about moments Take
20137 :ratioin Lengths 5
) 5 15 5
7 (b)
R F
θ
10 cm
N
100g
α
5
( ) ( ) ( )
( )( ) ( )( ) ( )
N 784or 80
20.010020.08.015.06.0
20.010020.0sin15.0cos
gF
gFF
gFF
=
=+
=+ θθ
5,5 5 25 5
Page 26 of 27
10(b)
( ) ( )( )( )
( )
( )
m/s 57.56
240
2201
280240 0
0 speed maximumFor
220
1280
240
220
1 240
1240240
240240
0240240ln220
240240ln220
3200
3200
32001 )(
220
2
220
2
2202
220
v
0
0 2
0
2
=⇒
=⇒
−=
=
−=
=−
−+−−
=−+
=−⎟⎟⎠
⎞⎜⎜⎝
⎛
−+
=⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛
−+
=−
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
∫∫
v
v
e v
dtdv
e v dtdv
e dtdv
v
vv
e vv
t vv
t vv
dt dv v
v dtdvii
t
t
t
t
tv
5
10 5
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Applied Maths Cvr 15/8/07 19:31 Page 2