Embed Size (px)
Citation preview
Coimisiún na Scrúduithe Stáit State Examinations Commission
Scéimeanna Marcála Scrúduithe Ardteistiméireachta, 2005 Matamaitic Fheidhmeach Ardleibhéal Marking Scheme Leaving Certificate Examination, 2005 Applied Mathematics Higher Level
M32
Coimisiún na Scrúduithe Stáit
State Examinations Commission
Scéimeanna Marcála Scrúduithe Ardteistiméireachta, 2005
Matamaitic Fheidhmeach Ardleibhéal
Marking Scheme Leaving Certificate Examination, 2005
Applied Mathematics Higher Level
2
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 Mark all answers, including excess answers and repeated answers whether cancelled
or not, and award the marks for the best answers.
3 Mark scripts in red unless candidate uses red. If a candidate uses red, mark the script
in blue or black.
4 Number the grid on each script 1 to 10 in numerical order, not the order of answering.
5 Scrutinise all pages of the answer book.
6 The marking scheme shows one correct solution to each question. In many cases there
are other equally valid methods.
3
1. (a) Car A and car B travel in the same direction along a horizontal straight road.
Each car is travelling at a uniform speed of 20 m/s.
Car A is at a distance of d metres in front of car B.
At a certain instant car A starts to brake with a constant retardation of 6 m/s2.
0.5 s later car B starts to brake with a constant retardation of 3 m/s2.
Find
(i) the distance travelled by car A before it comes to rest
(ii) the minimum value of d for car B not to collide with car A.
33.3or 3
100
)6(2020
2asu v (i)
2
22
=
−+=
+=
s
s
( )
( )
3.43
3.33106.66 of valueminimum
66.6or 6
400
32200
2
:applied are brakes When
10
05.020
Car B (ii)
2
22
2
21
=
−+=
=
−+=
+=
=
+=
+=
d
s
s
as u v
s
s
atuts
5
5
5
5
5
25
4
1 (b) A mass of 8 kg falls freely from rest. After 5 s the mass penetrates sand.
The sand offers a constant resistance and brings the mass to rest in 0.01 s.
Find
(i) the constant resistance of the sand
(ii) the distance the mass penetrates into the sand.
( )
( )
( ) ( )
N 4.39278
490088.98
4900
01.0490
Sand
49
58.90
s 5after Speed (i)
=
−=−
=−
−=
+=
+=
=
+=
+=
R
R
maRmg
a
a
atu v
v
v
atuv
( ) ( )
245.0
49002490
2 (ii)
2
22
=
−+=
+=
s
s
asuv
5
5
5
5
5
25
5
2. (a) A woman can swim at u m/s in still water.
She swims across a river of width d metres. The river flows with a constant
speed of v m/s parallel to the straight banks, where v < u.
Crossing the river in the shortest time takes the woman 10 seconds.
Find, in terms of u and v, the time it takes the woman to cross the river by the
shortest path.
v v d
u u 22 vu −
.
10
speed
distance time
speed
:pathshortest
10or 10
:imeshortest t
22
22
22
vu
u
vu
d
vu
udu
d
−=
−=
=
−=
==
5
5
5
5 20
6
α
X
2 (b) Two straight roads intersect at an angle
of 45°. Car A is moving towards the intersection with a uniform speed of p m/s.
Car B is moving towards the intersection with
a uniform speed of 8 m/s.
The velocity of car A relative to
car B is jirr
102 −− .
At a certain instant car A is 220 2 m from the intersection and car B is 136 m from the intersection.
(i) Find the value of p.
(ii) How far is car A from the intersection at the instant when the cars are nearest to each
other?
Give your answer correct to the nearest metre.
B 40 A
P VAB
( )
( )
( )
( )m. 11 Ans
m 10.8or 769.0210 travelsA time thisIn
0.769or 104
7.844
V
AXe tim
844.7or 1961.004
0cos4 AX
69.7810
2tan90
onintersecti thefrom m 40now is and
m 176or 228 travels B time thisIn
s 22or 210
2220 in onintersecti thereaches A (ii)
210
228102
822
102
(i)
AB
1
==
=
=
°=
−=
=⇒
−
−=−−
−−
−−=−−
−=
−
α
α
p
j p
i p
j i
i j p
i p
j i
V V V BAAB
rrrr
rrrrr
rrr
5
5
5
5
5
5 30
B
A
45°
7
3. (a) A ball is projected horizontally from
a point q above a smooth horizontal plane
with speed 2 m/s.
The ball first hits the plane at a point whose
horizontal displacement from q is 0.4 m.
The ball next strikes the plane at a
horizontal displacement of 1 m from q.
The coefficient of restitution between the ball
and the plane is e.
Find the value of e.
( )
( )
( )
( ) ( )
43
2
21
2
21
0 3.03.0 96.1
0 96.1
0
96.12 velocity Rebound
3.0
6.0 2 0.6 ir stage Second
96.1
2.08.90
2.0
4.0 2 0.4 ir stageFirst
=
=−
=−
=
+=
=⇒
=⇒=
−=
−=
+=
=⇒
=⇒=
e
ge
gtte
jr
j e i
t
t
at u jv
t
t
r
rr
r
r
r
5
5
5
5 20
q 2 m/s
0.6 m 0.4 m
8
3 (b) A plane is inclined at an angle β to the horizontal. A particle is projected up
the plane with initial velocity u at an angle α to the inclined plane. The plane of projection is vertical and contains the line of greatest slope.
(i) Find the range of the particle on the inclined plane in terms of u, α and β .
(ii) Show that for a constant value of u the range is a maximum when
245
βα −°=
.
{ }
( ){ }
( )
254
902
02cos 0
2coscos
2
sin2sincos2coscos
2
cossin2cos
sin22cos2
cos
sincos
sin22sin
cos
cos
sin2sin
cos
sin2cos
sincos Range
cos
sin2
0 cossin
0
2
2
2
2
2
22
2
2
22
2
21
2
21
2
21
βα
βα
βαα
βαβ
βαββ
ααββα
αββ
ββ
β
−°=⇒
°=+⇒
=+⇒=
+=
−=
−=
−=
−
=
−=
=⇒
=−
=
d
dR
g
u
α.g
u
.g
βu α.
g
u
d
dR
.g
βu α.
g
u
g
αuβ..g
g
αuα.u
β.tg α.t u
g
αu t
β.tg α.t u
r j
5
5
5
5
5
5 30
9
4. (a) A particle of mass 4 kg rests on a rough
horizontal table. It is connected by a
light inextensible string which passes over
a smooth, light, fixed pulley at the edge of
the table to a particle of mass 8 kg which
hangs freely under gravity.
The coefficient of friction between the 4 kg mass and the table is 4
1.
The system starts from rest and the 8 kg mass moves vertically downwards.
Find
(i) the tension in the string
(ii) the force exerted by the string on the pulley.
3
10
310
228
4
88 (i)
gT
Tg
gT T g
fgT
f T g
=
=
−=−
=−
=−
N 2.46or 3
210
2
Force (ii) 22
g
T
TT
=
=
+=
5
5
5
5
5
25
4
8
10
4 (b) Two particles of masses 3 kg and 5 kg are connected by a light inextensible
string, of length 4 m, passing over a light smooth peg of negligible radius.
The 5 kg mass rests on a smooth horizontal table. The peg is 2.5 m directly
above the 5 kg mass.
The 3 kg mass is held next to the peg and is allowed to fall vertically a distance
1.5 m before the string becomes taut.
(i) Show that when the string becomes taut the speed of each particle is
3 3
8
g m/s.
(ii) Show that the 3 kg mass will not reach the table.
( )( )
( )
table thereach
not willmass kg 3 1 As
0.84or 32
27
42
8
3g3 0
2 mass kg 3
45.2or 4
55
33 (ii)
8
3g3
5333
3
5.120
2 (i)
2
22
1
22
1
⇒<
=
−+
=
+=
−−=
=−
=−
=
+=
=
+=
+=
s
s
sg
asuv
ga
agT
aTg
v
vvg
gv
g
asuv
5
5
5
5
5
25
11
5. (a) Three identical smooth spheres P, Q and R, lie at rest on a smooth horizontal
table with their centres in a straight line. Q is between P and R.
Sphere P is projected towards Q with speed 2 m/s. Sphere P collides directly
with Q and then Q collides directly with R.
The coefficient of restitution for all of the collisions is 3
4.
Show that P strikes Q a second time.
( )
( )
timeseconda Q strikes P32
7
4
1 As
32
7
04
7 4
3 v v NEL
mvmv )0m( 4
7mM PCQR
4
7
4
1
02 4
3 v v NEL
mvmv )0m( 2m PCM PQ
1
21
21
2
1
21
21
>
=
−−=−
+=+
=
=
−−=−
+=+
v
v
v
25
5
5
5
5
5
12
5 (b) A smooth sphere A, of mass m, moving
with speed u, collides with an identical
smooth sphere B moving with
speed u.
The direction of motion of A, before impact,
makes an angle 45°with the line of centres at impact.
The direction of motion of B, before impact,
makes an angle 45°with the line of centres at impact. The coefficient of restitution between the spheres is e.
(i) Find, in terms of e and u, the speed of each sphere after the collision.
(ii) If 2
1=e , show that after the collision the angle between the directions
of motion of the two spheres is
−
3
4tan 1 .
( )
3
4
01
3
40
tan
2tan180tan1
2tan180tantan
2180
3
4
41
22
tan1
tan22tan
22 and
22 (ii)
12
Bof Speed
1222
A of Speed
2 and
2
22 NEL
22 PCM (i)
2
2121
2
2
22
21
21
21
=+
−−
=
+
−=
−=
−=−
=−
=
=−=⇒=
+=
+=
+
−=
=−=⇒
+−=−
+=−
θ
α
αθ
αθ
α
αα
uv
uve
eu
euueu
eu v
eu v
uu e v v
mv mv u
m u
m
5
5
5
5
5
25
αθ
α
45° 45°
u u
A B
13
6. (a) A conical pendulum consists of a light inelastic
string [pq], fixed at the end p, with a particle
attached to the other end q .
The particle moves uniformly in a horizontal circle
whose centre o is vertically below p.
If hpo = , find the period of the motion in terms of h.
T
mg
g
h
g
h
ω
mg
mrω
h
r
mg
mrω θ
mrωθ T
mgθ T
π
ω
π
2
2 Period
1
tan
sin
cos
2
2
2
=
=
=
=
=
=
=
5
5
5
5
5
25
θ
p
o
q
h
14
6 (b) A light elastic string of natural length a and elastic constant k is fixed at one
end to a point o on a smooth horizontal table. A particle of mass m is attached
to the other end of the string.
Initially the particle is held at rest on the table at a distance 2a from o, and is
then released.
Show that the time taken for the particle to reach o is 12
m
k
π +
.
+=
=
+=
+=
==
==
=
=
−=
−=
−=
21 timeTotal
0
s stage Second
2 e timstageFirst
Velocity
Amplitude
SHM with
onaccelerati
Force
2
21
41
π
π
ω
k
m
k
mt
t m
kaa
atut
k
mT
m
kaa
a
m
k ω
xm
k
kx
T
5
5
5
5
5
25
15
7. (a) A particle of weight 100 N lies on a plane.
The plane is inclined at 30° to the horizontal. A horizontal force F is applied to the particle.
The coefficient of friction between the particle
and the inclined plane is 5
3.
Find the least value of F that will move the particle up the plane.
R °30
F
100
R53
( )
80.11
30sin30cos10000sin301 cos30
00sin301 cos30
30sin30cos100
53
53
=
++=
+=
+=
F
FF
RF
FR
25
5
5
5
5
5
30°
F
16
7 (b) Two uniform rods, [ab] and [bc], of equal
length, are smoothly jointed at b.
They rest in a vertical plane with a and c on
rough horizontal ground and |∠abc| = θ2 . The weight of the rod [ab] is 2W and the weight of
the rod [bc] is W.
The coefficient of friction at both a and c is µ.
Find the least value of µ , in terms of θ , necessary for equilibrium.
R1 R2 2W W
F1 F2
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( )
( )
θµ
µθ
µ
θ
θθ
θθθ
tan5
3
4
5tan
4
3
tan 4
3
tan 4
5 tanW
sin sinW cos
:for about moments Take
4
7
4
5
4312
:systemfor about moments Take
3 Vert
22
12
2
21
2
221
2
1
2
2
21
=⇒
=
=⇒
<
=
=+
=+
=
=
=+
=+
WW
RF
RR
W F
W F
R F
bcb
WR
WR
R W W
a
WRR
lll
25
5
5
5
5
5
a c
b
θ2
a c
b
17
8. (a) Prove that the moment of inertia of a uniform rod of mass m and length 2l
about an axis through its centre perpendicular to the rod is 2
3
1ml .
{ }
{ }
2
31
3
32
3
-
2
2
m
M
3
x M
dx x M rod theofinertia ofmoment
x dx Melement theofinertia ofmoment
dx Melement of mass
lengthunit per mass Let M
l
l
l
l
l
l
=
=
=
=
=
=
=
−
∫
5
5
5
5
20
18
8 (b) A uniform rod [pq], of mass 9m and length 2l ,
has a particle of mass 2m attached at q.
The system is free to rotate about a smooth
horizontal axis through p. The rod is held in
a horizontal position and is then given an
initial angular velocity 3
2
g
l downwards.
The diagram shows the rod [pq] when it makes
an angle θ with the horizontal.
(i) Show that when the rod makes an angle θ below its initial horizontal position, its angular velocity is
( )15 13sin
10
g
l
θ+ .
(ii) Hence, or otherwise, show that the rod performs complete revolutions
about p.
( ) ( )( )
( ) ( ) ( ) ( )
( )
( )
srevolution complete
performs rod
0
10
1315
2
3 (ii)
10
sin1315
sin22sin9 2
320 20
PEin Loss KEin Gain
20
22 9inertia ofmoment (i)
2
212
2
2
21
2
1212
221
2
22
34
⇒
>
−=
=
+=
+=
−
=−
=
=
+=
l
l
lll
ll
l
ll
g
θg
mgmg g
mm
mghII
m
mm
ω
πθ
ω
θθω
ωω
30
5
5,5
5
5
5
p θ
q
19
9. (a) An alloy is made of iron and aluminium.
A piece of the alloy has a mass of 0.441 kg and a volume of 75 cm3.
The relative density of iron is 8 and the relative density of aluminium is 2.7.
Find
(i) the volume of iron in the piece of alloy
(ii) the mass of aluminium in the piece of alloy.
( )
( )
( ) ( )( ) ( )
( )kg 081.0
10752700 alluminium of ass M(ii)
cm 45
3.55.238
757.28441
107527001080004410
alum. of Mass iron of Massalloy of Mass
10752700 aluminium of Mass
108000 iron of ass M (i)
6
3
66
6
6
V
V
V
V V
VV .
V
V
-
--
-
-
=
−=
=
=
−+=
−+=
+=
−=
=
5
5
5
5
5
25
20
9 (b) A uniform rod of length l and relative density s
can turn smoothly about its upper end which is θ
fixed at a height h above the surface of water.
The rod is inclined at an angle θ to the vertical and is partially immersed in the water.
The rod is at rest.
Show that cos1
h
sθ =
−l, where 1s < .
r
x B
W
( ) ( ) ( )
( )
( )
( ) ( )
( )
( )
( )sh
x
hc
sx
sx
sx
Wx
s
xW
Wx
B
Wx
B
s
xW
s
Wx
s
sW LI
−=
=
−=
−=
=−
=
+
−
=
+
=
−−
−
−
=
=
1
os
1
1
2
1
2
2
1
2
sin2
1 sin
2
:rabout moments Take
or
1
BBuoyancy
22
222
l
l
l
ll
ll
l
l
ll
ll
l
l
ll
l
θ
θθ
25
5
5,5
5
5
θ
h
21
10. (a) Solve the differential equation
0dy
x xy ydx
− − =
given that y = 1 when x = 1.
( )
11ln or y
1C 1 ,1
lnln
11
1
0
−−+=
−=⇒==
++=
+=
+=
=−−
∫∫
xxx xee
xy
Cx x y
dxxy
dy
x
xy
dx
dy
yxydx
dyx
5
5
5
5
20
22
10 (b) A mass of 9 kg is suspended at the lower end of a light vertical rope.
Initially the mass is at rest. The mass is pulled up vertically with an initial pull
on the rope of 137.2 N.
The pull diminishes uniformly at the rate of 1 N for each metre through
which the mass is raised.
(i) Show that the resultant upward force on the mass when it is x metres
above its initial position is
49 x− .
(ii) Find the speed of the mass when it has been raised 15 metres.
(iii) Find the work done by the pull on the rope when the mass has been
raised by 15 m.
( )
( )5.622
09
K.E.in Change doneWork (iii)
76.11
5.112735 2
9
249
2
9
49 9
49 9 (ii)
49
937.21 Force (i)
2
21
2
15
0
2v
0
2
15
0
0
=
−=
=
=
−=
−=
−=
−=
−=
−−=
∫∫
v
v
v
xx v
dxx dv v
x dx
dvv
x
gx
v
5
5
5
5
5
5
30
23
Some Alternative Solutions
2(b) A
p 104 B
8
( )
( )
( )
( )
m. 11 Ans
onintersecti thefrom m 25081.60now is and
m 81.60or 77.5210 travelsA time thisIn
5.77or 104
58.8349
V
AXe tim
8.83495
69.33co205 AX
69.3310
2tan45
onintersecti thefrom m 250now is and
m 2170or 17210 travelsA time thisIn
s 17or 8
136 in onintersecti thereaches B(ii)
210
04028
2
18264041 (i)
AB
1
2
2
−
==
=
=
°=
−°=
=⇒
=−−
−+=
−
s
p
pp
p p
θ
5
5
5
5
5
5 30
45
250
X
ABV
θ
24
2(b)
O B C
X
( )
m. 11 Ans
on.intersecti from m 87.102220322.0now is and
m 322.0or 769.22210 travelsA time thisIn
22.769or 104
232.2
V
AXe tim
2.232845.7357.224
357.22431.101sin
45sin2220
845.769.78cos40
40
0.176
31.101sin
2220
sin33.69
OC
69.3310
2 tan45 (ii)
AB
1
=−
==
=+=+=
==
==
=
=
=
°=
−= −
CXACAX
AC
CX
CB
OC
θ
5
5
5 15
2220
ABV
θ
45
25
2(b)
A (220,220)
O (0,0) B C
X
( ) ( )
( )
m. 11 Ans
on.intersecti from m 87.102220322.0now is and
m 322.0or 769.22210 travelsA time thisIn
22.769or 104
232.2
V
AXe tim
32.22
69.722046.174220 AX
)69.7 (174.46, X of Coords
1365 : BXof equation
8805 :AX of equation
5176-220
0-220 AX of Slope
0) (136, Bof coords and (176,0) C of Coords
0.176
31.101sin
2220
sin33.69
OC
69.3310
2 tan45 (ii)
AB
22
1
=−
==
=
++−=
−
=+
=−
==
=
=
°=
−= −
yx
yx
OC
θ
5
5
5 15
2220
ABV
θ
45
26
3(b)
( )
( ){ }
( )
254
902
12sin whenmaximuma is This
cos
sin2sin
cos
sincos2
cos
sinsincoscossin2
cos
sinsin
cos
cossin2
cos
sin2sin
cos
sin2cos
sincos Range
cos
sin2
0 cossin
0
2
2
2
2
2
2
2
2
2
21
2
21
2
21
βα
βα
βα
β
ββα
β
βα
β
βαβ
β
βα
β
ββ
β
−°=⇒
°=+⇒
=+
−+=
+=
−=
−=
−
=
−=
=⇒
=−
=
g
u
g
αu
α.
g
αu
α.
g
αu
g
αuβ..g
g
αuα.u
β.tg α.t u
g
αu t
β.tg α.t u
r j
5
5
5
5
5
5
30
27
8 (b)
( ) ( ) ( ) ( )
srevolution complete
performs rod
0
5
229 202
320
PEin Gain KEin ss Lo
270 throughrotated has rod theWhen (ii)
2
2
2
212
21
2
2212
121
⇒
>
=
+=−
=−
=
°
l
llll
l
g
mgmgm g
m
mghII
ω
ω
ωω
10
5
5
