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