2012 AL- Combined Maths II-Model Answers New

- 1 - Done By : Chandima Peiris (B.Sc - Special)

General Certificate of Education (Adv. Level) Examination, August 2012

Combined Mathematics II - Part B

Model Answers

11. (a) A particle P is projected at a point O vertically upwards under the gravity with velocity u. After time

gu

2, another particle Q is projected at the point O vertically upwards under the gravity with velocity

v ( )u> . Let A be the highest point that the particle P reaches. The particles P and Q meet at the point A. Draw the velocity-time graphs for the complete motions of the particles P and Q in the same figure. Using these velocity-time graphs show that

(i) g

uOA2

2

= ,

(ii) 4

5uv = and the velocity of the particle Q at the point A is 4

3u .

(iii) when the particle Q reaches the highest point the height of the particle P from the point O, is g

u327 2

(b) A car of mass M kg is travelling on a level road against a resistance R of the motion which is a constant

at all speeds. If the maximum power of the engine is H kW and the car has a maximum speed of -1ms von a level road, find the resistance R in terms of M, H and v. Find the acceleration of the car in terms of M, H, v, g and α when it is moving

(i) at speed 1-ms 3v directly up,

(ii) at speed 1-ms 2v directly down

along a straight road inclined at an angle α to the horizontal. If the acceleration of the car in case (ii) is twice that in case (i), find αsin in terms of M, H, v and g. In this case find the maximum speed in terms of v that can be made by the car when moving directly up the road.

Answer (a)

g=θtan

From the triangle BFG, From the triangle GHA,

J

K

Q

P

v1/ −msv

st /

u

1v

1u

2u

θθ

θ θθ

1T 2Tg

u2

O A

B

E

CD

H

F G


gu

uu

2

tan 1−=θ

1

1tanTu

=θ

12uu

gug −=×⇒

1

2T

u

g =⇒

21uu =⇒

guT

21 =⇒

∴ =OA Area of the triangle ABO

⎟⎟⎠

⎞⎜⎜⎝

⎛+××= 122

1 Tg

uu

⎟⎟⎠

⎞⎜⎜⎝

⎛+×=

gu

guu

222

guu ×=

2

(i) =OA g

u2

2

From the triangle EDC,

1

1tanT

uv −=θ

)1(221 −−−−−=×=−⇒u

gugvv

When the particles P and Q meet at the point A, Area of ECAH = Area of the triangle ABO

( ) ug

ug

uvvT ×⎟⎟⎠

⎞⎜⎜⎝

⎛+×=+××⇒

2221

21

11

( ) uguvv

gu ×=+×⇒ 12

)2(21 −−−−−=+⇒ uvv

)1(21 −−−−−=− uvv

( ) ( );21 + 4

52

52 uvuv =⇒= ( ) ( );12 −4

32

32 11uvuv =⇒=

(ii) 4

5uv =

Velocity of the particle Q at the point A 4

3u=


From the triangle EHJ, From the triangle AJK,

21

tanTT

v+

=θ 2

2tanTu

=θ

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⇒

224

5

Tg

uug

gugu

43

2 ×=⇒

ugTgug 542

2 =+×⇒ 4

32

uu =⇒

ugT 34 2 =⇒

guT

43

2 =⇒

∴The height of the particle P from O when Q reaches the highest point 221 21

221 Tu

guTu ××−⎟⎟⎠

⎞⎜⎜⎝

⎛+××=

⎥⎦

⎤⎢⎣

⎡×−×=

guu

guu

43

43

21

⎥⎦

⎤⎢⎣

⎡−=

gu

gu

169

21 22

g

u167

21 2

×=

=g

u327 2

(b) Let F is the tractive force of the car. Power =Velocity × Tractive force of the engine FvH ×=⇒1000

v

HF 1000=⇒

vF

RM

For A/L Combined Maths (Group/Individual) Classes 2013/ 2014 English and Sinhala Mediums Contact :0772252158 P.C.P.Peiris B.Sc (Maths Special) University of Sri Jayewardenepura


Applying → amF = to the motion of the car, ( )0MRF =−

vHFR 1000==⇒

∴ R =v

H1000

Let 1a - Acceleration of the car 1F - Tractive force of the car,

when the car is move at speed 1-ms 3v directly up.

31000 1

vFH ×=

vHF 3000

1 =⇒

Applying amF = to the motion of the car,

11 sin MaMgRF =−− α

1sin10003000 MaMgv

Hv

H =−−⇒ α

1sin2000 MaMgv

H =−⇒ α

(i) ∴ =1a αsin2000 gMv

H −

Let 2a - Acceleration of the car

2F - Tractive force of the car,

when the car is move at speed 1-ms 2v directly down.

21000 2

vFH ×=

vHF 2000

2 =⇒

Applying amF = to the motion of the car,

22 sin MaRMgF =−+ α

21000sin2000 Ma

vHMg

vH =−+⇒ α

2sin1000 MaMgv

H =+⇒ α

R

1F

Mg

1a

3v

α

R

2F

2a2v

α Mg


(ii) =∴ 2a αsin1000 gMv

H +

If 12 2aa = ,

αα sin24000sin1000 gMv

HgMv

H −=+

MvHg 3000sin3 =⇒ α

=∴ αsinMgv

H1000

Let 1v be the maximum speed of the car, when it move directly up the road and 3F is the tractive force of the car.

Applying amF = to the motion of the car, ( )0sin3 MMgRF =−− α

αsin3 MgRF +=∴

Mgv

HMgv

H 10001000 ×+=

=3Fv

H2000

Power = 13vF

120001000 v

vHH ×=⇒

=∴ 1v2v

12. (a) A particle is projected under the gravity in a vertical plane with a velocity u at an angle θ to the

horizontal, at a point C which is at a height k from O. Consider a rectangular Cartesian system of coordinates by taking horizontal and vertical lines through the point O in the plane of projection as Ox and Oy axes respectively. If at time t the particle is at the point ( )yx, , show that

2

2

2sectanu

gxxky θθ −+= .

A particle P is projected under the gravity in the vertical plane at the point ( )hA ,0 , where h is positive, with a velocity v at an angle α to the horizontal. At the same instant another particle Q is projected

under the gravity in the same vertical plane at the point ⎟⎠⎞

⎜⎝⎛

2,0 hB with a velocity w at an angle ( )αβ > to

the horizontal. If the two particles P and Q meet at a point whose horizontal distance is d, show that βα coscos wv = and ( )αβ tantan2 −= dh .

Show also, that the time taken for the two particles to meet is ( )αβ sinsin2 vwh−

.

R

3F

Mg

1v

α


(b) One end of a light inextensible string is attached to a ceiling which is at a height of 3 metres from a horizontal floor. The string passes under a smooth light movable pulley P to which a particle of mass m is fixed and then over a smooth light pulley fixed to the ceiling. A particle Q of mass M ( )m> is

attached to the other end of the string. When the movable pulley P and the particle Q are at heights 21

metres and 1 metres respectively from the floor and the portions of the string not in contact with pulleys are vertical, the system is released from rest. Find the acceleration of the particle Q and the tension in the string.

Show that the particle Q will reach the floor after time ( )gmMmM

−+

24 seconds and the pulley P will rise

to height mM

M+

+4

321 metres from the floor.

Answer (a)

Applying PCatuts →+=→ 21 2

tux cosθ=

θcosuxt =⇒

Applying PCatuts →+=↑ 21 2

2

2 sin tgtuky −=− θ

θθθ

22

2

cos2cossin

uxgx

uuky −+=⇒

∴ 2

22

2sectanu

gxxky θθ −+= ( )1−−−−−−

u

θcosu

θsinu( )yxP ,

k

x

y

O

θC


Let the particles P and Q meet at the point ( )ydK , , when time t=

Applying K 21 2 →+=→ Aatuts Applying K

21 2 →+=→ Batuts

to the particle P to the particle Q

)2( cos −−−−= tvd α )3( cos −−−−= twd β From (2) and (3); twtv cos cos βα = ⇒ βα coscos wv =

When vu = , αθ = , hk = and dx = , from the equation (1);

)4(2sectan 2

22

−−−−−−+=v

gddhy αα

When wu = , βθ = , 2hk = and dx = , from the equation (1);

)5(2sectan

2 2

22

−−−−−−−+=w

gddhy ββ

( ) );5(4 − ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−+−−=

αβαβ 2222

2

sec1

sec1

2tantan

20

vwgddh

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛ −+−−=⇒βα

βααβ 2222

22222

secsecsecsec

2tantan

20

wvwvgddh

Since βα secsec wv = , 0secsec 2222 =− βα wv .

( )αβ tantan2

0 −−=∴ dh

⇒ ( )αβ tantan2 −= dh

From (2); ( )αβαα tantancos2cos −=⇒=

vht

vdt

y

( )ydK ,

O

v

( )hA ,0

d

α

⎟⎠⎞

⎜⎝⎛

2,0 hB

β

x

w

y

Q

P


⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⇒

αα

ββα

cossin

cossincos2v

ht

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⇒

ααα

βββ

cossincos

cossincos2 vw

ht ( )βα coscos wv =∴

∴ =t ( )αβ sinsin2 vwh−

(b)

Initial position General position Let x be the distance to the particle Q from the fixed line and y be the distance to the movable pulley P from that line. Since the string is inextensible, lyx =+ 2 , where l is a constant.

( )1202 −−−−−−=⇒=+ yxyx &&&&&&&& Applying amF =↓ to the motion of the particle Q,

( )2−−−−−=− xMTMg && Applying amF =↑ to the motion of the movable pulley P,

( )32 −−−−−−=− ymmgT && ( ) ( );223 ×+ ymxMmgMg &&&& −=− 22

⎟⎠⎞

⎜⎝⎛−−=−⇒

222 xmxMmgMg

&&&&

( )xmMmgMg &&+=−⇒ 424

=⇒ x&& ( )( )mM

gmM+−

422

P

Q

m

M

m1

m21

yx

Q

Mgmg

TT

P

y&&−x&&


From (2); ( )( )mM

gmMMMgT+−−=

422 From (1); ( )

( )mMgmMy

+−−=

42

&&

( ) ⎥⎦

⎤⎢⎣

⎡+

+−+=⇒mM

MmMMmMgT4

244 22

( )mMMmgT

+=⇒

43

∴Acceleration of the particle Q = ( )( )mM

gmM+−

422

Tension of the string = ( )mMMmg

+43

Let t be the time taken to the particle Q to reach horizontal floor.

Applying 2

21 atuts +=↓ to the particle Q

2

2101 tx&&+=

xt

&&

22 =⇒

( )( )gmM

mMt−+=⇒

22422

( )( )gmM

mMt−+=⇒

242

( )( )gmM

mMt−+=⇒

24

∴The particle Q reaches the floor after time ( )( )gmM

mM−+

24 seconds.

Let h and v be the height from the initial position and the velocity of the movable pulley P when the particle Q reaches the floor.

Applying 2

21 atuts +=↑ to the movable pulley P Applying asuv 222 +=↑ to the movable pulley P

( )( )gmM

mMyh−+−=

24

20

&&

21202 ××−= yv &&

( )( )

( )( ) 2

124

42

21 =

−+

+−=⇒

gmMmM

mMgmMh ( )

( ) gmMmMv

+−=⇒

422

21=∴ h m

When the particle Q reaches the floor, the string becomes slack then the movable pulley P moves under

gravity until it attains its maximum height 1h from the point where it above ⎟⎠⎞

⎜⎝⎛ + h

21 metres from the

floor.


We have 0=v , ( )( ) g

mMmMu

+−=

422 , ga = and 1hs =

Applying asuv 222 +=↑ to the movable pulley P ( )( ) 12420 ghg

mMmM −

+−=

( )( )mM

mMh+−=⇒

42

21

1

∴Height to the movable pulley P from the floor 121 hh ++=

( )( )mM

mM+−++=

42

21

21

21

⎟⎠⎞

⎜⎝⎛

+−+++=

mMmMmM

424

21

21

⎟⎠⎞

⎜⎝⎛

++=

mMM

46

21

21

= mM

M+

+4

321

13. A and B are two points on a smooth horizontal table at a distance 8l apart. A smooth particle P of mass m

lies at a point on AB in between the points A and B. The particle P is attached to the point A by a light elastic string of natural length 3l and modulus of elasticity λ4 and to the point B by a light elastic string of natural length 2l and modulus of elasticity λ .

If the particle P is in equilibrium at a point C, show that lAC1142= .

The particle P is held at the mid-point M of AB and then is released from rest. When the particle P is at a distance x from the point A along AB, obtain the tension of the two strings.

Write down the equation of motion of the particle P for lxl 41140 ≤≤ and show that, in the usual notation,

that 01142

611 =⎟

⎠⎞

⎜⎝⎛ −+ lx

mlx λ&& .

By writing lxy1142−= , show that 0

611 =+ y

mly λ&& .

Assuming that the solution of the above equation is of the form tBtAy ωω sincos += , find the constants A, B and ω .

Find the velocity of the particle P when it is at a point, distance l1141 from the point.



Answer

Equilibrium position Let a and b are the extensions of the strings AC and BC respectively. Let 1T and 2T are the tensions of the strings AC and BC respectively. Since the length lAB 8= , ( )13823 −−−−−−=+⇒=+++ lballbal From the Hooke's law,

laT

34

1λ=

lbT

22λ=

When the particle is in the equilibrium at the point C,

lb

laTT

234

21λλ =⇒=

234 ba =⇒

( )2038 −−−−−−−=−⇒ ba ( ) ( );231 +× la 911 =

119la =⇒

1124lb =∴

=∴ AC11933 llal +=+

=⇒ AC l1142

From the Hooke's law,

( )l

lxT3

343

−= λ ( )l

lxlT2

284

−−= λ

( )l

xlT2

64

−=⇒λ

Applying amF =→ to the particle xmTT &&=− 34

( ) ( ) xml

lxl

xl&&=−−−

⇒3

342

6 λλ

( ) xmlxxll

&&=+−−⇒ 2483186λ

A BCl8

1T 2Tl2l3 a b

CA B

M

A B3T 4T

x


( )xll

xm 11426

−=⇒λ

&&

01142

611 =⎟

⎠⎞

⎜⎝⎛ −+⇒ lx

lxm λ&&

⇒ 01142

611 =⎟

⎠⎞

⎜⎝⎛ −+ lx

mlx λ&&

Let xylxy &&&& =⇒−=1142

∴ 0611 =+ y

mly λ&&

This is of the form 02 =+ yy ω&& . Therefore the particle executes simple harmonic motion with ml6

11λω =

and the centre given by lxy11420 =⇒=

Let O be the centre of the above SHM, lAO1142= .

Since the particle is released from M, where lAM 4= , at lx 4= the velocity of the particle is zero.

If p is the amplitude of this motion, lpllpAOAMp112

11424 =⇒−=⇒−= .

Velocity v of the particle is given by ( )2222 ypv −= ω , where lp112= .

When lx1141= ,

111142

1141 llly −=−=

Velocity of the particle P when it is at a point, distance l1141 from A ⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛=

22

11112 llω

1213

611 2l

mlλ=

1216311 2

××=

mllλ

=ml

22λ

Let us assume that the solutions of the equation 0611 =+ y

mly λ&& is of the form tBtAy ωω sincos += , where A

and B are constants andml6

11λω = .

tBtAy ωω sincos +=

Differentiating with respect to t tBtAy ωωωω cossin +−=&

Since the particle is released from rest, when 0=t , 0=y& . ( ) ( )0cos0sin0 ωω BA +−=⇒


0=⇒ ωB =⇒ B 0 ( )0≠ωQ

tAy ωcos=∴

When 0=t , lx 4= . i.e. llly112

11424 =−=

( )0cos112 Al =⇒

=⇒ A 112l

=ωml6

11λ

Therefore the values of the constants are112lA = , 0=B and

ml611λω = .

14.

(a) Let A and B be two distinct points not collinear with a point O. Let the position vectors of the points A and B with respect to the point O be a and b respectively. If D is the point on AB such that DABD 2= ,

show that the position vector of the point D with respect to the point O is ( )ba +231 .

If akBC =→

, ( )1>k and the point O, D and C are collinear, find the value of k and the ratio DCOD : .

Express →

AC in terms of a and b. Further, if the line through the point O parallel to AC meets AB at E, show that ABDE =6 .

(b) The coordinates of the points A, B and C with respect to a rectangular Cartesian axes Ox and Oy, are

( )0,3 , ( )1, 0 − and ⎟⎟⎠

⎞⎜⎜⎝

⎛1 ,

332 respectively. Forces of magnitude 6P, 4P, 2P and P32 newtons act along

OA, BC, CA and BO respectively in the directions indicated by the order of the letters. Find the magnitude and the direction of the resultant of these forces. Find the position at which the line of action of the resultant cuts the y-axis. Hence, find the equation of the line of action of the resultant. Another force of magnitude P36 newtons is introduced to the system along AB in the direction indicated by the order of the letters. Show that the system is reduced to a couple of magnitude P10newton metre.

Answer (a)

1:2:2 =⇒= DABDDABD

2

O x

yB

D

A

1b

a


aOA =→

, bOB =→

→→→

+= OBAOAB ab −=

Since ABAD31= and A, D and B are collinear i.e.

→→= ABAD

31

( )abAD −=→

31

→→→+= ADOAOD

( )aba −+=31

3

3 aba −+=

=∴→

OD ( )ba +231

∴ The position vector of the point D with respect to the point O ( )ba += 231

Let akBC =→

( )1>k

bakOCBCOBOC +=⇒+=→→→→

( )baOD +=→

231

Since O, D and C are collinear, →→

= ODOC λ , where ℜ∈λ .

( )babak +=+⇒ 23λ

03

13

2 =⎟⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ −⇒ bak λλ

03

2 =−⇒λk and 0

31 =− λ

32λ=⇒ k and 3=λ

2=⇒ k and 3=λ ∴ =k 2

Since 2=k , baOC +=→

2 . →→→

+= OCDODC

( ) baba +++−= 2231

3

362 baba ++−−=


3

24 ba +=

( )ba += 232

→→=⇒ ODDC 2

∴ ODDC 2= ⇒ 2:1: =DCOD

→→→+=∴ OCAOAC

baaAC ++−=⇒→

2

⇒ baAC +=→

Since , →→

= ACOE μ where μ is a constant.

( )baOE +=∴→

μ →→→

+= OEAOAE

baaAE μμ ++−=⇒→

→→→

+= ODAOAD

( )3

231 abbaaAD −=++−=⇒

→

Since A, D and E are collinear→→

= AEAD γ , where ℜ∈γ .

( )baaab μμγ ++−=−3

031

31 =⎟

⎠⎞

⎜⎝⎛ −−+⎟

⎠⎞

⎜⎝⎛ −⇒ ab γμγγμ

031 =−⇒ γμ and 0

31 =−− γμγ

O

b

a

D

Eba +

C

By

x

A

AC OE


31=⇒ γμ and 0

31

31 =−−γ

31=⇒ γμ and

32=γ

31

32 =⇒

μ and 32=γ

21=⇒ μ and

32=γ

( )baOE +=∴→

21

( ) ( )babaOEDODE +++−=+=→→→

212

31

( )babaDE 332461 ++−−=⇒

→

( )abDE −=⇒→

61

→→=⇒ ABDE

61

ABDE61=∴ (QA, B, E and D are collinear)

⇒ DEAB 6=

(b)

Let R be the resultant of the system of forces and it makes an angle θ with the horizontal.

31

232tan =×

=φ

°=⇒ 30φ

31tan =α

°=⇒ 30α

323

3

323

1tan =−

=−

=β

°=⇒ 60β

Resolving βφθ sin2cos432sin PPPR −+=↑

232

23432 ×−×+= PPP

( )133sin −−−−−−= PR θ

R

θ ( )0,3A

B

C

32

P6

P32

P4

( )1,0 −

1 P2

O

1

φ

βα

( )yP −,0


Resolving φβθ sin4cos26cos PPPR ++=→

214

2126 ×+×+= PPP

( )29cos −−−−−−= PR θ

( ) ( ) ;21 22 + ( ) 3233333 222222222 ×××=⇒×+×= PRPR

PR 36=∴

( )( ) ;21

31

933tan ==θ

°=∴ 30θ Magnitude of the resultant = P36

Direction of the resultant = °30 with the horizontal Let ( )yP −,0 be the point of intersection of the line of action of the resultant and the y-axis. Taking moments about O

( )3

2sin21cos21sin490sin ×−×−×=×−° ββφθ PPPyR

32

2321

2121

214cos ××−××−××=×⇒ PPPyR θ

PPPPy 229 −−=⇒ 19 −=⇒ y

91−=⇒ y

∴The point at which the line of action of the resultant cuts the y-axis = ⎟⎠⎞

⎜⎝⎛ −

91, 0

Let cmxy += be the equation of the line of action of the resultant.

Here 3

130tantan =°== θm and 91−=c .

∴The equation of the line of action of the resultant ≡91

31 −= xy

O

P36

( )0,3A

B

C

32

P6

P32

P4

( )1,0 −

1 P2

O

1

°30

°60°30


PPPPPPPPX 9262336

212

2146 −++=×−×+×+=

→

0=⇒→X

PPPPPPPPY 33332322136

232

23432 −−+=×−×−×+=↑

0=⇒↑ Y

PPPP 922 −−−=

Here 0=→X , 0=↑ Y and , the system of forces is reduced to a couple.

Magnitude of the couple = P10

Direction of the couple =Clockwise

15. (a) Two equal uniform rods AB and AC each of weight W are freely jointed at A and the ends B and C are

connected by a light inextensible string. The rods are kept in equilibrium in a vertical plane with the ends B and C are on two smooth planes each of which inclined at an angle α to the horizontal; BC being horizontal and A being above BC. Find the reaction at B.

If αθ tan2tan > , where θ2ˆ =CAB , then show that the tension of the string is ( )αθ tan2tan21 −W .

Find the reaction at the joint A.

(b) Five light equal rods OA, OB, AC, AB and BC are smoothly jointed at their ends to form a framework as shown in the figure. The framework is smoothly hinged at O and carries a weight 35 newtons at C. The framework is held in a vertical plane, with OB horizontal by a horizontal force P newtons at A.

(i) Find the value of P. (ii) Find the magnitude and the direction of the

reaction at O. (iii) Using Bow's notation, draw a stress diagram for the framework and find the stresses in all rods,

distinguishing between tensions and thrusts.

oG 32136

32

2321

2121

214 ××−××−××−××= PPPP

P10−=oG

oG 0≠

O

A C

B

N35

P


Answer (a)

Let length of the rod AB is 2a.

Taking moments about B for the rod BA

( )10sinsin2cos2 −−−−−−=×−×+× θθθ aWaYaX

Taking moments about C for the rod AC

( )20sinsin2cos2 −−−−−−=×+×+×− θθθ aWaYaX

( ) ( );21 + 00sin4 =⇒= YYa θ ( ) ( );21 − θθ sin2cos4 WaXa =

θtan2

WX =⇒

Resolving ↑ for the rod AB

WRB =αcos

αsecWRb =⇒

∴Reaction at the point B = αsecW

A

B C

cRBR

X X

Y

Y

W WTT

α α

αα

B

θ θ

C



Resolving → for the rod AB

XRT B =+ αsin

θα

tan2cos

WWT =+⇒

( )αθ tan2tan2

−=⇒WT

∴The tension in the string = ( )αθ tan2tan2

−W

Magnitude of the reaction at the joint A = θtan2

W

Direction of the reaction at the joint A =Horizontal direction

(b)

Since the frameworks is in equilibrium,

Resolving → , ( )1−−−−= PX Resolving↑ , ( )235 −−−−= NY

Taking moments about O

aaP 33560sin2 ×=°× , where 2a is the length of the rod OA.

aaP 335232 ×=××⇒

NP 15=⇒

(i) The value of P = N15

Form (1); NX 15=

(ii) Magnitude of the reaction at O = 22 YX +

3515 22 ×+=

325 22 ××=

= N310

O

X O

A C

B

N35

P

Y1 2

3

45

6

°60 °60

°60


Direction of the reaction at O ⎟⎠⎞

⎜⎝⎛= −

XY1tan

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

1535tan 1

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

31tan 1

= °30 with the horizontal

( ) ( ) ( ) ( ) N105356254 22 =+×=−=−( ) ( ) N1065 =−

( ) ( ) N563 =−

( ) ( ) N1052 =−

Rod Magnitude Nature

OA 10N Thrust

OB 10N Thrust

AC 5N Tension

AB 10N Tension

BC 10N Thrust

5 2

641 3=

°30

°60

°60

°60°30

35

15

5



16. Show that the centre of mass of a uniform solid right circular

cone of height h is on its axis of symmetry at a distance h41

from the base of the cone. A uniform solid composite body consists of a right circular cone of base radius 3r and height h and a right circular cylinder of radius r and height 2h fixed together as shown in the figure.

Show that the centre of mass of the composite body is on its axis of symmetry at a distance h45 from the

vertex of the cone. The composite body is hanged freely in a vertical plane by a light inextensible string, one end of which is fixed to a ceiling and the other end to a point A on the circumference of the circular base of the cone. If the axis of symmetry of the composite body makes an angle α with the downward vertical, show that,

hr12tan =α .

By applying along the axis of symmetry of the composite body, a force P at the vertex of the cone, the composite body is kept in equilibrium so that the axis of symmetry of the composite body is horizontal. Find the force P and the tension of the string in terms of W and α , where W is the weight of the composite body. Answer

αtan=r

Elementary particle

Since the cone is symmetrical about the x-axis, its centre of gravity should lies on the x-axis. Then let

( )0 , xG ≡ . Consider the elementary particle in the shape of the cylinder with height dx and radius r, at a distance x from the vertex. Let dm ne its mass and ρ be its density.

ρπ dxrdm 2= dxxdm απ 22 tan=⇒ When x varies from 0 to h the whole object can be obtained. By the definition of centre of gravity,

dx

r

dx

αα x

x

r


∫

∫= h

h

dm

xdmx

0

0

∫

∫

∫

∫==⇒ h

h

h

h

dxx

dxx

dxx

dxxx

0

22

0

32

0

22

0

23

tan

tan

tan

tan

αρπ

αρπ

αρ

αρπ

∫

∫=⇒ h

h

dxx

dxxx

0

2

0

3

h

h

x

x

x

0

3

0

4

3

4

⎥⎦

⎤⎢⎣

⎡

⎥⎦

⎤⎢⎣

⎡

=⇒

3

4 34 h

hx ×=⇒

∴ =x 4

3h

∴The centre of gravity of a uniform solid right circular cone of height h is on its axis of symmetry at a distance

h41 from the base of the cone.

This composite body is symmetrical about x-axis. ∴The centre of gravity of it should lies on the x-axis. Let ρ be the density of the composite body.

y

y2GG1G

1W21 WW + 2W

r


Body Weight Coordinates of the centre of gravity

Cone ghr ρπ ×× 29

31

ghrW ρπ 21 3=

⎟⎠⎞

⎜⎝⎛−≡ 0 ,

41hG

Cylinder ghr ρπ 22 ×

ghrW ρπ 22 2=

( )0 , 2 hG ≡

Composite body ghrWW ρπ 221 5=+ ( )0 , xG ≡

hghrhghrxghr ×+⎟⎠⎞

⎜⎝⎛ −×=× ρπρπρπ 222 2

435

452

435 hhhx =+−=⇒

=∴ x4h

∴Distance to the centre of gravity of the composite body from the vertex of the cone =4hh +

=4

5h

∴The centre of mass of the composite body is on its axis of symmetry at a distance h45 from the vertex of

the cone. When the composite body is hang freely in a vertical plane by a light inextensible string, one end of which is fixed to a ceiling and the other end to a point A on the circumference of the circular base of the cone. Letα is the angle made by the axis of symmetry of the composite body with the downward vertical.

4

3tanhr=α

hr12tan =α

α

A

G

W

α4h

r3


When the composite body is in equilibrium, the point G is also in equilibrium under the action of three forces T, W and P, where T is the tension in the string. Applying Lami's theorem to the point G

( ) αα sin90sin90sinWTP =

°=

+°

αα sin1cosWTP ==⇒

αcotWP =⇒ and αcosecWT = The value of the force P = αcotW Tension in the string = αcosecW

17. (a) An urn contains 5 white, 3 black and 7 red similar balls. Three balls are taken from the urn at random

without replacement. Find the probability that

(i) all three balls are black, (ii) none of the three balls is white,

(iii) at least one ball is white, (iv) the balls are of different colours, (v) the three balls are taken in the order black, red then white.

(b) Students in a certain class were given a question paper in Statistics. The marks obtained by these

students are given in the following grouped frequency table. Range of Marks Number of students

00-20 14

20-40 1f

40-60 27

60-80 2f

80-100 15 The frequencies of the marks ranges 20-40 and 60-80 are missing in the table. However, the mode and the median of the grouped frequency distribution are known as 48 and 50 respectively. Calculate the two missing frequencies in the table. Hence, obtain the total number of students who sat for the Statistics paper. Find the mean and the standard deviation of the grouped frequency distribution.

PT

T

G

W

α


Answer (a)

Let W - The event that getting white ball. B - The event that getting black ball. R - The event that getting red ball.

(i) Probability that all three balls are black ( )BBBP=

131

142

153 ××=

=4551

5−W

3−B

7−R

W

W

W

W

W

W

W

W

W

W

W

W

W

B

B

B

B

B

B

B

B

B

B

B

B

B

R

R

R

R

R

R

R

R

R

R

R

R

R

WWW→

WWB→WWR→

WBW→

WBB→WBR→WRW→

WRB→WRR→BWW→BWB→BWR→BBW→BBB→BBR→

BRW→BRB→BRR→RWW→RWB→RWR→

RBW→

RBB→

RBR→

RRW→

RRB→

RRR→

155

153

157

144

143

147

145

142

147

145

143

146

133

133

137

134

132

137

134

133

136

134

136

137

135

131

137

135

132

136

134

133

136

135

132

136

135

133

135


(ii) Probability that of none of the three balls is white ( ) ( ) ( ) ( )BRRPBRBPBBRPBBBP +++=

( ) ( ) ( ) ( )RRRPRRBPRBRPRBBP ++++

131415673

131415273

131415723

4551

××××+

××××+

××××+=

131415567

131415367

131415637

131415237

××××+

××××+

××××+

××××+

455

352121721771 +++++++=

455120=

=9124

(iii) Probabilty that at least one ball is white = 1 - Probability that none of the three balls is white

91241−=

=9167

(iv) Probability that the balls are of different colours ( ) ( ) ( ) ( )BRWPBWRPWBRPWRBP +++=

( ) ( )RWBPRBWP ++

131415753

131415735

131415375

××××+

××××+

××××=

131415357

131415537

131415573

××××+

××××+

××××+

6131415

715 ×××

×=

132

6×

=

=133

(v) Probability that the three balls are taken in the order black, red and white ( )BRWP=

=××=135

147

153

261


(b) Range of Marks Number of students Cumulative frequency

00 - 20 14 14

20 - 40 1f 114 f+

40 - 60 27 141 f+

60 - 80 2f 2141 ff ++

80 - 100 15 2156 ff ++ Since the mode is 48, the model class is 40 - 60.

Mode ⎟⎟⎠

⎞⎜⎜⎝

⎛Δ+Δ

Δ+=

21

1CL , in the usual meaning.

Mode = 48, 40=L , 20=C , 11 27 f−=Δ , 22 27 f−=Δ

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

−+=

21

1

272727204048

fff

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=⇒

21

1

5427208

fff

( ) ( )121 275542 fff −=−−⇒

121 513522108 fff −=−−⇒ ( )12723 21 −−−−−−=−⇒ ff

Since the median is 50, the class which contains the median is 40 - 60.

Median ⎟⎠⎞

⎜⎝⎛ −+= −12 m

m

m FNf

CL , in the usual meaning.

Median = 50 , 40=L , 20=mC , 27=mf , 2156 ffN ++= , 11 14 fFm +=−

( )⎟⎠⎞

⎜⎝⎛ +−

+++= 1

21 142

5627204050 fff

⎟⎠⎞

⎜⎝⎛ −−++

=⇒2

22856272010 121 fff

122827 ff −+=⇒ ( )2121 −−−−−−=−⇒ ff

( ) ( );221 ×− 2523 11 =− ff 251 =⇒ f and 242 =f The value of =1f 25

The value of =2f 24 Total number of students who sat for the Statistics paper 242556 ++= = 105


Interval ix - Mid value if 20

50−= i

ix

u 2iu 2

iiuf iiuf

00 - 20 10 14 -2 4 56 -28

20 - 40 30 25 -1 1 25 -25

40 - 60 50 27 0 0 0 0

60 - 80 70 24 1 1 24 24

80 - 100 90 15 2 4 60 30

1055

1

=∑=i

if , 15

1

=∑=i

iiuf , 1655

1

2 =∑=i

iiuf

uCAx += , where 50=A , 20=C and ∑

∑

=

== 5

1

5

1

ii

iii

f

ufu

⎟⎠⎞

⎜⎝⎛+=⇒105

12050x

19.50=⇒ x ∴Mean of the grouped frequency distribution = 19.50

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−==∑

∑

=

= 25

1

5

1

2

2222 20 uf

ufC

ii

iii

ux σσ

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=⇒

222

1051

10516520xσ

⎟⎠⎞

⎜⎝⎛≈⇒1051652022

xσ ( 0105

1 2

≈⎟⎠⎞

⎜⎝⎛

Q )

57.120=⇒ xσ

06.25=⇒ xσ ∴Standard deviation of the grouped frequency distribution = 06.25

☺☺☺☺☺☺☺


2012 AL- Combined Maths II-Model Answers New

Documents

Transcript of 2012 AL- Combined Maths II-Model Answers New