COMBINED MATHEMATICS

2008 A/l Applied Maths

ksfrdaIa pdñkao
B.Sc(Engineering) University of Moratuwa
 Paper II

( 01 )
Solution

(a) Velocity - Time Graph

Velocity
~

o m
.c
da
5
5

in
5
V
am
d V
U 2d 3d
ch

t1 t2 t3 t
sh

...................................................................................................................................... 15
iro

t1
Displacement at uniform retardation is =2 ( V + U ) = d
.n

Displacement at uniform velocity U is = t2U = 2d

w

t3
Displacement at uniform acceleration =2 ( V + U ) = 3d 5
w
w

t1 + t2 + t3 = 2d 2d 6d
V+U + U + V+U 5
Total time taken 2d 2d ( 5U + V ) hours
=U ( U+ V ) { 4U + ( U+V) } = U ( U+ V ) 5

6d hours
Time taken if no track repair is done = V

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


01
Time lost due to track - repair
2d ( 5U + V ) 6d 5
= U(U+V) - V

2d { 5UV + V2 - 3U2 - 3UV }

= UV ( U + V )

2d { V2 + 2UV + 3U2 } 5
= UV ( U + V )

2d ( V - U ) ( V + 3U ) hours
= UV ( U + V )

m
...................................................................................................................................... 25

o
1(b)

.c


da
A  a
( B

in
am
ch
sh

D C
iro

Velocity of H , W = v
W,E =w
.n

5
Vector addition Vel . of H , E = v + w
w

Velocity triangles
w
w

P R

( ( S
 ) 

w v
v

Q
10

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


02
For the flight from A to B , in the direction AB , velocity of H , E = PR

For the flight from C to D , in the direction CD , velocity of H , E = SR

{
PR = v2 - w2 sin2  + w cos  5
From the two velocity - triangles ,
SR = v2 - w2 sin2  - w cos  5

a
Time from A to B = PR

m
a
Time from C to D = SR

o
.c
Their sum T1 = a ( SR + PR )
( PR )( SR )

da
2a . v2 - w2 sin2  5

in
= hours
v -w
2 2
am
...................................................................................................................................... 30

Sum of the time from B to C and from D to A

ch

5
( Replacing  by -  ) = 2a v - w cos  = T
2 2 2
sh

2 v2 - w2
2
iro

5
Total time T = T1 + T2 = A ( v2 - w2 sin2  + v2 - w2 cos2  ) 5
where A is a constant = 2a / ( v - w ).
.n

2 2

....................................................................................................................................... 15
w

dT
[
- w2 sin  cos  w2 sin  cos 
]
w

d = A +
v2 - w2 sin2  v2 -w2 cos2 
w

=
Aw2 sin  cos 
v2 - w2 sin2  v2 - w2 cos2 
[ v2 - w2 sin2 - v2 - w2 cos2  ]
5
When 0 < <  , dT > 0
4 d 5

 4 < <  2 , dT < 0

d
ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao

03
 T is a maxiumum when  = 4
5
....................................................................................................................................... 15
( 02 )
Solution

(a)
V
m
F
f M
05 marks for the FIGURE

o m
Acceleration of wedge M , T = F , acceleration of m , M = f

.c
Vector Addition => acceleration of m , T = ( F - f cos ) + f sin  5

da
For the system , m ( F - f cos ) - MF = 0 10

in
( M + m ) F = mf cos 
am
F m cos 
f = M + m = const ............... ( i ) 5
ch

....................................................................................................................................... 25
sh

For motion of the partcle , along the inclined face

mg sin  = m ( f - F cos ) 10
iro

g sin  = f - mf cos  , by making use of ( i ),

2
.n

M+m
w

f ( M + m sin2 )
=M+ m
w

( M + m ) g sin 
= M + m sin2 
w

5
5
This is the cons tan t retardation of m , relative to the wedge M
Using s = ut - 1 f t2 , for motion of the particle relative to wedge ,
2

time taken to come back to the ponit of projection is = 2V = 2V ( M + m sin )

2

f ( M + m ) g sin 

............................................................................................................... 5 ................ 25 ..

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


04
 02'(b).
B C
u A
 a b c

v 
1st collision  
y
2nd collision x 


1st collision 
5
Conservation of momentum :   av + bw = au
Newton` s law of restitution : w - v = eu

m
5
( a - be ) u ............( i ) 5

o
v= a+b

.c
( l + e ) au 5
w= ............ ( ii )

da
a+b

Similarly ,for second collision , replacing ( a , b ) by ( b , c ) , ( v , w ) by ( x , y ) and u by w ,

in
am
( b - ce ) w
x= .............. ( iii ),
b+c 5
ch

( l + e ) bw ................ ( iv ) 5
y= b+c
sh

abu ( l + e )2 ( l + e )2 u 5
from ( ii ) , ( iv ) , y = ( a + b )( b + c ) =
( )( )
iro

1+b 1+c
a b
.n

.......................................................................................................................................
35
a b
w

Given v = 0 , x = 0  = e =
b c 5
w

a : b : c : = e2 : e : 1
w

1
Initial KE T0 = au2
2
1
Final KE ( retained in the system ) T = c y2
2 5
T c y
( )
2
 T =a u
0

T c ( l + e )4 1 5
T0 = a = e4 = e2
( )( )
1+12 1+1 2
e e
e2
.............................. ..........................................................................................................15
ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao

05
( 03 )
Solution v

(i) R

 mg

m
5

o
For the figure

.c
u

da
Let m be the mass of the particle.

in
Conservation of energy :
am
1 mv2 + mga cos  = 1 mu2 - mga . 15
2 2
ch

v2 = u2 - 2ga ( 1 + cos  ). 5
sh

 v= u2 - 2ga (1 + cos  ). 5
iro

....................................................................................................................................... 30
For particle P
.n

R + mg cos  = mv2 10
w

a
w

u2 5
R - mg cos  + m [a 2g ( 1 + cos  )]
w

5
= m [ ua - g ( 2 + 3 cos  )].
2

....................................................................................................................................... 20

3 ( ii ) The particle will leave the bowl at the edge , provided that R > 0 , when
1
()
= cos-1 4 =  , an acute angle.
5

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


06
 i . e . provided that u2 - ga 2 + 3
( 4 ) >0
5
i . e . provided that u > 1 1 ga . 2

4
....................................................................................................................................... 10

3 ( iii ) When leaving the bowl at the edge where = , the speed V of the particle is given by

V2 = u2 - 2ga 1 + ( 1
4 ) 5
= u2 - 5ga .

m
2
In free motion under gravity

o
x - t . V cos  5

.c
0 = t . V sin  - 1 gt2
2

da
5
2V sin 

in
t= g
am
2V 2
5
x = g sin  cos 
ch

Particle will not fall back in to the bowl , provided that x > 2a sin  . 5
sh

i . e . provided that 2V sin  cos  > 2a sin 

2
5
g
iro

ga
i . e . V2 > cos  5
.n

u2 - 5 ga > 4 ga
w

2
13 ga . 5
i . e . u2 >
w

2
....................................................................................................................................... 40
w

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


07
 ( 04 )
Solution

l l

4a x T
T0

m
Q
mg

o
4mg

.c
da
T0 = 4 mg
}  = mg

in
=  4a
l l a
am
5 5
At time t
ch

x mg
( )
T= l = a x, x>0.
5
sh

..
mg - T = mx
iro

5
mg -
mg ..
a x = mx
.n
w

d2 x g 5
dt 2 + a ( x - a ) = 0 , x > 0 .
w

......................................................................................................................................... 25
w

Given Solution

g
x = a + b sin  t + c cos  t ...............( i ) , 2 = a .
dx 5
dt = b  cos t - c sin  t

When t = 0 , x = 4a 5

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


08
 4a = a + c
5
c = 3a

dx = 0 5
When t = 0 ,
dt
5
0=b b=0.
....................................................................................................................................... 25
When the string becomes slack x = 0 ,
 0 = a + 3a cos  t 5

 cos  t = - 1

m
5
3

o
.c
Moving under SHM , time t1 at which the string becomes slack is given by

da
1
t1 =  (  -  ), where a is the acute angle cos-1 3 . () 1 5

in
......................................................................................................................................... 15
am
Then , velocity of P , just before the string becomes slack is

dx 1-1 =2 2
ch

dt = - c sin t 1
, where sin  t 1
= 9 3
sh

5
dx = - 3a . ( g / a ) . 2 2
3 = -2 2ga = -V
iro

dt

5 5
.n

........................................................................................................................................ 15
w

In free motion under gravity ,maximum height reached is

w

V2 8ga
h = 2g = 2g = 4a . 5
w

Total maximum height , from the initial position

= 4a + 4a = 8a
5
....................................................................................................................................... 10
V
In free motion under gravity , time t2 = g 5
2 2ga a
= g = 2 2 g

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


09
 Total time to reach the highest position
= t1 + t2

1 1 5
=  (  -  ) +  2 2
........................................................................................................................................ 10

Alternative Method : Last Part of Q . 4

With X = x - a , the equation of motion reads
X - 2 X = 0
This equation has a solution in the form X2 = 2 ( A2 - X2 ).

m
When t = 0 , X = 3a , X = 0
5

o
=> A = 3a

.c
( Amplitude of SHM ) . 5

da
When
. the string becomes slack x = 0 , X = -a and
X2 = 2 { 9a2 - ( -a )2 } 5
. .
in
am
X = x = -a ( 2 2 ) = -V 5
......................................................................................................................................... 20
ch

In free motion under gravity , maximum heighr reached is

sh

V2 8ga 5
h = 2g = 2g = 4a . In
iro

Total maximum height , from the initial position

= 4a + 4a = 8a 5
.n

....................................................................................................................................... 10
w

Time taken by the particle in SHM ( using the circular are in the figure below )
w
w

1 5
T1 =  (  - )

FIGURE : ( 05 )

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


10
 V
In free motion under gravity , time t2 = g 5

2 2ga a
= g = 2 2 g

Total time to reach the highest position = t1 + t2

1 1 5
 (  -) + ( 2 2 )
......................................................................................................................................... 20
( 05 )

m
Solution
F

o
(a)
R

.c
m

da
w

in
am
ch

( 05 ) for the FIGURE .

sh

Resolving horizontally
iro

5
F cos  = R sin  ................( i )
.n

Also , F = R ......................( ii ) 5
w

Dividing ( ii ) by ( i ) , = tan  5

w

Resolving along AO : R = ( W + w ) cos  .............. ( iii ) 5

w

Taking moments about O Fa = wa ...........................( iv ) 5

w
Dividing ( iv ) by ( iii )  =W + w sec  5

(W+w
w ) 2
=
1 + 2
2

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


11
 ( W+w
) 1
2

w -1 = 2

 = w = w
5
( W + w )2 - w2 W ( W + 2w )
....................................................................................................................................... 40
5 ( b)


a -b

A b B

m
R x x R

o
.c
da
2 R cos  = 2W ; R = W sec 

in
am
X = R sin 
= W tan 
b
ch

= W
( a - b )2 - b2
sh

Wb
=
iro

a(a-2b)
........................................................................................................................................
20
.n
w

C
w

S
w

W
S
X X
R1 5
W
For the FIGURE

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


12
 2S 3 =W 5
2

2R1 cos  = 3W 5

tan  = b
a ( a - 2b )

S 5
X + 2 = R1 sin 

m
X = 3W tan  - W >0 5
2 2 3

o
.c
tan  > 1
3 3

da
b2 1 5

in
a - 2ab > 27
2
am
a2 - 2ab < 27b2
5
( a - b )2 < 28b2
ch

a - b < 2 7b
sh

5
a<b(1+2 7 )
........................................................................................................................................ 40
iro

( 06 )
Solution
.n

A B
(a) O

w

MC
w
w

C
D X
X
Y Y
Figure : ( 05 ) Marks

Let ( X , Y ) be the components of the reaction at C parallel and perpendicular to the string ,
respectively .

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


13
Taking moments about B for BC :
10
X . 2a sin  + Y . 2a cos = T . a sin  ............ ( 1 )
Taking moments about D for DC :
X . 2a sin  - Y . 2a cos = T .a sin  ............ ( 2) 10

( 2 ) - ( 1 ) => Y = 0 . The reaction at C is parallel to the string, 5

1
and X = 2 T 5
Taking moments about A for AD , DC :
M + X . 4a sin  = T . 3a sin  10
M
Eliminating X between the last two equations , we get M = T . ( 3 -2 ) a sin  => T =a sin  .

m
5
...................................................................................................................................... 50

o
6(b) c

.c
da
W
b

in
d
B
am
30 0
30 0
ch
sh

d
c 30 0
iro

30 0 D E C a
a A b
Stress diagram : ( 10 ) marks
[ ( 05 ) for each triangle of forces]
.n
w

From  abc , ab bc ca
0 = 0 =
sin 30 sin 45 sin 1050
w

=> bc = W ; Tension in AD . 5  , 5

w

Also => ca = W sin 750 ; Thrust in AB . 5  , 5

From  bdc , cb = bd = dc .
sin 1050 sin 600 sin 150

W 3 5  , 5
=> bd = ; Tension in DE .
2 2 sin 750

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


14
 Also => dc = W tan 150 ; Tension in DB .
5  , 5
2
....................................................................................................................................... 50
( 07 ) D
Solution

a
A

  
 O1 X

m
B

o
.c
da
Figure : ( 05 ) Marks C

in
Height of the frustum h = a ( l -  ) cos , where a is the semi - vertical angle .
am
= > a cot  = h 5
l-
a cot 
ch

Mass of the frustum M =  ( x tan  )2 dx 5

a cot 
sh

=  tan2 1 ( l - 3) a3 cot3 

3
iro

= 1
  ( l - 3) a3 cot  5
3
.n

=  1 ( l - 3) a2h / ( l - )

3
w

=  h ( l +  + 2) a2
w

3
20
.........................................................................................................................................
w

_
The_distance x of the centre of mass , G , from the vertex O of the cone is given by
M x = x .  ( x tan  )2 dx
5
=  tan2 1 ( l - 4) a4 cot4 
4
5
=  1 ( l - 4) a4 cot2 
4
_ 3 ( l - 4)
Hence x  . a cot 
4 ( l - 3) 5
ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao

15
 _
Distance of G from the centre O1 of the smaller face , O1G = x - a cot 

h
{
3 ( l - 4 )
= ( l -  ) 4 ( l - 3 ) -  } 5

h . 3 ( l - 4 ) - 4( l - 3 )
= 4 (l-)

h . 3 - 4 4
= 4 ( l -  )2 ( l +  + 2)

m
h . 3 + 2 2 5
=

o
4 l +  + 2

.c
........................................................................................................................................ 25
Letting 0+ , we get M = 1 a2 hp,

da
5
3

in
. and x = 3 h . 5
am
4
........................................................................................................................................ 10
ch

The distance O1G1 = x~ of the centre of mass of the body J is given by

sh

1 a2 hp . h ( 3 + 2 2 ) - 1 a22 1 hp . 1 . h

3 4 3 2 4 2 20
iro

x~ =
1 a2 hp . ( 1 +  2 ) - 1 a22 1 hp
3 3 2
.n

h [ 4 ( 3 + 2 2 ) - 2 ]
w

= 48 5
w

1 ( 2 +  2 )
6
w

= h . ( 12 +  2 )
2
5
8 ( 2 +   )
........................................................................................................................................ 30

Suppose G1  V so that ~x = h => 12 +  2 = 8 +  2 => 2 = 4 => = 2 > 1 .
2
Therefore G1 and V cannot coincide . 5

......................................................................................................................................
05
ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao

16
 When suspended from a point D , the axis of symmetry of J is inclined at an angle  to the
vertical DG1 , given by D

O2 C a 5
tan = =
O2 G1 h-~x O2
p
a = 8a . ( 2 + 2  2)
2
= A V

G1
h . ( 4 + 8  )
2
h ( 4 + 8  )
8 ( 2 + 2  )2

5 C

.......................................................................................................................................

m
B
10
( 08 )

o
Solution

.c
da
(a)
( i ) Two events A and B are said to be independent if P ( A  B ) = P ( A ). P ( B )

in
5
( ii ) Two events A and B are said to be mutually exclusive if A  B = 
am
5
( iii Two events A and B are said to be exhaustive if A  B = S
5
ch

......................................................................................................................................... 15
sh

Since the events A  B and A  B1 are mutually exclusive ,

5
iro

P [ ( A  B ) ( A  B1 )] = P [ ( A  B ) + P( A  B1 )]
Also , since ( A  B ) ( A  B1 ) = A ,
P [ ( A  B ) ( A  B1 )] = P ( A ) 5
.n

Hence P [ ( A  B ) + P( A  B1 )] = P ( A ) ..................... ( 1 )

w

....................................................................................................................................... 10
1 1
w

Using ( 1 ) , P ( A  B ) + 2 = 2 => P ( A  B ) = 0 5
w

Also , using ( 1 ) , P ( A  B ) + P( A  B ) = P ( B ) 1

5
1
0 + P ( A 1  B ) = 3
5
1
P ( A  B )
1
= 3
........................................................................................................................................ 15

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


17
P ( A1  B1 ) = P [( A  B )1]
= 1-P(AB)
= 1 - [ P ( A ) +P ( B ) ] 5

(
= 1- 1 + 1 = 1
2 3 ) 6 5
........................................................................................................................................ 10

8 ( b ) The conditional probability of A given B , is defined by

P(AB)
p ( A  B) = . 5
P(B)
Let C be the event : ‘ the student cycles to school ’

m
5
Then C1 is the event : ‘ the student goes to school by bus ’
Let T be the event : ‘ the student arrives at the school early or on time ’

o
Then T 1 is the event : ‘ the student arrives late ’

.c
da
Since P ( T ) = 19 it follows that P ( T 1 ) = 1 - 19 = 9 5
28 28 28

3 3 1
in
am
5
Since P ( T  C1) = , it follows that P ( T 1  C1) = 1 - =
4 4 4
ch

1 1 5
Hence P ( T 1  C ) = 2P ( T 1  C1) = 2 x =
4 2
sh

Now P ( T 1 ) = P ( T 1  C ) + P ( T 1  C1 )
= P ( T 1  C ) .P ( C ) + P( T 1  C1) . P( C1 ) 5
iro

1 1 9
= 2 P ( C ) + 4 { 1 - P ( C )} = 28 5
.n

Hence P ( C ) = 2 and P ( C1 ) = 5 .
7 7
w

5
........................................................................................................................................ 40
w

P ( C1  T 1 ) P( T 1  C1) . P ( C1 )
P ( C1  T 1 ) = = 5
P( T 1 ) P( T 1 )
w

( 1 / 4 )( 5 / 7 ) 5
= 5
( 9 / 28 ) 9

........................................................................................................................................ 10

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


18
Iternative Method for 8 ( b )

L
1\2
C
c
1\2
T L
l-c 1\4

m
B

o
3\4

.c
T

da
Tree Diagram 25

in
( with all probabilities expressed numerically , or in terms of C )
am
( i ) Here C denotes the probability of cycling to school , P ( C )
Then the probability of the student ’ s coming to school on time is
ch

1 3 19 5
sh

2 c + 4 ( l - c ) = 28
iro

c = 2 5
7
.n

........................................................................................................................................ 35
ace P ( B ) = 1 - c = 1 - 2 = 5 .
w

7 7
w

P ( B  L ) P ( L  B ) . P ( B ) ( 1 / 4) ( 5 / 7 ) 5 .
P(BL) =
w

= = =
P(L) P(B) ( 9 / 28 ) 9
5 5
........................................................................................................................................
10
Are the event B = C1 , and the event L = T 1 .

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


19
( 09 )
Solution

5

n
_ 1
(i) Mean x = n xi
i=1

_ 1 ,
n

 
n


n n
Using the given transformation , y1 ( ax1 + b ) a xi + b
i=1 i=1 i=1 i=1
= a n x + bn , 5
( since a , b are constants ).

_

n
_ 1 5
Hence y = n yi = a x + b

m
i=1
........................................................................................................................................ 15

o

n
_
( ii ) Standard deviation Sx = 1

.c
n ( x i
- x )2
i=1
5

da
_ _
 
n n
_
Also , ( yi - y )2 = ( axi + b - ax - b )2 = a2 ( xi - x )2 = a2 . n sx2

in
i=1 i=1
5
am

n
1 _ a2 n sx2
Hence Sy = n ( yi - y )2 = n = a Sx ,
i=1 5
ch

( since Sx > _ 0 ).
....................................................................................................................................... 15
sh

Consider Geography .
Let x1 denote the original mark of ith the candidate , and
5
iro

let y1 = ax1 + b be his / her scaled mark ( i = 1 , 2 , .... n ).

40 = 40 a + b ....................( 1 ) 5
Also , by result ( i ) 50 = a m + b .....................( 2 )
.n

5
and by ( ii ) 15 = 12 a => a = 5
w

4
5
w

(2)-(1) => 10 = a ( m - 40 );
w

10 5
=> m = 40 + a
m = 48 or 32 5
......................................................................................................................................... 30
Next consider History .
Let x1j denote the original mark of the jth candidate , and let yj 1 = c xj1 + d be his / her
scaled mark ( j = 1 , 2 , ....... N ).

Then 56 = 61 c + d ....................( 3 ) 5
Also , by ( i ) 50 = 53 c + d ....................( 4 ) 5
ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao

20
 3 5
( 3 ) - ( 4 ) => c=4
by result ( ii ) 15 = c s 5
s = 20 5
........................................................................................................................................ 25
Let N dente the total number of candidates who sat for History.

The sum of marks of all candidates before re - scrutiny = 53 N 5

The number of candidates whose marks were changed = N

1000

m
The sum of marks of all candidates after re - scrutiny = 53 N + N ( 68 - 65 )
1000

o
5

.c
53003 N
= 1000

da
in
The mean mark of all candidates after re - scrutiny = 53003 N
1000 N
am
= 53 . 003
5
ch

....................................................................................................................................... 15
sh
iro
.n
w
w
w

ksh; jYfhkau Tnj úIfhys olaIhl= lrkafkuq' ksfrdaIa pdñkao


21