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- 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
- 2 - Done By : Chandima Peiris (B.Sc - Special)
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=
- 3 - Done By : Chandima Peiris (B.Sc - Special)
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
- 4 - Done By : Chandima Peiris (B.Sc - Special)
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
- 5 - Done By : Chandima Peiris (B.Sc - Special)
(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
- 6 - Done By : Chandima Peiris (B.Sc - Special)
(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
- 7 - Done By : Chandima Peiris (B.Sc - Special)
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
- 8 - Done By : Chandima Peiris (B.Sc - Special)
⎟⎟⎠
⎞⎜⎜⎝
⎛ −=⇒
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&&
- 9 - Done By : Chandima Peiris (B.Sc - Special)
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.
- 10 - Done By : Chandima Peiris (B.Sc - Special)
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.
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
- 11 - Done By : Chandima Peiris (B.Sc - Special)
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
- 12 - Done By : Chandima Peiris (B.Sc - Special)
( )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 +−=⇒
- 13 - Done By : Chandima Peiris (B.Sc - Special)
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
- 14 - Done By : Chandima Peiris (B.Sc - Special)
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 ++−−=
- 15 - Done By : Chandima Peiris (B.Sc - Special)
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
- 16 - Done By : Chandima Peiris (B.Sc - Special)
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
- 17 - Done By : Chandima Peiris (B.Sc - Special)
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
- 18 - Done By : Chandima Peiris (B.Sc - Special)
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
- 19 - Done By : Chandima Peiris (B.Sc - Special)
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
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
- 20 - Done By : Chandima Peiris (B.Sc - Special)
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
- 21 - Done By : Chandima Peiris (B.Sc - Special)
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
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
- 22 - Done By : Chandima Peiris (B.Sc - Special)
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
- 23 - Done By : Chandima Peiris (B.Sc - Special)
∫
∫= 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
- 24 - Done By : Chandima Peiris (B.Sc - Special)
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
- 25 - Done By : Chandima Peiris (B.Sc - Special)
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
- 26 - Done By : Chandima Peiris (B.Sc - Special)
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
- 27 - Done By : Chandima Peiris (B.Sc - Special)
(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
- 28 - Done By : Chandima Peiris (B.Sc - Special)
(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
- 29 - Done By : Chandima Peiris (B.Sc - Special)
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
☺☺☺☺☺☺☺
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
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