Vectors
March 22, 2017 | Author: Sesha Sai Kumar | Category: N/A
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ELEMENTS OF VECTORS
PHYSICS - I A
VECTORS
3
CHAPTER
Leonhard Euler (15 April 1707 – 18 September 1783) was a pioneering Swiss mathematician and physicist. He made important discoveries in fields as diverse as infinitesimal calculus and graph theory. He also introduced much of the modern mathematical terminology and notation, particularly for mathematical analysis, such as the notion of a mathematical function. He is also renowned for his work in mechanics, fluid dynamics, optics, and astronomy. Euler was featured on the sixth series of the Swiss 10-franc banknote and on numerous Swiss, German, and Russian postage stamps. The asteroid 2002 Euler was named in his honor. He is also commemorated by the L eonhar d Euler Lutheran Church on their Calendar of Saints. 1707-1783
h
To differ entiate between scalar and vector quantites with examples.
h
To find the resultant of vectors, and resolve a vector into its rectangular components
3.1 CL ASSI FI CATI ON OF PHYSI CAL QUANTI TI ES All measurable quantities are called physical quantities. Most of the physical Quantities are classified into 'Scalars' and 'Vectors'. Scal ar: - Physi cal quant i t i es havi ng onl y magnitude ar e called Scalar s. Exampl es : Length, t ime, volume, density, temperature, mass, work, energy, electric charge, electric current, potential ,resistance, capacity, etc..... To describe a scalar quantity we require a)
The specific unit of that quantity
b)
The number of times that unit is contained in that quantity
Ex: A bag contains 100 kg of sugar. Here kg is the unit and 100 is the number of units of sugar present in the bag. Note3.1: Unit is not a compulsion to represent a scalar Ex : - Specific gravity, Refractive index Note3.2: Mathematical operations of scalar quantities yield scalar quantities and these quantities are manipulated by ordinary algebraic rules. Vector : - Physical quantities having both magnitude and dir ection and that obeys laws of vector addition ar e called vector s (or ) AKASH MULTIMEDIA
h
To explain the Relative Motion.
h
To compute the product of vectors.
The vector, as a mathematical obj ect, is defined as a dir ected line segment. I t should also obey the laws of vector addition. Examples : – Displacement, velocity, acceleration, force, momentum, angular momentum , moment of force, Torque, magnetic moment, magnetic induction field, Intensity of electric field, etc .... To describe a vector quantity we require. a)
The specific unit of that quantity.
b)
The number of times that unit is contained in that quantity.
c)
The orientation of that quantity.
Ex : - A plane is flying from west to east with a velocity of 50 ms–1. Here ms–1 is the unit, 50 is the number of units of velocity and west to east is the direction. Note 3.3: 1) A physical quantity having magnitude and direction but not obeying laws of vector addition is treated as a scalar. Ex : Electric current is a scalar quantity Electric current is always associated with direction, but it is not a vector quantity. It does not obey law of vector addition for its addition. i1 (i1 + i2)
q i2
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PHYSICS - I A
ELEMENTS OF VECTORS
The resultant of i1 and i2 is (i1 + i2) by Kirchoff’s current law. The resultant does not depend on angle between currents i1 and i2.
b) Axial Vector s : The vectors whose direction changes with the co-ordinate system in which it is defined changes is called axial vector.
Note3.4: Equations in vector form indicate both mathematical and geometrical relationships among the quantities. Physical laws in vector form are very compact and independent of choice of coordinate system.
Ex:Angular velocity, torque, angular momentum
3.2 GEOM ETRI CAL REPRESENTATI ON OF VECTORS Any vector A can be represented geometrically as a directed line segment shown in (an arrow), as fig. The magnitude of A is denoted by A , and the direction of A is specified by the sence of the arrow and the angle ‘ ’ that it makes with a fixed reference line.
c) L ike vector s (or ) par allel vector s : Two or more vectors (representing same physical quantity) are called like vectors if they are parallel to each other, however their magnitudes may be different. A
C
B
d) Unlike vectors (or) anti parallel vectors : Two vectors (representing same physical quantity) are called unlike vectors if they act in opposite direction however their magitudes can be different.
A
A
B
q
When using graphical methods, the length of the arrow is proportional to the magnitude of the vector, and the arrow head represents the direction. Ex:- The velocity vector v is represented by an arrow OP as shown in figure. The initial point of the vector is O, the final point of the vector is P. The length OP is the magnitude of the velocity and its direction is 600 north of east (or) 300 east of north . N
600 North of East or 300 Eash of North
P
N 30 0
NW
NE
v
O
A
B C
Ex : Suppose two trains are running on parallel tracks with same speed and direction. Then their velocity vectors are equal vectors. Note : If a vector is displaced parallel to itself its magnitude and direction does not change. C
450
q 60 0
W
e) Equal vect or s : Two or more vecto rs (representing same physical quantity) are called equal if their magnitudes and directions are same.
E
W
S
E SW
SE S
3.3 TYPES OF VECTORS a) Polar Vector s : The vector whose direction does not change even though the co-ordinate system in which it is defined changes is called polar vector. Ex : Force, momentum, Acceleration. AKASH MULTIMEDIA
A
Y X
B
A B C
f) Negative Vector : A vector having the same magnitude and opposite in direction to that of a given vector is called negative vector of the given vector A
A
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ELEMENTS OF VECTORS
PHYSICS - I A
g) Co-initial vector s : The vectors having same initial point are called co-initial vectors. B
A
C
O
E
D
h) Collinear vector s : Two or more vectors are said to be collinear when they act along the same line however their magnitudes may be different. Ex : Two vectors A and B as shown are collinear vectors. A B (or)
A
B
i) Coplanar vector s : A number of vectors are said to be coplanar if they are in the same plane or parallel to the same plane. However their magnitudes may be different. Y
A C B
O
X
j) Unit vector : A vector whosemagnitudeequals one and used to specify a convenient direction is called a unit vector. A unit vector has no units and dimensions. Its purpose is to specify the direction of given vector. In cartesian coordinate system, unit vectors along positive x, y and z axis are symbolised as i , j and k respectively. These three unit vectors are mutually perpendicular and their magnitudes ˆi ˆj kˆ 1 . If A is a non zero vector, then the unit vector in A ˆ the direction of A is given by A A
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3.4 REPRESENTATI ON OF ANGL E BETWEEN THE TWO VECTORS The angle between two vectors is represented by the smaller of the two angles between the vectors when they are placed tail to tail by displacing either of the vectors parallel to it self. Ex: The angle between A and B is correctly represented in the following figures.
B
B q
(or)
B
q
B
q
A
A
A
Fig.(ii)
Fig.(i)
a) If the angle between A and B is , then the angle between A and KB is also . Where ‘K’ is a positive constant. KB B q q A A b) If the anglebetween A and B is , then the angle between A and KB is (180 – ) . Where K is a positive constant. B q B q KB 0 A 180 q A
c) Angle between collinear vectors is always zero or 1800C Q
Q
q 00
P
or
q 1800
P
Problem : 3.1
Three vectors A, B, C are shown in the figure. Find angle between (i) A and B (ii) B and C (iii) A and C. A
300
x
300
B
x
450
x
C
Sol. To find the angle between two vectors we connect the tails of the two vectors. We can shift the vectors parallel to themselves such that tails of A,B and C are connected as shown in figure.
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ELEMENTS OF VECTORS y
2. A 300
x
300 450
B
C
Now we observe that angle between and B is 60, B A and C is 15 and between A and C is 75 Problem : 3.2 I f A, B, C represents the three sides of an equilateral triangle taken the angle in the same order then find between i) A and B ii) B and C iii) A and C . B
When two vector s are acting at some angle: First join the initial point of Q with the final point of P and then, to find the resultant of these two, draw a vector R from the initial pointof P to the final point of Q . This single vector R is the resultant of vectors P and Q . R
Q
P
Q
P
of and both in R represents the resultant P Q magnitude and direction. So, R P Q.
3.6 L AWS OF VECTOR ADDI TI ON a) Vector addition obeys commutative law.
1200
C
Addition of two vectors is independent of the orderof the vectors in which they areadded. If P and Q are two vectors then P Q Q P as shown in fig.
B
C
0
120
1200
A
A
Sol. From the diagram the angle between the vectors A and B is 1200 , the angle between and B C is 1200, the angle between A and C is 1200
Q
3. Now, draw an arrow from the initial point of the first vector to the final point of the second vector. This arrow represents the resultant of the two vectors. 1.
When two vector s ar e acting in the same direction. Let the two vectors P and Q be acting in the same direction. Q
P
Q
R PQ
(a)
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(b)
R PQ (c)
Q
A
b) Vector addition obeys associative law. While adding more than two vectors, the resultant is independent of the order in which they are added. If P,Q and R are three vectors, then P Q R P Q R as shown in fig.
P
(P Q) R
PQ
c)
Q R
Q
Examples :
P
P
O
2. Join the initial point of the second vector with the final point of the first vector by moving parallel to itself.
C
Q P R
P Q R
3.5 ADDI TI ON OF VECTORS (GRAPHI CAL M ETHOD) 1. To add two vectors geometrically represent the vectors by arrow head lines using the same suitable scale, with their proper directions in the choosen coordinate system.
P
B
R
P P Q R
Q
R
Vector addition obeys distr ibutive law.
If k, k1, k2 are scalars then k P Q kP kQ and P k1 k 2 k1 P k 2 P
Note 3.5 : Vector addition is possible only between vectors of same kind. 68
ELEMENTS OF VECTORS
PHYSICS - I A
3.7 PARAL L EL OGRAM L AW OF VECTORS Statement : I f two vector s ar e dr awn fr om a point so as to r epr esent the adj acent sides of a par allelogr am both in magnitude and the dir ection, the diagonal of the par allelogr am dr awn fr om the same point r epr esents the r esultant of the two vectors both in magnitude and dir ection. B
B
P
q
N
A
Let the two vectors P and Q , inclined at angle be acting on a particle at the same time. Let they be represented in magnitude and direction by two adjacent sides OA and OB of parallelogram OACB, drawn from a point O.
According to parallelogram law of vectors, their resultant vector R will be represented by the diagonal of the parallelogram. OC M agnitude of r esultant : The line OA is extended upto point N and draw a perpendicular from point C on to the extended line as shown in fig. From the parallelogram OACB.
R
a
q
P
O
a
C
Q
q
R
O
Dir ection of r esultant :
C
Q
q
The magnitude of the resultant R depends on the magnitudes of the vectors P and Q and also on the angle between them.
A
N
If the line of action of the vector P is taken as reference line, the resultant R makes an angle with it. This angle indicates the direction of R . Then from right angled triangle ONC, CN CN Q sin .....(4) ON OA AN P Qcos Note3.6: If is the angle between the resultant R
tan
P sin and the vector Q, then tan 1 , Q P cos Special Cases:
a) If the magnitude of P > Q then . i.e., the resultant is closer to the vector of larger magnitude. b) When angle between two vectors increases, the magnitude of their resultant decreases. c) W hen t wo vect or s ar e i n t he same direction(par allel).
BOA = CAN = and
Q
P
OA = P, OB = AC = Q, OC = R and
R PQ
In CNA, AN = AC cos = Q cos CN = AC sin = Q sin
then = 00, cos 00 = 1 and sin 00 = 0
From triangle ONC OC2
=
ON2
+
from eq.(3), R = P 2 Q 2 2PQ(1)
NC2
=
(OC)2 = (OA+AN)2 + (NC)2 R2 = (OA)2 + 2(OA)(AN) + (AN)2 +(NC)2 2
R2 = P2 + 2 PQ cos + QCos QSin R 2 P 2 2PQCos Q 2 R=
P 2 Q 2 2PQ cos
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........ (3)
2
(P Q)2 = (P + Q)
Q 0 = 0 or = 00. P Q(1) Thus for two vectors acting in the same direction, the magnitude of the resultant vector is equal to the sum of the magnitudes oftwo vectors and acts along the direction of P and Q . from eq.(4), tan =
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ELEMENTS OF VECTORS
(d) When two vector s ar e acting in opposite directions (antiparallel).
P + Q cos = 0
P
Q
R P Q
From eq. (3), R = P 2 Q 2 2PQ(1) (P Q)2 = (P–Q) or (Q–P) Q0 From eq. (4), tan = = 0 P Q(1) or = 00 or 1800.
=
Thus for two vectors acting in opposite directions, the magnitude of the resultant vector is equal to the difference of the magnitudes of the two vectors and its direction is along the vector of larger magnitude. (e) When two vectors are per pendicular to each other, = 900, sin = 1 R
R=
Q
P 2 Q 2 2PQ(0) =
a
ii) The resultant vector is perpendicular to the vector having smaller magnitude. Q sin P Q cos
Q sin tan 90 0 P Q cos
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R
R P 2 Q 2 2PQ cos
R P2 P2 2P2 cos q
[P=Q]
2 R 2P2 2P2 cosq , R 2P 1 c os
R 2P 2 2 cos2 , 2
2 and the resultant makes an angle ‘ ’ with P is Q sin P s in tan P Q cos P P cos tan
P2 Q2
i) The angle between the two vectors is greater than 900
f
h) If magnitudes of P and Q are equal and the angle between them is then their resultant is
P
f) If the resultant of two vectors is perpendicular to any one of the vectors, then
Q
Q2 P2 , q P R P P sin ,cos , tan . Q Q R The angle between P and Q is 900 R max P Q P R max R min g) R P Q and Q R R . min max min
R 2P cos
P sin s in P 1 cos 2 cos2 / 2
2 sin cos 2 2 tan tan 2 2 cos / 2 2
Q (1) Q From eq(4), tan = = P Q(0) P Q or = tan–1 P
we know tan
P and Q
R=
then = 1800, cos 1800 = –1 and sin 1800 = 0.
and cos = 0 From eq(3),
cos =
P
tan tan , 2
2
a)
If 600 , then R = 3P and 30 0
b)
If 90 0 , then R = 2P and 450
If 120 0 , then R = P and 60 0 The unit vector parallel to the resultant of two PQ vectors P & Q is nˆ . PQ c) I)
Note 3.7 : We can add a vector to another vector of same kind but we can’t add a vector quanity to a scalar quantity. 70
ELEMENTS OF VECTORS
PHYSICS - I A Problem : 3.3
Problem : 3. 7
The resultant of two forces whose magnitudes are in the ratio 3 : 5 is 28 N. I f the angle of their inclination is 600, then find the magnitude of each force. Sol : Let F1 and F2 be the two forces.
Then F1 = 3x; F2 = 5x; R = 28 N and 600 R F12 F22 2F1F2 cos
28
3x 2 5x 2 2 3x (5x)cos600
28 9x 2 25x 2 15x 2 7x
x
Theresultant of two vectors A and B is perpendicular to A and equal to half magnitude of B . Find of the the angle between A and B ? Sol. Since R is perpendicular to A . figure shows the three vectors A , B and R . angle between A and B is – B R R B 1 sin = = = p q q B 2B 2 A A = 30° angle between A and B is 150°.
Problem : 3.8
28 4 , 7
F1 = 3 × 4 = 12 N, F2 = 5 × 4 = 20 N. * Problem : 3.4
I f vectors A and B are 3i – 4j + 5k and 2i + 3j – 4k respectively then find the unit vector parallel to A B A B 5i j k nˆ (Ans : ) Hint : A B 27
What is the displacement of the point of a wheel initially in contact with the ground when the wheel rolls forward half a revolution? Take the radius of the wheel as R and the x-axis as the forward direction? Sol: From figure, during half revolution of the wheel, the point A covers AC R horizontal distance, and BC = 2R vertical distance.
B
* Problem : 3.5
y
I f magnitude of the resultant of two vectors equal of magnitude, is equal to the magnitude of either of the vectors, what is the angle between them ? (Ans : 1200) Hint : R 2P cos
B 7 2 24 2 25
A 3i 4j 1 3i 4j and A A 32 4 2 5
1 C 25 3i 4j 15i 20 j 5
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A
x
C
x = R and y = 2R; Displacement, s
2
Problem : 3.6 I f A 3 i 4j and B 7i 24j , find a vector having the same magnitude as B and par allel to A Sol. The vector parallel to A and having magnitude of B is A ^ C B B A A
q
x2 y 2
( R)2 (2R)2 R 2 4
y 2R 2 and Tan 1 Tan 1 Tan 1 with x R x-axis.
i.e., Displacement has magnitude R 2 4 and makes 2 an angle Tan 1 with x-axis.
3.8. TRI ANGL E L AW OF VECTORS " If two vectors are represented in magnitude and direction by the two sides of a triangle taken in one or der, their resultant vector is represented in magnitude and dir ection by the thir d side of the tr iangle taken in r ever " . se or der Let the two vectors P and Q , inclined at an angle be acting on a particle at the same time. Let 71
PHYSICS - I A
ELEMENTS OF VECTORS
they be represented in magnitude and direction by two sides OA and AB of triangle OAB, taken in the same order shown in fig. Then, according to triangle law of vector addition, the resultant R is represented by the third side OB of triangle, taken in opposite order.. B g R Q p q R P Q a q O P A N M agnitude of r esultant R : R P2 Q 2 2PQ cos Dir ection of r esultant R :
is r epr esented by the closing side of the polygon taken in reverse order in magnitudeand direction . D F F E C 4
3
FR
F2
F1
A
B
(or) I f a number of coplanar forces ar e acting simultaneously at a point keep the par ticle in equilibr ium, these forces can be r epr esented as the sides of a polygon taken in or der both in magnitude and dir ection. D F F3
4
E
C
Q sin Tan . P Q cos
F5
Statement of tr iangle law when thr ee for ces keep a par ticle in equilibr ium : When three forces acting at a point can keep a par ticle in equilibr ium the thr ee for ces can be r epr esented as the sides of a tr iangle taken in or der both in magnitude and dir ection.
F2
A B F1 F1 F 2 F 3 F 4 F5 0
Note 3. 8 : I f x is the side of a hexagon ABCDEF as shown in figur e E
D
Q F2
O F3
P
have equal ratio i.e., F1 F2 F3 . PQ PR QR 3.9 POLYGON L AW OF VECTORS Statement : If a number of vector s are r epresented by the sides of a polygon both in magnitude and direction taken in or der, their resultant
C
3x
x
F2
R Suppose three fo rece F1 , F2 , F3 are simultaneously acting at point O and the point is in equilibrium. Then the three forces can be represented as three sides of a triangle. The triangle PQR is constructed by drawing parallel lines to the directions in which the forces are applied. The magnitudes of the forces and the corresponding sides of the triangle
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0
90
2x
F
F3
x
3x
F1
F1
600
A
B
x
AB = x, AC = 3x , AD = 2x, AE = 3x , AF= x Problem : 3.9 ABCDEF is a regular hexagon with point O as centre. Find the value of AB AC AD AE AF .
E
D
O
F
C
B A Sol. From the diagram AB DE BC EF
72
ELEMENTS OF VECTORS
PHYSICS - I A AB AC AD AE AF = AB AB BC AD AD DE AD DE EF 3AD 3 2AO 6 AO
Application 3.1 : L ami's Theor em
3.10 NUL L VECTOR (OR) ZERO VECTOR A vercotr whose magnitude is equal to zero is called a null vector. Its direction is indeterminate. Examples of zer o vector : 1. The velocity of a particle at rest. 2. The acceleration of a particle moving at uniform velocity. 3. The displacement of a stationary object over any arbitrary interval of time. 4. The position vector of a particle at the origin Note 3.9 : A 0 A , A 0 0
" I f thr ee coplanar for ces acting at a point k eeps it in equilibr ium, t hen each f or ce is pr opor tional to the sine of the angle between the other two for ces" If F1 , F2 , F3 are the magnitudes of three forces and , , are the angle between forces F 2 and F 3 , F3 and F1 and F1 and F2 respectively, as shown in fig. Then according to lami’s theorem F1 F F 2 3 sin sin sin
F2
F3
a
g
b F1
Problem 3.10 : I n the given figure the tesion in the string OB is 30N. Find the weight ‘W’and the tension in the string OA. A
3.11 EQUI L I BRI ANT
30 0
A vect or havi ng same magnit ude and opposite in dir ection to that of the r esultant of a number of vector s is called the equilibr iant. (or ) Negative vector of the r esultant of a number of vector s is called the equilibr iant (E). If F1 , F2 & F3 are the three forces acting on a body, then their resultant force is FR F1 F2 F3 E FR , E F1 F2 F3 .
B
W
Sol: Let T1 and T2 be the tensions in the strings OA and OB respectively A T1
30 0 900 O
1500
Note 3.10 : i) Single force cannot keep the particle in equilibrium. ii) Minimum number of equal forces required to keep the particle in equilibrium is two. iii) Minimum number of unequal coplanar forces required to keep the particle in equilibrium is thr ee. iv) Minimum number of equal or unequal non coplanar forces required to keep the particle in equilibrium is four . AKASH MULTIMEDIA
O
900
T2
B
W
According to lami’s theorem T1 sin 90 0
T2 sin150 0
W sin120 0
(T2 = 30N)
on sloving w 30 3 N and T1= 60N
3.12 SUBTRACTI ON OF VECTORS The pr ocess of subtr acting one vector fr om another is equivalent to adding vector ially the negative of the vector to be subtr acted. Let P and Q be the two vectors as shown in figure. We want to find the difference P Q . Let a 73
PHYSICS - I A
ELEMENTS OF VECTORS
vector Q be added to the vector P by the laws of vector addition. Their resultant gives the value of PQ .
Q
P
q
O
A
R
180 q
O
B
C
C
B B
A B C
B C
P 2 Q 2 2PQ cos and
Q sin (1800 -) Q sin tan . 0 P Q cos (180 ) P Q cos Note 3.7 : If P Q then P Q 2P sin . 2 S 2P sin 2 Note 3.8 : When two vector s P and Q have same magnitude and is the angle between them
Resultant Difference 2Pcos / 2 2P sin / 2
00
2P
0
600
3P
P
900
2P
2P
1200
P
3P
1800
0
2P
C A
B PQ
P
B Q
A
A
Applications 3.2 : If v i is initial velocity of a particle and vf is its final velocity. Then change in its velocity is given by v v f v i . v v2i v2f 2v i vf cos Where ' ' is the angle between initial and final velocities. Application 3.3 : When a particle is performing uniform circular motion with a constant speed v, the magnitude of change in velocity when it describes an angle at the centre is V 2vsin . 2 V
q
Q
C Q
O
O
3.13 L AWS OF VECTOR SUBTRACTI ON : a) the vector subtractio n does not follow commutative law i.e. P Q Q P, But P Q Q P
A
R
C
S P 2 Q 2 2PQ cos (1800 )
P
P Q R
Q
A P Q
P
C
The magnitude of P Q is
S=
A
O
C B
P
Q
S PQ
P
B
Q
Q
b) the vector subtractio n does not follow associative law ie. P Q R P Q R
P
QP
q V
Application 3.4 : If velocity of a particle changes from v i to vf in time t then the acceleration vf v i . of the particle is given by a t Problem : 3.11 A particle is moving eastwards with a velocity of 5m/s. I n 10s the velocity changes to 5 m/s northwards. Find the average acceleration in this time
D
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ELEMENTS OF VECTORS
PHYSICS - I A Sol.
v vf v i a av t t v
v f 5m / s
N
5 2m / s
W
E
0
45
vi 5m / s
vi 5m / s
S
5j 5i 5 2 1 a av m/s2 10 10 2
north – west direction. * Problem : 3.12
Two vectors A and B haveprecisely equal magni A B to be larger than tudes. For the magnitude of the magnitude of A B by a factor n, what must be the angle between them ? Sol. A B n A B 2A cos tan
n2A sin 2 2
(A = B)
1 1 1 , tan 1 , 2 tan 1 n 2 n 2 n
3.14RESOL UTI ON OF A VECTOR I NTO RECTANGUL AR COM PONENTS Resolution of a vector is the pr ocess of obtaining the component vector s which when combined, accor ding to laws of vector addition, produce the given vector. j
Y Q
r
y
O
R
q x
P i
from right angled triangle OPR 2 2 OR OP 2 PR2 or r x y .... (6) The equation (6) gives the magnitude of the vector in terms of its rectangular components. Note 3.9 : Method involving resolution of vectors into components to find the resultant of the vectors is known as analytical method.
Note 3.10 : The components of a vector are independent of each other and can be handled separately. X
Consider a vector r in the xy plane and it makes an angle with x axis as shown in figure. i and j are unit vectors along X-axis and Y-axis respectivley. From t he point R, draw RP and RQ perpendicular to X-axis and Y-axis respectively. From the parallelogram law of vector addition, it follows that OR OP OQ ............ (1)
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If OP = x and OQ = y, then OP xi and OQ yj From equation (1) r xi yj ....... (2) In the above equation xi and y j are called the x-component and the y-component of the vector r , and x and y are called magnitudes of the two component vectors. from the triganle OPR. OP x cos x r cos ......... (3) OR r PR y and sin y r sin ......(4) OR r y ............. (5) from (3) and (4) tan x The equation (3) and (4) gives the magnitudes of the rectangular component vectors in terms of the magnitude of the given vector and its inclination with X-axis. from equation (2), r r cos ˆi r sin ˆj
Note 3.11 : Theoretically, a given vector can be made the diagonal of infinite number of parallelograms. Thus there can be infinite number of ways to divide a vector into components. Note 3.12 :
y
B
-x
o
A
D
C
x
-y 75
PHYSICS - I A
ELEMENTS OF VECTORS
a) b)
If the vector A is in first quadrant then it can be written as A A x ˆi A y j If the vector B is in second quadrant then B B ˆi B j x
Application 3.5 :
F
F sin q
y
q
c) If the vector C is in third quadrant then C Cx ˆi Cy j
d) If the vector D is in fourth quardant then D D x ˆi D y j
3.15 ADDITION OF TWO VECTORSIN TERMS OF COMPONENTS Let us consider the addition of two vectors P and Q in terms of their components. We have P P ˆi P ˆj, Q Q ˆi Q ˆj x
y
x
A block is placed on smooth horizontal surface and pulled by a force ‘F’ making an angle ‘ ’ with horizontal. The component of force along horizontal = Fcos . The component of force along vertical = Fsin . Application 3.6 :
y
m
y mg sin q
Q
Qy
P
Ry
x
Qx
Px
Rx
Let R be the resultant vector with component.
Rx and Ry along x and y axis respectively. Then R x R x ˆi and R y R y ˆi. From the diagram Rx = Px + Qx and Ry = Py+Qy R x Px Q x ˆi, R y Py Q y ˆj R R x ˆi R y ˆj; R Px Q x ˆi Py Q y ˆj
R
Px Qx 2 Py Q y
Tan
Ry Rx
, Tan
Py Q y Tan 1 Px Q x AKASH MULTIMEDIA
2
Py Q y Px Q x
mg cos q
q mg
q
R
Py
O
F cos q
A block of mass ‘m’ is placed on inclined plane of angle ‘ ’ then the component of weight parallel to the inclined plane is ‘mg sin ’, the component of weight perpendicular to the inclined plane is mgcos . Application 3.7 :
O
q
2 x 2
x
T q
T cos q
F
T sin q
mg A simple pendulum having a bob of mass ‘m’ is suspended from a rigid support and it is pulled by a horizontal force ‘F’. The string makes an angle q with the vertical as shown in figure. The horizontal component of tension = T sin The vertical component of tension = T cos 76
ELEMENTS OF VECTORS
PHYSICS - I A
When the bob is in equilibrium .............(1) T sin F, .............. (2) T cos mg
T
mg mgl cos l2 x 2
From equations (1) and (2)
Tan
F x F mgTan mg 2 mg l x2
T F 2 mg
2
Note- 3.13 : If a vector is rotated through an angle other than integral multiple of 2 (or 3600) it changes, but its magnitude does not change. B
Problem : 3.14 A weight mg is suspended from the middle of a rope whose ends are at the same level. The rope is no longer horizontal. What is the minimum tension required to completely straighten the rope is Sol. From the diagram q q 2Tsin q 2T sin = mg T T mg T q q 2 sin T cosq T cosq The rope will be staraight when = 0 mg T 2 sin 0 0 mg The tension required to completely straighten the rope is infinity. Problem : 3.15 The sum of magnitudes of two forces acting at a point is 16N. If their resultant is normal to the smaller force and has a magnitude of 8N. Then the forces are
A
q A B
F2
Ve cto
r
Note - 3.14 : If the frame of reference is rotated the vector does not change (though its components may change). S
F
q 900
F1
Sol. Let F be the resultant of two forces F1 and F2 as shown in figure with F2 > F1.
F2 sin = F1 ...................... (i)
O
F2 cos = F = 8 ...................... (ii) Squaring and adding Eqs (i) and (ii), we get F22 = F12 + 64 ...................... (iii)
Problem : 3.13 A vector 3 iˆ j rotates about its tail through an angle 300 in clock wise direction then the new vector is Sol: The magnitude of 3 ˆi j is 3 1 = 2 The angle made by the vector with x - axis is Tan
Ay Ax
y
3i j
1 300
3
o
300
x
When the given vector rotates 300 in clock wise its direction changes along x - axis but its magnitude does not change.
y
The new vector is 2iˆ AKASH MULTIMEDIA
o
2i
x
Given F1 + F2 = 16 ...................... (iv) Solving Eqs. (iii) and (iv), we get F1 = 6N and F2 = 10N
* 3.16 RESOL UTI ON OF A VECTOR I NTOTHREE RECTANGUL AR COM PONENTS : Let us consider a vector A be represented by OP as shown in fig. With O as origin, construct a rectangular parallelopiped with three edges along the three rectangular axes which meet at O. A becomes the diagonal of the parallelopiped. A x , A y and A z are three vector intercepts along x, y and z axes respectively. These are the three rectangular components of A . 77
PHYSICS - I A
ELEMENTS OF VECTORS
Applying triangle law of vectors, OP OK KP .................... (1)
Squaring and adding
Applying parallelogram law of vectors,
A 2x
cos 2 cos 2 cos 2 =
A2
Y S
A 2y A2
A 2z
A2
A 2x A y2 A z2 A2
A2 A2
1
and sin 2 sin 2 sin 2 2
Ay A
P
Ax
O
Q
Az
3.18 POSI TI ON VECTOR Position vector of any point, with r espect to an ar bitr ar ily chosen or igin, is defined as the vector which connects the or igin and the point and is dir ected towar ds the point.
X
K
T Z
OK OT OQ ....................... (2) from (1) and (2) OP OT OQ KP But KP OS, OP OT OQ OS A Az A x A y , or A Ax i Ay j Az k
Again, OP2 = OK2 + KP2 = OQ2+OT2+KP2 or A2 = Ax2 + Az2+Ay2 KP OS A y 2 2 2 or A A x A y A z This gives the magnitude of A in terms of the magnitudes of components A x , A y and A z .
* 3.17 DI RECTI ON COSI NES : The dir ection cosines , m and n of a vector ar e the cosines of the angles , and which a given vector makes wit h x-axis, y-axis and z-axis respectively. Z A kˆ
g
O
b
p x, y , z
Y
r
O
iˆ
X
Z Position vector of a point helps in locating the position of the point. Its magnitude gives the distance between origin and the point. Consider a point ‘P’ having co-ordinates (x,y,z) as shown in figure. If ‘O’ is the origin, then OP is called the position vector (r), r xi yj zk Magnitude of r is, r x 2 y 2 z 2 The unit vector along r is given by r xiˆ yjˆ zkˆ ˆr r x2 y 2 z2
Application 3.8 Displacement vector : Y
r1
a
P
X O
A x1 , y1, z1 s B x ,y ,z 2 2 2
r2
X
ˆj Y If , , are angle made by A with x, y, z axis and cos ,cos ,cos are called direction cosines. Ay A A cos x ,m cos , n cos z A A A
AKASH MULTIMEDIA
If r1 is initial position vector of the particle and r2 is final position vector of the particle then the dis placement of the particle is given by s r2 r1 . s x 2 x1 ˆi y 2 y1 ˆj z 2 z1 kˆ 78
ELEMENTS OF VECTORS
PHYSICS - I A
The magnitude of the displacement vector is
S AB
2
2
x2 x1 y2 y1 z2 z1
2
Application 3.9 Condition for collision : Consider two particles, 1 and 2 move with con stant velocities v1 and v 2 . At initial moment (t=0) their position vectors are r1 and r2 . If particles then collide at the point ‘P’ after time y ‘t’ sec. s v t 1 P From the diagram r s v t r r r1 s1 r2 s2 2 r r1 v1t r2 v2 t r1 x O r1 r2 v2 v1 t (or) r1 r2 v2 v1 t ... (1) r r t 1 2 v2 v1 2
2
2
1
Problem : 3.16 Find the resultant of the vectors shown in figure. Sol. y
A m 5c
370
C 4cm B
R OR AB BC R 5 cos37iˆ 5sin 37 ˆj 3iˆ 4 ˆj R 4iˆ 3ˆj 3iˆ 4ˆj R = 7i + 7j, R 7 2 cm and
450 with horizontal
Problem : 3.17
Find the resultant of the vectors OA, OB, OC asshown in figure. Theradiusof thecircleisR. Sol : R OA OB OC
AKASH MULTIMEDIA
B
R ˆ R ˆ R R i R j 2 2 R = 2R+R along OB .
450 450
O
A
Problem : 3.18 A man walks 40m north, then 30m east and then 40m south. The displacement from the starting point is N Sol. S 40 ˆj 30iˆ 40 ˆj E W S 30iˆ S displacement is 30 m east Problem : 3.19 Vector A is 2 cm long and is 600 above the x – axis in the first quadrant, vector B is 2cm long and is 600 below the x – axis in the fourth quadrant. Find A B . Sol: A
Y
x
A
600
-X
O
X
600
By
B
-Y
Ay
Bx
R AB ˆ R 2 cos60 0 ˆi 2 sin 60 ˆj +2cos60 ˆi – 2sin60 j R 4 cos60iˆ 2iˆ 2cm, along x – axis * Problem : 3.20
x
O
C
1
substitute ‘t’ in equation ........(1) r1 r2 r1 r2 v 2 v1 v2 v1 r1 r2 v 2 v1 r1 r2 v 2 v1
3cm
R Riˆ R cos 45i R sin 45ˆj Rjˆ
A vector has x component of -25.0 units and y component of 40.0 units find the magnitude and direction of the vector. Sol. Consider a vector A A x ˆi A y ˆj A 25iˆ 40 ˆj 2 2 2 A A 2 x A y 25.0 40.0
= 47.16 47.2 units. and tan
Ay Ax
40.0 1.6 25.0
an (-1.6)= – 58.00with – ve x - axis.
(This is in clock wise direction). This is equivalent to (1220 ) in anticlockwise direction with the x – axis. 79
PHYSICS - I A
ELEMENTS OF VECTORS y
* Problem : 3.21
y1
Find the resultant of the vectors shown in fig by the component method
P
rps
y b
c
5
-X
rs| s
3
370
1 0
53
6
O
a
x
If the position of frame S1 relative to S at any time is rs|s .
R y 3jˆ 3jˆ 4.8ˆj R y 1.2ˆj
6.6 2 1.22 y
Rx1.2
x
R
R x6.6
Then from the above diagram rps rps| rs|s 6.7
R y1.2
O
x
10.30
y
tan
Ry Rx
1.2 0.18; tan 1 0.18 10.30 6.6
with –ve x axis in clock wise direction or 169.70 with + ve x–axis in anti clock wise direction. * Problem : 3.22
Two vectors a and b have equal magnitudes of 12.7 units. These vectors are making angles 28.20 and 133.20 with the x axis respectively. Their sum is r. Find (a) the x and y components of r, (b) the magnitude of r, (c) the angle r makes with the +x axis. Ans : 2.5, 15.3, 15.5, 6.120 Hint : same as 3.21
3.19 REL ATI VE VEL OCI TY When we consider the motion of a particle, we assume a fixed point relative to which the given particle is in motion. For example, if we say that water is flowing or wind is blowing or a person is running with a speed ‘v’, we mean that these all are relative to the earth.(Which we have assumed to be fixed) AKASH MULTIMEDIA
x
Now to find the velocity of a moving object relative to another moving object, consider a par ticle P whose position relative to frame S is rps and relative to S| is rps| .
-Y R x 1iˆ 5 cos 370 ˆi 6 cos 53iˆ R x 1iˆ 4iˆ 3.6iˆ R x 6.6iˆ R y 3jˆ 5sin 37ˆj 6 sin 530 ˆj
R R 2x R 2y
x1
S1
S
d
Sol.
rps|
Differentiating the above equation with respect to time d rps d rps1 d rs1s dt dt dt vps v ps1 vs1s ; v ps1 v ps v s1s ...... (1) Thus, if the velocities of two bodies (here particle and frame S| ) are known with respect to a common frame (s), we can find the velocity of one body relative to the other. Note 3.15 : I f vA and vB are velocities of two bodies A and B relative to earth, then the velocity of A relative to B is v AB v A v B . a) If two bodies are moving along the same line in the same direction with velocities vA and vB relative to earth then the magnitude of velocity of A relative to B is given by vAB = vA – vB If vAB is positive the direction of vAB is that of A and if negative the direction of vAB is opposite to that of A (or) along the direction of B. b) If two bodies are moving towards each other or away from each other with velocities vA and vB. Then the magnitude of velocity of A relative to B is given by vAB = vA – (– vB ) = vA + vB and directed towards A (or) away from A 80
ELEMENTS OF VECTORS
PHYSICS - I A
c) If two bodies A and B are moving with velocities vA and vB in mutually perpendicular directions, then the magnitude of velocity of A relative to B is given by v AB v 2A v B2 . The direction of vAB with vA is tan
If the boy is running on the train in a direction opposite to the motion of train, then VBG = VBT – VTG Problem : 3.23
vB vA
d) If two bodies A and B are moving with ve locities vA and v B , and is the angle between the velocities, then the magnitude of velocity of A relative to B is given by v AB v A2 v B2 2v A v B cos
and the direction of vAB with vA is given by
An object A is moving with 5 m/s and B is moving with 20 m/s in the same direction. (Positive x-axis). Find a) velocity of B relative to A. b) Velocity of A relative to B. Sol. a) VB 20iˆ m/s ; VA 5iˆ m/s ; VB VA 15iˆ m/s ˆ / s, V 5iˆ m/s ; b) VB 20im A V V V = – 15 ˆi m/s AB
v B sin v A v B cos
Two object A and B are moving each with velocities 10 m/s. A is moving towards East and B is moving towards North from the same point as shown. Find velocity of A relative to B (VAB ) .
Application 3.10: If two bodies A and B are mov ing with accelerations a A and a B with respect to ground the acceleration of A relative to B is
aAB aA aB
a AB a 2A a B2 2a A a B cos q
Where is the angle between a A and a B .
N VB
W
450 0
Sol:
45
Differentiating equation (1) with respect to time d | d d | v PS v PS vS S dt dt dt a PS | a PS aS |S ............. (2)
B
Problem : 3.24
v B sin 180 Tan v A v B cos 180
Tan
A
O
VA
S VAB VA VB = 10i – 10j V AB 10 2 10 2 10 2 ms1 along SE
Problem : 3.25 Find the speed of two objectsif when they move unifomly towards each other, they get 4.0 metres closer each second and when they move uniformly in the same direction with original speeds, they get 4.0 metres closer each 10 sec. Sol. Let their speeds be v1 and v2 and let v1 > v2 .
Application 3.11:
I n Fir st Case:
Relative M otion on a moving tr ain : If a boy is running with velocity VBT relative to train and train is moving with velocity VTG relative to ground , then the velocity of the boy relative to ground VBG will be given by:
Relative velocity, v1 + v2 = 4
VBG VBT VTG
So, if the boy is running along the direction of train VBG = VBT + VTG AKASH MULTIMEDIA
E
VAB
......
(1)
I n Second Case: 4 = 0.4 ...... (2) 10 Solving eqns. (1) and (2), we get v1 = 2.2 m s–1, v2 = 1.8 m s–1
Relative velocity, v1 – v2 =
Problem : 3.26 A person walks up a stationary escalator in time t1. I f he remains stationary on the escalator, then it can take him up in time t2. How much time would it take him to walk up the moving escalator ?
81
PHYSICS - I A
ELEMENTS OF VECTORS Sol. Let L be the length of escalator. Speed of man w.r.t.
escalator is vme
L . t1
L Speed of escalator ve = . t2 Speed of man with respect to ground would be 1 1 vm vme ve L t1 t 2
The desired time is t
tt L 12 . v m t1 t 2
A
Two ships A and B are 10km apart on a line running south to north. Ship A farther north is streaming west at 20 km/h and ship B is streaming north at 20 km/h. What is their distance of closest approach and how long do they take to reach it ? Sol.
N
VB
V B 20 km / h
E
W
VA
VBG = VBW + VWG The time taken for the boat to move to a distance ‘d’ along the direction of flow of water is d .................(1) t1 VBW VWG b) If the boat is moving opposite to the direction of flow of water (up stream) VBW A
A C VAB 450 B
0
45 VA 20km / h
Velocity of B relative to A is VBA VB VA
20 2 202
= 20 2 km/h
i.e., VBA is 20 2 km/h at an angle of 450 from east towards north. Condition A is at rest and B is moving with VBA in the direction shown in figure. Therefore, the minimum distance between the two is 1 Smin = AC = AB sin 450 = 10 km = 5 2 km 2 BC 5 2 1 and the time is t 20 2 4 VBA
h= 15 min
AKASH MULTIMEDIA
VWG
B
d
S
VBA 20 2km / h
VBA
Suppose a boat moves relative to water with velocity VBW and water is following relative to ground with velocity VWG, velocity of boat relative to ground VBG will be given by : VBG VBW VWG a) If the boat is moving along the direction of flow of water (down stream) VBW
Problem : 3.27
1Gm1m2 2 d2
3.20M OTI ON OF A BOAT I N THE L I NE OF RI VER FL OW
d
B
VWG
VBG = VBW – VWG The time taken for the boat to move to a distance ‘d’ opposite to the direction of flow of water is d t1 .........(2) VBW VWG t1 VBW VWG From equations (1) and (2) t V V 2 BW WG c) If the engine of boat is switched off , then the boat will be carried by the river current. VBG = VWG
(VBW = 0)
The time taken for travel a distance ‘d’ is d t3 VWG time taken by person to go down streams a distance ‘d’ and comes back T td t u d d T VBW VWG VBW VWG 82
ELEMENTS OF VECTORS
PHYSICS - I A
3.21 M OTI ON OF A BOAT ACROSS A RI VER a) To cross the r iver over the shotest distance: V WG
C
V BW
d
V BW
d
V BG
q
q
Suppose a boat starts from a point ‘A’ on one bank of the river of width ‘d’. To cross the river over the shortest distance, the boat should move by making an angle q with the normal to the flow of water as shown in above figure. Let V WG is the velocity of the water with respect to ground, V BW is the velocity of boat in still water (or) velocity of boat relative to water. Velocity of boat with respect to ground is given by VBG VBW VWG from the triangle ABC 2 2 .......... (1) VBG VBW VWG
V and cos BG ......... (2) VBW V sin WG ......... (3) VBW To cross the river in shortest path the angle made by the velocity of boat with the flow of water is 90 + VWG = 90 + sin–1 VBW The component of velocity of boat normal to the flow of water is = VBW cos . The time taken for the boat to cross the river is d ......... (4) t VBW cos d d t V VBG VBW BG VBW 2 2 VBW VBG
C
V BG
A To cross the river in shortest time the boat should be rowed along the normal to flow of water (or) by making an angle of 900 with the flow of water The minimum time taken to cross the river is
t min
d .... (5) VBW
[from (4) 0 ]
The velocity of boat with respect to ground is 2 2 ... (6) VBG VBW VWG
In this case the boat reaches the other bank at point ‘C’, due to the flow of water. The displacement of boat along the direction of flow of water (or) drift is given by
x = VWG (tmin) , x VWG d .... (7) V BW The displacement of the boat is
s d2 x2 Note 3.16 : The time can be obtained by dividing the distance in a particular direction by velocity in that direction. AB BC AC t VBW VWG VBG c) Suppose the boat starts at point ‘A’ on one bank with velocity V BW and reaches the other bank at point ‘D’ as shown in the diagram. B x D C
d
V BW
........... (5) [ from (1)]
AKASH MULTIMEDIA
V WG
B
B
A
d
Shor test Time :
x
900
t
b)
VBW sinq
q
VBW cosq
A
V BE
VWG
83
PHYSICS - I A
ELEMENTS OF VECTORS
The componentof velocity of boat parallel to the flow of water is V BW
Substitute (1) and (3) in (2) we get t1 = t2
The component of velocity of boat normal to the flow of water is of VBW Cos
T = 2t1 = 60 min and the distance (AB+BC) travelled by the boat before meets the cork is D = 2d – x
The boat meets the cork again after
The time taken for the boat to cross the river is t
D 2 VB VW t1 VW 2t1
d VBW Cos
D 2VB t1 2VW t1 2VW t1
The horizontal displacement of the boat (or) drift is x VWG VBW sin t
d x VWG VBW sin VBW Cos
D 2VB t1 2 5
Problem : 3.29 A man wishe 00000s to cross a river flowing with velocity u jumps at an angle with the river flow. Find the velocity of the man with respect to ground if he can swim with speed v. Also find how far from the point directly opposite to the starting point does the boat reach the opposite bank. I n what direction does the boat actually move. I f the width of the river is d.
Note 3.17 : i) If VWG > VBWsin , the boat reaches the other end of the river to the right of B. ii) If VWG < VBWsin , the boat reaches the other end of the river to the left of B.
Sol.
C
v d
Problem : 3.28
Sol.
C
B
iii) If VWG = VBWsin , the boat reaches the exactly opposite point on the other bank. i.e, at B.
v
A boat ismoving with a velocity vbw = 5 km/hr relative to water. At time t = 0, the boat passes through a piece of cork floating in water while moving downstream. If it turns back at time t1=30 min, a) when the boat meet the cork again ? b) The distance travelled by the boat during this time.
30 5km 60
fq
A V MG V MW V WG
u
x
and is the angle between V MG and V WG
VMG V 2 u2 2Vu cos
t=0
drift = u V cos A
B
tan
C vw
vw
Liet AB=d is the distance travelled by boat along down stream, in ‘t1’ sec and it returns back and it meets the cork at point C after ‘t2’ sec.
Let AC=x is the distance travelled by the cork dur-ing (t1 + t2) sec. d VB Vw t1 .................. (1) d x VB Vw t 2 ........... (2) and x Vw t1 t 2 ............. (3)
AKASH MULTIMEDIA
d V sin
V sin u V cos
Problem : 3.30 A swimmer crosses a flowing stream of width ‘d’to and fro in time t1. The time taken to cover the same distance up and down the stream is t2. If t3 is the time the swimmer would take to swim a distance 2d in still water, then Sol. Let v be the river velocity and u the velocity of swimmer in still water. Then d t1 2 ..... (i) u2 v2
84
ELEMENTS OF VECTORS
PHYSICS - I A t2
Sol : Let the velocity of the boat with respect to water be u, the velocity of the water current with respect to the banks is v, and the width of the river ' '.
d d 2ud 2 .... (ii) u v u v u v2
2d ........ (iii) u from equations (i), (ii) and (iii)
and t3 =
B
v
t12 t 2 t 3 t1 t 2 t 3
u
a
Problem : 3.31 Two persons P and Q crosses the river starting from point A on one side to exactly opposite point B on the other bank of the river. The person P crosses the river in the shortest path. The person Q crosses the river in shortest time and walks back to point B. Velocity of river is 3 kmph and speed of each boat is 5 kmph w.r.t river. I f the two persons reach the point B in the same time, then the speed of walk of Q is Sol : For per son (P) : B C
For person (Q) :
VB
tP
VB2 VW 2
d 52 32
Then = ut1 ........(2) Now we consider the case when the boat moving with velocity u makeing an angle ' ' with the line AB.
VW
d 4
The relative velocity of the boat with respect to the bank along the river is v – u sin = 0, or v = u sin
d d x , 4 5 Vman
d But x VW VB
VW d d d , 4 5 VB Vman
d d 3d 4 5 5 Vman
1 1 3 , 4 5 5Vman
1 3 20 5Vman
320 12kmph 5
* Problem : 3.32 A man in a boat attempts to cross a river from point A. I f he rows the boat perpendicular to the banks, he reaches point C at a distance s = 120m down stream, from point B, in 10 minutes shown in figure. I f the man rows his boat at an angle to the straight line AB he reachesthe point B in 12.5 minutes. Find a) the width of the river, (b) velocity of the boat relative to the water, (c) velocity of the curent and (d) the angle . Assume the velocity of the boat relative to the water to be constnat and of the same magnitude in both cases.
AKASH MULTIMEDIA
Then BC = s = vt1 = 120m............(1)
A
d d tQ , t P t Q t VB 5
Vman
When boat is rowed perpendicular to the banks the resultant velocity of the boat with respect to the banks is along AC. The boat moves along the river with velocity v. Suppose the boat moves a distnace BC = s along the river in time t1. Exactly in this time t1 boat moves across the river and travels a distnace' ', (width of the river), with velocity u.
VW
d
A
B x C VB
A
C
........(3)
The relative velocity of the boat with respect to the bank and perpendicular to it is ucos a . If the time taken to reach the point B in the case is t2 then (ucos )t2 = ......... (4) From (1), (2), (3) & (4) t1 = s/v; t2 = /ucos sin = s/ ; cos = t1/ t2 We know that sin2 + cos2 = 1. From this identity
s / 2 t1 / t 2 2 1 Solving for ' ' we get
= st2 t 22 t12 and substituting the values s = 120 m, t1 = 10 minutes, t2 = 12.5 minutes, we get = 200m. using (1) and (2) we get u = 19.2 mmin; v = 12 m/min and sin–1(v/u) = sin–1(12/19.2) = sin–1(0.6667) = 410.49’ 85
PHYSICS - I A
ELEMENTS OF VECTORS
3.22 A M AN I N RAI N Let a veloc us consider the rain is falling with ity VR and a man moves with a velocity VM relative to ground, he will observe the rain falling with a velocity VRM V R V M . Since VR is assumed as a constant (for the rain falling at a constant rate), the man will record different velocities of the rain if he moves with different velocities relative to ground. i) If rain is falling vertically with a velocity VR and an observer is moving horizontally with velocity , VM the velocity of rain relative to observer will be :
V RM1 V R V M1
From the diagram tan and tan
x v M1 y
.......... (2)
iii) If the man speeds up, at a particular velocity V M 2 , the rain will appear to fall vertically with V RM2 V R V M2 as shown in figure. VRM 2 VRM 2
VR
VM 2
VR
b
VRM
V M 2
VR
a
VM
x ......... (1) y
VM
iv) If the man increases his speed further, he will see the rain falling with a velocity as shown in figure.
VM
VRM VR VM
V RM 3
The magnitude of velocity of rain relative to man is VRM VR2 VM2 If is the angle made by the umbrella with V V horizontal, then tan R (or) sin R (or) VM VRM VM cos . VRM If is the angle made by the umbrella with V V vertical, then tan M (or) sin M (or) VRM VR VR cos VRM ii) When the man is moving with a velocity VM1 relative to ground towards east, and the rain is falling with a velocity VR relative to ground by making an angle with vertical.Then the velocity of rain relative to ground VRM1 is as shown in figure. N
VRM1
q
y
VR
b
W
VRM1
VM
1
VM 1
x
AKASH MULTIMEDIA
VM 2
VM 1
S
E
V RM 3
b
q
VR
y
VM 3
VM3
x
VM 3
vM x x V RM3 V R V M3 tan ; tan 3 y y (where x and y are assumed unknowns involved in the problem and their value can be either obtained or eliminated). From the above threee cases we understand that sometimes the man obseves the rain falling forwards, sometimes vertically downwards, and sometimes the rain appears to fall backwards. Hence the velocity of rain relative to man depends upon the velocity of man relative to ground. * Problem : 3.33 A pipe is fixed to a moving cart such that the axis of the pipe makes an angle with the floor of the cart. The cart moves uniformly along a horizontal path with velocity v1 = 2ms-1. I f rain drops which are falling with a velocity v2 = 6ms-1 vertically down reach the bottom of the pipe without touching the walls of the pipe find the value of . Assume that the air resistance is negligible and velocity v of the drops is constant.
86
ELEMENTS OF VECTORS
PHYSICS - I A Sol.
v1
v1
From the fig . v 6 tan 2 3 . v1 2
v21
a
tan (3)
Now, Velocity of Rain relative to man = VRM VM
v2
ˆ (5i) ˆ ˆ 15i) (20K Vertical
v1
a
Problem : 3.34
ˆ 20K
To a man in car which is moving due east with a speed of 20 m/s, the rain appears to fall at an angle of 300 with vertical. When he reduces the speed to 8 m/s, again the rain appears to fall at the same angle with vertical. Find the velocity of rain.
Vr m 300 y
VR
Tan
300
y
x Tan 300 =
Tan 1
1 2
To a man walking at the rate of 3km/h the rain appears to fall vertically. When he increases his speed to 6 km/ h it appearsto meet him at an angle og 450 with vertical. Find the angle made by the velocity of rain with the vartical and its value. Sol.
VR
VRM
1 2
Problem : 3.36
x Vm 20
20 x .... (i) y
10iˆ
ˆ 10iˆ 20K
Sol:
Tan300
a
VR
VRM
VRM
6
From the diagram Tan450
x = 14 and y 6 3 Hence V rg 14i 6 3 j
From (1) and (2) 450 and sin 450
Resultant velocity of rain and wind = ˆ 15i VRM 20K
AKASH MULTIMEDIA
3 ....... (1) y
3 .............. (2) y
and Tan
Rain is falling vertically with a speed of 20ms–1. A person is running in the rain with a velocity of 5 ms–1 and a wind is also blowing with a speed of 15ms–1 (both from the west). The angle with the vertical at which the person should hold his umbrella so that he may not get drenched is: Sol. VRain VR 20(K) VMan VM 5i Vwind VW 15i
3
3
VM 3kmph
Solving eq. (i) and (ii), we have
Problem : 3.35
VR
y
-VM = 8
x8 ...............(ii) y
q
450
3 , VR
1 2
3 VR
VR 3 2 kmph
3.23 M ULTI PL I CATI ON OF A VECTOR BY A SCAL AR Consider a vector P as shown in figure (a). When it is multiplied by a scalar ‘m’, the result is another vector R such that R mP So, whenever a vector is multiplied by a scalar ‘m’ the result is another vector having magnitude ‘m’ times that of the first and directed along the first vector. 87
PHYSICS - I A
ELEMENTS OF VECTORS
P
B 2A C 2 A
a
b c
An illustration of this for m = + 2 and m= –2 is shown in figure (b) and figure (c )respectively. If m = 0, the result is a null vector. If m
1 then R is a unit vector.. P
3.24 SCAL AR PRODUCT OR DOT PRODUCT Dot pr oduct of two vector s is defined as the pr oduct of their magnitudes and the cosine of the angle between the two. (or) The dot pr oduct is the pr oduct of the magnitude of one of the vector quantities and the scalar component of second vector in the dir ection of the fir st vector Q
q (i)
(d) Scalar product of two vectors will be maximum when cos = max = 1, i.e., = 0°, i.e., vectors are parallel. (P.Q)max PQ (e) P.Q P (Q cos ) = Q (P cos ) (f) If the scalar product of two non zero vectors vanishes, t hen the vect ors are orthogonal. (perpendicular). i.e., = 900, P.Q PQ cos 90 0 0 (g) The scalar product of a vector by itself is termed as self dot product and is given by 0 P.P = PP cos = P2 ( 0 ) (h) In case of unit vector nˆ , nˆ . nˆ = 1 x 1 x cos = 1 nˆ . nˆ = ˆi . ˆi = ˆj . ˆj = kˆ . kˆ = 1 In case of orthogonal unit vectors ˆi , ˆj and kˆ ; ˆi . ˆj = ˆj . kˆ = kˆ . ˆi = 0 If P P ˆi Py ˆj Pz kˆ and Q Q ˆi Q y ˆj Q z kˆ P.Q P ˆi Py ˆj Pz kˆ . Q ˆi Q y ˆj Q z kˆ
(i)
(j) P
The Scalar product of the two vectors, P and Q is defined as P.Q P Q cos . Here is the angle between P and Q . The angle between the two vectors P and Q is P.Q P.Q 1 given by cos cos P Q P Q
PROPERTI ES OF SCAL AR PRODUCT : (a) The Dot product between two vectors obeys commutative law. i.e., P.Q Q.P (b) The Dot product obeys distributive law, i.e., P.(Q R) P.Q P.R (c) It is always a scalar which is positive if angle between the vectors is acute (i.e. < 90°) and negative if angle between them is obtuse (i.e. 90° < 180°) AKASH MULTIMEDIA
= (PxQx + PyQy + PzQz) (k) The product of vectors is possible between vectors of different kind. Application 3.12 : Geometrically, Q cos is the projection of Q onto P and P cos is the projection of P onto Q Q as ofthe magnitude of shown. So P . is the product Q and the component of along P P and vice versa. Q
Q
Q
q
q P
Q P
Q cos q
sq co
q
P
P.Q (a) Component of Q along P = Q cos = P ˆ P.Q 88
ELEMENTS OF VECTORS
PHYSICS - I A
(b) The vector component of Q along P = P.Q ˆ Bcos Pˆ Pˆ P.Q Pˆ P P.Q (c) Component of P along Q = P cos = Q ˆ P.Q (d) The vector component of P along Q = P.Q ˆ ˆ P.Q ˆ Q ˆ P cos Q Q Q e) The component of P perpendicular to Q and Q in the same plane is C P P.Q f) The component of Q perpendicular to P and P in the same plane is D Q Q.P
Examples of Scalar Pr oduct a) The scalar product of force and displacement, is called workdone by that force. If a force F acts on a body and displaces it through a displacement S , the work done by the force W F.S F(Scos ) (F cos )S Workdone is maximum when force and displacement are in the same direction (the angle between them = 0) and is equal to FS. If force and displacement are perpendicular to each other, the workdone by the force is zero. b)
Dot product between force and velcity vecotors F.S gives power i.e., power = F.V t c) A block of mass 'm' is lifted to a vertical height 'h' from the surface of the earth. The work done (W) against the gravitational force (F=mg) is W F.h Fh cos 0 W = Fh h
W = mgh
mg
AKASH MULTIMEDIA
If h decreases, potential energy of the body will decreases. d) The dot product of the magnetic field induction and area of the coil gives the magnetic flux. d B.ds (or) B.ds. ds
n
B
q O
e) The dot product of current density J and area ds gives electric curret i.e., dI J.ds (or) I J.ds f) Electric potential is the scalar product of electric field intencity E and position vector dr . dV = E.dr or V E.dr. g) If a magnetic dipole ofmoment of M is in a magnetic field of induction B, the potential energy of the dipole is given by U B M.B.
3.25 CROSS PRODUCT OF TWO VETORS The vector pr oduct of two vetor s is a vector whose magnitude is equal to the pr oduct of magnitudes of the two vector s and sine of the angle between them. T he pr oduct vect or i s i n a pl ane per pendicular to the plane of two vector s. The dir ection of pr oduct vector can be known either with ‘r ight hand scr ew r ule’or with ‘r ight hand thumb r ule’. PQ nˆ
q
Q P
According to vector product P Q PQ sin nˆ nˆ is a unit vector perpendicular to plane of P and Q as shown in fig. The direction of P Q is given by the following rules. 89
PHYSICS - I A
ELEMENTS OF VECTORS
(a) Right handed scr ew r ule : Imagine a righ handed screw to be placed along the normal of the plane (ABCD) containing P and Q . Rotate the cap of the screw from P to Q through the smaller angle between them. The direction of motion of the tip of the screw gives the direction of P Q . P Q
D
C
Q q
O
P B
A
c) The vector product of two parallel vectors is a null vector i.e., P Q PQ sin nˆ = PQ isn 0 nˆ = 0 d) The vector product of a vector by itself is a null vector (zero) P P = P P sin nˆ = PP sin 0 nˆ = 0 e) The magnitude of the vector product of two vectors mutually at right angle is equal to the product of the magnitude of the vectors. P Q = P Q sin nˆ = PQsin900 nˆ = PQ nˆ f) The vector product of unit orthogonal vectors, ˆi, ˆj and kˆ have the following relations.
QP
It is clear from fig. that tip of the screw moves up when cap is rotated from P to Q . Hence P Q directed upwards. Similarly, it can be proved that Q P is directed downward along the normal. (b) Right hand thumb r ule : Imagine the normal to the plane (ABCD) containing P and Q to be held in the right hand with the thumb erect. If the fingers curl in the direction from P to Q , through smaller angle between them, then the direction of thumb gives the direction of P Q .
ˆi ˆi ˆj ˆj kˆ kˆ 0 ˆi ˆj ˆj ˆi k, ˆ ˆj kˆ kˆ ˆj ˆi, and kˆ ˆi ˆi kˆ ˆj
They follow the right hand screw rule as shown in fig. j
i
P Q
k
D
Q
O A
q
QP
C
P
B
Fig. shown as that P Q is directed along the normal in the up ward direction while Q P is downward. Pr oper ties of vector pr oduct : Following are the properties of vector product : a) The vector product is not commutative PQ QP. b) The vector product is distributive P QR PQPR .
AKASH MULTIMEDIA
g) The vector product of two vectors interms of their x, y and z components can be expressed in the form of determinant. If P P ˆi P ˆj P kˆ x
y
z
Q Q x ˆi Q y ˆj Q z kˆ
ˆi P Q Px Qx
ˆj Py Qy
kˆ Pz Qz
Py Q z Pz Q y ˆi Px Q z Pz Q x ˆj Px Q y Py Q x kˆ
h) The unit vector perpendicular to both P and P Q Q is n PQ 90
ELEMENTS OF VECTORS
PHYSICS - I A
ˆ P Px ˆi Py ˆj Pz k, Q Q x ˆi Q y ˆj Q z kˆ
i) If
are parallel then
Py Px P z = constant Q x Qy Qz
Note 3.18 :
1) The angle between P and P Q is 900 2) The angle between Q and P Q is 900 3) The angle between (P Q) and (P Q) is 900 4) The angle between (P Q) and (P Q) is 900 5) The angle between P Q and Q P is 1800 Application 3.13 : Area of pr ar allelogr am : If P and Q are the two adjacent sides of a parallelogram OABC as shown in figure.
From the diagram d1 P Q and d 2 P Q d1 d 2 P Q P Q = PPPQQP Q Q = OPQ P Q O = 2(P Q) d1 d 2 2 P Q
B
Q
q
O
q O
P
C
A
1 (base)(height) 2 1 = (OA)(BC) 2 From the triangle OBC BC Q sin 1 Area of the triangle = P Q sin 2 1 1 PQ sin P Q 2 2 Area of triangle is equal to half the magnitude of the cross product of two vectors representing two sides of the triangle .
Area of the triangle =
P
N
A
The area of parallelogram = OA X perpendicular distance BN = OA X OB sin = PQ sin = PQ Application 3.14 : Area of par allelogr am in termsof diagonals. If d1 and d 2 are the diagonals of a parallelogram of OABC as shown in fig. B
Q
O
1 d1 d 2 . 2
Application 3.15 : Area of tr iangle : Let ‘OAB’ be a triangle formed with P and Q as its two sides as shown in figure.
C
Q
Area of the pralleogram =
BN is the perpendicular drawn from B on OA. B
d2 d1
P
AKASH MULTIMEDIA
A
C
Examples of Vector Product : a) Rel at i on bet ween L i near Vel oci t y V, Angular Velocity and Position Vector r . a) Consider a particel is moving in a circular pat h about an axis passing thro ugh OO 1 , perpendicular t o plane of t he path, wit h a constant angular velocity . P is the position of the 91
PHYSICS - I A
ELEMENTS OF VECTORS
body at any time given by the position vector r described with 'O' as the origin. The radius of the circular path is O1 P = r sin . The linear velocity v = (rsin ). The direction of the velocity is tangential to the path at 'P' and it is perpendicular to the plane contained by and r . So vectorially v r . z
v
w O
p
r sin q r
q
O
b)
Torque (M oment of force) : Turning effect of a force is called torque, and is equal to the pr oduct of the for ce and the per pendicular distance between axis of r otation and the line of action of for ce. Let F be the force acting on a particle at a point P in XY plane. The position vector of the particle be r . The torque “ ” acting on the particle with respect to the origin “O” is given by ON F r sin Fnˆ ON r sin The torque acts perpendicular to XY plane i.e., along Z – axis. Magnitude of torque is r sin F Z
Consider a particle of mass m at a point Q in XY plane. The position vector of Q is r . If it is moving with a velocity ‘v’ in that plane, its momentum P mv. Angular momentum = perpendicular distance from origin to the line of action of momentum x momentum L = ON(P) = (rsin )P L r p, L rp sin nˆ The direction of angular momentum is along Z–axis. d) For ce on a char ged par ticle q moving with velocity v in a magnetic field B is given by F q(v B) e) Tor que on a magnet i c moment M i n magnetic field B is given by (M B) f) The for ce on a cur r ent car rying conductor of length in a magnetic field B is given by F i( B) g) If ' ˆi ' denotes a unit vector along incident r ay rˆ a unit vector along refracted ray into a medium of r efr active index and nˆ is unit vector normal to boundar y of medium dir ected towar ds the incident medium, the law of r efr action is ˆi nˆ (rˆ n) ˆ * Problem : 3.37
O
Find the angle between two vectors A 2i j k and B = i – k. A.B Hint : cos , Ans : 300 A B
X r
q Y
c)
F
P
N r
q
Angular momentum : T he moment of moment um is cal led angular momentum.
Show that the vectors a 3iˆ 2 ˆj kˆ, b iˆ 3 ˆj 5kˆ and c 2iˆ ˆj 4kˆ from a right angled triangle. Sol.
Z
O
Y
r Q
X
AKASH MULTIMEDIA
q
b
c
L rp
N
Problem : 3.38
p
a
a 9 4 1 14, b 1 9 25 35 and c 4 1 16 21.
92
ELEMENTS OF VECTORS
PHYSICS - I A which gives b2 = a2 + c2 ˆ (2iˆ ˆj 4k) ˆ 3iˆ 2ˆj kˆ a b c (iˆ 3ˆj 5k) Hence, a,b, c are coplanar.. So they form a right angled triangle.
Problem : 3.44 A particle of massm is moving with velocity 'v' parallel to x axis along line y = b. I ts angular momentum w . r to origin after time (t) will be
Problem : 3.39 I f a1 and a2 are two non collinear unit vectors and if a1 a2 3, then the value of ( a1 a2 ).(2a1 a2 ) is
Sol: Shortest distance between line of momentum and origin is 'b'.
mv
(0,b)
Sol. a1 = a2 = 1 and
a12 +a22 + 2a1a2cos
3
(0,0)
2
3 or
Hence angular momentum of the particle with respet ˆ . to origin is L mvb( k)
1 + 1 + 2 cos = 3 or cos = 1/2 Now (a1 a2 ).(2a1 a2 ) 2a12 a 22 a1a 2 cos 1 1 = 2 1 2 2
Problem : 3.40 If a and b are two unit vectors such that a 2b and 5a 4b areperpendicular to each other, then theangle between a and b is Sol. (a 2b).(5a 4b) 0
5a 2 4ab cos 10ab cos 8b2 0 = 5 – 4 cos + 10 cos – 8 = 0 [a = b = 1] = – 3 + 6 cos = 0 cos = 1/2, = 600 Find the component of 3 iˆ + 4 ˆj along iˆ ˆj ? Sol. Componant of A along B is given by A.B (3iˆ 4 ˆj).(iˆ ˆj) 7 2 2 B
135
.
Sol. Unit vector in the direction of iˆ 2 Jˆ 2 kˆ 1 ˆ iˆ 2 ˆj 2kˆ is nˆ i 2 Jˆ 2 kˆ 3 9 Angular velocity of rigid body 1 3 iˆ 2 ˆj 2kˆ , iˆ 2 ˆj 2kˆ 3 and position vector r 2iˆ 3 ˆj 4 kˆ 2iˆ ˆj kˆ , r 4 Jˆ 3kˆ
= – 17i+13j+5k * Problem : 3.43
AKASH MULTIMEDIA
134
1
Fi nd the uni t vector per pendi cul ar A 2i 3 j k and B i j k. AB 4i j 5k ˆ Hint : n , Ans : AB 42
, = sin–1
I f a rigid body is rotating about an axis passing thr ough the poi nt 2iˆ ˆj kˆ and paral l el to iˆ 2 ˆj 2kˆ with an angular velocity of 3rad/sec. Find the velocity of the point of the rigid body whose position vector is 2iˆ 3 ˆj 4kˆ
i j k r F 1 1 6 = i(1–18)–j(–1–12)+k(3+2) 3
5 27
Problem : 3.46
* Problem : 3.42 I f F 2i 3 j k and r = i–j+6k find r × F..
2
134
sin =
Problem : 3.41
Sol.
Problem : 3.45 I f A = ( i j ) and B = ( i j + 5 k ). Find angle between A and B . i j k Sol. A B = 2 1 0 1 1 5 = i (5 + 0) – j (10 – 0) + k (–2 –1) A B = 5 i – 10 j – 3 k 2 2 52 10 3 | AB| sin = AB 22 1 1 1 25
to
Linear velocity of the point 'P' is ˆj i v r 1 2 0
4
k
2 3
v 2i 3 j 4 k 93
PHYSICS - I A
ELEMENTS OF VECTORS Problem : 3.47
* Problem : 3.49 Find the angle between the diagonals of a cube with edges of length a.
Fi nd the vector components of vector A =2ˆ i + 3 ˆj along the directions of B = ˆ i + ˆj . Sol.
Sol.
A.B B ˆ C= Acosq B= B B
z F
A
y
E
O
c
D
x
* Problem : 3.48
From the diagram OG d1 ai+aj+ak, CF d2 –ai+ aj + ak
A particle is in equilibrium under the simultaneous action of three forces. Prove that each force bears a constant ratio with the sine of the angle between the other two. Sol : Let, P+Q and R be the three forces acting at a point O
shown in figure. Since the particle is in equilibrium. ............... (1) P+Q +R=0 (or) (P+Q)=-R ................ (2) Taking cross product, with P on both sides of (2), P× P+Q =-P×R (or) P×P+P ×Q=R×P (or) P ×Q=R×P ............... (3) Taking cross product, with Q on both sides of (2)
G
B
2+3 i+j 5 ˆ ˆ C= = i+j . 1+1 1+1 2
The angle between the daigonals is d1d2 1 1 Hint : cos Ans : cos 3 d d 1
1.
2
L ong Answer Questions
Define scalar and vector products and explain their properties. Give two examples for each
Shor t Answer Questions
1.
Define scalar product. Give its properties and two examples.
2.
Define vector product. Give its properties and two examples
3.
State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector.
From (3) and (4) P×Q=Q×R=R×P
4.
State triangle law of vectors and explain it.
5.
What is relative motion. Explain it?
Taking their magnitudes
6.
Show that a boat must at an angle of 900 with respect to river water in order to cross the river in minimum time ?
7.
Define unit vector, null vector and position vector. If a b a b prove that the angle between a and b is 900.
Q× P+Q =- Q×R
Q
P
Q×P+Q ×Q=-Q×R Q×P= Q×R (or) P ×Q=Q×R
g b
a R
PQ sin = QR sin RP sin Dividing throughout by PQR
PQ sin QR sin RP sin PQR PQR PQR (or)
P Q R sin sin sin
AKASH MULTIMEDIA
8.
94
ELEMENTS OF VECTORS
PHYSICS - I A
Ver y Shor t Answer Questions 1.
2.
3.
4. 5.
6. 7.
The vertical component of a vector is equal to its horizontal component. What is the angle made by the vector with x–axis? A vector V make an angle with the horizontal. The vector is rotated through an angle . Does this rotation change the vector V ? Two forces of magnitudes 3 units and 5 units act at 600 with each other. What is the magnitude of their resultant? A ˆi ˆj. What is the angle between the vector and x–axis ? When two right angled vectors of magnitude 7 units and 24 units combine, what is the magnitude of their resultant? If P = 2i + 4j + 14k and Q = 4i +4j + 10k find the magnitude of P Q . A force 2i + j –k newton acts on a body which is initially at rest. At the end of 20 seconds the velocity of the body is 4i + 2j + 2k ms–1. What is the mass of the body?
8.
Can a vector of magnitude zero have a nonzero components?
9.
What are the units of a unit vector?
10. Can two vectors of unequal magnitude add up to give the zero vector? Can three unequal vectors and upto give the zero vector? 11. During practice a player throws a ball high in to the air, and then runs in a straight line and catches it. Which had and greater displacement, the player or the ball ? 12. In the below fig. there are four forces that are equal in magnitude. Identify any three of them which act on a body simultaneously and keep it with unchanged velocity. C
A
D
B AKASH MULTIMEDIA
13. The mass of a body is 5kg and its velocity is 4iˆ 2 ˆj kˆ ms–1. What is the momentum of the body? 2 2 14. Find the values o f a.b a b and 2 2 a.b a b . 15. Does vector subtraction obey commutative law? Assess Your self 1)
Identify the vector and scalar quantities among the following. a) temperature b) force c) volume of the milk in a bottle d) bus fare to Hyderabad from your village/town e) height of Charminar f) velocity of horse - cart you travel, the age of the Earth.
Ans : 1) Force, velocity of horse - cart are vectors. 2) Temperature, volume of the milk, bus fare, height of charminar, the age of the earth are Scalars. 2. Can we add a vector quantity to a scalar quantity ? Ans : We can add a vector to another vector of same kind but we can’t add a vector quantity to a scalar quantity. 3. The magnitudes of two vectors A and B are 8 units and 5 units respectively. What are the maximum and minimum possible values for the magnitude of the resultant vecor R A B ? Ans. is maximum when A and B are in the same directions. The magnitude of R is equal to the magnitude of A and B. Rmax = A + B = (8+5) unit = 13 units. R has minimum magnitude when A and B are opposite (antiparallel) to each other.
4.
Rmin = A – B =(8–5) units = 3 units. If A and B are two vectors, under what conditions A B A B . Under what condition the resultant vector equals to zero ? 95
PHYSICS - I A
ELEMENTS OF VECTORS
The angle made by this vector with X–axis is 180 + 66 = 2400 in anti clock wise
Ans. when these vectors have same direction. A + B = 0 when they are oppositely directed. A displacement vector in x – y plane has magnitude 25m and direction 300 with x – axis. What are its x and y components? Y
y X
3 x 25 cos30 25 21.65N 2
Ans. All their components may or may not be equal. If one component of a vector is not zero, can the magnitude of the vector be zero ? Ans. The magnitude of vector A is given by A 2x +A 2y +A 2z . Hence if any component is non zero, A can never be zero. 9.
1 y 25sin 30 25 12.5N 2
6.
Ans :
A certain force has magnitude of 200N, its x component is – 80N. What is its y component and its direction. Two possible answers exist find them. in first case
80N
Under what circumstances can the component of a vector be equal to the magnitude of the vector ?
Ans : If vector is directed along positive direction of one of the three axes x, y or z then the magnitude of the vector is the same as the component associated with the direction, other components are zero. 10. Write the following vectors in their unit vector form in cartesion coordinate system in two dimensions (i.e., consider x – y plane only.)
Y 200N -X
a) A velocity vector of 10 ms–1 at an angle of elevation of 600;
X
O
Y
-Y
Ans :
a) F Fx2 Fy2
200
183
90 + 24 = 1140 in anti–clock wise direction. b)
F
s m 10 600 O Vx Q
200 80
183 tan Fx 80 66 0
AKASH MULTIMEDIA
-X
-80
200N
b) A vector A of magnitude A = 5m and angle = 2250
Fy2
Y
Ans :
in second case Y
Fy 183N
Fy
2
2250 -X
O Fy -Y
X
X
ˆ ˆ V=10cos60i+10sin60j V 5iˆ 8.66 ˆj
= 240
The angle made by the force with X–axis is Fy2
-1
Vy
Fx 80 and tan Fy
P
R
2 80 Fy 2 Fy 183N
Fx2
Vx 45
0
X
s -1
m 25 30 O x
y
If A B , what can you conclude about the components of A and B.
8.
P
Ans :
7.
10 m
5.
Vy
-Y
96
ELEMENTS OF VECTORS
PHYSICS - I A
V 10 cos 45iˆ 10 sin 45 ˆj V 3.54 iˆ 3.54 ˆj
Ans : The boat which heads straight across reaches the opposite side first because it takes where
d
c) A displacement from the origin to the point x = 14m, y = – 6m Ans : S S ˆi 5 ˆj , S 14iˆ 6 ˆj
minimum time to cross i.e., t
11. If the resultant of three vectors is zero, must they all be in the same plane ? Ans. Yes, the vectors must be in the same plane. Resultant of two vectors lies in the plane in which the vectors lie. Hence the third vector which is equal to the resultant of the two and opposite in direction should lie in the same plane.
and VWG is river velocity. The boat which heads upstream at an angle crosses the river following the shortest path. The condition for that is
1
2
12. Two boats travelling at the same speed set off across a river at the same time. One heads straight across and is pulled downstream somewhat by the currentThe other one heads upstream at an angle so as to arrive at a point opposite the starting point. Which boat reaches the opposite side first ?
AKASH MULTIMEDIA
2 2 VBW VWG
is width of the river and VBW is boat velocity
V = sin–1 WG . VBW 13. Prove that P.P = Px2+Py2+Pz2 = P2. ˆ Ans. Consider P Px ˆi Py ˆj Pz k.
P.P Px ˆi Py ˆj Pz kˆ . Px ˆi Py ˆj Pz kˆ
P.P Px2 Py2 Pz2 P 2
97
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