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Chapter Notes: Motion in Two and Three Dimension Physics Class 11


Notes for motion in a plane chapter of class 11 physics. Dronstudy provides free comprehensive chapterwise class 11 physics notes with proper images & diagram.

 

VECTORS

How to Locate Points in Space?
Positions in space are designated relative to coordinate systems. The Cartesian coordinate system is a particularly convenient coordinate system in which position are designated by distances (x, y, z) along three perpendicular axes that intersect at a point called the originIn a polar coordinate system positions in a plane are designated by a length r from the origin, and an angle q usually measured from the positive x - axis. From simple trigonometry we see that the relationships between the polar coordinates and the Cartesian coordinates are

x = r\cos \theta
y = r\sin \theta

A frame of reference is another name for the particular coordinate system with respect to which we are making observations of physical phenomena.
A scalar quantity requires only a number for its complete description. Mass, volume, density, pressure and temperature are all examples of scalar quantities. The mathematics of scalar quantities is the ordinary algebra of numbers.
Vector quantities require  both magnitude and direction for its complete description. Velocity, acceleration, force and momentum are the examples of vector quantities. A vector can be represented graphically by a directed line segment. The length of the line segment represents the vector's magnitude and its angle with respect to some coordinate system specifies its direction.
We will represent vectors by  bold face type letters, with an arrow over the letter such as {\vec a}. When written by hand the same representation may be used.
The magnitude of a vector will be represented by italic type letters such as a, b, c etc.
If two vectors have the same direction, they are parallel. If they have the same magnitude and the same direction, they are equal, no matter where they are located in space.
If two vectors have the same direction, they are parallel. If they have the same magnitude and the same direction, they are equal, no matter where they are located in space.
The negative of a vector is defined as a vector having the same magnitude as the original vector but the opposite direction.
When two vectors  and  have opposite directions, whether their magnitudes are the same or not, we say that they are antiparallel.

Vector Algebra

Addition of Vectors
(i) Geometrical Method
Two vectors  {\vec a} and {\vec b} may be added geometrically by drawing them to a common scale and placing then head to tail. The vector connecting the tail of the first to the head of the second is the sum vector {\vec c} .
Vector addition is commutative and obeys the associative law.

(ii) Analytical Method (Parallelogram law of vector addition)
If the two vectors {\vec a} and {\vec b} are given such that the angle between them is \theta . The magnitude of the resultant vector {\vec c} of their vector addition is given by

c = \sqrt {{a^2} + {b^2} + 2ab\,\cos \theta }
and its direction is given by angle \phi  with
vector {\vec a}
\tan \phi = {{b\sin \theta } \over {a + b\cos \theta }}
It is a common error to conclude that if \vec c = \vec a + \vec b the magnitude of {\vec c} should be just equal to the magnitude of {\vec a} plus the magnitude of {\vec b} In general, the conclusion is wrong; one can see that c < a + b. The magnitude of the vector sum \vec a + \vec b depends on the magnitudes of {\vec a} and of {\vec b} and on the angle between {\vec a} and {\vec b} Only in the special case in which {\vec a} and {\vec b} are parallel; the magnitude of \vec c = \vec a + \vec b equal to the sum of the magnitudes of {\vec a} and {\vec b} By contrast, when the vectors are anti-parallel the magnitude of {\vec c} equals the difference of the magnitudes of {\vec a} and {\vec b}.

Subtraction of Vectors
To subtract {\vec b} from {\vec a}, reverse the direction of {\vec b} to get -{\vec b} ; then add -{\vec b} to {\vec a}

Resolution of a Vector
The (scalar) components, ax and ay of any two dimensional vector {\vec a} are found by dropping perpendicular lines from the ends of {\vec a} onto
the coordinate axes. The components are given by
{a_x} = a\,\cos \theta
and {a_y} = a\,\sin \theta
where \theta  is the angle from the positive direction of the x axis to the direction of {\vec a} The algebraic sign of a component indicates its direction along the associated axis.

Given the components, we can reconstruct the vector from
a = \sqrt {a_x^2 + a_y^2}
and \tan \theta = {{{a_y}} \over {{a_x}}}

Application 1

Find the components of the vectors {\vec a} and {\vec b} as shown in the figure, if a = 2 units  and b = 3 units.

Solution

{a_x} = a\,\cos \theta = 2\cos {30^ \circ } = 1.73 units
{a_y} = a\,\sin \theta = 2\sin {30^ \circ } = 1 units
{b_x} = b\,\cos \theta = 2\cos {45^ \circ } = 2.12 units
{b_y} = - b\,\sin \theta = - 3\sin {45^ \circ } = - 2.12 units

Application 2

Find the magnitude and direction of the resultant vector \vec c = \vec a + \vec b  where {\vec a} and {\vec b} are vectors shown in figure.

Solution

{c_x} = {a_x} + {b_x} = 1.73 + 2.12 = 3.85 units
{c_y} = {a_y} + {b_y} = 1 - 2.12 = - 1.12 units
c = \sqrt {c_x^2 + c_y^2} = \sqrt {{{(3.85)}^2} + {{( - 1.12)}^2}} = 4 units
\tan \theta= {{{c_y}} \over {{c_x}}} = - {{1.12} \over {3.85}} = - 0.29 units
\theta = {\tan ^{ - 1}}( - 0.29)

Note
There is one slight complication in using equation \tan \theta = {{{a_y}} \over {{a_x}}} to find \theta Suppose {a_x} = 2 m and {a_y} = - 2 m; then \tan \theta = - 1 But there are two angles having tangents of -1, namely 135° and 315° (or -45°). In general, any two angles that differ by 180° have the same tangent. To decide which is correct, we have to look at the individual components. Because ax is positive and ay is negative, the angle must be in the fourth quadrant; thus \theta = 315° (or -45°) is the correct value. But if instead we have ax = -2 m and ay = 2 m, then the correct angle is 135°. Similarly, when ax and ay are both negative, the tangent is positive, but the angle is in the third quadrant. One should always draw a sketch to check which of the two possibilities is the correct one.

Unit Vectors
A unit vector is a vector that has a magnitude of 1, with no units. Its only purpose is to point, that is, to describe a direction in space. Unit vectors provide a convenient notation for many expressions involving components of vectors

In an x-y coordinate system we can define a unit vector {\hat i} that points in the direction of the positive  x - axis and a unit vector {\hat j} that points in the direction of the positive y-axis. Then we can express the relationship between component vectors and components as
{{\vec a}_x} = {a_x}\hat i
The vector {\vec a} can be written in terms of its components as
{{\vec a}_x} = {a_x}\hat i + {a_y}\hat j
When two vectors {\vec a} and {\vec b} are represented in terms of their components, we can express the vector sum {\vec c} using unit vector as follows
\vec a = {a_x}\hat i + {a_y}\hat j
\vec b= {b_x}\hat i + {b_y}\hat j
\vec c = ({a_x}\hat i + {a_y}\hat j) + ({b_x}\hat i + {b_y}\hat j)= \vec c = ({a_x} + {b_x})\hat i + ({a_y} + {b_y})\hat j
In three dimensional coordinate system a vector {\vec a} may be expressed as
\vec a = {a_x}\hat i + {a_y}\hat j + {a_z}\hat k
where {\hat k} is the unit vector along the z-axis.

Application 3

If \vec a = 2\hat i + \hat j and \vec b= 4\hat i +7 \hat j
(a) Find the components of \vec c = \vec a + \vec b
(b) Find the magnitude of {\vec c} and its angle with x-axis.

Solution

(a) \vec c = \vec a + \vec b  = (2\hat i + \hat j) + (4\hat i + 7\hat j)
\vec c = (2 + 4)\hat i + (1 + 7)\hat j
Thus {c_x} = 6; {c_y} = 8
(b) c = \sqrt {c_x^2 + c_y^2} = \sqrt {{{(6)}^2} + {{(8)}^2}} = 100
\tan \theta = {{{c_y}} \over {{c_x}}} = {8 \over 6} = {4 \over 3}
\theta = {\tan ^{ - 1}}\left( {{4 \over 3}} \right) = {53^ \circ }

Multiplication of Vectors
(i) Multiplication of a vector by a scalar
The product of a scalar b and a vector {\vec v} is a new vector whose magnitude is {\vec v} and whose direction is the same as that of {\vec v} if b is positive, and opposite to that of {\vec v} if b is negative. To divide {\vec v} by b, multiply {\vec v} by (1/b).

(ii) Multiplication of a vector by another vector
(a) Scalar or Dot Product
The scalar (or dot) product of two vectors {\vec a} and {\vec b} is written \vec a.\vec b and is the scalar quantity c given by

c = \vec a.\vec b = ab\,\cos \theta
In which \theta  is the angle between the directions of {\vec a} and {\vec b}. The scalar product may be positive, zero, or negative, depending on the value of \theta . A scalar product is the product of magnitude of one vector and the component of the second vector along the direction of the first vector.

Some properties of Dot product
1) It is commutative, i.e. \vec a.\vec b\vec b.\vec a
2) It is distributive over addition and subtraction i.e. \vec a.(\vec b \pm \vec c) = \vec a.\vec b \pm \vec a.\vec c
3) If {\vec a} and {\vec b} are perpendicular \vec a.\vec b= 0.
4) If {\vec a} and {\vec b} are parallel then \vec a.\vec b =a.b
5) Square of a vector is defined as \vec a.\vec a= a2 (scalar) and not \vec a \times \vec a
\hat i.\hat i = \hat j.\hat j = \hat k.\hat k = (1)(1)cos{0^ \circ } = 1
\hat i.\hat j = \hat j.\hat k = \hat k.\hat i = (1)(1)cos{90^ \circ } = 0
6) In unit-vector notation we have, 

\vec a.\vec b = ({a_x}\hat i + {a_y}\hat j + {a_z}\hat k).({b_x}\hat i + {b_y}\hat j + {b_z}\hat k)
Thus, \vec a.\vec b = {a_x}{b_x} + {a_y}{b_y} + {a_z}{b_z}

Application 4

If \vec a = 2\hat i + 3\hat j; \vec b = 4\hat i + 2\hat j Find c = \vec a.\vec b

Solution

\vec c = \vec a.\vec b = (2\hat i + 3\hat j) + (4\hat i + 2\hat j) = (2)(4) + (3)(2) = 14

(b) Vector or Cross product
The vector (or cross) product of two vectors {\vec a} and {\vec b} is written as \vec a \times \vec b and is a vector {\vec c} whose magnitude c is given by

c = ab\,\sin \theta

In which \theta is the smaller of the angles between the direction of {\vec a} and {\vec b}. the direction of {\vec c} is perpendicular to the plane defined by {\vec a} and {\vec b} and is given  by right hand rule.
\vec c = \vec a \times \vec b
\left| {\vec c} \right| = \vec a \times \vec b= ab\,\sin \theta
and \vec a \times \vec b = \left| {\vec a} \right|\left| {\vec b} \right|\sin \theta \,\hat n
where {\hat n} is the unit vector.
The vector {\vec c} is directed perpendicular to the plane formed by {\vec a} and {\vec b} The direction of vector {\vec c} may be obtained by using the Right Hand Thumb Rule. Stretch all the fingers and thumb of your right hand such that they are perpendicular to each other.  Now align  your hand such that its plane is perpendicular to the plane formed by vectors {\vec a} and {\vec b}. Align the stretched fingers along the direction of the vector written first in order i.e., {\vec a} in this case . Curl the fingers of your hand towards the second vector through the smaller angle. Then, the direction of the thumb gives the direction of the cross product.

Properties of Cross Product

1) It is not commutative \vec a \times \vec b \ne \vec b \times \vec a. In fact \vec a \times \vec b = - \vec b \times \vec a
2) It is distributive over addition and subtraction
\vec a \times (\vec b \pm \vec c) = \vec a \times \vec b \pm \vec a \times \vec c
3) If \vec a\parallel \vec b, then \vec a \times \vec b = 0 and if \vec a \bot \vec b\left| {\vec a \times \vec b} \right| = ab
4) If {\hat i}{\hat j}{\hat k} be the unit vectors along in the positive directions of x, y, and z axes then 
\hat i \times \hat i = \hat j \times \hat j = \hat k \times \hat k = 0
\hat i \times \hat j = \hat k\hat j \times \hat k = \hat i\hat k \times \hat i = \hat j (maintain cyclic order) \hat i \times \hat j = - \hat j \times \hat i
5) The cross product may also be expressed by the determinant
\vec a \times \vec b = \left| {\matrix{{\hat i} & {\hat j} & {\hat k} \cr {{a_x}} & {{a_y}} & {{a_z}} \cr {{b_x}} & {{b_y}} & {{b_z}} \cr} } \right|
where \vec a = {a_x}\hat i + {a_y}\hat j + {a_z}\hat k
and \vec b = {b_x}\hat i + {b_y}\hat j + {b_z}\hat k

Application 5

If \vec a = 2\hat i + 3\hat j; \vec b = 4\hat i + 2\hat j Find \vec d = \vec a \times \vec b

Solution

\vec d = (2\hat i + 3\hat j) \times (4\hat i + 2\hat j)
\vec d = (2)(4)(\hat i \times \hat i) + (2)(2)(\hat i \times \hat j) + (3)(4)(\hat j \times \hat i) + (3)(2)(\hat j \times \hat j)
Since \hat i \times \hat i = 0\hat j \times \hat j = 0
\hat i \times \hat j = \hat k\hat j \times \hat i = - \hat k
\vec d = 4\hat k - 12\hat k = - 8\hat k



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