Scalars and Vectors; Multiplication of Vectors by Real Numbers
Scalars vs. Vectors
Scalars Vectors A scalar quantity has only magnitude. A vector quantity has both magnitude and direction. Scalars can be added, subtracted, multiplied, and divided just as the ordinary numbers i.e., scalars are subjected to simple arithmetic operations. Vectors cannot be added, subtracted, and multiplied following simple arithmetic laws. Arithmetic division of vectors is not possible at all. Example: Mass, volume, time, distance, speed, work, temperature Example: Displacement, velocity, acceleration, force
Position Vector Position vector of a point in a coordinate system is the straight line that joins the origin and the point.
Magnitude of the vector is the length of the straight line and its direction is along the angle θ from the +vex-axis.
Displacement Vector Straight line joining the initial and final positions
Equality of Vectors Two vectors and are said to be equal if and only if they have the same magnitude and the same direction.
Multiplication of Vectors by Real Numbers
Multiplication of a vector with positive number k only changes the magnitude of the vector keeping its direction unchanged.
if k > 0.
Multiplication of a vector with negative number −k gives a vector in the opposite direction.
Addition and Subtraction of Vectors
Addition of Vectors: Triangle Method
The given vectors and have to be arranged head to tail, keeping their directions unchanged.
The line , joining the starting point of and the end point of , represents a vector that is the sum (resultant vector) of the vectors and , i.e., = + .
Vector addition obeys commutative law and associative law, i.e.,
+ = + And
Subtraction of Vectors
Difference between two vectors and is defined as the sum of two vectors and.
Parallelogram Method of Vector Addition
The given vectors and have to be arranged, keeping their directions unchanged, such that their starting point is a common point O.
If a parallelogram OQSP is drawn with these two vectors as the two sides, then the diagonal OS is the sum (resultant vector) of the given two vectors.
Length of the diagonal is the magnitude of the resultant vector, and its direction is along the diagonal OS.
Resolution of Vectors
Unit Vector
A unit vector is a vector of unit magnitude and points towards a particular direction.
Unit vector can be expressed as
are three special unit vectors along x, y, and z axes respectively.
Resolution of Vector in Rectangular Components (in Two Dimensions)
The process of splitting a vector into rectangular components is called resolution of vector.
The components of a vector are found by projecting the vector on the axes of a rectangular coordinate system. The coordinate system can be considered according to our convenience.
and are the components of vector along x-axis and y-axis respectively. From triangle law of vector addition, we have
Let and be the unit vectors along x-axis and y-axis respectively.
Hence
From right-angled triangle ORP, ax = a cos q and ay = a sin q Thus, the magnitudes of the components are
Therefore, if the components of a vector are known, then its magnitude and direction can be determined by using the following equations.
Rectangular Components of a Vector in Three Dimensions
Using triangle law of vectors,
Using parallelogram law of vectors,
If, α, β, and γ are the angles which makes with x, y, and z axes respectively, then
And,
Addition of Vectors by Analytical Method
Let and represent the two vectors and , making an angle θ between them.
For right-angled triangle ONS,
However,
Similarly,
Combining equations (ii) and (iii), we obtain
Using equation (iv), we obtain
Where ‘R’ is given by equation (i)
Equation (i) gives the magnitude of the resultant and equation (v) and (vi) its directions.
Equation (i) is known as the law of cosines and equation (iv) as the law of sines.
Direction cosines of vectors
If a vector A makes angles α, β and γ with positive directions of X-axis, Y-axis and Z-axis, then
Motion in a Plane
Displacement
Suppose the particle is at point P at time t and P' at time. The displacement is
In component form,
Velocity
The instantaneous velocity is given by the limiting value of the average velocity as the time interval approaches zero i.e.
Acceleration
The instantaneous acceleration is the limiting value of the average acceleration as the time interval approaches zero i.e,
Motion in a Plane with Constant Acceleration and Relative Velocity in Two Dimensions
Suppose that an object is moving in x-y plane and its acceleration is constant.
Let be the inital velocity of the object at time t = 0 and be the final velocity of the object at time t = t
Then,
In terms of components,
Let be the position vector of the particle at time 0 and be the position vector of the particle at time t.
The displacement is the average velocity multiplied by the time interval.
In component form:
and
Motion in a plane can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions.
Relative Velocity in Two Dimensions
Suppose that two objects A and B are moving with velocities and . Then, velocity of object A relative to that of B is
And the velocity of object B relative to that of A is
A special case − Relative velocity of rain with respect to the moving man Let Velocity of man walking West represented by Velocity of the rain falling downwards Relative velocity of rain with respect to man From the given figure, it is evident that represents the relative velocity of rain with respect to the man.
Note If the man wants to protect himself from rain, then he should hold an umbrella at an angle
towards his motion with the vertical.
Example −
Projectile Motion
Projectile Given Angular Projection
Equation of path of projectile − Suppose at any time t, the object is at point P (x, y). For motion along horizontal direction, the acceleration is zero. The position of the object at any time tis given by,
Here, x0 = 0, ux = u cos θ, ax = 0 [Velocity of an object in the horizontal direction is constant] Putting these values in equation (i),
= ut cos
⇒ For motion along vertical direction, the acceleration ay is −g. The position of the object at any time t along the vertical direction is given by,
∴ Putting the value of t from equation (ii),
⇒ This is an equation of a parabola. Hence, the path of the projectile is a parabola.
Time of flight −
Total time for which the object is in flight It is denoted by T. Total time of flight = Time of ascent + Time of descent ∴ T = t + t = 2t [Time of ascent = Time of descent = t] ⇒ At the highest point H, the vertical component of velocity becomes zero. For vertical motion of the object (from 0 to H), ∴
Maximum height − Maximum height ‘h’ reached by the projectile For vertical upward motion from 0 to H, Using the relation we obtain ⇒
That is,
Horizontal Range − Horizontal distance covered by the object between its point of projection and the point of hitting the ground. It is denoted by R. ‘R’ is the distance travelled during time of flight T. ∴ ⇒ ⇒ For the maximum horizontal range, sin 2θ = 1 = sin 90° ⇒ 2θ = 90°
⇒ θ = 45° ∴ Maximum horizontal range (Rm) is
Uniform Circular Motion
Angular displacement (θ) − Angle traced out by the radius vector at the centre of the circular Path in a given time:
Angular velocity (ω) − Rate of change of angular displacement
Relation between linear velocity and angular velocity
Let ω→ Uniform angular velocity of point object moving along PQ v → Linear speed r → Radius of circular path t → Time at which the object is at point P t + Δt → Time at which the object is at Q Let POQ = Δθ and It means that an object describes an arc PQ of length Δl in time interval Δt. Also, From equations (i) and (ii),
Direction of − Velocity at any point in circular motion is directed along the tangent to the circle at that point in the direction of motion.
Angular acceleration (α) − Rate of change of angular velocity
Relation between linear acceleration and angular acceleration
We know that v = rω Differentiating with respect to time, we obtain
Centripetal acceleration −
Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration. Consider a particle of mass m, moving with a constant speed v and uniform angular velocity ω. Let at any time, the particle is at point P, where and at time t + Δt, the particle be at Q, whereand POQ = Δθ Also, Now, Let and be the velocity vectors of the particle at locations P and Q respectively. We can represent and in magnitude and direction by the tangents and. Since the particle is moving with a uniform speed v, the length of the tangents is equal i.e, To find the change in velocity in time interval t and t + Δt, take an external point. Draw and representing the velocity vectors and . Clearly, From triangle law of vectors, As, lies close to . Then,can be taken as an arc of circle of radius When represents the magnitude of centripetal acceleration at P, which is given by, Thus Direction of centripetal acceleration − Centripetal acceleration vector acts along the radius of the circular path and is directed towards the centre of the circular path.