In physics, we can classify quantities as scalars or vectors. Basically, the
difference is that a direction is associated with a vector but not with a
scalar.
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Scalars and Vectors
A scalar quantity is a quantity with magnitude only. It is specified completely
by a single number, along with the proper unit. Scalars can be added,
subtracted, multiplied and divided just as the ordinary numbers. Examples :
current, speed, pressure etc.
Vectors
Quantities which require both magnitude and direction to describe a situation
fully are known as vectors. For example, displacement and velocity are vectors.
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Scalars
A unit vector has a magnitude of one and hence, it actually gives just the
direction of the vector.
A unit vector can be determined by dividing the original vector by its
magnitude
â = a/|a|
Unit vectors along different co–ordinate axis are as shown below:
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Unit vector
Dot Product of Two Vectors
The dot product, also known as the scalar product or inner product, is an
operation that takes two vectors and returns a scalar.
It is defined as the product of the magnitudes of the two vectors and the
cosine of the angle between them. The dot product of two vectors a and b is
denoted as a · b or <a, b>.
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Dot and Cross Product of Two Vectors
where |a| and |b| are the magnitudes of the vectors, and θ is the angle between
them.
Dot product is called scalar product as A, B and cosθ are scalars. Both
vectors have a direction but their scalar product does not have a direction.
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→ →
A
· B = |A| |B| cos(θ)
Dot product is distributive
A . (B + C) = A . B + A . C
Dot product of a vector with itself gives square of its magnitude
A . A = AA cosθ = A
A . (λB) = λ(A . B)
where λ is a real number
The Cross Product of Two Vectors
The cross product, also known as the vector product, is an operation that
takes two vectors and returns a third vector that is perpendicular to both of
the input vectors.
It is defined as the product of the magnitudes of the two vectors, times the sine
of the angle between them, and a unit vector perpendicular to the plane
containing the two vectors. The cross product of two vectors a and b is denoted as
a × b.
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A.B=B.A
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Dot product is commutative
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Properties OF Dot Product of Two Vectors
Properties of The Cross Product of Two Vectors
The vector product is do not have Commutative Property.
A×B = – (B×A)
The following property holds true in case of vector multiplication
(kA)×B= k(A×B) =A×(kB)
If the given vectors are collinear then
A×B= 0
Following the above property, We can say that the vector multiplication of
a vector with itself would be
^ =0
A×A= |A||A|sin0 n
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Where A & B are magnitudes of vectors A and B respectively and θ is the
smaller angle between them. Cross product is called vector product as A, B and
sinθ are scalars. Both vectors have a direction and their vector product has a
same direction.
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A · B = |A| |B| sin(θ)
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→ →
Position and Displacement Vectors
Position and displacement vectors are concepts used in physics and mathematics
to describe the location and movement of objects in space.
Position vector:A position vector is a vector that points from the origin of a coordinate
system to a particular point in space. It represents the position of the
point relative to the origin, and its magnitude is the distance from the
origin to the point.
Position vector of the object is the vector joining the origin to the point where
→
the object lies directed from origin to the point. It is usually denoted by r.
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From the above discussion it also follows that
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^ × ^k=0
^i × ^i=j × ^j=k
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Also in terms of unit vector notation
When an object is displaced from its position at point P to a new position at
-→
point P' (say), then the vector PP' having its tail at P and head at P' is called the
displacement vector of the object corresponding to its motion from P to P'.
Types of Vector
Equal vectors
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Two vectors having same direction and equal magnitude are said to be
equal vectors. This is the necessary and sufficient condition for any two
vectors to be equal.If two vectors P and Q are equal, we can write P=Q
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Displacement vector:-
A vector with zero magnitude and an arbitrary direction is called a zero
vector. It is presented by →
0 and also known as Null vector.
Negative of a Vector
The vector whose magnitude is same as that of A but the direction is opposite
to that of vector A is called a negative of A and is represented by -A.
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Zero vector
Vectors are said to be coplanar if they lie in the same plane or they are
parallel to the same plane, otherwise they are said to be non-coplanar
vectors.
Addition of Vectors
Triangular Law of Vector Addition
According to this law, if two vectors are represented by two adjacent sides of a
triangle, then the third side of the triangle, taken in the opposite direction,
represents the resultant vector.
In the figure, vectors A→ and →
B have an angle θ between them. Let the resultant
be R making an angle α with A. SN is perpendicular dropped from S on the line
→ →
→
→ →
OP. A, B and R are taken as magnitude of A , B and R.
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Coplanar Vectors
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A and B are said to be parallel vectors if they have same direction,and may or
may no thave equal magnitude (A॥B). If the directions are opposite, then A is
anti-parallel to B.
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Parallel vectors
2
2
2
2
OB = (OA +AC) + BC
2
In the right triangle ABC, we have
cos θ = AC/AB and sin θ = BC/AB
⇒ AC = AB cos θ and BC = AB sin θ
⇒ AC = Q cos θ and BC = Q sin θ --- (2)
Substituting values from (2) in (1), we have
2
2
R = (P + Q cos θ) + (Q sin θ)
2
⇒ R = P + Q cos θ + 2PQ cosθ + Q sin θ
⇒ R = P + 2PQ cos θ + Q (cos θ + sin θ)
⇒ R = P + 2PQ cos θ + Q [cos θ + sin θ = 1]
⇒ R = √(P + 2PQ cos θ + Q ) → Magnitude of the resultant vector R
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
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2
OB = OC + BC
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In triangle OCB,
Triangle law of vector addition is used to find the sum of two vectors when
the head of the first vector is joined to the tail of the second vector.
Magnitude of the resultant sum vector R: R2 =
√(P + 2PQ cos θ + Q )
2
2
Direction of the resultant vector R: ϕ = tan-1 [(Q sin θ)/(P + Q cos θ)]
Ques: Two vectors A and B have magnitudes of 4 units and 9 units and make
an angle of 30° with each other. Find the magnitude and direction of the
resultant sum vector using the triangle law of vector addition formula.
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Important Notes on Triangle Law of Vector Addition
If the resultant vector R makes an angle ϕ with the vector P, then the
formulas for its magnitude and direction are:
|R| =
√(P + Q + 2PQ cos θ)
2
2
β = tan-1 [(Q sin θ)/(P + Q cos θ)]
Parallelogram Law of Vector Addition Proof
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If two vectors are represented in magnitude and direction by two adjacent sides
of a parallelogram drawn from a point, then the diagonal of the parallelogram
passing through that point will represent their resultant in magnitude and
direction.
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Parallelogram Law of Vector Addition
2
2
OC = OD + DC
2
⇒ OC = (OA + AD) + DC --- (1)
2
2
2
In the right triangle CAD, we have
cos θ = AD/AC and sin θ = DC/AC
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In right-angled triangle OCD, we have
⇒ AD = AC cos θ and DC = AC sin θ
⇒ AD = Q cos θ and DC = Q sin θ --- (2)
Substituting values from (2) in (1), we have
2
⇒ R = P + Q cos θ + 2PQ cos θ + Q sin θ
⇒ R = P + 2PQ cos θ + Q (cos θ + sin θ)
⇒ R = P + 2PQ cos θ + Q [cos θ + sin θ = 1]
⇒ R = √(P + 2PQ cos θ + Q ) → Magnitude of the resultant vector R
2
2
2
2
2
2
2
2
2
2
2
2
2
2
Next, we will determine the direction of the resultant vector. We have in right
traingle ODC,
tan β = DC/OD
⇒ tan β = Q sin θ/(OA + AD) [From (2)]
⇒ tan β = Q sin θ/(P + Q cos θ) [From (2)]
⇒ β = tan-1[(Q sin θ)/(P + Q cos θ)] → Direction of the resultant vector R
Important Notes on Parallelogram Law of Vector Addition
To apply the Parallelogram Law of Vector Addition, the two vectors are
joined at the tails of each other and form the adjacent sides of a
parallelogram.
The triangle law and the parallelogram law of vector addition are
equivalent and give the same value as the resultant vector.
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2
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2
R 2 = (P + Q cos θ) + (Q sin θ)
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cos θ = AD/AC and sin θ = DC/AC
Multiplication of Vectors by Real Numbers
→
When a vector A is multiplied by a real number n, the quantity obtained is a vector
→
→
→
n A whose magnitude is n times that of the original vector. |n A | = n| A |. Its
direction might be the same or opposite to that of the original vector depending
upon whether n is positive or negative.
→
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Ques:Two forces of magnitudes 4N and 7N act on a body and the angle between
them is 45°. Determine the magnitude and direction of the resultant vector with
the 4N force using the Parallelogram Law of Vector Addition.
→
→
→
If n is zero, the magnitude of n A is also zero. Such a vector, whose magnitude is
→
zero is called a zero vector or a null vector and is denoted by 0.
Since the
magnitude of a null vector is zero, its direction cannot be specified.
If n is a scalar quantity rather than just being a pure number, then the
→
dimension of n A is the product of dimensions of n and A.
MOTION IN 2D (PLANE)
Position vector and Displacement
The position vector r→ of a particle P, located in a plane with reference to
the origin of on xy–coordinate system is given by r→ = x ^i +y ^j , as shown
below.
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→
If n is a negative number, n A and A have opposite directions.
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If n is a positive number, nA and A have the same direction.
Average velocity
Average velocity is given by
→
vavg = Δ r /Δt =(Δx ^i + Δy ^j )/Δt
vavg = vx ^i + vy^j
Instantaneous velocity
Instantaneous velocity is given by,
V = lim
Δv/Δt = dr→ /dt
Δt-0
v= vx ^i + vy^j
Here
vx = dx/dt and vy =dy/dt
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r1 = x 1^i +y1 j^
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Now, if the particle moves along the path as shown to a new position P1 with
→
the position vector r1 ;
√(v +v )
2
y
2
x
Also,
tan θ=(vx /vy)
-1
θ=tan (vx /vy )
Average acceleration
-1
θ=tan (vx /vy ) Average acceleration is given by,
^
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v =
^
Instantaneous acceleration is given by,
a = dv/dt = (dvx /dt) i^ + (dvy /dt)j^
^
^
a = a xi +ayj
Projectile Motion
Projectile motion refers to the motion of an object that is launched into the
air and then moves under the influence of gravity. The object, known as a
projectile, follows a curved path called a parabola.
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Instantaneous acceleration
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a avg= ax i +ay j
If x and y are the coordinates of particle after time t,
x=(u.cosθ).t .....(i)
y=(u.sinθ)t−1/2g.t 2.....(ii)
From the equation (i) by putting the value of t, as a function of x, in equation
(ii), we get
2
y=(u.sinθ).x/(u.cosθ)−1/2g(xu.cosθ)
y=x.tanθ−(g/u2 .cos 2θ)x2
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Equation of Trajectory
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The motion of the projectile is affected by various factors, including its initial
velocity, the angle at which it is launched, and the force of gravity.
The time taken by a projectile to return its initial elevation after projection
is known as time of flight. It is denoted by (T) and given by
T=2u.sinθ/g
Maximum Height Attained
The maximum vertical height traveled by the projectile during its journey is
called the maximum height attained by the projectile.
It is denoted by Hmax and given by
Hmax =u2 .sin2 θ/2g
Horizontal Range
The maximum horizontal distance between the points of projection
and the point of horizontal plane where the projectile hits is called
horizontal range.
It is denoted by R and give by
R=u2.sin2θ/g
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Time of Flight (T)
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If any three quantities, as mentioned, are known then the fourth quantity can be
solved directly.
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The above equation is called the equation of trajectory. As the equation represents
a parabola. Thus, the trajectory (or the path) of a projectile is a parabola. Here,
u, θ, x and y are four variables.
A reference frame is a physical object to which we attach our coordinate
system.
If you observe the motion of a flying kite while standing on the ground, your
reference frame is the ground and if you observe the motion of kite from
inside a car moving on the ground, your reference frame is the car.
Uniform Circular Motion
When an object follows a circular path at a constant speed, the
motion of the object is called uniform circular motion.
v = ωr
Centripetal Acceleration
Centripetal acceleration is the acceleration experienced by an object
moving in a circular path.
the acceleration of an object moving with speed v in a circle of radius R has
a magnitude v2 /R and is always directed towards the centre.
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The velocity of a particle depends on the reference frame from where the
particle is observed.
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Relative Velocity
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Note : The range of projectile will be maximum if θ = 45
→
At an instant the angle θ made by the position vector r of the particle
with the positive direction of x-axis is called the angular position of
particle.
Angular Velocity
The angular velocity of particle at any instant is defined as the rate of
change of angular position θ. That is,
→
→
ω = dθ / dt
This means that the angular velocity is a vector in the direction of dθ.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity with respect to
time.
α = dω / dt
Importent Question
Ques :A boy throws a ball in the air at 60⁰ to the horizontal along a road with a
speed of 10 m/s (36 km/h). Another boy sitting in a passing by car observes the ball.
Sketch the motion of the ball as observed by the boy in the car, if the car has a speed
of (18 km/h). Give an explanation to support your diagram
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Angular Position
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2
a = v2/ R = ω r
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This is why this acceleration is called centripetal acceleration.
sol:
Velocity of car is in the direction of u x and equal to
=18*5/18
=5m/s
Now since they have equal horizontal velocity hence they will cover equal
distances when the ball comes down. But, the ball also has a vertical velocity
of
uy =ucos30
=10.
=5
√3/2
√3m/s
hence it travels upwards in a parabolic trajectory.
Ques : In dealing with the motion of a projectile in the air, we ignore the effect of
air resistance on the motion. This gives trajectory as a parabola as you have studied.
What would the trajectory look like if air resistance is included? sketch such a
trajectory and explain why you have drawn it that way.
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u x =ucos60
=10.1/2
=5m/s
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That gives,
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Given that, the initial velocity of the ball u=10m s
Sol:
The velocity of the fighter plane at a height h=1.5 km=1500 m is given as
u=720km/h=200m/s
Let the plane drops the bomb t seconds before the target is exactly below
the plane
Then in t seconds, the bomb must cover the vertical distance of 1500m under free
fall with initial velocity zero.
Hence,
2
h=ut+1/2gt
That gives,
1500 = 0+1/2.10t
√
or, t= 300=10
2
√3s
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Ques : A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720
km/h. At What angle of sight (w.r.t horizontal) when the target is seen, should the
pilot drop the bomb in order to attack the target?
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If the air resistance is included in a parabolic trajectory then both the horizontal
and vertical speed of the particle will decrease due to that resistance. This will result
in a decrease in the maximum attained height and in the range achieved by the
particle. These two conditions can be compared by the figure drawn below:
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sol:
hence,
tanθ=1500/2000
√3/4
√3
=
√3/4
-1
θ=tan
Ques :A hill is 500 m high. Supplies are to be sent across the hill using a canon that
can hurl packets at a speed of 125m/s over the hill. The canon is located at a distance
of 800 m from the foot of the hill and can be moved on the ground at a speed of 2
m/s; so that its distance from the hill can be adjusted. What is the shortest time in
which a packet can reach on the ground across the hill?
Sol:
We are given that, the speed of packets is 125 m/s and the height of hill is
500 m.
Now, in order to cross the hill by packet, the vertical components of the speed
of packet must be reduced so that it can attain a height of 500 m and the
distance between the hill and the cannon must be half of the range of the
packet
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Hence,
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√3s
ut=200x10
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Now distance covered by the plane is
v2= u2+ 2gh
or,
0 = u2 + 2gh
That gives,
√2gh
=√(2x10x500)
u2 =
=100m/s
Now, we have
u2 = u2x + uy2
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Hence,
or
2
=25-125
=75m/s
t s =10
Hence, time of half flight and time to reach the top of the hill is 10
seconds. Therefore the cannon must be at the half range i.e. at the
horizontal distance in 10 sec
Hence the distance between the hill and the canon is given by
u x 10
=75 x 10
=750
Therefore the required distance for the cannon is 800 750 50 − = m
Therefore, the total time taken by the packet from 800 m away from the hill
to reach the other side of the hill is equal to 25 + 10 + 10 =45 s .
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2
u2x = 125 - 100
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Hence,
(a) Since the football is kicked vertically upwards and the only acting
force on the ball is the gravitational force. Hence, the acceleration at
every point of trajectory of the ball is g .
*After studying from
this notes
NOTE : Worksheet (Important questions of all typology with
answers) is provided as a seperate PDF on website padhleakshay.com
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(b) The velocity of the football at the highest point will be zero because both
components of its velocity will be equal to zero at the highest point.
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Ques:A football is kicked into the air vertically upwards. What is its
(a) acceleration, and (b) velocity at the highest point?