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# Motion In A Plane-CBSE CLASS 11

Motion in a plane -CBSE CLASS 11

CONTENT LIST

## Motion in plane

Motion in a plane means motion of object in two dimensional plane. Two describe motion in plane two coordinate axes are taken into consideration & they are generally X-axis and Y-axis.

## Scaler and Vector

Physical quantities that have only magnitude and not related to any direction is called a scalar quantity. For example, the quantities such as mass, length, time, temperature, area, volume, speed, density, work etc. are measured by their magnitude only. They can be added  and subtracted by arithmetic rule

Physical quantities that have magnitude and direction both and can be added by triangle rule are called Vector physical quantity.

Characteristics of vectors:
• Vector possess both magnitude and direction.
• Vectors do not obey the ordinary laws of Algebra.
• Vectors change if either magnitude or direction or both change.

## Representation of Vectors

Vectors are denoted by a directed line segment .Length of line segment show the magnitude of vector quantity and arrow show the direction.

Point A is called tail of vector and B is called head of vector. Length of directed line AB =4m signifies magnitude of vector AB and It is directed from point A to point B.

## Position vector

Position vector is a vector in a coordinate system that describe the position of a point in space with respect to origin.

The direction of the position vector starts from the origin and points toward the given point.

In a two-dimensional space, OA represents position vector of A (2, 3) and OB represents the position vector of B (6,5)  with reference to origin O = (0,0)

Position vector of A = OA = 2î + 3

Position vector of B = OB = 6î + 5

## Displacement vector

Let an object move from A (x1,y1) to B (x2,y2) in x-y plane.

Vector AB is called the displacement vector of the object.

Position vector of point A = OA = x1î + y1j

Position vector of point B = OB = x2î + y2j

According to triangle law, we get

OA + AB = OB

=> AB = OB - OA = (x2î + y2j) - (x1î + y1j) = (x2 - x1)î + (y2 - y1)j

Addition of two or  more vectors is done based on by using  Triangle, Parallelogram and Polygon Law of Vectors.

```Triangle Law of Vector AdditionLet we have two vectors, a and b.Draw a line AB representing vector a ( A as the tail and B as the head). Draw another line BC representing vector b ( B as the tail and C as the head)  by keeping head of vector a with tail of b close. Join the line AC with A as the tail and C as the head. The line AC represents the resultant (The resultant is the vector sum of two or more vectors.) sum of the vectors a and b.
Parallelogram Law of Vector Addition ```

Let we have two vectors, a and b.

• Draw a line AB representing vector a using the same scale in
given direction.
• Draw the second vector line AC representing vector b using the
same scale from the tail of the first vector a ie. point A.
• Considering these vectors as the adjacent sides and complete
the parallelogram ABDC.
• The diagonal AB represents the resultant vector (a+b) in both
magnitude and direction.

`What is the formula of magnitude of resultant of two vectors at an angle .The diagrammatic representation of the vectors and the resultant is`
Let a and b be two vector at angle θ
Resultant vector of sum of vector a and b is a + b.
Magnitude of a + b can be determined by considering right angled triangle AED.
AE = AB + BE = a + bcosθ
DE = bsinθ
By applying Pythagoras Theorem, we get
AD² = AE² + DE²
| a + b | = √[(a + bcosθ)² + (bsinθ)²]
=> √[a² + b²cos²θ + 2abcosθ + b²sin²θ]
=> √[a² + 2abcosθ + b²(cos²θ + sin²θ)]
=> √[a² + 2abcosθ + b²]
`             => √[a² + b² + 2abcosθ]`

### How do you find the magnitude of an angle between two vectors?

Now we need to find the angle between the base vector and resultant vector.
If angle between resultant and vector a is α, then
`tan α = bsinθ/(a + bcosθ)`

### How do you find the resultant of two vectors at 90 degrees?

`Let a and b be two vector at angle 90°.We know that,Magnitude of resultant vector of a and b = √[a² + b² + 2abcosθ]=> √[a² + b² + 2abcos90°]=> √[a² + b²]            (cos90° = 0)Angle of resultant from vector a = tanα = bsinθ/(a + bcosθ)=> bsin90°/(a + bcos90°)=> (b x 1)/[a + (b x 0)]=> b/atanα = b/a`

Question: Addition of 2î + 7ĵ and î + ĵ gives ______
a) 3î + 8ĵ
b) î + 35ĵ
c) î + 8ĵ
d) 2î + 7ĵ
Answer: a) 3î + 8ĵ
Question: Subtraction of 2î + 7ĵ and î + ĵ gives ______
a) -î – 6ĵ
b) 3î + 8ĵ
c) î + 6ĵ
d) 7ĵ
Answer: c) î + 6ĵ
Question: When two vectors in the same direction are added, the magnitude of resulting vector is equal to _______
a) Sum of magnitudes of the vectors
b) Difference of magnitudes of the vectors
c) Product of magnitudes of the vectors
d) Sum of the roots of magnitudes of the vectors
Answer: a) Sum of magnitudes of the vectors
Question: A vector, 5 units from the origin, along the X axis, is added to vector 11 units from the origin along the Y axis. What is the resultant vector?
a) 3î + 8ĵ
b) 5î + 11ĵ
c) 11î + 5ĵ
d) 2î + 7ĵ
Answer: b) 5î + 11ĵ
Question: Unit vector which is perpendicular to the vector 4î + 3ĵ is _____
a) ĵ
b) î
c) 4î + 3ĵ
d) k̂
Question: How many variables are required to define the motion of a body in a plane?
a) 1
b) 2
c) 3
d) 4
Question: Two vector having magnitude 2 units and 8 units, the magnitude of resultant of the two vector is :
a) Exactly 6 unit
b) Exactly 10 unit
c) 2 unit or 10 unit or between 2 and 10 unit
d) None of these
Answer: c) 2 unit or 10 unit or between 2 and 10 unit

## Rectangular components of a vector

To determine the component of any vector along a given axis, we drop a perpendicular from head of vector to the given axis .It is also called projection of vector along the axis.

For example OA is the given vector. We have to determine its component along the the horizontal axis (x-axis). If OA makes angle θ with the horizontal axis.

Consider triangle OAC,
Cos θ=OC/OA
=> OC = OA Cos θ
OC is called horizontal component of OA or projection of OA on OB.

For example OA is the given vector. We have to determine its component along the the horizontal axis (x-axis). If OA makes angle θ with the horizontal axis.

Consider triangle OAC,
Sin θ=AC/OA
=> AC = OA sin θ
=> OE = OA sin θ
OE is called vertical component of OA or projection of OA on OD.

## Product of Vectors

Product of vectors are of two types:
i) Cross Product or Vector Product
ii) Dot Product or Scalar Product

Cross product:- When two vector quantity are multiplied and result is also a vector than it is said to be cross product or vector product.

Dot product:- When two vector physical quantity are multiplied and result is scalar than it is said to be dot product or scalar product.

Cross Product or Vector Product

Vector quantity x Vector quantity = Vector quantity

Let two vector a and b are lying in x-y plane, then the cross product |a x b| = |a||b|sinθ
and the direction of resultant vector is perpendicular to both a and b.

Where θ is the angle between a and b.

### Dot Product or Scalar Product

Vector quantity x Vector quantity = Scalar quantity

Let two vector a and b are lying in x-y plane, then the dot product |a x b| = |a||b|cosθ
resultant of dot product is a scalar quantity. Hence they have no direction.

## Right Hand Rule

Direction of a x b is determined with the help of right hand rule as follow.
i) Index finger pointing towards 1st vector.
ii) Middle finger pointing towards 2nd vector.
iii) Thumb pointing towards the direction of cross product.

How do you find the cross product of two vectors?

Let us consider two vectors such as,
A = 2î + 3ĵ + 4k̂
B = 5î + 6ĵ + 7k̂

As we known that,
i × j = k ; j × i = -k ; j × k = i ; k × j = -i ; k × i = j ; i × k = -j
i × i = j × j = k × k = 0

Now,
A × B = (2i + 3j + 4k) × (5i + 6j + 7k)
=> 10(i x i) + 15(j x i) + 20(k x i) + 12(i x j) + 18(j x j) + 24(k x j) + 14(i x k) + 21(j x k) + 28(k x k)
=> 0 - 15k + 20j + 12k + 0 - 24i - 14j + 21i + 0
=> -3i + 6j -3k

Cross product of two vectors using matrix method

Let us consider two vectors such as,
A = 2î + 3ĵ + 4k̂
B = 5î + 6ĵ + 7k̂

Create a matrix as shown below

Now select i column and row, then multiply 3 and 7 and subtract product of 6 and 4. By this step we get coefficient of i = -3.

Now select j column and row, then multiply 4 and 5 and subtract product of 2 and 7. By this step we get coefficient of j = 6.

Now select k column and row, then multiply 2 and 6 and subtract product of 3 and 5. By this step we get coefficient of k = -3.

Now, A x B = -3i + 6j - 3k

How do you find the dot product of two vectors?

Let us consider two vectors such as,
A = 2î + 3ĵ + 4k̂
B = 5î + 6ĵ + 7k̂

Now multiply the coefficients of i, j and k and simply add them, as shown below.

A.B = (2 x 5) + (3 x 6) + (4 x 7) = 10 + 18 + 28 = 56

Question:Vector a has magnitude 4, vector b has magnitude 5, the angle between a and b is 30° and n is the unit vector at right angles to both a and b .What is a × b ?
a) 20n
b) 10n
c) 5n
d) None of these

Question: Vector a has magnitude 3√2, vector b has magnitude 4. The angle between a and b is 135° and n is the unit vector at right angles to both a and b.What is the value of a × b?
a) 12n
b) 12√2n
c) 6√2
d) None of these

Question: Vector a has magnitude 1/√3, vector b has magnitude 10. The angle between a and b is 30° and n is the unit vector at right angles to both a and b.What is the value of a × b?
a) 10√3n
b) 5√3n
c) 5n
d) None of these

Question: Vector a = 5î and vector b = 3ĵ . What is the value of a × b?
a) 15î
b) 15ĵ
c) 15k̂
d) None of these

Question: Vector a = 5î and vector b = 3k̂. What is the value of a × b?
a) -15ĵ
b) 15ĵ
c) 15k̂
d) None of these

## Parallel vectors

• Two vectors are said to be parallel if the angle between them is 0 or 180 degrees.
• If angle between two vectors is zero than they are parallel and have same direction.
• If angle between two vectors is 180 than they are parallel and in opposite direction.They are also called anti parallel vectors.
• Cross product of two parallel vector is always zero.
• ## Projectile motion

Let  an object is projected from point A with velocity u at an angle θ from horizontal. Object acquires maximum height PO in time t second and hit the ground at point B. AB  is called horizontal range of projectile.

## Time of flight:

Time of ascend: t second (time to move object from A to P)
Initial velocity in vertical direction: usinθ
Final velocity in vertical direction: 0  (At point P)
Acceleration: -g
Using equation of motion v = u + at
0 = usinθ - gt
usinθ = gt
t = usinθ/g

Total time taken to reach the maximum height = usinθ/g.

Time of descend: t second (time to move object from P to B)

It will take same time to come down from its maximum height which is usinθ/g.

Total time of flight = time of ascend + time of descend

Total time of flight = t + t = 2t = 2usinθ/g

Total time of flight = 2usinθ/g

## Maximum height of projectile:

Let maximum height of projectile be H.

Initial velocity in vertical direction: usinθ

Acceleration: -g

Time of ascend: usinθ/g

Using equation of motion s = ut+ (1/2)at²

H = usinθ(usinθ/g) + (1/2)(-g)(usinθ/g

H = u²sin²θ/g - u²sin²θ/2g

H = u²sin²θ/2g

Maximum height of projectile = u²sin²θ/2g

## Horizontal range of projectile:

In horizontal direction velocity of projectile = ucosθ (constant)

Time of flight: 2usinθ/g

For uniform velocity, distance travelled in horizontal direction= velocity x time

Distance travelled in horizontal direction = ucosθ(2usinθ/g)

Distance travelled in horizontal direction = 2u²sinθcosθ/g

Distance travelled in horizontal direction = u²sin2θ/g

Horizontal range of projectile = u²sin2θ/g

## Equation of trajectory:

Let any point Q (x, y) at the trajectory at any time t means projectile covers horizontal distance x and vertical distance y in time t.

Horizontal motion of projectile is with uniform velocity .

x= (ucosθ)xt

=> t=(x/ucosθ) ---Equation1

Vertical motion of projectile is with accelerated velocity .

y=(usinθ)xt-(1/2)gt² ------Equation2

Putting value of t in Equation 2

y=(usinθ)x(x/ucosθ)-(1/2)g(x/ucosθ)²

y=(usinθ)x(x/ucosθ)-(1/2)g(x/ucosθ)²

y=xtanθ-(1/2)g(x/ucosθ)²

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