Chapter 3 Motion2D

8/3/2019 Chapter 3 Motion2D

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CHAPTER 3

KINEMATICS IN TWODIMENSION


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ObjectiveAble to define and differentiate between scalar and

vector quantities.

Able to solve addition of vector problems using the

graphical method.

Able to solve addition of vector problems using the

component method.

Able to solve subtraction and multiplication of a vectorby scalar problems.

Able to explain, analyze and solve projectile motion.


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KINEMATICS IN TWO DIMENSION

Scalars & VectorsAddition of Vectors

Subtraction & multiplication of vectors by a scalar

Projectile motion


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What is a Vector???

A vector is a mathematical object possessing, and fully

described by, a magnitude and a direction.

The vectors magnitude is equal to the length of the arrow, and

its direction corresponds to where the arrow is pointing.

Its tip represent the point of a vector and the base as its tail.

There are a number of ways to label vectors such as or


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Scalar vs vector

Each physical quantity can be categorized as

either ascalar quantityor a vector quantity.

Scalar quantities are physical quantities that

have magnitude only.

It does not depend on direction

(Examples :mass, distance and time)


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Vector quantities

Vector quantities are physical quantities that possess both

magnitude and direction.

(Examples : velocity, force and momentum)

Example:

Suppose a particle, moves from some point A to some point B

The direction of the tip (arrowhead) represents the direction of the

displacement and the length of the arrow represent the magnitude of the

displacement.

so displacement depends only on the initial and final positions,

the displacement vector is independent of the path taken between these

two point


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Scalar

Quantities

Distance Charge

Power

Work Speed

Vector

Quantities

Displacement Electric field

Force

MomentumVelocity

Example of scalar quantities and vector quantities


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Equality of two vectors

Two vector A and B may be defined to be

equal if they have the same magnitude and

point in the same direction

Example

All the vector in figure are equal even though they

have different starting points


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2 - Addition of Vectors one dimension

For vectors in onedimension, simpleaddition and

subtractionare all thatis needed.

You do need to be

careful about thesigns, as the figureindicates.


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Adding vector

In two dimensions, the situation is somewhat

more complicated.

The easiest way to learn how vector addition

works is to look at it graphically.

There are several way to add vectors

graphically:

1. tip-to-tail

2. Parallelogram


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Useful method for determining the result of adding two vectorwhich make a right angle to each other.

The method not applicable for adding more than two

vectors or for adding vector which are not at 90 deqrees to

each other.

Pythagorean theorem


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Example


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Adding the vectors in the opposite order gives thesame result:


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Tip-to-Tail method

We can add any two vector A and B by placing

the tail so that it meets the tip of A. the sum

A+B, is the vector from the tail of A to the tip

of B.


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Even if the vectors are not at right angles, theycan be added graphically by using the tail-to-tipmethod.


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Parallelogram method

To add A and B using the parallelogram method, place the tail

ofB so that it meets the tail ofA. Take these two vectors to be

the first two adjacent sides of a parallelogram, and draw in

the remaining two sides.

The vector sum, A + B, extends from the tails of

A and B across the diagonal to the opposite corner of the

parallelogram.

If the vectors are perpendicular and unequal in magnitude,

the parallelogram will be a rectangle. If the vectors are

perpendicular and equal in magnitude, the parallelogram willbe a square


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The parallelogram method may also be used; here againthe vectors must betail-to-tip.


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Law of addition

1. Commutative law of addition

when two vector are added, the sum is

independent of the order of the addition

2. Associative law of addition

when three or more vectors are added,their sum is independent of way in which

the individual vectors are qroup together.


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Adding Vectors by Components

Any vector can be expressed as the sum of two other

vectors, which are called its components. Usually theother vectors are chosen so that they are perpendicularto each other.


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If the components areperpendicular, they can befound using trigonometricfunctions.


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Adding vectors:

1. Draw a diagram; add the vectors graphically.

2. Choosexand yaxis.

3. Resolve each vector into xand ycomponents.

4. Calculate each component using sines and cosines.

5. Add the components in each direction.

6. To find the length/magnitude and direction of the vector,use:

3 S bt ti f V t d


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3 - Subtraction of Vectors, andMultiplication of a Vector by a Scalar

In order to subtractvectors, we definethenegativeof a vector, which has thesame magnitude but points in theoppositedirection.

Then we add the negative vector:


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A vector V can be multiplied by a scalar c; the result is avectorcV that has the same directionbut amagnitude cV. If

c is negative, the resultant vector points in the oppositedirection.


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EXAMPLE 1

Suppose object moves from r1 = 4i + 3j to r2 = -6i + 6j in 5

seconds. Find the displacement of the object. Find theaverage velocity of the object.ANSWER :

If object moves from one position (r1) to another(r2) then displacement given by

r = r2

r1 that isr = (x2i + y2j) (x1i + y1j) orr = (x2 x1)i + (y2 y1)j

Therefore,

r = r2 r1

= (-6i + 6j) (4i + 3j)

= (-6 4)i + (6-3)j

= -10i + 3j


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r1 = 4i + 3j to r2 = -6i + 6j

3j

4i-6i

6j

r1

r2

r

-10i

3j


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5 Projectile Motion


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5-Projectile MotionAprojectileis an object moving in two dimensions under theinfluence of Earth's gravity; its path is a parabola.

The speed in the x-direction isconstant; in the y-direction theobject moves with constantacceleration g.

This photograph shows twoballs that start to fall at the sametime. The one on the right hasan initial speed in the x-

direction. It can be seen thatvertical positions of the two ballsare identical at identical times,while the horizontal position ofthe yellow ball increases

linearly.


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It can be understood by analyzing the horizontal and verticalmotions separately.


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If an object is launched at an initial angle of 0with thehorizontal, the analysis is similar except that the initial velocityhas a vertical component.

Solving Problems Involving Projectile


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Solving Problems Involving ProjectileMotion

Projectile motion is motion with constant acceleration in two dimensions,where the acceleration is gand is down.

1. Readthe problem carefully, and choose the object(s) you are going toanalyze.

2. Drawa diagram.

3. Choosean origin and a coordinate system.

4. Decideon the time interval; this is the same in both directions, and includesonly the time the object is moving with constant acceleration g.

5. Examine the xand ymotions separately.

6. List known and unknown quantities. Remember that vx

never changes, and that

vy= 0 at the highest point.

7. Planhow you will proceed. Use the appropriate equations; you may have to

combine some of them.


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J CT T N


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PROJECTILE MOTION

Horizontalcomponent

Verticalcomponent

Kinematic

formula(v = u + at)

vx = ux + axt

( ax = 0 )

vy = v0y + ayt

(ay = -g)

Velocity vx = (uxcos )( ax = 0 )

vy = (v0sin ) gt(ay = -g)

Kinematicformula

(s = ut + at2)

x = uxt + axt2( ax = 0 )

y = v0yt + ayt2(ay = -g)

Distance Sx = (uxcos )t Sy = (uysin )t gt2


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EXAMPLE 2

A movie stunt driver on motorcycle speeds horizontally off a 50.0 m highcliff. How fast must the motorcycle leave the cliff top to land on levelground below, 90.0 m from the base of the cliff where the cameras are?

Ignore the air resistance.

Ans :


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Answer :Sy = 50.0 mSx = 90.0 mUy = 0

Ux = ?t = ?

Use, s = ut + at2

Sy = uyt - gt2

-50.0 = 0 - (9.8)t2

t = 3.19 s

To calculate the initial velocity ;Sx = uxt + axt

2

90.0 = ux(3.19) + 0Ux = 28.2 ms

-1


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EXAMPLE 3

A football is kicked at an angle 37 with a velocity of 20 ms-1 as shown in the figureabove. Calculate

a) The time of travel at the maximum heightb) The maximum heightc) How far away it hits the groundd) The velocity vector at the maximum heighte) The acceleration vector at maximum height.Assume that the ball leave the foot at ground level and ignore the air resistance.

Answer :


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Answer :

= 37u = 26 ms-1

ux = u cos = (20)cos 37 = 16.0 ms-1

uy = u sin = (20)sin 37 = 12.0 ms-1

a) vy = uy gt0 = (12.0) (9.8)tt = 1.22 s

b) Sy = (uysin )t gt2= (12.0)(1.22) ()(9.8)(1.22)2

= 7.35 m

c) Sx = (uxcos )t why times two?= (16)(1.22x2)= 39.04 m

d) Vy = 0vx = ux = 16.0 ms

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e) The acceleration vector is the same at the highest point as it isthroughout the flight, which is 9.8 ms-2 downward.


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ASSIGNMENT 3

A car stuntman is moving horizontally takes off from a point 15.0

m above the ground and lands 60.0 m away as shown in the

figure.

Calculate

a) The time taking off and landing.

b) The speed of the car at take-off

15.0 m

60.0 m
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Thank YouComing up next :

NEWTONS LAW OFMOTIONThank You

Coming up next :NEWTONS LAW OF

MOTION

Y

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Chapter 3 Motion2D

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