Motion in two dimetion

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Lesson 3: Motions in two dimension (Contd.)

Contents:

Vectors and Scalars

Projectile Motion

Circular Motion

Introduction

Quantities like mass or volume are scalar quantities because they have magnitude buthave no directions. Addition or subtartion of scalar quantities are easy as they can be

simply added or subtracted.

e.g. 40kg + 60kg = 00kg

!00ml "#0 ml = #0ml

Addition of vector Quantities

Parallel Vectors

$hen the vectors are parallel vector addition is easy. $hen both vectors are in same

direction take magnitude of all vectors positive. $hen any one of the vectors is in

opposite direction take the magnitude negative.

Eam!le "

Ahmed %alks 0m from his house due &ast and reach Aisha's house then he %alksanother 4m &ast to (ariyam's house. $hat is his displacement)

*isplacement is 4 + 4 = 4 m &

Eam!le #

Ahmed %alks 0 m from his house due &ast and reaches Aisha's house then he %alks 4

m $est to ameedh's house. $hat is his displacement from his house)


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*isplacement = 0 + ,"4- = 6 m &

Vectors at $i%&t an%le

$hen t%o vectors are at right angle %e can use ythagoras theorem to find the magnitude

of the resultant vectors. /ts direction ,angle- can be found using trigonometry.

Eam!le 3

A boat travels m1s north%ards %hile the ocean current and %aves move it 6m1s

east%ards. $hat %ill be resultant velocity of the boat)

Using Pythagoras Theorem

(agnitude of the 2esultant 3elocity = = 00 = 0 m1s

Using Trigonometry

*irection ,earing-5 tan = pposite side

Ad7acent side


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8an = 61

= tan " ,0.9#-

= !9o& of :

8herefore the boat %ill travel 0 m1s in the direction of !9 o&ast of :orth

Com!onents (at ri%&t an%le) of a vector

A vector such as a force can be resolved into t%o components at right angle.

/f the magnitude of the vector is ;< then

3ertical component< ;y = ; sin

oriontal component< ;> = ; cos

&.g. ?alculate the oriontal and vertical components of the follo%ing force vector.

3ertical ?omponent ;y = 4 sin !6o = @.@ :

oriontal ?omponent ;> = 4 cos !6o = !. :

'utractin% two vectors w&ic& are not !arallel or at ri%&t an%le

. ;irst change the direction of one vectors

@. econd< find the components of t%o vectors!. 8hird< add the vertical components and horiontal components of the vectors

separately.

4. ;ind the magnitude and direction of the resultant vector using the resultantcomponents

$hen %e add t%o vectors %e do not need to change the direction of any vector.


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Circular Motion

$hen an ob7ect is moving in a circle it is in circular motion. Bou may %hirl an ob7ectaround %ith a constant speed. o%ever the velocity changes because the ob7ect'sdirection of travel is changing. 8herefore %e say the ob7ect is accelerating. 8he

centripetal force %hich is to%ards the centre keeps the ob7ect in a circular path. 8herefore

the acceleration is to%ards the centre and it is called centripetal acceleration. 8heinstantaneous velocity is directed at a tangent to the circle.


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8hree e>amples of centripetal force are sho%n belo%.

As a car makes a turn< the

force of friction acting upon

the turned %heels of the car

provides centripetal force

required for circular motion.

As a bucket of %ater is tied to a

string and spun in a circle< the

tension force acting upon the

bucket provides the centripetal

force required for circular

motion.

As the moon orbits the

&arth< the force of

gravity acting upon the

moon provides the

centripetal force

required for circular

motion.

Average speed = distance = @2

8ime 8

?entripetal Acceleration = v@

2

= 4@2

8@$here 8 is period< 2 is radius of the circle and v is average velocity

Proectile Motion

A pro7ectile is an ob7ect upon %hich the only force acting is gravity and also that once

projectedor dropped continues in motion by its o%n inertiaand is influenced only by thedo%n%ard force of gravity. 8here are a variety of e>amples of pro7ectiles. /n all these

e>amples the air resistance is negligible.

An ob7ect dropped from rest is a pro7ectileAn ob7ect %hich is thro%n vertically up%ard is also a pro7ectile

And an ob7ect is %hich thro%n up%ard at an angle to the horiontal is also a pro7ectile
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/newtlaws/u2l1b.htmlhttp://www.glenbrook.k12.il.us/gbssci/Phys/Class/newtlaws/u2l1b.html


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E*uations of Motion

(any problems involve in moving ob7ects can be solved using the follo%ing equations.

v = u + at

s = (v + u)t

s = ut + at2

v2= u2+ 2as

%hereC

u= initial velocity ,m1s-

v= final velocity ,m1s-a= acceleration ,m1s@-

s= displacement ,m-t= time taken ,s-

Example:A ball %as thro%n up%ards from the ground %ith a velocity of !0 m1s. After ho% many

seconds %ill it strike the ground again) Assume g = 0 m1s@ and air resistance is

negligible.

s = 0 m since the ball came back to the starting point

u = "!0 m1s as initially the ball is moving up%arda = +0m1s as acceleration of free fall is do%n%ards.

= ut + D at@

0 = "!0 > t + D > 0t@

0 = "60t + 0t@

0 = 6t + t@

t,t + 6- =0< 8herefore t = 0 s or t =6 s


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Practice Question

Question "

Question #

Question 3


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Question +

Question ,


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Question -

Question

8he boy in the picture above is pushing on the handle of the la%nmo%er %ith a force of

00:.a- ;ind the vertical and horiontal components of the force.

b- /f the la%nmo%er %eights !00 :< %hat is the total do%n%ard force on the ground

%hen the la%nmo%er is being pushed)


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c- /f the la%nmo%er is pulled rather than pushed using the same force on the handle

Motion in two dimetion

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