PPU 960 Physics Note [Sem 1 : Chapter 2 - Kinematics]

00

STPM

2012

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Facebook: www.facebook.com/groups/josh.lrt Email: [email protected] [Mr. Josh]

Contact No: +6018-397 6808 [Mr. Josh]

Chapter 2 – Kinematics By : Josh, LRT

2012 © LRT Documents Copyrighted. All rights reserved. Page 1 of 14

Chapter 2 – Kinematics 2.1 Linear motion

Kinematics is the branch of classical mechanics that describes the motion of points, bodies (objects)

and systems of bodies (groups of objects) without consideration of the forces that cause it. It deals

with the concepts that are needed to describe motion.

Dynamics deals with the effect that forces have on motion. Together, kinematics and dynamics

form the branch of physics known as Mechanics.

1D – Kinematics Basics:

a. Displacement

𝑥 = final position

∆𝑥 = 𝑥 − 𝑥0 = displacement

𝑥0 = initial position



SAMPLE QUESTION:



b. Velocity & Speed

Speed is a scalar quantity and defined as the rate of change of distance.

=

SI units for speed → meters per second ( )

=

Velocity is a vector quantity and defined as the rate of change of displacement.

=

SI units for velocity → meters per second ( )

=



SAMPLE QUESTION:

Instantaneous velocity, v is defined as the instantaneous rate of change of displacement. (Velocity

at a particular instant)

= lim∆ 0∆

∆

=



c. Acceleration & Deceleration

Acceleration is a vector quantity and defined as is defined as the rate of change of velocity.

=

SI units for velocity → meters per second ( )

Deceleration is a negative Acceleration which decreasing it speed with time.

= −



Let’s understand the graph and the concept of S, V and A.

** This is the way to get the equation!

Going Down, using differentiates function, opposite for the integrate function.

Example:

1. To find V,

=

or = ∫

2. To find A,

=

or =

3. To find S,

= ∫ or ∬



** This is the way to find values from graph!

Going Down, using differentiates function, opposite for the integrate function.

Example:

1. To find V,

=

or = ∫

Graph of v against t, gradient = , Area =

2. To find A,

=

or =

Graph of a against t, gradient = , Area =

3. To find S,

= ∫ or ∬

Graph of s against t, gradient = , Area =



** This is the formulas to calculate 1D! Using Linear Motion formulas:

This only applied for a body that travels along the straight line.

Defining symbol:

=

=

=

=

From acceleration,

=

Acceleration = rate change of velocity

=

= (

) Displacement = Product of average velocity and time

= (

)

=

From, = and =

( − )( ) =

=



2.2 Projectile

Projectile, is when a particle is projected under gravity at a velocity u at an angle θ to the horizontal

(neglecting air resistance, due to what we learn now is VACUUM PHYSICS) it follows the

curve of a parabola.

This motion – Projectile is a 2D motion due to exist of 2 components in the action of kinetic

projectiles. The components are vertical (y-axis) and horizontal (x-axis).

Diagram below is Oblique Projectile:

At Fy, the motion should be constant

acceleration (due to GRAVITY).

At Fx, the motion should be constant

velocity (due to linear).

What we can saw in the diagram is:

1. The 𝒗𝒙 is constant, because there is no

any force [horizontal] acting on it.

2. The 𝒗𝒚 is changing, because the height

of motion per second is different.

[Decreasing towards maximum point and

increasing towards same level of initial

point]

3. At highest point of the trajectory:

𝑣𝑦 = 0 but 𝑣𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

4. Acceleration is constant and vertically

downwards. Therefore, 𝑎 = −𝑔.



Projectile Equations and Formula:

Symbol define:

1. = accele ation

2. = g avit

3. = initial velocit

4. = angle of p ojectile

5. = ma imum height of p ojectile

6. = ange of p ojectile

Since, the gravity is equal to acceleration and it is always towards to the earth. Therefore the

equation is derived as: = − .

To calculate Y component, information’s needed is:

a. = sin

b. = −

c. =

Given that,

=

= −

Since, maximum point is reached, final velocity of = 0.

0 = ( sin ) −

=( )

. → used to calculate MAXIMUM height for component Y.

𝜃

𝑎 = −𝑔

𝑢

𝐻

R



d. Since, an object is projected from a starting time and end with a ending time.

Therefore, =

Given that,

= −

0 = sin −

=

→ used to calculate time of HALF projectile for component Y.

e. Instantaneous can calculate at any time by using the formula below:

Given that,

= −

= ( sin ) −

→ used to calculate any time of projectile for component Y.

f. To calculate FULL TIME of an object projectile, just simply take answer of

⟨ =

⟩ = → time of FULL projectile.

To calculate X component, information’s needed is:

a. = cos

b. Instantaneous horizontal displacement at any time is

= ( cos )

c. To find Range, R which is the total distance from start point X to end point.

= ( cos )

= ( cos ) ( sin

)

= ( sin

)



Let

= = cos 0

= = sin 0

The displacement of the ball in horizontal axis,

=

0 0 = ( cos 0)

ituation =

= 0 0

0

= ( sin 0)

= ( 0 − 00)

=

= −

( 0 ) = ( sin 0) ( 0 0

cos 0) −

( ) (

0 0

cos 0)

0 = ( 0 tan 0) −

( ) [

0 0

(cos 0)

]

0 − ( 0 tan 0)

− ( )

= [ 0 0

(cos 0)

]



(cos 0) =

[

0 0

0 − ( 0tan 0)

−

( )]

=

{

[

0 0

0 − ( 0tan 0)

−

( )]

(cos 0)

}

=

= √

= 0

= 0 [** DUE TO DIRECTION, MOVING TO RIGHT!]

0

PPU 960 Physics Note [Sem 1 : Chapter 2 - Kinematics]

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