Saturday, September 21, 2013

KONVOKESYEN USM 2013 (ke 48)




Sidang Kelima / Fifth Session
Sabtu, 21 September 2013, 3.00 petang
Bachelor of Applied Science / Bachelor of Engineering / Bachelor of Technology 
Tahniah dan syabas  untuk anakanda. Semoga menjadi jurutera  berwibawa berbakti untuk alam dan manusia dan mendapat keredhaan Yang Maha Esa.

Friday, September 13, 2013

Fitnah Di Alam Siber : Penghakiman
Hakim berkata "....mahkamah mendapati perkara memuatkan artikel yang berunsur fitnah dalam sesawang internet telah berleluasa dan ada beberapa pihak yang tidak bertanggungjawab seperti defendan yang bermaharajalela di alam siber, bersembunyi di belakang tirai nama samaran atau ‘anonymity’ untuk menulis apa sahaja yang mereka mahu tanpa mempedulikan kesan akibat penulisan mereka".

"... mahkamah mengambil serius terhadap gejala negatif ini yang telah menular tanpa batasan sejak akhir-akhir ini dan sekiranya tidak dibendung dan dibanteras akan merosakkan ‘social fabric’ masyarakat yang berpotensi menggugat ketenteraman awam dan keselamatan negara".

"... “Dengan ini, saya mahu penghakiman yang diberikan diambil sebagai satu teladan kepada defendan dan juga semua pengguna internet supaya mereka tidak menyalahgunakan kemudahan ini".

"... Sekiranya disalahgunakan, mahkamah akan mengambil pandangan yang amat serius, seperti pepatah “sepandai-pandai tupai melompat, akhirnya jatuh ke tanah juga” dan sepandai-pandainya defendan menyembunyikan diri dengan nama samaran, akhirnya telah didedahkan melalui ‘footprint’ yang beliau tinggalkan di alam siber,” tegasnya.

Pengajarannya : Fikir-fikirkanlah apabila mahu "membaling batu sembunyi tangan" , ada orang yang cedera terkena batu itu sedangkan dia tidak sepatutnya dibaling batu.

Sunday, July 7, 2013

PROJECTILE MOTION : Q & A FOR UNDERSTANDING


1.      A roadrunner runs directly off a cliff with an initial velocity of 3.5 m/s
a)      What are the components of this velocity?
Vx = 3.5 m/s                            Vy = 0 m/s

b)      What will be the horizontal velocity 2 seconds after the bird leaves the cliff?
3.5 m/s – horizontal velocity is unchanging

c)      If the cliff is 300 m high, at what time will the roadrunner reach the ground?
h = dy = ½ x 10 x t2 = 300
300 x 2 / 10 = t2 = 60
t = 7.75 s

d)     How far from the cliff will this bird land?
dx = 3.5 x 7.75 = 27.125 m

e)      If there is a small pond which begins 25m away from the cliff and extends 2.5 meters from there; will the roadrunner land in the pond?
Yes, the pond is from 25 m to 27.5 m, so the roadrunner will land in the pond.

f)       What is the final vertical velocity at which the roadrunner is traveling? [The vertical velocity at the time when the bird reaches the ground]
Vy = 10 x 7.75 + 0 = 77.5 m/s

g)      What is the final horizontal velocity at which the roadrunner is traveling? [The horizontal velocity at the time when the bird reaches the ground]
Vx = 0 + 3.5 = 3.5 m/s

h)      What is the total final velocity of this motion? [magnitude and direction]
V2 = 77.52 + 3.52 = 6018.5
V = 77.579 m/s
q = tan-1 (77.5 / 3.5) = 87.41o below the horizontal


2.      An object (any object) is dropped from a height of 300m
a)      How long does it take this object to fall to the ground?
h = dy = ½ x 10 x t2 = 300
300 x 2 / 10 = t2 = 60
t = 7.75 s

b)      Compare this answer with you answer from question 1, part c). What are the reasons for any similarities or differences?
They are the same because their vertical motions are identical. All objects fall with the same gravitational acceleration, so two objects at the same height with the same initial vertical velocity will reach the ground at the same time.



3.      The intent of a bean bag toss game is to get your bean bag to land on the ‘bull’s-eye’ of a target. The target is set up parallel to the ground and is the same height above the ground as your hand is when you let go of the bean bag. The game’s rules further require you to be 5 m from the center of the target when you release the bag.
a)      Evaluate the following questions for both an angle of 32o and an angle of 58o if the bean bag is thrown with an initial velocity of 6 m/s:
                  32o                                                                   58o
i.        What are the components of velocity?
            Vx32: Cos 32o = Vx32/6                                                Vx58: Cos 58o = Vx58/6           
            6 x Cos 32o = Vx32 = 5.09 m/s                                    6 x Cos 58o = Vx58 = 3.18 m/s

            Vy32: Sin 32o = Vy32/6                                                 Vy58: Sin 58o = Vy58/6
            6 x Sin 32o = Vy32 = 3.18 m/s                                     6 x Sin 58o = Vy58 = 5.09 m/s


ii.      What is the maximum height of the bean bag’s motion?
            tTOP32 = Vy/10 = 3.18/10 = 0.318s                   tTOP58 = Vy58/10 = 5.09/10 = 0.509s
            hMAX32 = ½ x 10 x 0.3182 = 0.506 m               hMAX58 = ½ x 10 x 0.5092 = 1.295 m

iii.    How long will the bean bag be in the air?
            tTOTAL32 = 2 x tTOP32 = 0.636 s                        tTOTAL58 = 2 x tTOP58 = 1.018 s


iv.    How far away from you will the bag land?
            dx32 = 5.09 x 0.636 =   3.24 m                         dx58 = 3.18 x 1.018 = 3.24m

v.      If the center of the bull’s-eye ranges from 4.9 m to 5.1 m away from you, does your bean bag win?
                        No                                                                   No

4.      A stunt driver drives a red mustang convertible up a ramp and off a cliff. The car leaves the ramp at a velocity of 60 m/s at an angle of 45o to the horizontal; the cliff and ramp combined cause the car to begin its projectile motion at a height of 315m above the ground. If you were coordinating this stunt, how far away would you put a landing surface so that your stunt driver was not injured?
In order to find horizontal distance we need horizontal velocity and time. We can find both horizontal and vertical velocity from the initial conditions, but we’ll have to calculate the time it will take for the car to reach the ground. So first we’ll find the components of velocity:

            Vx: Cos 45o = Vx/60                                       Vy: Sin 45o = Vy/60
            60 x Cos 45o = Vx = 42.43 m/s                       60 x Sin 45o = Vy = 42.43 m/s

Horizontal velocity is constant, but the vertical velocity we calculated above is only the initial vertical velocity.
We use the initial vertical velocity to find the time it takes the car to reach the top of its path and fall to the ground. Let’s think about this in two parts; the time it takes to reach hMAX first:

tTOP = 42.43/10 = 4.243 s

Now what about the time it takes to fall from the maximum height? Well first we need to know the maximum height:

hTOP = ½ x 10 x 4.2432 = 90.02 m

hMAX = hTOP + ho =  90.02 + 315 = 405.02 m

Now we calculate the time it takes to fall from a height of 405.02 m:

405.02 = ½ x 10 x tDOWN2

tDOWN2 = 81.004

tDOWN = 9.000 s

Putting these two times together, we have the total time it takes the car to travel up to its maximum height and then fall back down. This is the total time in the air and this is the time we will want to use to solve for horizontal distance.

tTOTAL = tTOP + tDOWN = 4.243 + 9.000 = 13.243 s

dx = 42.43 x 13.243 = 561.9 m

The landing surface is centered 561.9 m from the base of the cliff.

Friday, July 5, 2013

DEFINITION OF SIMPLE HARMONIC MOTION (SHM)

Definition of Simple Harmonic Motion

  1. Simple harmonic motion occurs when the force F acting on an object is directly proportional to the displacement x of the object, but in the opposite direction.
  2. Mathematical statement F = - kx
  3. The force is called a restoring force because it always acts on the object to return it to its equilibrium position.

An object is undergoing simple harmonic motion (SHM) if;
  1. the acceleration of the object is directly proportional to  its displacement from its equilibrium position.
  2. the acceleration is always directed towards the equilibrium position.
- See more at: http://physicsnet.co.uk/a-level-physics-as-a2/further-mechanics/simple-harmonic-motion-shm/#sthash.iGsH4mWo.dpuf
Descriptive terms
  1. The amplitude A is the maximum displacement from the equilibrium position.
  2. The period T is the time for one complete oscillation. After time T the motion repeats itself. In general x(t) = x (t + T).
  3. The frequency f is the number of oscillations per second. The frequency equals the reciprocal of the period.  f = 1/T.
  4. Although simple harmonic motion is not motion in a circle, it is convenient to use angular frequency by defining w = 2pf = 2p/T.
An object is undergoing simple harmonic motion (SHM) if;
  1. the acceleration of the object is directly proportional to  its displacement from its equilibrium position.
  2. the acceleration is always directed towards the equilibrium position.
- See more at: http://physicsnet.co.uk/a-level-physics-as-a2/further-mechanics/simple-harmonic-motion-shm/#sthash.iGsH4mWo.dpuf
An object is undergoing simple harmonic motion (SHM) if;
  1. the acceleration of the object is directly proportional to  its displacement from its equilibrium position.
  2. the acceleration is always directed towards the equilibrium position.
- See more at: http://physicsnet.co.uk/a-level-physics-as-a2/further-mechanics/simple-harmonic-motion-shm/#sthash.72NJe4OB.dpuf

DISTANCE-TIME GRAPH

Interpreting Distance - Time Graph
Generally, time is plotted on the x axis and distance is taken on the y axis of the coordinate system. 

PROJECTILE MOTION GRAPH

Interpreting Projectile Motion Graph
Projectile motion follows the two dimensional kinematics. The trajectory of the projectile is always a parabola. It is not a linear graph.
In the figure below the projectile motion graph is shown.
Vx is showing the Horizontal Component of the velocity of the projectile.
Vy is showing the Vertical Component of the velocity of the projectile.  


 Horizontal Component of the Velocity (Vx)
The horizontal component of the velocity does not changes and the displacements covered by the horizontal components of the velocity are uniform. Horizontal velocity component is equal to the initial velocity component. Gravity force does not affect or does not make any change in the horizontal velocity component of the velocity.

Vertical Component of the Velocity (Vy): 
Vertical component of the velocity does not remain constant during the projectile motion. Gravity force acts on it and changes the vertical component of the velocity of the projectile. The displacements covered by the vertical component of the velocity are not uniform. 

1.During the projectile motion, both the magnitude and direction changes.
2.If the projectile is moving in the upward direction, the vertical component of the velocity is in the upward direction and decrease in its magnitude.
3.When the projectile moves in the downward direction, the direction of the vertical components of the velocity is in the downward direction and increase in the magnitude.

PROJECTILE MOTION EQUATIONS

Projectile Motion Equation is used to find the distance, velocity and time taken in the projectile motion. The Curved path along which the projectile travels is what is known as trajectory. Projectile Motion is the free fall motion of any body in a horizontal path with constant velocity. 

Where Vx is the velocity along x-axis,
       Vxo is the initial velocity along x-axis,
          Vy is the velocity along y-axis,
          Vyo is the initial velocity along y-axis.
  g is the acceleration due to gravity and
         t is the time taken.

 Where Vo is the initial Velocity
 sin θ is the component along y-axis,
 cos θ is the component along x-axis.

Solved Examples 

Question 1:  

A body is projected with a velocity of 20 ms-1 at 50o to the horizontal. Find
 (i) Maximum height reached
(ii) Time of flight and
(iii) Range of the projectile.
Solution:
Initial Velocity Vo = 20 ms-1,
θ = 50o

Time of flight, t = 2V0sinθg
                           = 2×20×sin5009.8
                           = 3.126 s.

Maximum Height reached, H = V2osin2θ2g
                                                    = (20)2sin25022×9.8
                                                    = 11.97 m.

Horizontal Range R = v20sin2θg
                 
= 202sin100o9.8
                                    = 40.196 m.


Question 2:
John is on top of the building and jack is down. If john throws a ball at an angle of 60o and with initial velocity 20 m/s. At what height will the ball reach after 2 s?
Solution: Given: Vyo = 20 m/s,
           Δ t = 2s,
The Vertical velocity in y direction is given by Vy = Vyo sin 60o
                                                                                     = 20 × 32
                                                                                     = 17.32 m/s.
 Vertical distance, y = Vyo t - 12 gt2
                                   = 20 × 2 - 0.5 × 9.8 × 4
                                   = 20.4 m.

Saturday, June 22, 2013

PROJECTILE'S TRAJECTORY

Characteristics of a Projectile'sTrajectory

The force acting is only gravity.

Two components of the projectile's motion - horizontal and vertical motion.

Horizontally Launched Projectiles
Consider a cannonball projected horizontally by a cannon from the top of a very high cliff

In the absence of gravity, the cannonball would continue its horizontal motion at a constant velocity. This is consistent with the law of inertia.

If merely dropped from rest in the presence of gravity, the cannonball would accelerate downward, gaining speed at a rate of 9.8 m/s every second. This is consistent with our conception of free-falling objects accelerating at a rate known as the acceleration of gravity.

 

Cannonball is projected horizontally in the presence of gravity, it would maintain the same horizontal motion as before - a constant horizontal velocity.

The force of gravity will act upon the cannonball to cause the same vertical motion as before - a downward acceleration. The cannonball falls the same amount of distance as it did when it was merely dropped from rest

 




 
 
 

Forces

Horizontal


Motion

No
 Vertical 


Motion

Yes The force of gravity acts downward


Acceleration

No Yes "g" is downward at 9.8 m/s/s


Velocity

Constant Changing (by 9.8 m/s each second)

 

Non-Horizontally Launched Projectiles


 



PROJECTILE

Projectile
A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity such as
i. an object dropped from rest
ii. an object that is thrown vertically upward
iii. an object which is thrown upward at an angle to the horizontal.


Projectile would show a single force acting downwards and labeled force of gravity (or simply Fgrav).
 

Projectile Motion and Inertia

 

 

 

Friday, June 21, 2013

PERHIMPUNAN HARAM....TINGGALKANLAH YANG BURUK

Presiden Pertubuhan Muafakat Sejahtera Masyarakat Malaysia (Muafakat), Ismail Mina Ahmad menegaskan 
"umat Islam perlu mengelak daripada menyertai perhimpunan yang jelas bertentangan dengan hukum syarak kerana cuma mahu menunaikan impian seseorang melalui rusuhan jalanan......"

Pengerusi Sekretariat Ilmuwan Malaysia (Ilmu), Prof. Madya. Dr. Fadlan Mohd. Othman berkata, 
"...selain solat hajat umat Islam juga digalakkan berdoa supaya Allah menggagalkan segala perancangan jahat untuk menimbulkan keadaan huru-hara melalui perhimpunan haram itu. Pegangan Ahli Sunnah Wal Jamaah sendiri mewajibkan umat Islam untuk menyokong serta patuh kepada pemerintah Islam sedia ada. Kalau pemerintah tidak baik sekalipun, dalam Islam ada etika atau adab yang digariskan oleh Nabi Muhammad SAW sendiri iaitu dengan menasihati bukannya dengan berdemonstrasi. Saya menyeru kepada semua umat Islam tidak menyertai perhimpunan tersebut dan sebaliknya bersama-sama kita mendoakan supaya kesejahteraan serta keamanan dinikmati sekarang tidak tergugat...".

Ikutlah nasihat yang baik, tinggalkanlah perbuatan yang merosakkan diri.