PROBLEM SET on MOMENTUM for BSCS2.1A, BSIT2.1A, BSIT2.1C, ASCT1.1B,



BSIT2.1A,  BSIT2.1C, ASCT2.1B, BSCS2,1A

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           Note: Please answer with solution on  short bond paper with border and presented                         neatly

            Date of Submission:
                   BSIT2.1 C - September 2, 2015 (Wed)
                   BSIT2.1A - September 2, 2015 (Wed)
                   BSCS2.1A  - September 2, 2015 (Wed)
                  ASCT2.1B  - September 3, 2015 (Thurs)

             Failure to submit on time (during your Physics Time) will be automatic Zero (0) on Seatwork no. 1.
              Follow accordingly,

                                                                                                    Sir George Mata 
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WORK, POWER AND ENERGY

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WORK, POWER AND ENERGY
       Work (W) is energy transferred to or from an object by means of force acting on the object.
                 
                                                                W = F ∙ d
                  * dot product
Note: This is the multiplication of two vectors in which the product  is a scalar quantity.
To obtain the magnitude of work , the following formula is used:
                                                W = Fd cos θ
        W = work
          F = force
          d = displacement
         θ  = angle
       The commonly used units of work is:
                                joules (J) = 1J = 1 N-m = 1 kg m2/s2
                                calorie (cal): 1 cal = 4.18J
                                electron-volt (eV): 1eV = 1.6x 10-19 J
       Work done by a single force
1.       A box was pushed with a 10N force causing it to move by 5m to the same direction as the force. What is the work done?

2.       A 1.0kg ball was thrown upward; reaching up to a 4.50m above the point it was released. What is the work done by the gravity on the ball?


3.       A 100.0 kg crate is being pushed across the factory floor covering a distance of 24m. What is the work done by the normal force exerted by the floor on the crate?

4.       A 100.0N force is applied at an angle of 40° above the horizontal on a 5.0kg object, moving it with a displacement of 15.0m. What is the work done
ENERGY
Energy  - the ability to do work
Mechanical, heat, light... Etch

MECHANICAL ENERGY
      Is the energy possessed by a physical object due to its motion or state.
       FORM
      Kinetic Energy – is the energy an object possessed due to its motion
      Potential Energy – is the energy due to state  of the object; state could mean the position of the object or the extent it was compressed or stretched.
       Kinetic Energy
      Motion of an object
      Depends on its mass and speed
Kinds of Kinetic Energy
Vibrational – energy due to vibrational motion
   Rotational – energy due to rorational motion
   Translational – energy due to motion from one location to another.
KE  = ½ mv2
                                                where
                                                KE-  Kinetic Energy
                                                m= mass
                                                 v = speed
1.       What is the kinetic energy of a 72kg man running at 3.0m/s




Potential Energy
                                is the energy due to the state of the object; state could mean the position of the object or the extent it was compressed.
2 Form
                 Gravitational Potential Energy  PEg
                Elastic Potential Energy PE

       Gravitational Potential Energy (PEg)
      Is the potential energy associated with an object due to the position of the object.
                                                Peg = mgh
                                PE = Potential Energy
                                g -  gravity
                                h - height
1.       What is the potential energy of a 5.4kg vase on the balcony 3.5m above the ground

2.       At what height above the ground should a 4.0kg bag of rice be placed to have a potential energy of 196J?

       Elastic Potential Energy
      Is the potential energy in a stretched or compressed elastic object
Hooke’s Law
                                                Fspring  = -kx
                where k = the spring constant
               x = the amount of compression
PEspring  = 1/2kx2

       LAW OF CONSERVATION OF ENERGY

       FORMS OF ENERGY
      Electrical Energy – energy due to the movement of electrons
      Radiant Energy – is the energy that is carried by electromagnetic waves
      Thermal Energy – or heat, is the energy that causes an increase in temperature
      Sound Energy – is the energy transmitted through substance in longitudinal waves
      Chemical Energy – is energy stored in the chemical bonds of atoms and molecules
      Nuclear Energy – is the energy stored in the nucleus of an atom. Its an energy that holds the nucleus together.
       SOURCES OF ENERGY
      Non-renewable energy source – are those energy sources that cannot be replenish
      Renewable Energy – those that can be replenish in a short period of time
Ex. Hydropower, solar, wind, geothermal and biomass

POWER
       POWER
      The rate at which work is done
      Defines how fast work is done
               
                                P  = W/t
                                where P = power
                                                W = work
                                                t = time
               
       The SI unit for power
      1 Watt   = 1J/s
Another unit used is
1Hp = 746watts
P = W/t
                                P = Fd/t
                                P = F x d/t                           
                                P = F v
                                1 Watt = 1 N m/s

Little Nellie Newton lifts her 40kg body a distance of 0.25m in 2 sec. Find the force she exerts, the work done, and the power she delivered.






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GRAVITY, FREE FALL AND PROJECTILE

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Free Fall Motion

       An object is said to be in the state of free fall if gravity is the only force acting on it
       Assumption
       The motion is along the vertical
       Gravity is the only force
       Acceleration  is always 9.8m/s2
       Air resistance is neglected
       Scenarios
       Dropped
       Viy = 0
       Acceleration is 9.8m/s2 as it falls
       Thrown Downward
       Non zero initial velocity
       Acceleration is 9.8m/s2 as it falls
       Thrown upward
       Non zero initial velocity
       Slow down on its way up
       Zero velocity at the peak
       Accelerate at 9.8m/s2


Example
       A ball was dropped from a very high building
      A. How far below its dropping point will it have a speed of 1960m/s2 . How many seconds after it was dropped will have this speed?   Ans.
      B. What is the speed of the ball and how far is it below its dropping point 5.0s after it was dropped?
      C. How many seconds since the ball was dropped will it be 3,136m below its dropping point? What will be the velocity at the moment?
      What is the acceleration at the peak
      F. What is the acceleration 38.0sec after it was fired
      G. What is the velocity 50s after it was fired?
      I.  At what time is it 3926 m above the ground?
A ball is dropped from rest from the top of a building. Find:
a)      The instantaneous velocity of the ball after 6 sec.
b)      How the ball fell.
c)       The average velocity up to that point.
PROJECTILE MOTION
-          A projectile motion is a case of free falling motion where velocity of the object has a horizontal component.
-          It is also a free falling motion and Gravity is the only force acting on it
-          Has no air resistance
Example
                Volleyball, cannon , baseball
Path of the projectile: Like a parabola

Acceleration
Since there is only acceleration in the vertical direction, the velocity in the horizontal direction is constant, being equal to Vi cos θ. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, being equal to  g .[1] The components of the acceleration are:
ax = 0
a= -9.8m/s or  32.2 ft/s2
Note: Acceleration is entirely vertical along the path of projectile
     

Velocity

The horizontal component of the velocity of the object remains unchanged throughout the motion. The vertical component of the velocity increases linearly, because the acceleration due to gravity is constant. The accelerations in the  x  and  y  directions can be integrated to solve for the components of velocity at any time  t ,
Note: Velocity is entirely horizontal at the peak of the projectile
          Velocity remain the same at the entire horizontal.
Initial Velocity (Vi) is the beginning of the projectile motion,
-          Cannot be zero
Angle (θ)  - angle of elevation with respect to horizontal ground

There are two components of Velocity
Vix = Vi cos θ    (horizontal Component)
Viy = Vi sin θ    (Vertical component)

Maximum Height (dymax) – highest altitude the object reach during the projectile motion
            Vy = 0
dymax =  Viyt + ½ gt2
dymax = Vfy – ½ gt2
2gdymax = Vfy2 – Viy2
2dymax = (Vfy + Viy)t
 Time to reach the maximum height  (t)
      t =   (Vfy – Vi y)/g
TOTAL TIME OF FLIGHT  (T)  = 2 x t
MAXIMUM RANGE – Rmax 

            Rmax  =  Vix T

Example:
An arrow leaves a bow at 30m/s at angle of 30° from the horizontal.
a)    What is the max Range (R)  
b)    Find the maximum Height (dymax)
c)    How long it would take the arrow to land on the other side = (t = time)

Solve first for the Horizontal and Vertical Component of initial velocity
Viy = ?
Vix =?

Then solve for Max. Height  (dymax)   Answer: 11.5m
Then solve for time at Max height : answer: 1.5s
Since max height is at halfway of the projectile, therefore
            Total time of flight (T) = 2t     Answer: 3.0s
Solve for Max range:
            R =  Vix t   Answer: 78m

2. Supposed that a coin rolls off the edge of a table 1.5m high at a horizontal speed of 1.0m/s
a. Find the time it would take before the coin land in the floor
b. Find the Maximum range
Since Vi is already in horizontal, Vix = 1.0m/s

To solve for the time: use 
dymax =  Viyt + ½ gt2    Answer: 0.55s
Given:
     dymax =  -15m (since it is downward)
     g = -9.8m/s2
                           
Solve for Maximum Range:

R =  Vix t        Answer: 0.55m
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NEWTON's LAW


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POINTERS TO REVIEW IN PHYSCIS 1


1. Physics as a science
a. Scientific method
b. Branches of Science
c. What is Physics
d. Theory, model, law and hypothesis
e. SI unit and Si derived unit
f. Prefixes
g. Significant figure
h. Conversion
   in to cm
kg to lbs
km to miles
km to meter
i. scientific notation
2. Scalar and Vectors
a. Sketching
b. Splitting of vector
c. negative vector
d. addition of vector
e. subtraction of vector
f. resultant vector
3. General Concept of Motion
a. total distance
b. total displacement
c. average speed
d. average velocity
e. acceleration

in one dimension and two dimension (please study)

Example: (please study this)
A car is heading to the north and is traveling at 40km/hr and then smoothly made a westward turn and then travels at 25km/hr; hence, it is now heading to the west. What is the change in velocity of the car (∆V) What is the acceleration of the car (a)? The total travel time of the car as it smoothly changed its direction is 15min.


Given: Vi = 40km/hr  due north
           Vf = 25km/hr   due west
           t = 15min    = 0.25hr
           ∆V = ?
           a = ?




            





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LECTURE NO.4 GENERAL CONCEPT OF MOTION

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GENERAL CONCEPTS OF MOTION
Speed -  scalar quantity that refers to how fast an object is moving.
-          Amount of distance travel d per unit time t.
Speed (v) =  d/t
Velocity (v) – a vector quantity
-          The rate at which object changes its position
-          m/s (SI unit)
-          amount of displacement traveled (d) per unit time t
Velocity (v) =  d/t
Average speed and Average velocity
Average speed – the total distance traveled divided by the total time, independent of the direction
                vave =     dT/t
Average Velocity – total displacement divided by time
                vave = d/t
Example:
A car traveled 400.okm to the north. It turned and traveled 100.0km  due east. Then, it turned northwest and traveled 250.0km. Finally, it turn and traveled 200.0km to the south. If the entire trip is 10.0 hrs:  (a) What is the car’s average speed? (b) What is the average velocity
Given:
d1 = 400.0km due north  
d2= 100.0km due east
d3= 250.0km due northwest
d4= 200.0km due south
t=10.0h

(a)    Average speed
vave = dT/t

compute for the total distance (dT)
                dT  =  d1+d2+d3+d4  
                    =  400.0 + 100.0 + 250.0km +200.0km
                    = 950.0km
Compute for the average speed (vave )

vave = dT/t   =   950.0km/10hr   = 95.0km/hr


(b)   Average Velocity

Since  it is a vector quantity, we need to use addition of vector

vave = dT/t  ,  first ween need to compute for total displacement

Total Displacement: (dT)

Sketch:


Get summation of each axis


X
Y
d1
0
400km
d2
100km
0
d3
-250cos45˚ = -176.8km
250sin45˚ = 176.8km
d4
0
-200km

 ∑dx = -76.8km
∑dy= 376.8km



Constant and instantaneous speed and velocity

·         An object is saild to have no constant speed if it neither speeds-up or slows-down
·         An object is said to have constant velocity if it has constant speed and constant direction


Instantaneous speed  - the speed of an object at one particular moment in time
Instantaneous velocity – the velocity of an object at one particular moment in time

Example:
A car entered a bridge at 2:45pm and exited at 5:15pm. If the bridge is 105km long and the car traveled at constant speed across the bridge, what is the speed of the car at 4:00pm?

Solution:
Given:
                D = 105km
                t=   2:45pm to 5:15pm  = 2h 30 min = 2.5hr

Get the constant speed

        v = d/t    =  105km/2.5hr    =  42km/hr

        v4:00 =  42km/hr



There is an acceleration if:
a.       The speed is changing
b.      The speed is constant but the direction is changing
c.       The speed and direction are both changing


Example:
An object uniformly accelerates from 3.0m/s to 15.0m/s in 4.0s. There is no change in the direction of motion. Find (a) the acceleration and (b) the displacement traveled.
In two dimensions

A car is heading to the north and then smoothly made a westward turn; hence, it is now heading to the west. During the travel, the speed of the car remains constant at 1.5km/h. What is the acceleration of the car? The total travel time of the car as it smoothly changed its direction is 15min.



 


To add the velocity vectors:

               

X
Y
vf
0
-1.5km/h
vi
-1.5km/h
0

-1.5km/h
-1.5km/h


DECELERATION:
                Is an acceleration that is directed towards the direction opposite that of velocity

Example:
                An object slow down from 23.0m/s due east to 5.0m/s due east. What is the acceleration? Indicate its direction.




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