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					NAME         : MOHD HAZWAN BIN MD KHALID

MATRIX NO.   : MT0921533103

CLASS        : 1K02

LECTURER     : Mr. Amri Bin Ab. Rahman

DATE         : 18 JUNE 2009
                  Experiment 2: Free Fall and Projectile Motion



Objective :   To determine the acceleration due to gravity, g using two methods:
              a.     Free fall motion
              b.     Projectile motion




Theory :

A) Free fall motion




                                          Figure 2.1
When a body of mass, m falls freely from a certain height, h above the ground, it experiences a
linear motion. The body will obey the usually equation of motion, that is

                                           1
                               s = ut + at2……………. 2.1
                                           2



By substituting,

s = -h = downward displacement of the body from the falling point to the ground

u = 0 = the initial velocity of the body

a = -g = the downward acceleration due to gravity

t = time taken for the body to reach the ground



We obtain the displacement of the body, h as



                                      1
                                h = gt2……………… 2.2
                                      2



B) Projectile motion

From the law of conservation of energy, the potential energy of a body of mass, m equal its
kinetic energy that is



                                      1
                            mgh = mvx2………………. 2.3
                                      2



where      m = the mass of the body

           g = the acceleration due to gravity

           h = the height of the body

           Vx = the velocity of the body
A body which is moving in a projectile motion with the velocity of vx will have a range of

                             R = vxt………………….. 2.4


where vx is the horizontal component of the velocity.




Combining equation 2.3 and 2.4, we obtain

                                   ���� 2
                             h=             ……………….. (2.5)
                                  2�������� 2



where t is the time taken for the body from the end of the curved track to reach the ground
Apparatus

     Apparatus                    Range                 Sensitivity           Uncertainty
        Timer           (0.000000 – 100.000000)s        0.000001s             ±0.000001s
     Metre rule              (0.0 – 100.0)cm               1mm                  ±1mm
    Retort stand
  Free fall adaptor
  Horizontal table
      Steel ball
   1 carbon paper
  1 drawing paper
  Cellophane tape
      Plasticine
       Scissors
        String
   Pendulum bob
    Curved railing


Procedures

A. Free-fall Motion
   1) The apparatus was set up as shown in Figure 2.1.
   2) The height, h of the electromagnet base was adjusted above the point of impact, and
       begun with a small value of h.
   3) The value of h was recorded.
   4) The circuit was switched on and the steel ball was attached onto the upper contact.
   5) The circuit was switched off and the ball was let fell.
   6) 8 sets of reading were taken at different values of h and t.
   7) A graph of h against t2 was plotted to determine the value of g.


B. Projectile Motion

       1) The apparatus was set up as shown in Figure 2.2.
       2) The steel ball was slide on the curvature railing from 8 different heights, h and the
          values of R were recorded.
       3) The height, H was measured from the edge of the railing to the landing surface. By
          referring to the graph of h against t2 from experiment A, the value of t2 for H was
          obtained by using extrapolation.
       4) A graph of h against R2 was plotted and the value of g was calculated.
       5) The value of g obtained from both experiments was compared with standard value.
          As so comments were given.
Observation

A. Free fall motion

  No.               Height, h (±0.1)cm            Time, t (0.000001)s         t2,s2
   1                         5.0                      0.096777             0.009
   2                        10.0                      0.136659             0.019
   3                        15.0                      0.165403             0.027
   4                        20.0                      0.197505             0.039
   5                        25.0                      0.222394             0.049
   6                        30.0                      0.204749             0.058
   7                        35.0                      0.264239             0.070
   8                        40.0                      0.280573             0.079
                                                                          2
average                   ħ=22.5                                         ŧ =0.044


       (5.0+10.0+15.0+20.0+25.0+30.0+35.0+40.)
ħ=
                             8

  = 22.5 cm


       (0.009+0.019+0.027+0.039+0.049+0.058+0.070+0.079)
ŧ2 =
                                 8

   = 0.044 ���� 2



Centroid point is (ŧ2 , ħ)

(ŧ2 , ħ) is (22.5 , 0.044)



CALCULATION

    A. Free fall motion

                                                    1
                                         ���� = �������� + �������� 2
                                                    2
S = -h = downward displacement of the body from falling point to the ground
u = 0 = initial velocity
a = -g =downward acceleration due to the gravity
hence :

                                                   1
                                       −���� = 0 ���� − �������� 2
                                                   2
                                                1
                                         −���� = − �������� 2
                                                2
                                               1
                                           ���� = �������� 2
                                               2



refer to graph :

                                           ���� = ���� ����������������

                                          ���� 2 = ���� ����������������



                                    1
                                   ∴ ���� = ����(��������������������������������           )
                                    2
                                              1
                                       ���� =     ����
                                              2 (��������������������������������   )




Gradient for the best line:
                                                   ����2 − ����1
                                         �������� =
                                                   ����2 − ����1

                                                  22.5 − 0
                                        �������� =
                                                 0.044 − 0
                                       22.5
                              �������� =         = 511.36 �������� ���� −2
                                       0.044
                                                 = 5.11 m s-2



Gradient for maximum line:
                                                      ����2 − ����1
                                        ���������������� =
                                                      ����2 − ����1
                                              30.0 − 10.0
                                        =
                                             0.058 − 0.019
                                             20.0
                                         =         �������� ���� −2
                                             0.039
                                           = 5.13 ���� ���� −2




Gradient for minimum line:


                                                      ����2 − ����1
                                         ���������������� =
                                                      ����2 − ����1

                                              35.0 − 10.0
                                         =
                                             0.07 − 0.019
                                          = 490.19�������� ���� −2

                                          = 4.90 ���� ���� −2



                                   ������������������������������������ = �������������������� ± ∆����
                                               ���������������� − ����������������
                                     ∆���� =
                                                        2
                                              5.13 − 4.90
                                          =
                                                   2
                                           = 0.16 ���� ���� −2



������������������������������������ = 5.11 ± 0.16

∴ ���� = 2����

     = 2 (0.5.11)

     = 10.22 m s-2
B. Projectile Motion



   No.         Height, h             R,cm2                            R2,cm2
              (±0.1)cm
   1              5.0                24.80                               615.04
   2             10.0                38.10                              1451.61
   3             15.0                46.00                              2116.00
   4             20.0                51.80                              2683.24
   5             25.0                57.50                              3306.25
   6             30.0                62.00                              3844.00
   7             35.0                65.80                              4329.64
   8             40.0                72.10                              5198.41
Average=        ħ=22.5                                         2
                                                             ���� ���������������������������� = 2943.02


                         5.0 + 10.0 + 15.0 + 20.0 + 25.0 + 30.0 + 35.0 + 40.0
                  ���� =
                                                  8
                                             = 22.5 ��������

         615.04 + 1451.61 + 2116.00 + 2683.24 + 3306.25 + 3844.00 + 4329.64 + 5198.41
���� 2 =
                                               8
                                         = 2943.02 ��������2



Centroid point (R2 , H) is (2943.02,22.5)




CALCULATION

Gradient for the best line.
                                                     ����2 − ����1
                                            �������� =
                                                     ����2 − ����1

                                                 22.5 − 0
                                             =
                                                 2943 − 0
                                                 = 0.007
Gradient for mmax

                                                             35.0 − 15.0
                                              ���������������� =
                                                            4400 − 1800
                                                      = 0.007



Gradient for mmin

                                                             29.0 − 9.0
                                               ���������������� =
                                                            4000 − 800
                                                      = 0.006


         ���� ������������ −���� ������������
∆���� =              2

    0.007 − 0.006
=                 = 0.0005
          2



������������������������������������ = ���� ± ∆����

               = 0.007 ± 0.0005



                                                              ���� 2
                                                      ���� =
                                                              2��������

                                                             1 2
                                                    ���� =          ����
                                                            2��������



                                    85−0
From graph 1.1            ���� 2 =
                                    0.007

                            ���� = 12142.85

                                = 110.19 ����
 1
      = ������������������������������������
2��������

                 1
���� =
       2���� ������������������������������������

              1
   =
       2 110.19 0.007

   = 0.65 �������� ���� −1

���� = 0.0065 ���� ���� −1



Discussion:

          The probability of doing error was very high.
          The error might be done by the student when the student adjust the height of the retord
           stand because the meter rule position might not 900 to the horizontal table which cause
           many error in record the value for height.
          The experiment was carried out in open ventilation system which cause the air resistance
           to the steel ball when falling.
          For experiment B, which is using projectile motion, the error might be made by the
           student is higher than free fall motion experiment. Moreover, in experiment B there will
           be more resistance than experiment B, there were more resistance which is air resistance
           and frictional force between curvature railing with the steel ball.
          The experiment was failed because the acceleration due to gravity is 1.44ms-2 for free
           fall motion experiment and 0.46ms-2 for projectile motion.
          The theoretical value for acceleration due to gravity is 9.81ms-2
          The experiment was failed because the difference between theoretical value of
           acceleration due to gravity was too big and unexpected.
          To reduce the error, our eyes must be perpendicular to the scale in order to avoid parallax
           error. Besides that, we must make sure the curvature railing was straight to ensure that
           the ball can slide in a straight line.
          When taking or record the value of range, we must draw a vertical or horizontal line as
           the reference line and the meter ruler must 900to the reference line.
Conclusion:

      Both experiment must use the same height because when the height is same, so the time
       taken for whole flight time will be also the same.
      To calculate the acceleration due to gravitational for free fall motion using equation
       below:

                                                             1
                                                         ���� = �������� 2
                                                             2
                ∴ ���������������������������������������������������� ������������������������������������������������ = 2(�������������������������������� �������� ������������ ��������������������)

      From the graph of free fall motion, we can determine the flight time that we used it to
       find the gravitational acceleration of projectile motion.
      The equation to find acceleration due to gravitational for projectile motion as shown
       below:



                                                                 ���� 2
                                                          ���� =
                                                                 2��������

                                         1
                                              = �������������������������������� �������� ������������ ��������������������
                                        2��������

                                                               1
                                   ���� =
                                            2 ���� (�������������������������������� �������� ������������ ��������������������)

				
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