# physic matriculation 1

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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
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.
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 (��������������������������������   )

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

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

����2 − ����1
���������������� =
����2 − ����1
30.0 − 10.0
=
0.058 − 0.019
20.0
=         �������� ���� −2
0.039
= 5.13 ���� ���� −2

����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

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

22.5 − 0
=
2943 − 0
= 0.007

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

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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