Phy 433 Experiment 2

PHY433
EXPERIMENT 2 – MOTION IN ONE DIMENSION
LECTURER’S NAME : PROFESOR MADYA DR SAIDATULAKMAR SHAMSUDDIN
GROUP : RAS2031A
GROUP MEMBER’S NAME:
NO. NAME STUDENT ID

1 HARIS ISKANDAR BIN ZEIFERI IDZLIN 2022827876
2 MOHAMMAD AIDIL IKHWAN BIN ZULKIFLI 2022899608
3 MUHAMMAD AMIN ASRAFF BIN KAMARUDIN 2022498818
4 MUHAMMAD NA’IMULLAH BIN ABD AZIZ 2022659716
Title : Motion In One Dimension
Objectives : Determine g from the given system moving in one dimension.
Apparatus : i) Linear air track

ii) Glider with flag
iii) Photo gate timer
iv) pulley and string
v) slotted mass
vi) electronic balance
Theory :
Figure 1
By applying Newton’s 2nd Law to the system in Figure 1;
(1)
Where;
a= acceleration of mA and mB
mA = mass of glider and flag
mB = mass of slotted mass
g =gravitational acceleration
The acceleration of mA and mB is equal by assuming the string connecting mA and mB is massless and
the friction between the string and pulley is zero.
The instantaneous final horizontal velocity v of mA after travelling a horizontal distance s starting
from initial velocity u = 0, is determined by dividing its body length (L) to the time it takes to pass
through the photogate timer (t)
𝐿
𝑣= (2)
𝑡
Here, the velocity v is assumed to be constant while passing through the photogate timer in time t.
Hence the acceleration of the system, can then be determined by calculating the acceleration of mA
by using an equation of motion v2 = u2 + 2as, which gives
𝑣2
𝑎= (3)
2𝑠
Procedure
1. Mass mA and mB are determined by using an electronic balance.

2. mA is placed on a linear air tract and then connected to mB by a string through a pulley as shown in
Figure 2.
3. Distance s, between mA and a photogate timer is fixed at 31.0 cm.
4. Both masses (mA and mB) are released and the time interval (t) for the flag of length L to pass
through the photogate is recorded. This step is repeated 3 times so that an average value (tave)
could be calculated.
5. mB is increased by adding a slotted mass on it and Step 4 is repeated.
6. Step 4 and 5 is repeated for 5 different values of mass mB.
7. The data is recorded in Table 1.
8. The instantaneous velocity v of mA passing through the photogate is calculated by dividing length
𝐿
L to tave, [𝑣 = 𝑡 ].
𝑎𝑣𝑒
9. The acceleration a of mA is then calculated by using an equation of motion v2 = u2 + 2as.
𝑣2
Since mA starts from rest u = 0, therefore 𝑎 = 2𝑠.
10. The acceleration due to gravity g is determined by calculating the gradient of the best drawn
line of a versus graph.
11. The uncertainty in the measured value of g is determined.
Data And Results
Glider mass with flag, mA = (203.90 + 0.01 g)

Length, L = (4.0 +0.1cm)
Distance from mA at rest to photogate timer, s = (31.0 + 0.1 cm)
mB Time taken ,t Tavg 𝐿 v2 𝑣2

𝑣= 𝑎=
(+ 0.01 ( + 0.0001)x10-1s (+0.0001) 𝑡𝑎𝑣𝑒 2𝑠
x10-1 s (m2/s2) mB/(mA+mB)
g)
t1 t2 t3 (m/s) (m/s2)
20.00 0.6162 0.6213 0.6202 0.6192 0.65 0.42 0.68 0.089
30.00 0.5212 0.4993 0.5113 0.5106 0.78 0.61 0.98 0.128
40.00 0.4450 0.4566 0.4462 0.4493 0.89 0.79 1.27 0.164
50.00 0.3922 0.3960 0.4000 0.3961 1.01 1.02 1.65 0.197
60.00 0.3737 0.3783 0.3535 0.3685 1.09 1.19 1.92 0.227

Calculations
𝑚𝐵
∑
̅̅̅̅̅̅̅̅̅̅̅̅
𝑚𝐵 𝑚𝐴 + 𝑚𝑏 ∑ 𝑎
𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 = ( , 𝑎̅) = ( , )
𝑚𝐴 + 𝑚𝑏 5 5
0.805 6.5
=( , )
5 5
= (0.161,1.3 𝑚/𝑠 2 )
(1.875−0.500)
𝑚𝑏𝑒𝑠𝑡 𝑙𝑖𝑛𝑒 = 0.225−0.070
= 8.87 𝑚/𝑠2 = 𝑔
2.150 − 0.625
𝑚𝑚𝑎𝑥 = = 9.84 𝑚/𝑠 2
0.250 − 0.095
2.475 − 0.750
𝑚𝑚𝑖𝑛 = = 8.21 𝑚/𝑠 2
0.300 − 0.090
𝑚𝑚𝑎𝑥 − 𝑚𝑚𝑖𝑛 9.84 − 8.21

∆𝑚 = = = 0.82 𝑚/𝑠 2 = ∆𝑔
2 2
|𝑔𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡 − 𝑔𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑 |
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝑥 100%
𝑔𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑
|8.87−9.80|
= 𝑥100%
9.80
= 9.49%
Conclusions
The acceleration due to gravity, 𝑔 = 8.87 ± 0.82 𝑚/𝑠 2
Even though, the difference between the value of g determined in this experiment and the accepted
value is 9.49 %, it is still within the calculated experimental error. That is, the g accepted = 9.80 m/s2
lies between g minimum = 8.87 – 0.82 = 8.05 m/s2 and g maximum = 8.87 + 0.82 = 9.69 m/s2.
Possible sources of uncertainties.
1. The estimation of the best line from the plotted graph.
2. The possible inconsistency in the distance from which mA is released.
3. The assumption that the velocity v with which mA of length L passes through the photogate
timer is the instantaneous velocity of mA.
Post Lab Questions
1. State an example of a systematic and a random error involved in this experiment?
Systematic error may come from the delay or faster recording of time by the photogate timer
and random error may be derived from the measurement of distance s on the linear air tract.
The error in estimating the best straight line from the graph may also be random in nature.
2. What source of error do you believe, affected the measurement of g most?
The estimation of the best straight line from the plotted graph.
3. What would the graph of v versus s look like? Why didn’t we use this graph to find a?
𝑣 2 = 2𝑎𝑠 ⇒ 𝑣 ⇒ √2𝑎𝑠
It would be a curve like a graph of 𝑦 = √𝑥.
Acceleration, a cannot be easily determined by finding the gradient of the graph.

Phy 433 Experiment 2

Uploaded by

Copyright:

Available Formats

Phy 433 Experiment 2

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Phy 433 Experiment 2

Uploaded by

Copyright:

Available Formats

PHY433

EXPERIMENT 2 – MOTION IN ONE DIMENSION

LECTURER’S NAME : PROFESOR MADYA DR SAIDATULAKMAR SHAMSUDDIN

GROUP MEMBER’S NAME:

NO. NAME STUDENT ID

Title : Motion In One Dimension

Objectives : Determine g from the given system moving in one dimension.

Apparatus : i) Linear air track

By applying Newton’s 2nd Law to the system in Figure 1;

1. Mass mA and mB are determined by using an electronic balance.

Data And Results

Glider mass with flag, mA = (203.90 + 0.01 g)

mB Time taken ,t Tavg 𝐿 v2 𝑣2

20.00 0.6162 0.6213 0.6202 0.6192 0.65 0.42 0.68 0.089

30.00 0.5212 0.4993 0.5113 0.5106 0.78 0.61 0.98 0.128

40.00 0.4450 0.4566 0.4462 0.4493 0.89 0.79 1.27 0.164

50.00 0.3922 0.3960 0.4000 0.3961 1.01 1.02 1.65 0.197

60.00 0.3737 0.3783 0.3535 0.3685 1.09 1.19 1.92 0.227

𝑚𝑚𝑎𝑥 − 𝑚𝑚𝑖𝑛 9.84 − 8.21

The acceleration due to gravity, 𝑔 = 8.87 ± 0.82 𝑚/𝑠 2

Possible sources of uncertainties.

1. The estimation of the best line from the plotted graph.

2. The possible inconsistency in the distance from which mA is released.

Post Lab Questions

1. State an example of a systematic and a random error involved in this experiment?

2. What source of error do you believe, affected the measurement of g most?

It would be a curve like a graph of 𝑦 = √𝑥.

Acceleration, a cannot be easily determined by finding the gradient of the graph.

You might also like