Acceleration Slope

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Title: The acceleration of a ball down a slope Criteria: Dn, DCP, CE Aim: i.

ii. iii.

To determine the dependance of the distance moved from rest down the ramp, s, on the time it takes to cover this distance, t. To determine the acceleration of the motion To determine a ÷ sinθ and compare it to the accepted value of g

Raw Data Vertical Height of Ramp /cm ±0.05cm Length of Ramp / cm ±0.05cm Distance moved from rest down the ramp, S / cm ±0.05m 140.00 120.00 100.00 80.00 60.00 40.00 20.00

25.20 153.00 Time Taken1 , t1 / s ± 0.01

Time Taken2 , t2 /s ± 0.01

2.66 2.38 2.18 2.03 1.78 1.41 1.12

2.46 2.44 2.35 2.00 1.69 1.50 1.10

Observation As the distance moved from rest down the ramp decreased, the time taken for it to fall down increases but it did not increase uniformly. Also as the distance moved from rest down the ramp decreased, it became more difficult to stop the stop clock as soon as the ball reached the end of the ramp because the ball covers less distance.

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Processed Table of Results

Vertical Height of Ramp /cm ±0.05cm

25.20

Length of Ramp /cm ±0.05cm

153.00

Angle Of Incline 9.48 Of Slope/ o

Distance Distance moved from moved from Time rest down rest down Taken1 the ramp, S the ramp, S , t1 / s / cm /m ± 0.01 ±0.05cm ± 0.0005m

Time Avg / Relative Taken2 t Uncertaint s 2 , t / s ± 0.01 y in (tAvg) ± 0.01

(tAvg) 2 / s2

Uncertainties in (tAvg) 2 / ± s2

140.00

1.4000

2.66

2.46

2.56

0.00391

6.55

0.05

120.00

1.2000

2.38

2.44

2.41

0.00415

5.81

0.05

100.00

1.0000

2.18

2.35

2.27

0.00441

5.15

0.05

80.00

0.8000

2.03

2.00

2.02

0.00495

4.08

0.04

60.00

0.6000

1.78

1.69

1.74

0.00575

3.03

0.03

40.00

0.4000

1.41

1.50

1.46

0.00685

2.13

0.03

20.00

0.2000

1.12

1.10

1.11

0.00901

1.23

0.02

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Calculation of values and uncertainties in Processed Table Of Results ∆S

= Limit of reading on ruler ÷ 2 = 0.001m ÷ 2 = 0.0005m

∆ t1 or ∆t2

= Smallest unit of measurement on stop watch = 0.01s

tAvg

= ( t1 + t2) ÷ 2

∆tAvg

= (∆ t1 + ∆t2) ÷ 2 = (0.01s + 0.01s) ÷ 2 = 0.01s

Relative Uncertainty in tAvg = (tAvg) 2 = tAvg × tAvg ∆(tAvg)2 =(Relative Uncertainty in tAvg+Relative Uncertainty in tAvg)× (tAvg)2

= To find the angle, θ of the incline of the slope The height of the slope, h = 25.2cm The Length of the slope , s = 153.00cm 153.00 cm 25.20cm θ Sin θ = Opposite ÷ Hypotenuse θ = Sin-1(25.20 ÷ 153.00) θ = 9.48o

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To find the dependence of s on t , a graph of t against s was plotted y

1.8

1.6

1.4

1.2

1

0.8

istanceovrdbyl,/m D

0.6

0.4

0.2

x -0.2

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

Time /s Series 1

As the graph is parabolic, the graph shows that the dependence of s on t isf(x)=0.1634509*x^2.269321; that R²=0.9965 n n s is directionally proportional to t (s α t ) where n > 1. Now to see whether s is directly proportional to t2 where n = 2 a graph of t2 graph of t2 against s will be plotted.

Graph of (time) 2 against s

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2 /s 2 ) vgA (t

5 4.5 4 3.5 3 2.5

etakn, Tim

2 1.5 1 0.5

x 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Distance fallen by ball, s /m

1.1

1.2

1.3

1.4

Comments on Graph Error bars were not used because the error bars obtained from the uncertainties Series 1 were negligible when compared with the appropriate scales used. f(x)=4.5428571*x+0.36285714; R²=0.9955 This graph shows that s is directly proportional to the square of the time taken as the graph of t2 against s is a straight line that almost passes through the origin. This is proven by the formula relating the acceleration, a, of a ball moving down a ramp of length, s, with in a time, t, a = 2s ÷ t2 When t2 is made the subject substitute of the above equation t2 = 2s ÷ a Hence this equation shows that s is directly proportional to the square of the time taken. Therefore, the relationship between s and t obtained by the experiment agrees with that of the equation that relates the s and t.

1.5

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Determining the of acceleration of the ball from the graph of t2 against s and formula t2 = 2s ÷ a From this equation The gradient, m = 2 ÷ a

1

The software automatically applied and generated the calculation for the gradient using the following formulae; Gradient, m = =

Change in Y = Change in t2 Change in X Change in s

The gradient was outputted as 4.5429 s 2m1 from the graphing software. The gradient will be expressed to 4 significant figures since the least significant figure used in the calculation, t2was to 4 S.F. The gradient, m = 4.5429 s 2m-1

2

Equating equation 1 with equation 2 2 ÷ a = 4.5429 a = 2 ÷ 4.5429 a = 0.4402 m s-2 Therefore the acceleration of the ball is 0.4402 m s-2 To find the value of g using the results Force Applied on Ball = mg sin θ Also Force Applied on Ball = ma Therefore equating these two equations mgsin θ = ma gsin θ = a g = a ÷ sin θ Substituting a = 0.4402 and θ = 9.48 o g = 0.4402 ÷ sin (9.48 o) = 2.67 m s-2

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Therefore the acceleration due to gravity is 2.67 m s-2 Percentage Discrepancy of g

= 72.78 % Conclusion And Evaluation Conclusion Distance moved from rest down the ramp, s is related to the time taken, t for it to cover this distance by a proportional relationship. S is directly proportional to t2 as the graph of t2 against S showed a straight line almost passing through the origin. • The acceleration of the ball was 0.4402 m s-2. This acceleration was not accurate because of the stated errors in the experiment. The acceleration due to gravity was 2.67 m s-1 with a percentage error of 72.78 % from its literal value of 9.81 m s-1. This result was highly inaccurate. Although the experimental procedure itself was inaccurate, the main reason for this inaccurate result in the value of g was that the formula for g , g = a ÷ sin θ, does not take in to account the frictional force but only the force provided by the acceleration to due to gravity. •

Comment on Results Looking at the values of t1 and t2, the maximum difference in these values were 0.20 seconds. This is a significant difference in values considering that the times for t1 and t2were 2.46s and 2.64s. Hence, this shows that the errors in the time taken were not precise. The experiment was very inaccurate as shown by the large percentage error of 72.78 % for the value of g. Therefore there must be some significant systematic of errors in the experiment. Sources of error, Limitations and Weaknesses of experiment 1. The ranges of values of s were very small. Therefore, the uncertainties

associated with the experiment had a larger impact on acceleration and the value of g obtained.

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2. Also it was very difficult to stop the stop clock when the ball reached the end of the ramp because a. The ramp was very short b. The ball was moving quite fast This could have caused a systematic error in the timing. 3. For the value of g, the main reason for the inaccurate value is that the frictional force between the ball and wooden ramp was not accounted for in the calculations for g using the formula, g = a ÷ sin θ.

Realistic Improvements to the investigation The ramp should be made longer to about 400cm so that the range of values of s would be higher making the uncertainties associated with the experiment have less impact on the final results 2. The angle of the slope should be lowered so that the ball will not move very fast allowing for accurate timing of the ball. 3. A separate experiment should be carried out to determine the frictional force between the ball and wooden ramp. This could be used to calculate the resultant force on the ball. Resultant Force = Force provided by g – Frictional Force Resultant Force = Force provided by g - (Coefficient of Friction × Normal Force) Resultant Force = mg sin θ - (mgcos θ×µ) Resultant Force = mg (sin θ -µcos θ) Also Resultant force(R) = ma Therefore mg (sin θ -µcos θ) = ma g (sin θ -µcos θ) = a g = a ÷ (sin θ -µcos θ) Therefore this formula should be used to find the value of g. This will give a far more accurate value of g than the previous one the frictional force was not accounted for in the acceleration. 1.