Mathematics SL Exploration Modelling the fall of an object

Transcription

1 Mathematics SL Exploration Modelling the fall of an object Candidate Number: glt548

2 Introduction I was fascinated by the equations modelling the movement of falling objects as a physics student. The SUVAT equations model the projectile motion of a falling object (Hamper, 2014). However, they limited because they do not factor in factors such as air resistance (Hamper, 2014). As a result, I was inspired to set out to verify how well the formulae worked. A meter rule was erected perpendicular to the floor using a retort stand. A ball bearing was dropped parallel to the meter rule. This process was repeated 5 times. A camera was set up perpendicular to the system running at 60 frames per second to film the fall of the ball. Therefore, the smallest increment of time was in 1 of a second. The ball s displacement was 60 recorded frame by frame until it hit the floor. FIGURE 1: How displacement was measured Displacement was measured with a meter rule Retort stand held ruler perpendicular to ground The displacement of the falling ball was recorded until it hit the floor 1

3 Table 1 lists the displacement and time for each experiment. TABLE 1: Data table showing displacement, mean displacement and time for each trial Displacement (S)/m Frame number Time (t)/s Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Mean The data may reflect some inaccuracies due to; 1. Parallax error between the camera and ruler may arise in inaccuracies for the displacement of the ball. This may be rectified by having the camera follow the fall of the ball. 2. The low quality of the frames led to high imprecision because the smallest readings were in 0.01m. To improve upon this a higher-quality resolution camera may be used. 3. The framerate of the camera should be increased to more accurately track the displacement and time of the ball. The following are the definitions for the exploration: 2

4 Displacement will be defined as S and is measured in m. Velocity will be defined as V and is measured in ms 1. Acceleration will be defined as A and is measured in ms 2 Time will be defined as t and is measured in s. Exploration Using the data from Table 1, a scatter plot was plotted to s the relationship between S and t. Graph 1 shows the scatter plot of the mean displacement over time. GRAPH 1: Scatter plot of data Through inspection of the scatter plot in graph 1, the function of S over t starts with a small gradient before increasing towards the end. Therefore, the shape of the scatter plot suggests a quadratic curve, an exponential curve or a sinusoidal curve. These three curves will be considered in this exploration. The relationship between S and t may fit these three curves in reality because; 1. The quadratic nature of the graph may be due to constant acceleration the ball experiences due to gravity on Earth, therefore the displacement of the ball is quadratic 2. The exponential model of the curve may fit the curve because while gravitation may be constant, air resistance acting on the ball will impact the displacement of the ball over time because it slows the ball down, therefore the relationship of S and t may be exponential 3

5 3. The sinusoidal model may fit the scatter plot because the ball bounces up and down within a time period, thus the displacement over time increases before decreasing and therefore forming a sinusoidal curve. Based on these three assumptions, linear regression combined with Pearson s productmoment correlation coefficient will be used to verify the type of curve. Due to the large number of calculations involved for linearising each value of t with respect to quadratic, exponential and sinusoidal models, a spreadsheet was used. As such, the working for linearising the quadratic curve and the following correlation will only be shown. The equation Y 1 = m 1 X 1 + c 1 will be used for the quadratic model, where Y 1 = S, X 1 = t 2, m 1 is the gradient and c 1 is the Y intercept. TABLE 2: Values of S, t, t 2, e t and sin(t) S t t 2 e t sin(t)

6 The values of S and t 2 were then plotted to find m 1 and c 1 with a line of best fit as seen in Graph 2. GRAPH 2: S against t 2 As seen in Graph 1, c 1 = This is noteworthy because the experiment started with S = 0, therefore suggesting systemic error from parallax error in the experiment, thus resulting in c 1 > 0. m 1 is the rate of change of S over t 2, therefore the rate of change of S = S 1 S 2, while the rate of change of t 2 = t 1 2 t 2 2, hence m 1 = S 1 S 2 t 1 2 t 2 2. Therefore, to find m 1 ; m 1 = S 1 S 2 t 1 2 t 2 2 = = Substituting the value of m 1 and c 1 into the equation Y 1 = m 1 X 1 + c 1 gives the full linearised equation, thus; Y 1 = X Substituting Y 1 for S and X 1 for t now puts linearised equation back into the context of displacement over time, therefore the relationship between S and t is: S = t To find the strength of this regression, Pearson s product moment correlation coefficient will be used to measure the strength of linear dependence between S and t 2, where r 1 is Pearson s product moment correlation coefficient for the linearised relationship between S and t 2, where r 1 = S St 2 S S S t 2 and; 5

7 S St 2 = St 2 ( S)( t2 ), S n S = S 2 ( S)2, S n t 2 = t 22 ( t2 ) 2 n r 2 and r 3 will be Pearson s product moment correlation coefficient for S against e t and S against sin(t) respectively. For S S in the relationship between S and t 2 ; S S = (S) 2 ( S)2 = = (5s. f) For S t 2 S and t 2 ; S t 2 = (t 2 ) 2 ( t2 ) 2 = = (5s. f) For S St 2 in the relationship between S and t 2 ; S St 2 = St 2 ( S)( t2 ) = ((0 0) + ( ) + ( )) (8.442)(1.5349) = = (5s. f) For r 1 in the relationship between S and t 2 ; r 1 = = (5s. f) In comparison, r 2 = (5s. f) and r 3 = (5s. f). Since r 1 = and was the highest value of r, this shows that the strongest positive correlation was between S and t 2, hence the quadratic model is the most appropriate model. The lower correlation 6

8 values are also reflected in the more dispersed scatter plots of S against e t and S against sin(t) as seen in Graph 3 and Graph 4. GRAPH 3: Scatter plot of S against e t GRAPH 4: Scatter plot of S against sin(t) As seen in Graph 3, the exponential model is less suitable because the scatter plots after linearisation suggest a relationship that is not linear due to the curve implied by the scatter plots. Therefore, while r 2 indicated a very strong positive linear correlation, the actual shape of the curve as suggested by the scatter plots indicates otherwise. Likewise, for the sinusoidal model, the scatter plots suggest a non-linear relationship because they suggest a curve, with an even weaker r as seen in r 3.Therefore, both the exponential model and the sinusoidal 7

9 model is rejected. Therefore, the quadratic equation S = t will be used to model the fall of the ball bearing. Graph 5 shows the model. GRAPH 5: S = t with data plots This demonstrates that the quadratic model is valid because the model is able to fit in all of the error bars as created by uncertainty in the measurement of S and t. This shows that the model is able to fit the data with an excellent degree of certainty. To further analyse the fall of the object, differentiation will be used to find V. Differentiation gives the rate of change of the gradient. Since the gradient rate of change of S with respect to t, therefore differentiating S will give V because V is the rate of change of displacement over time. ds dt = (2)t ds dt = V V = t Graph 6 shows V against t. 8

10 GRAPH 6: V against t The fact that V is directly proportional to t, as seen by V = 0 when t = 0, is significant because it reflects the reality that the ball had no velocity when it was released. Moreover, the constant gradient shows that A is uniform. Differentiating V will give A because A is the rate of change of V over t. dv dt = t0 dv dt = A A = ms 2 This therefore verifies that A is a constant because there are no variables. However, this is higher than the actual acceleration due to gravity where A = 9.81ms 2. By the process of integration, we reverse the process of differentiating and find the area under the curve. Thus A dt is the area under A, which is the velocity of the ball, otherwise known as V. Likewise, V dt gives the area under V, which is S.Therefore, when integrating that actual value of A, this shows that the model of displacement of the ball should be lower. S = V dt = A dt S = 4.905t 2 + c It will be assumed that c = 0 in the process on integrating A back to S because the ball should have started at S = 0 in the experiment, thus c = 0. Graph 7 compares the model S = t versus S = 4.905t 2 (as predicted from A = ms 2 ). 9

11 GRAPH 7: S = t versus S = 4.905t 2 As seen in Graph 7, the model S = 4.905t 2 shows that the displacement of the ball over time should be lower than model created in the investigation, S = t The error in predicting S increases greatly because both graphs are exponential, thus the distance between S for the 2 graphs increases greatly as t increases. Moreover, the model S = 4.905t 2 gradually moves out of the error bars for t, thus indicating increasing error as t increases. However, the model of S = 4.905t 2 still fits within majority of error bars, the model S = t is only useful in modelling the fall of the ball for small values of t. In conclusion, this investigation has found that SUVAT works to a reasonable degree of accuracy. As a result of this investigation, this has broadened my view on SUVAT because this shows that models do not necessarily conform to our observations. A possible real life application to this model would be to predict the fall of objects from greater heights to investigate the safety of people exposed to falling objects. Bibliography Hamper, C. (2014). Pearson Baccalaureate: Physics Standard Level for the IB Diploma. Essex: Pearson Education Limited. 10