Lab 6: Physics 4A Trajectories

                                                            Trajectories Laboratory 
                                                                 I-Shou Lin  & Kirk Paderes
                                                                        March  15th, 2017'
'
Explore the behavior of projectile motion in two dimension in order to determine the impact point of a ball on an inclined board.

When a projectile is in motion across space, the horizontal and vertical components of its motion may seem related and indistinguishable. However, the horizontal and vertical components of its velocity are independent of each other, and the only quantity relating them is time. When a projectile motion is in free fall, only its vertical motion is affected by the acceleration due to gravity. Its horizontal component of velocity remains constant. The horizontal and vertical components of its velocity can be combined together through vector addition to yield the overall velocity at a given point in time.


The laboratory procedure consisted mainly of setting up the apparatus                    
                            
The black cover at the floor is carbon paper that leaves a mark when the ball lands on it. The ball starts accelerating and falls on the floor on top of the carbon paper. From the measurements, determine the ball's horizontal velocity. Then, attach a board such that it couches the end of the lab table and the floor. Place carbon paper to determine where the ball lands on the paper which is on the board. 

                                             Data Table
                 

Horizontal distance	Height	Distance	Angle
67.3+/- 0.1 cm	89.1 +/- 0.1 cm	70 +/- 2 cm	46 +/- 0.5

     

                                                           Calculated Results

                         velocity= ( 1.578 m/s to 1.579 m/s to 1.581 m/s) 

Distance= ( 0.737 m to 0.758 m to 0.780 m)


Conclusions 

This trajectory laboratory was really simple and straightforward. Assembling the apparatus was self-explanatory and easy. For the first part of the lab where the horizontal velocity needs to be determined from the height and displacement, measuring the height is easy, but for the displacement, it's where the uncertainty values come to play with the impact points of the steel ball on the carbon paper. Since the steel ball is dense, has a small surface area, and drops from a low height, the air resistance's effect on the ball is negligible. The second part of the lab was more interesting because now there was a board that was elevated at an angle to the ground in which the ball would have to strike. The derivation of the formula for the distance is in the photo. Then we performed the experiment and compared the theoretical to the experimental results for distance.  The propagated uncertainty for the distance was 0.22 m due to the uncertainty in the displacement and subsequently the velocity of the ball. Differences between the experimental and theoretical values for distance is 5.8 cm a intermediate range, 5.7 cm for lower bound. and 6 cm for the upper bound. These significant differences is likely the result of propagated uncertainty values from measurements and setup of apparatus that affected later results. 

Lab 4: Propagated Uncertainty in Measurements

                                        Propagated Uncertainty in Measurements
                                                  Ian Lin 
                                              March 6, 2017

           To measure the density of metal cylinders taking into account propagated uncertainty


The density of any object can be calculated through its simple formula of mass divided by volume. In the case of a metal cylinder, the mass is determined through the use of a balance, and the volume is calculated through the formula of v=  πr2h due to it being a right cylinder. However, inherent in these measurements of the mass, radius, and height are uncertainties that are transmitted into the density value through propagated uncertainty.

The first step in measuring the density is to mass the metal cylinder on an electronic balance. Then the volume formula for a right cylinder is applied in order to find the volume. Finally, the density value is calculated through mass divided by volume.
       ( electronic balance)


                                          Data Table

Type	Values
Mass	70.58 g
Height	8.4 cm
Diameter	1.8 cm
Density	3.3 g /cm^3

This was a simple and straightforward lab. All I had to do was measure the mass of the metal cylinder and calculate the volume of that cylinder through the formula. Then the density of the cylinder could be determined by its formula. The uncertainty values for each of the measurement types were in the last digit to the right of the decimal point. The total propagated error in the density value is +/- 0.332 g/cm^3. That is within 10% of the density value calculations. 

Lab 3: Non-Constant Acceleration Lab

                             Non-Constant Acceleration Lab
                                        Ian Lin March 8, 2017
This experiment is trying to analyze the numeric and analytical approach to solving when an object in question reaches zero velocity when the acceleration is not constant.

Most acceleration underwent by objects in the world are not constant. In fact, even objects accelerating under the influence of gravity isn't constant due to air resistance. Therefore, when determining the three crucial variables of acceleration, velocity, and position, one has to use calculus. However, some functions that express acceleration as a function of time is complex and hard to integrate. This gives rise to the numeric approach of using an excel spreadsheet to measure accurately measure acceleration, velocity, and position.

We entered the relevant information onto the excel spreadsheet and let the computer compute the values. The initial variables are the mass, velocity, burn rate, force, and change in time. Then the columns were filled down with time, acceleration, average acceleration, change in velocity, velocity, average velocity, displacement, and position. That's basically all there is to it.

Mo	6500	kg		a given by	F / (Mo - bt)
Vo	25	m/s			(-8000) / (6500 - (1500-20)t)
b	20	kg/s
F	-8000	N
Δt	1	s
t	a	a_avg	Δv	v	v_avg	Δx	x
0	-1.230769231			25			0
1	-1.234567901	-1.232668566	-1.232668566	23.76733143	24.38366572	24.38366572	24.38366572
2	-1.238390093	-1.236478997	-1.236478997	22.53085244	23.14909194	23.14909194	47.53275765
3	-1.242236025	-1.240313059	-1.240313059	21.29053938	21.91069591	21.91069591	69.44345356
4	-1.246105919	-1.244170972	-1.244170972	20.04636841	20.66845389	20.66845389	90.11190745
5	-1.25	-1.24805296	-1.24805296	18.79831545	19.42234193	19.42234193	109.5342494
6	-1.253918495	-1.251959248	-1.251959248	17.5463562	18.17233582	18.17233582	127.7065852
7	-1.257861635	-1.255890065	-1.255890065	16.29046613	16.91841117	16.91841117	144.6249964
8	-1.261829653	-1.259845644	-1.259845644	15.03062049	15.66054331	15.66054331	160.2855397
9	-1.265822785	-1.263826219	-1.263826219	13.76679427	14.39870738	14.39870738	174.6842471
10	-1.26984127	-1.267832027	-1.267832027	12.49896224	13.13287826	13.13287826	187.8171253
11	-1.27388535	-1.27186331	-1.27186331	11.22709893	11.86303059	11.86303059	199.6801559
12	-1.277955272	-1.275920311	-1.275920311	9.951178622	10.58913878	10.58913878	210.2692947
13	-1.282051282	-1.280003277	-1.280003277	8.671175346	9.311176984	9.311176984	219.5804717
14	-1.286173633	-1.284112458	-1.284112458	7.387062888	8.029119117	8.029119117	227.6095908
15	-1.290322581	-1.288248107	-1.288248107	6.098814781	6.742938834	6.742938834	234.3525296
16	-1.294498382	-1.292410481	-1.292410481	4.8064043	5.45260954	5.45260954	239.8051392
17	-1.298701299	-1.29659984	-1.29659984	3.509804459	4.158104379	4.158104379	243.9632435
18	-1.302931596	-1.300816447	-1.300816447	2.208988012	2.859396236	2.859396236	246.8226398
19	-1.307189542	-1.305060569	-1.305060569	0.903927443	1.556457727	1.556457727	248.3790975
20	-1.31147541	-1.309332476	-1.309332476	-0.405405034	0.249261204	0.249261204	248.6283587
21	-1.315789474	-1.313632442	-1.313632442	-1.719037475	-1.062221254	-1.062221254	247.5661375

     
                     
                                                                                 Acceleration
                                         acceleration= F( net) / m(t)= -8000 n /6500 kg- 20 kg/s * T
                                         acceleration (1)= -8000 n /6500-20* 1 = -1.234567901 m/s^2

                                                                               Change in Velocity
                                           Δv= average acceleration *Δt
                                        Δv= -1.234567901 m/s^2 * 1s=-1.234567901 m/s

                                                                       Velocity
                                                              velocity= Δv+ v ( initial)
                                            Velocity (1 s)= -1.23 m/s + 25 m/s=  23.77 m/s 
    
                                                                         Δ x
                                                             Δx =velocity average *ΔT 
                                                           Δx ( 1 s ) = 24.38 m/s * 1 s = 24.38 m

                                                       
  This lab didn't involve any physical measurement of any processes at all. All of the lab was on paper and excel spreadsheet. The problem was to figure out how far the elephant that had a variable acceleration with time would go before slowing down to a stop. The analytical approach is to use integral calculus to figure out velocity and position and then setting velocity of the elephant equal to zero and then plugging that time into the position function to find the displacement. The excel spreadsheet method allowed the computer to calculate numerically to great precision I might add, the distance that the elephant travels before coming to rest. The two answers from the two methods agree closely to a tenth of a meter. 

1. The two answers agree closely: analytical answer: 248.7 m , numerical answer: 248.9 m 

2. The time interval for doing the integration is small enough when there're a bunch of velocity values that are close to 0 by 0.1 m increment, and the distance values that correspond to the velocity values differ by 0.1 m increment. When the distance values appear to differ by smaller and smaller values so that the precision is sufficiently small, then the numerical result is good enough.

3. The elephant would go about 129 meters given the conditions. 
                   

Lab 2: Free Fall Lab

                                                                       Free Fall Lab
                                            Ian, Mo, and Caesar 
                                                   3/1/17

The fact that in the absence of any air resistance and other external forces, any object in free fall will accelerate at a constant rate of 9.81 m/s^2

When objects are relatively close to the surface of the earth, they undergo a constant acceleration of g when in free fall. In order to measure physically the effects of g on objects, a measuring apparatus must be used. This apparatus is a 1.5 m long column that is anchored to a heavy tripod base with an electromagnet and a spark sensitive tape. When the object is released from rest, its displacement is recorded with on the tape at 60 frames/ second. Recording and graphing this data gives us the position vs time graph for the object. Then, in the velocity vs time graph is obtained by using the formula of  Δx / the time due to the frequency = 1/60th of a second. This will give us the desired result. Finally, the slope of the velocity vs time graph as determined through the " equation on chart" on excel is the value of g because acceleration is change in velocity per time.

The measurement/ observation portion of the lab consisted of setting up the column with the tripod base with the tape. Since all of the actual processes of recording the spark on the tape are already completed. This initial step is just for reference. With the tape already marked, the next step is to measure the distance of each mark from the 0-cm mark. Then, we proceed to find displacement, mid-interval time, and mid-interval speed. All of these data are recorded on an excel spreadsheet.

TIME(s)	DISTANCE (cm)	∆x (cm)	Mid-Interval time (s)	Mid-Interval Speed               (cm/s)
0	0	0	0.008333333
0.01666667	1.2	1.2	0.025	72
0.03333333	2.8	1.6	0.041666667	96
0.05	4.6	1.8	0.058333333	108
0.06666667	6.7	2.1	0.075	126
0.08333333	9	2.3	0.091666667	138
0.1	11.5	2.5	0.108333333	150
0.11666667	14.4	2.9	0.125	174
0.13333333	17.3	2.9	0.141666667	174
0.15	20.9	3.6	0.158333333	216
0.16666667	24.4	3.5	0.175	210
0.18333333	28.6	4.2	0.191666667	252
0.2	32.6	4	0.208333333	240
0.21666667	37	4.4	0.225	264
0.23333333	41.7	4.7	0.241666667	282
0.25	46.7	5	0.258333333	300
0.26666667	52	5.3	0.275	318
0.28333333	57.3	5.3	0.291666667	318
0.3	63.2	5.9	0.308333333	354
0.31666667	69.3	6.1	0.325	366
0.33333333	75.8	6.5	0.341666667	390
0.35	82.3	6.5	0.358333333	390

Mid-Interval Time Sample Calculations

                                                  Mid-Interval Time=time+(1/120);  

                                                  0.025 s= 0.01666667 s+(1/120)s

                                                 

                                                   Mid-Interval Speed

                                                  

                                                  Mid-interval speed= Δx/(1/60 s) 

                                             

                                             72 cm/s= 1.2 cm/ 0.025 s

The two graphs above depicts velocity vs time and distance vs time. From the slope of the basic vs time graph is the velocity, and from the slope of the velocity vs time graph is acceleration. Therefore, the graphs allow the data to display visually the relationship between distance, velocity, and acceleration. These graphs were obtained through the use of scatter plot on excel document, and the line and equation were further obtained through the " display equations on chart" and " Display R-squared value on chart."

This lab was fairly straightforward with only the " hard " part on measuring the distance of each of the marks from  the 0-cm mark. The rest of the lab was filling in an excel spreadsheet and graphing the results of the position vs time and velocity vs time. The only major source of error in this lab is the precision of the measurement of the distances from the 0-cm mark. This human error includes the discrepancy in measurement of the uncertainty value of 0.1 cm. This initial uncertain value of distance can affect the value of the mid-interval speed. 

Questions/Analysis:

1. In the case of constant acceleration, the average velocity during that time interval is equal to the mid-time interval velocity because the graph of velocity vs time is a straight line, and the average of the initial velocity and final velocity just so happens to lie right on the midpoint between initial and final. 

2. I can get the acceleration due to gravity from my velocity/time graph by calculating the slope of the line or by looking at the equation for the line of best fit. My result is 9.63 m/s^2 which is 0.18 m/s^2 below the accepted value of 9.81 m/s^2. That is about 1.8% relative difference. 

3. I can get the acceleration due to gravity from my position/ time graph by taking the derivative of the slope of the line. Since the slope of the position vs time graph is velocity, and the derivative of velocity with respect to time is acceleration. The result I get is 9.67 m/s^2 which is 0.14 m/s^2 less than the accepted value of 9.81 m/ s^2. That's about 1.4 % relative difference. 

Additional Questions

1. There aren't any particular or noticeable pattern in the values of the values of g. 

2. The class' average value is below the accepted value by 0.18 m/s^2 or 1.8% relative difference. 

3. There aren't any patterns in the class' values of g. 

4. The factors of measurement of the distance on the ruler and air resistance. The human measurement of the distance on the ruler is a random error, and the air resistance is a systematic error. 

5. Part 2 of the free fall lab is to investigate and analyze the average of each group and the average of the class as well as to calculate the standard deviation of the mean for the class data. Introduction to the concept of the deviation of the mean is with the coffee machines that delivered variable grams of coffee. The two coffee machines exhibited different variability in each trial of the amount of instant coffee delivered. This gives rise to the theory of deviation from the mean value and how to calculate the mean so that it's value is positive. Then the formula of the standard deviation of the mean is introduced with the sigma symbol. The standard deviation of the mean describes the spread of the data. We were supposed to learn the definition and concept of the mean and the various deviations of the mean. In addition, the graph ofthe standard deviation of the mean was also introduced as a bell curve.