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Student ID Number_901071640__________ CRN Section____90366______________ Summary Assignment: I. Use of your knowledge of Newton’s Laws of Motion from your lab activities to generate a proof of the conservation of momentum for a collision of two objects in one dimension. Explain the rationale for each step. Remember that momentum is defined as p = mv.

II. Search for and find the commonly accepted value for the acceleration due to gravity (a g ) near the Earth’s surface (refer to your lab activities). Cite your source using The Chicago Manual of Style Notes and Bibliography Style. Evaluate its reliability; for example, did you use a government or university website? Perform a per cent difference calculation between the value you measured, and the value obtained from your reference source. The acceleration due to gravity was determined by using the Free Fall Apparatus, to compare the height of the ball drop within seconds and the distance the ball dropped was varied. The acceleration due to gravity a g at the Catonsville lab location was determined by the longitude, latitude, and local density. According to Local Gravity Calculator, the local density of Catonsville is 9.80063 ms -2 . From the given graph, slope = 0.0021 Acceleration due to gravity, g = 2/slope = 2/0.0021 = 9.5238 m/s 2 Accepted value = 9.80063 m/s 2 Percentage difference = [ |9.5238 - 9.80063| / 9.80063] * 100 = -2.86% Mass of ball = 15.6 grams Citation “Local Gravity Calculator.” SensorsONE. Accessed September 25, 2023. https://www.sensorsone.com/local-gravity-calculator/. III. In the space below, insert a copy of your graph of the kinetic, potential, and total energies of a glider from your Conservation of Mechanical Energy lab report. Below that, insert a copy of your graph from Newton’s Second Law. Check that correct axis labels and units are present, as well as a legend if appropriate. Newton second’s law graph

Conservation of mechanical energy graph y = 4.2815x + 0.0077 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Accerlation (m/s^2) Force (Newtons) Acceleration vs Force 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 1 2 3 4 5 6 7 Energy (Joules) Position Energy (J) vs Position K.E. P.E. E tot

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IV. Analyze your graph from Newton’s Second Law of Motion (refer to the lab activity). There are two important ideas to be extracted from your data. The first is that the resulting acceleration is proportional to the applied force. Does the graph support the first concept, or, does it disprove it? How? The second concept is that the proportionality constant between the force and acceleration should be the reciprocal of the object’s mass. Does the graph support the second concept, or disprove it? How? These acceleration over accelerating force line is aligned with the origin in a straight, that means acceleration and accelerating force are directly proportional to each other. The value of our total masses was also inversely proportional to the values of the forces and from newton’s second law of motion m=f/a. therefore the experiment was successful, and the objective was achieved. y = 4.2815x + 0.0077 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Accerlation (m/s^2) Force (Newtons) Acceleration vs Force

Lab 103- Acceleration Due to Gravity September 25 th , 2023 Objective of the Experiment : To determine the acceleration a g due to gravity at surface of the earth. List of the Equipment : Free fall apparatus, Stopwatch, Meter stick Procedures: Same as in the lab manual. Results and Analysis: The acceleration due to gravity was determined by using the Free Fall Apparatus, to compare the height of the ball drop within seconds and the distance the ball dropped was varied. The acceleration due to gravity a g at the Catonsville lab location was determined by the longitude, latitude, and local density. According to Local Gravity Calculator, the local density of Catonsville is 9.80063 ms -2 . From the given graph, slope = 0.0021 Acceleration due to gravity, g = 2/slope = 2/0.0021 = 9.5238 m/s 2 Accepted value = 9.80063 m/s 2

Percentage difference = [ |9.5238 - 9.80063| / 9.80063] * 100 = -2.86% Mass of ball = 15.6 grams Conclusion These findings show that the acceleration is constant because of the straight line to the origin in the excel graph. The straight line also shows that the height is proportionate to the time squared. The result we derived from the graph was a -2.86% different from the actual value which is 9.80063 m/s 2 . Citation “Local Gravity Calculator.” SensorsONE. Accessed September 25, 2023. https://www.sensorsone.com/local- gravity-calculator/.

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Lab 105- Newtons 2 nd Law of Motion October 8, 2023 Objective of the Experiment : To gain confidence that newtons second law of motion is valid. List of the Equipment : Pasco box, Air track, Glider, Small masses, Hanger, String, Smart wheel Procedures: Same as in the lab manual. Results and Analysis: A scatter plot graph was plotted with the accelerating force on the x-axis and the acceleration on the y- axis to give; Mass of hanger = 229g Converting to kg; 229g x 1kg/1000g= 0.229kg. These values were converted to kg and multiplied by 9.8 to give the accelerating force. Accelerating force Gm h Acceleration m/s 2 y = 4.2815x + 0.0077 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Accerlation (m/s^2) Force (Newtons) Acceleration vs Force

0.0167 0.0809 0.1147 0.496 0.2127 0.919 0.3107 1.33 The linear equation from the graph: y = 4.2815x + 0.0077 Therefore, slope of best fit line= 4.2815 The mass of the system was gotten by using the relation; 1/slope= 1/4.2815= 0.2336g The accepted value of the mass was given as 0.229kg Percentage difference = [|0.2336 - 0.229 / 0.229|] * 100 = 0.75% Conclusion These acceleration over accelerating force line is aligned with the origin in a straight, that means acceleration and accelerating force are directly proportional to each other. The value of our total masses was also inversely proportional to the values of the forces and from newton’s second law of motion m=f/a. therefore the experiment was successful, and the objective was achieved.

Lab 108- Conservation of Mechanical Energy October 28 th , 2023 Objective of the Experiment : To verify the conservation of mechanical energy. List of the Equipment : Air track, Glider, Ruler, Pasco interference, Photogate, Rubber band Procedures: Same as in the lab manual. Results and Analysis:

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Calculations: Kinetic energy, Potential energy and mechanical energy was derived using the formulas: KE = ½ x m x v 2 PE = mgh M.E = K.E + P.E Average total energy value is 0.610. Largest diff between E tot and E tot Ave is 0.00725 Percentage difference = [ |Largest difference| / Total energy ave ] * 100 = [ |0.00725| / 0.610469] * 100 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 1 2 3 4 5 6 7 Energy (Joules) Position Energy (J) vs Position K.E. P.E. E tot

= 1.1867% Conclusion The objective of this lab is to verify the Law of Conservation of Mechanical Energy. The total energy relatively constant in the graph, it was constant within to 1.1876%. The law notion of conservation of mechanical energy was supported because the beam was set at different altitudes. As the altitude decreased, the kinetic energy of the hanger increases, and the potential energy of the hanger decreases but the mechanical energy of the hanger stays the same. This can be proven using the trendline in the graph. The Kinetic trendline goes up, the potential trendline goes down and the mechanical trendline remains in an almost straight line. The trendline for the mechanical energy should be perfectly straight but there are multiple sources of error affecting the graph. Some errors were the improper positioning of the photogate not capturing the velocity correctly, inaccurate measurements of the altitude, human error when releasing the glider.

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