Data sheet final

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PHYSICS LAB Data Sheet: Activity 4 – Conservation of Momentum Name Course Date Sileola Akinjolire Physics Setup Elastic 1 Elastic 2 Elastic 3 Inelastic Ball 1 Mass (kg) 0.150 0.250 0.250 0.250 Ball 2 Mass (kg) 0.250 0.250 0.100 Unknown Activity Date Code: RBZ Elastic 1 Procedure Data & Analysis (Mass of Ball 1 < Mass of Ball 2) Complete the tables and questions below using your data plot and information found under the Background and Activity Form tabs. Note: 1.00 cm = 0.0100 m, 100 mJ = 0.100 J Table 1 – Determine Ball 1 Initial Velocity Spring PE (J) Ball 1 Initial KE (J) Ball 1 Mass (kg) Ball 1 Initial Velocity (m/s) 146.25 mJ = 0.14625 J 0.14625 0.150 1.39642 Sample Calculation : Show your Ball 1 initial velocity calculation. For Conservation of Energy KE = Spring PE = 0.14625 1 2 m v 2 = 0.14625 v 2 = 0.14625 × 2 m = 0.14625 × 2 0.150 = 1.95 v = √ 1.95 = 1.39642 m / s Table 2 – Determine Initial Momenta

Ball 1 Initial Momentum (kg ∙m/s) Ball 2 Initial Momentum (kg∙m/s) Combined Initial Momentum (kg∙m/s) 0.20946 0 0.20946 Sample Calculation : Show your combined initial momentum calculation. For mass 1: initial momentum = mass, m x initial velocity = 0.15 x 1.39642 = 0.20946 kgm/s For mass 2: initial momentum = mass, m x initial velocity = 0.25 x 0 = 0 kgm/s Combined initial momentum: 0.20946 kgm/s + 0 kgm/s = 0.20946 kgm/s Table 3 – Data from Position vs Time Plot Blue Line Data Point Time (s) Position (m) Red Line Data Point Time (s) Position (m) 1 0.2863 35.29cm = 0.3529m 1 0.2863 51.95cm = 0.5195m 2 0.4530 29.47cm = 0.2947m 2 0.4530 69.41cm = 0.6941m Table 4 – Determine Slope, Final Velocity, and Final Momentum of Ball 1 Blue Line Slope Ball 1 Final Velocity (m/s) Ball 1 Mass (kg) Ball 1 Final Momentum (kg∙m/s) -0.3491 -0.3491 0.150 -0.052365 Sample Calculation: Show your blue line slope calculation. Blue line slope at the two points(0.2863, 0.3529) and (0.4530, 0.2947) Then, slope = ∆ y ∆x = y 2 − y 1 x 2 − x 1 = 0.2947 − 0.3529 0.4530 − 0.2863 = − 0.0582 0.1667 =− 0.3491 m / s Table 5 – Determine Slope, Final Velocity, and Final Momentum of Ball 2 Red Line Slope Ball 2 Final Velocity Ball 2 Mass (kg) Ball 2 Final

(m/s) Momentum (kg∙m/s) 1.04679 1.04679 0.250 0.26170 Sample Calculation: Show your Ball 2 final momentum calculation Red line slope at the two points (0.2863, 0.5196) and (0.4530, 0.6941) Then, slope = ∆ y ∆x = y 2 − y 1 x 2 − x 1 = 0.6941 − 0.5196 0.4530 − 0.2863 = 0.1745 0.1667 = 1.04679 m / s Final momentum calculation: mv = 0.25 x 1.04679 = 0.2617 kgm/s Table 6 – Determine Final Momenta Ball 1 Initial Momentum (kg ∙m/s) Ball 2 Initial Momentum (kg∙m/s) Combined Initial Momentum (kg∙m/s) -0.052365 0.26170 0.209335 Sample Calculation: Show your combined final momentum calculation. Combined final momentum = Ball 1 final momentum + Ball 2 final momentum = 0.052365 + 0.26170 = 0.209335 kgm/s Observations and Questions [1] Describe the general behavior of the balls for this collision. Assuming that you did a few different trials runs, was this general behavior affected by changing initial speed of the projectile ball? For this procedure there is a difference in the mass of the two balls. Based on your observations what do you think the behaviour of the balls would be if the mass difference was very, very large?

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Answer – Through all trial runs, the momentum is observed to be the same, that is, the initial momentum is equal to the final momentum. Also, it can be said that the collision is elastic even though there was mass variation in the trial runs. [2] Based on your results for this data run, is momentum conserved in this collision? Provide specific results to support your answer. Answer – Yes, momentum is conserved since the initial momentum is 0.209 kgm/s and also the final momentum is 0.209 kgm/s approximately which predicted them being same. [3] Based on your results for this data run, is kinetic energy conserved in this collision? Provide specific results and/or a calculation to support your answer. Answer - Summation of kinetic energy before collision is 0.14625 J And summation of kinetic energies after collision is = 1 2 ( 0.15 ) ( − 0.3491 ) 2 + 1 2 ( 0.25 ) ( 1.04679 ) 2 = 0.14611 J Since, summation of kinetic energy before collision is equal to summation after collision, then, yes, kinetic energy is conserved. Elastic 2 Procedure Data & Analysis (Mass of Ball 1 = Mass of Ball 2) Complete the tables and questions below using your data plot and information found under the Background and Activity Form tabs. Note : 1.00 cm = 0.0100 m, 100 mJ = 0.100 J Table 7 – Determine Ball 1 Initial Velocity Spring PE (J) Ball 1 Initial KE (J) Ball 1 Mass (kg) Ball 1 Initial Velocity (m/s) 0.21060 0.21060 0.250 1.2980

Sample Calculation: Show your Ball 1 initial velocity calculation. Initial K.E = Spring PE = 0.14625 1 2 m v 2 = 0.21060 v = √ 0.21060 × 2 0.250 = √ 0.4212 0.250 = 1.2980 m / s Table 8 – Determine Initial Momenta Ball 1 Initial Momentum (kg ∙m/s) Ball 2 Initial Momentum (kg∙m/s) Combined Initial Momentum (kg∙m/s) 0.3245 0 0.3245 Sample Calculation : Show your combined initial momentum calculation. For mass 1: initial momentum = mass, m x initial velocity = 0.25 x 1.2980 = 0.3245 kgm/s For mass 2: initial momentum = mass, m x initial velocity = 0.25 x 0 = 0 kgm/s Combined initial momentum: 0.3245 kgm/s + 0 kgm/s = 0.3245 kgm/s Table 9 – Data from Position vs Time Plot Blue Line Data Point Time (s) Position (m) Red Line Data Point Time (s) Position (m) 1 0.2920 0.3853 1 0.2920 0.5169 2 0.4247 0.3853 2 0.4247 0.6891 Table 10 – Determine Slope, Final Velocity, and Final Momentum of Ball 1 Blue Line Slope Ball 1 Final Velocity (m/s) Ball 1 Mass (kg) Ball 1 Final Momentum (kg∙m/s) 0 0 0.250 0 Sample Calculation: Show your blue line slope calculation.

Blue line slope at the two points(0.2920, 0.3853) and (0.4247, 0.3853) Then, slope = ∆ y ∆x = y 2 − y 1 x 2 − x 1 = 0.3853 − 0.3853 0.4 247 − 0.2920 = 0 0.1327 = 0 m / s Table 11 – Determine Slope, Final Velocity, and Final Momentum of Ball 2 Red Line Slope Ball 2 Final Velocity (m/s) Ball 2 Mass (kg) Ball 2 Final Momentum (kg∙m/s) 1.2977 1.2977 0.25 0.3244 Sample Calculation: Show your Ball 2 final momentum calculation Red line slope at the two points (0.2920, 0.5169) and (0.4247, 0.6891) Then, slope = ∆ y ∆ x = y 2 − y 1 x 2 − x 1 = 0.6891 − 0.5169 0.4247 − 0.2920 = 0.1722 0.1327 = 1.2977 m / s Final momentum calculation: mv = 0.25 x 1.2977 = 0.3244 kgm/s Table 12 – Determine Final Momenta Ball 1 Initial Momentum (kg ∙m/s) Ball 2 Initial Momentum (kg∙m/s) Combined Initial Momentum (kg∙m/s) 0 0.3244 0.3244 Sample Calculation: Show your combined final momentum calculation. Combined final momentum = 0 + 0.3244 = 0.3244 kgm/s Observations and Questions [1] Describe the general behavior of the balls for this collision. Assuming that you did a few different trials runs, was this general behavior affected by changing initial speed of the projectile ball?

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Answer – Through all trial runs, the momentum is observed to be the same, that is, the initial momentum is equal to the final momentum. Also, it can be said that the collision is elastic even though there was mass variation in the trial runs. [2] Based on your results for this data run, is momentum conserved in this collision? Provide specific results to support your answer. Answer – Yes, momentum is conserved since the initial momentum is 0.324 kgm/s and also the final momentum is 0.324 kgm/s approximately which predicted them being same. [3] Based on your results for this data run, is kinetic energy conserved in this collision? Provide specific results and/or a calculation to support your answer. Answer – Summation of kinetic energy before collision is 0.21060 J And the summation of kinetic energies after collision is = 1 2 ( 0.25 ) ( 0 ) 2 + 1 2 ( 0.25 ) ( 1.2977 ) 2 = 0.21050 J Since, summation of kinetic energy before collision is equal to summation after collision, then, yes, kinetic energy is conserved. Elastic 3 Procedure Data & Analysis (Mass of Ball 1 > Mass of Ball 2) Complete the tables and questions below using your data plot and information found under the Background and Activity Form tabs. Note : 1.00 cm = 0.0100 m, 100 mJ = 0.100 J Table 13 – Determine Ball 1 Initial Velocity Spring PE (J) Ball 1 Initial KE (J) Ball 1 Mass (kg) Ball 1 Initial Velocity (m/s) 0.28665 0.28665 0.250 1.51433

Sample Calculation: Show your Ball 1 initial velocity calculation. 1 2 mv 2 = 0.28665 v = √ 0.28665 × 2 0.250 = 1.51433 m / s Table 14 – Determine Initial Momenta Ball 1 Initial Momentum (kg ∙m/s) Ball 2 Initial Momentum (kg∙m/s) Combined Initial Momentum (kg∙m/s) 0.37858 0 0.37858 Sample Calculation : Show your combined initial momentum calculation. For mass 1: initial momentum = mass, m x initial velocity = 0.25 x 1.51433 = 0.37858 kgm/s For mass 2: initial momentum = mass, m x initial velocity = 0.100 x 0 = 0 kgm/s Combined initial momentum: 0.37858 kgm/s + 0 kgm/s = 0.37858 kgm/s Table 15 – Data from Position vs Time Plot Blue Line Data Point Time (s) Position (m) Red Line Data Point Time (s) Position (m) 1 0.2443 0.4162 1 0.2443 0.5231 2 0.3247 0.4684 2 0.3247 0.6969 Table 16 – Determine Slope, Final Velocity, and Final Momentum of Ball 1 Blue Line Slope Ball 1 Final Velocity (m/s) Ball 1 Mass (kg) Ball 1 Final Momentum (kg∙m/s) 0.6493 0.6493 0.250 0.1623 Sample Calculation: Show your blue line slope calculation. Blue line slope at the two points(0.2443, 0.4162) and (0.3247, 0.4684)

Then, slope = ∆ y ∆x = y 2 − y 1 x 2 − x 1 = 0.4684 − 0.4162 0.3247 − 0.2443 = 0.0522 0.0804 = 0.6493 m / s Table 17 – Determine Slope, Final Velocity, and Final Momentum of Ball 2 Red Line Slope Ball 2 Final Velocity (m/s) Ball 2 Mass (kg) Ball 2 Final Momentum (kg∙m/s) 2.1617 2.1617 0.100 0.21617 Sample Calculation: Show your Ball 2 final momentum calculation Red line slope at the two points (0.2443, 0.5231) and (0.3247, 0.6969) Then, slope = ∆ y ∆ x = y 2 − y 1 x 2 − x 1 = 0.6969 − 0.5231 0.3247 − 0.2443 = 0.1738 0.0804 = 2.1617 m / s Final momentum calculation: mv = 0.100 x 2.1617 = 0.21617 kgm/s Table 18 – Determine Final Momenta Ball 1 Initial Momentum (kg ∙m/s) Ball 2 Initial Momentum (kg∙m/s) Combined Initial Momentum (kg∙m/s) 0.1623 0.21617 0.3785 Sample Calculation: Show your combined final momentum calculation. Combined final momentum = 0.1623 + 0.21617 = 0.37847 kgm/s Observations and Questions [1] Describe the general behavior of the balls for this collision. Assuming that you did a few different trials runs, was this general behavior affected by changing initial speed of the projectile ball? For this procedure there is a difference in the mass of the two balls. Based on your observations what do you think the behaviour of the balls would be if the mass difference was very, very large?

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Answer – Through all trial runs, the momentum is observed to be the same, that is, the initial momentum is equal to the final momentum. Also, it can be said that the collision is elastic even though there was mass variation in the trial runs. [2] Based on your results for this data run, is momentum conserved in this collision? Provide specific results to support your answer. Answer – Yes, momentum is conserved since the initial momentum is 0.3785 kgm/s and also the final momentum is 0.3785 kgm/s approximately which predicted them being same. [3] Based on your results for this data run, is kinetic energy conserved in this collision? Provide specific results and/or a calculation to support your answer. Answer - Summation of kinetic energy before collision is 0.28665 J And the summation of kinetic energies after collision is = 1 2 ( 0.25 ) ( 0.6493 ) 2 + 1 2 ( 0.100 ) ( 2.1617 ) 2 ¿ 0.286346 J Since, summation of kinetic energy before collision is equal to summation after collision, then, yes, kinetic energy is conserved. Inelastic Procedure Data & Analysis (Mass of Ball 2 Unknown) Complete the tables and questions below using your data plot and information found under the Background and Activity Form tabs. Note : 1.00 cm = 0.0100 m, 100 mJ = 0.100 J Table 19 – Determine Ball 1 Initial Velocity Spring PE (J) Ball 1 Initial KE (J) Ball 1 Mass (kg) Ball 1 Initial Velocity (m/s)

0.0936 0.0936 0.250 0.8653 Sample Calculation: Show your Ball 1 initial velocity calculation. Initial KE = Spring PE 1 2 mv 2 = 0.0936 v = √ 0.0936 × 2 0.25 = √ 0.7488 = 0.8653 m / s Table 20 – Determine Initial Momenta Ball 1 Initial Momentum (kg ∙m/s) Ball 2 Initial Momentum (kg∙m/s) Combined Initial Momentum (kg∙m/s) 0.2163 0 0.2163 Sample Calculation : Show your combined initial momentum calculation. For mass 1: initial momentum = mass, m x initial velocity = 0.25 x 0.8653 = 0.2163 kgm/s Combined initial momentum: 0.2163 kgm/s + 0 kgm/s = 0.2163 kgm/s Table 21 – Data from Position vs Time Plot Blue Line Data Point Time (s) Position (m) Red Line Data Point Time (s) Position (m) 1 0.4773 0.4741 1 0.4773 0.5120 2 0.8373 0.6471 2 0.8373 0.6850 Table 22 – Determine Slope, Final Velocity, and Final Momentum of Ball 1 Blue Line Slope Ball 1 Final Velocity (m/s) Ball 1 Mass (kg) Ball 1 Final Momentum (kg∙m/s) 0.4806 0.4806 0.250 0.1201 Sample Calculation: Show your blue line slope calculation.

Blue line slope at the two points(0.4773, 0.4741) and (0.8373, 0.6471) Then, slope = ∆ y ∆x = y 2 − y 1 x 2 − x 1 = 0.6471 − 0.4741 0.8373 − 0.4773 = 0.173 0.36 = 0.4806 m / s Table 23 – Determine Slope and Final Velocity of Ball 2 Red Line Slope Ball 2 Final Velocity (m/s) 0.4806 0.4806 Sample Calculation: Show your red line slope calculation Red line slope at the two points (0.4773, 0.5120) and (0.8373, 0.6850) Then, slope = ∆ y ∆ x = y 2 − y 1 x 2 − x 1 = 0.6850 − 0.5120 0.8373 − 0.4773 = 0.173 0.36 = 0.4806 m / s Observations and Questions [1] Describe the general behavior of the balls for this collision. Assuming that you did a few different trials runs, was this general behavior affected by changing initial speed of the projectile ball? Answer – It was observed that both balls stuck together after collision and also, energy was not conserved. [2] The mass of Ball 2 is unknown for this data run. Assume that momentum is conserved in this collision and use that information to determine the mass of Ball 2. Show your calculation below. Answer – If momentum is assumed to be conserved, let the unknown mass be m. Then, summation of initial momentum = summation of final momentum. ( 0.250 × 0.0936 ) + ( m× 0 ) = ( 0.250 × 0.4806 ) +( m× 0.4806 ) ( 0.250 × 0.0936 ) = 0.4806 ( 0.250 + m )

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0.250 + m = 0.0234 0.4806 0.250 + m = 0.04 m = 0.04 − 0.25 m =− 0.21 kg Since mass cannot be negative, then it is 0.21 kg.

Data sheet final

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