OMAR MOHAMED - L_Collisions Nov 2023

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Jan 9, 2024

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Analyzing Collisions in Two Dimensions (L2) lab performed on Oct 27, 2023 Omar Mohamed SPH4U Mr.Robert Finn Partners: Hassan Ahmad
Purpose : To determine the type of collision between two air pucks; To compare the total momentum of two air pucks before and after the collision; Materials : air pucks air table apparatus timing device ruler large format paper protractor level Procedure : 1. To ensure the air table's surface is level, a water level or clinometer is used, and the end supports are adjusted accordingly. 2. A large piece of format paper is cut to fit onto the air table's surface. If necessary, adhesive tape is used to attach the paper to the table's edge, not to the carbon paper itself. 3. With their group partner, the individuals turn on the air table without the timing device. They practice creating a collision with two air pucks by holding them at the bottom right and left of the table and launching them at an angle and speed that would result in a collision near the middle of the paper. They make sure to remove the external force acting on the pucks (whether their hand or another device) as quickly as possible once the pucks are in motion. This collision technique is practiced several times. 4. The timing device is turned on, and it is set to a reasonable frequency, approximately 50 Hz. The frequency is noted for later use in determining time intervals. 5. If the masses of the air pucks haven't been measured, they are weighed, and the masses are recorded.. 6. The air pucks are positioned at the bottom right and left of the air table. Someone with better timing skills operates the timing pedal of the timing device. The air pucks are launched in the same way as during practice, and the timing pedal is pressed and held instantly once the pucks are launched. The timing pedal is released before the air pucks hit the table's borders. The air table and timing device are turned off. 7. The format paper is carefully removed from the air table. In each corner of the paper, it is indicated whether the points closer to it were recorded by the pucks before or after the collision. Each set of points is paired with the mass of the air puck that generated it. 8. Four groups of points which represent air puck positions are obtained on the paper. Two of them belong to the velocities of air puck 1 before and after the collision, and the other two belong to the velocities of air puck 2 before and after the collision. 9. The first 4-5 points in the two trajectories before the collision and the first 4-5 points in the two trajectories after the collision are identified. A displacement vector is drawn and measured for each of these four trajectories. To calculate the time interval between each
point, the frequency of the timing device is divided by one. 10. A set of axes is drawn on the format paper. The y-axis is made to be exactly parallel to the displacement vector of one of the two trajectories before the collision to eliminate the need for measuring that trajectory as a fourth angle. 11. Using a protractor and a ruler or set square, the angles of the three remaining displacement vectors are measured in terms of the cardinal directions (with the positive y-axis being [N] and the positive x-axis being [W]). If necessary, extensions to the length of the vectors are carefully drawn until they intersect the y-axis for simplified angle measurements. Observations : Puck Mass (g) Distance before Collision(cm ) Distance After Collision(cm ) Time Elapsed Before Collision (s) Time Elapsed After Collision (s) Direction Before Collision Direction After Collision 1 188.3 g 8.95 cm 0.0895m 6.62 cm = 0.0662 m (7/50.0)s = 0.140s (7/50.0)s = 0.140s N [W 2° S] 2 187.2 g 6.21 cm = 0.0621 6.71 cm = 0.0671m (5/50.0)s = 0.100s (5/50.0)s = 0.100s W [N 13° W]
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Analysis : 1. Calculate the speed of each air puck before and after the collision. Include ONE sample calculation in the body of the report. Put all of the results in the form of a table. Create two additional columns in the table and leave them blank for the results of Analysis #2 and Analysis #3. Δd = 8.95 cm Δt = 0.140 s ∴ The initial velocity of puck 1 was 64 cm / s [ N ] . Puck Velocity (cm/s) Momentum (cmg/s) Kinetic Energy (J) 1 Initial v 1 = 64.0 cm/s [N] 12100 cmg / s [ N ] 0.0386 J Final v 1 ' = 47.2 cm/s [W 2° S] 8890 cmg / s [W 2° S] 0.0210 J 2 Initial v 2 = 62.1cm/s [W] 11700 cmg / s [W] 0.0363 J Final v 2 ' = 67.1 cm/s [N 13° W] 12600 cmg / s [N 13° W] 0.0424 J 2. Calculate the amount of momentum of each air puck before and after the collision. Include ONE sample calculation in the body of the report. Put all of the results in the first empty column from the table created in the previous step. m 1 = 188.3 g v 1 = 64.0 cm / s [ N ] p 1 = m 1 v 1 =( 188.3 g )( 64.0 cm / s [ N ]) 12100 cmg / s [ N ] ∴ The initial momentum of puck 1 was 12100 cmg / s [ N ] . 3. Calculate the amount of kinetic energy (in joules) of each air puck before and after the collision. Include ONE sample calculation in the body of the report. Put all of the results
in the last column from the table created in Analysis #1. m 1 = 188.3 g v 1 = 64.0 cm / s E K = ? Theinitial kinetic energyof puck 1 was 0.0386 J 4. Calculate the total kinetic energy for the system before and after the collision (in joules). Show all your work in the space below. Let be Total Initial kinetic energy and be Total Final kinetic energy. Theinitial totalkinetic energy of the systemwas 0.0749 J ¿ final kinetic energy was 0.0634 J 5. Determine the total momentum of the air pucks before and after the collision. Show all your work, including sketches of vector diagrams that helped to find the answers. Initial total momentum: m 1 = 188.3 g v 1 ' = 64.0 cm/s [N] m 2 = 187.2 g v 2 ' = 62.1cm/s [W]
P T = P 1 + P 2 m 1 v 1 = (188.3g * 64.0 cm/s [N]) = 12100 gcm/s m 2 v 2 = (187.2g * 62.1cm/s [W]) = 11600 gcm/s p T = ) p T = ) = 16800 gcm/s tan θ = 11600 gcm / s ¿ 12100 gcm / s 11600 gcm / s θ = ta n 1 ¿ ¿ 12100 gcm / s ) = 43.8 ° N 43.8 Theinitial totalmomentumof the puck is 16800 gcm / s ¿ ° W ¿ Final total momentum: m 1 = 188.3 g v 1 ' = 47.2 cm/s [W 2° S] m 2 = 187.2 g W + N + S E
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v 2 ' = 67.1 cm/s [N 13° W] P T ' = ¿ P 1 ' + P 2 ' p T ' = + c 2 -2bc cos A) p T ' = + 12600 2 -2(8890 *12600) cos 101 ° ) = 16700 gcm/s = 31.5 ° ∴ the final total momentum is 16700 cmg / s [ W 29.5 N ] . 13 ° 2 ° 88 ° P T ' 2 ° W + E N + S θ
Discussion : 1. Use the percentage difference calculator to determine the percent difference between the magnitudes of the answers for Analysis #5. Express the answer to the correct number of significant digits and display the calculation in GRASP format in the space below. 0.00597015 100 0.6% \ ∴ The percentage difference between the magnitudes is 0.6% 2. Was momentum conserved for this collision? Comment on whether the results were as expected, with respect to the magnitude and direction of the total momentum before and after the collision. Momentum conservation states that the total momentum before a collision should be equal to the total momentum after the collision. In this particular experiment, the initial total momentum was calculated to be 16800 gcm/s [N 43.8°W], while the final total momentum after the collision was found to be 16700 gcm/s [W 29.5 N]. Although the final total momentum showed a slight decrease compared to the initial total momentum, resulting in a small percentage difference of 0.6%, which indicates a minor loss of momentum in the system. The change in direction of the final total momentum, along with the change in angle was expected due to the nature of the collision and the forces involved. Considering potential sources of error in real-world experiments, the observed results align reasonably well with the expectations of momentum conservation.
3. Using evidence from the observations and analysis of the collision, which type of collision must have occurred? Justify your answer with specific values from the lab as well as observations made during the experiment. Types of collisions include elastic, inelastic or completely inelastic. The observations and analyisis indicate that an inelastic collision must have occurred. This was due to few key factors: Kinetic Energy Calculation: The initial kinetic energy of puck 1 was calculated as 0.0386 J, and the final kinetic energy was calculated as 0.0210 J. For puck 2, the initial kinetic energy was calculated as 0.0363 J, and the final kinetic energy was calculated as 0.0424 J. In an elastic collision, the total kinetic energy of the system should remain the same before and after the collision, which was not the case here. Momentum Change: The initial total momentum of the system was calculated as 16800 gcm/s [N 43.8°W], and the final total momentum was calculated as 16700 cmg/s [W 29.5 N]. In an elastic collision, the total momentum of the system should remain constant. The change in both direction and magnitude of momentum indicates that it was an inelastic collision. 4. Provide three logical sources of error in this experiment to explain any discrepancy between experimental results and theory. There are more than three sources of error, so please pick the ones that would have the most negative impact on the result. It is inappropriate to note the possibility of a preventable human error. Sources of error: 1. Friction and Air Resistance : One of the major sources of error in this experiment could be the presence of friction and air resistance. Even though an air table is used to minimize these effects, it is nearly impossible to completely eliminate them. These forces can still exert a small but significant influence on the motion of the air pucks, leading to discrepancies between the observed and theoretical outcomes. 2. Measurement and Alignment of Displacement Vectors: Drawing and measuring displacement vectors can lead to errors, especially if they are not perfectly aligned or if the measurements are not taken accurately. The angles measured with the protractor may
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have slight inaccuracies, which can affect the calculation of velocities and the subsequent analysis of momentum and kinetic energy. Small errors in vector measurements can accumulate and lead to significant discrepancies in the final results. 3. Timing Device Accuracy: The accuracy of the timing device used to measure the time intervals can be a critical source of error. If the timing device lacks precision, it can lead to inaccuracies in the calculation of velocities and in the determination of momentum and kinetic energy. This could result in discrepancies between the experimental and theoretical values. Conclusion : In conclusion, the experiment aimed to determine the type of collision between two air pucks and compare their total momentum before and after the collision. It was observed that the initial velocity of puck 1 was 64.0 cm/s [N] and puck 2 was 62.1 cm/s [W], while the final velocities after the collision were 47.2 cm/s [W 2° S] for puck 1 and 67.1 cm/s [N 13° W] for puck 2. These values led to an initial total kinetic energy of the system of 0.0749 J, which decreased to 0.0634 J after the collision. Additionally, the initial total momentum of the system was found to be 16800 cmg/s [N 43.8°W], which subsequently transformed into 16700 cmg/s [W 29.5°N] post-collision. The percentage difference between the magnitudes of the initial and final total momentum was calculated to be 0.6%. These results indicate that the purpose of the experiment was achieved, as the observed changes in velocities and momentum align with what we learned based on the principles of conservation of momentum and energy. The slight discrepancy in percentage difference can be attributed to experimental minor sources of error. Overall, the experiment successfully provided valuable information regarding the conservation of momentum and energy.