lOMoARcPSD|17093303 Copy of 2DCollisions SE COLLEGE PHYSICS Physics for Electronics (Seneca College) StuDocu is not sponsored or endorsed by any college or university Downloaded by kasra mehrinejad (kasra.mh.zh@gmail.com) lOMoARcPSD|17093303 Name: Daniella Bafas Date: 21/11/21 Student Exploration: 2D Collisions Directions: Follow the instructions to go through the simulation. Respond to the questions and prompts in the orange boxes. Vocabulary: center of mass, conservation of energy, conservation of momentum, elasticity, kinetic energy, momentum, speed, vector, velocity Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Gizmo Warm-up Objects collide all the time, but often with very different results. Sometimes colliding objects will stick together. Other times, they will bounce off each other at an angle. What determines how objects will behave in a collision? You can use the 2D Collisions Gizmo to find out. Note the arrows, or vectors, on each puck. Click Play ( ). 1. How does the direction and length of its vector relate to the motion of a puck? The direction and the length of the vector determine the velocity of the puck. 2. The velocity (speed and direction) of each puck is described by components in the i and j directions. The symbol for velocity is v. (Vector quantities are shown in bold.) A. Which component represents movement in the east-west direction? Compon ent “i” B. Which component represents movement in the north-south direction? Compon ent “j” Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved Downloaded by kasra mehrinejad (kasra.mh.zh@gmail.com) lOMoARcPSD|17093303 3. The speed (v) of a puck is equal to the length of its velocity vector. To calculate the speed of a puck with a velocity of ai + bj, use the Pythagorean theorem: Set the velocity of the blue puck to 12.00i + 5.00j m/s. What is its speed? v= Activity A: Elastic collisions 13.00m/s Get the Gizmo ready: ● Click Reset. Make sure Elasticity is set to 1.0. ● Set the blue puck’s velocity to v = 4.00i + 3.00j and the gold puck’s velocity to v = 0.00i – 4.00j. Introduction: An object’s elasticity describes how readily it returns to its original shape after it has collided with another object. In a perfectly elastic collision (in which elasticity equals 1), the two colliding objects return to their original shape immediately after the collision takes place. Question: What is conserved during an elastic collision? 1. Calculate: The kinetic energy (KE) of an object is a measure of its energy of motion. The equation for kinetic energy is: KE = mv2 ÷ 2, and the unit for kinetic energy is the joule (J). In the equation, m represents an object’s mass and v represents its velocity. A. Calculate the kinetic energy of each puck. (Note: The mass of the pucks can be found on the CONTROLS pane, and the magnitude of the pucks’ velocities (v) can be found at the bottom of the SIMULATION pane.) Blue puck KE = 62.5 J Gold puck KE = 24.0 J B. Add the kinetic energy of the blue puck to that of the gold puck to find the total kinetic energy for the system. Total system KE = 86.5 J 2. Compare: Turn on Velocity vectors during motion. Click Play and observe the pucks. A. Calculate the final kinetic energy of the two pucks and the total system. Blue puck KE = 11.45 J Gold puck KE = 75.0 J Total system KE = 86.5 J Use the CALCULATION tab to check your work. B. How did the kinetic energies of the two pucks change, and how can you explain these changes? The kinetic energy of the blue puck decreased whereas, the kinetic energy of golden puck increased after the collision. C. How did the total system kinetic energy before the collision compare to that of after Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved Downloaded by kasra mehrinejad (kasra.mh.zh@gmail.com) lOMoARcPSD|17093303 the collision? The total kinetic energy before the collision equaled to the kinetic energy after the collision. The kinetic energy was conserved during the collision. 3. Make a rule: Complete the sentence: During an elastic collision, the total kinetic energy of the system is conserved. . This rule is part of the law of conservation of energy. 4. Calculate: It takes force to deflect or stop a moving object. Momentum (p) is a measure of an object’s tendency to continue moving in a given direction. The formula for momentum is p = mv and the unit is newton-seconds (kg•m/s). Click Reset. Select the CONTROLS tab. Because momentum has direction, it can be described in both the i direction and j direction. Calculate the initial momentums (pay attention to +/- signs): Blue puck: p in i direction = 20.0 kg m/s p in j direction = 15.0 kg m/s Gold puck: p in i direction = 0.00 kg m/s p in j direction = 12.0 kg m/s Total system: p in i direction = 20.0 kg m/s p in j direction = 3.00 kg m/s 5. Calculate: Click Play and observe the pucks collide. Calculate the final momentums: Blue puck: p in i direction = 0.25 kg m/s p in j direction = 10.7 kg m/s Gold puck: p in i direction = 19.8 kg m/s p in j direction = -7.71 kg m/s Total system: p in i direction = 20.0 kg m/s p in j direction = 3.00 kg m/s Use the CALCULATION tab to check your answers. 6. Compare: Look at the momentum values you calculated for before and after the collision. A. What did you notice about the total system momentum in the i direction? The total system momentum in the “i” direction is the same before and after collision. B. What did you notice about the total system momentum in the j direction? The total system momentum in the “j” direction is the same before and after collision. During an elastic collision, the total momentum in both the i direction and the j direction remains the same. This rule is part of the law of conservation of momentum. 7. Compare: Click Reset. Select the MOMENTUM tab. Set up several different collisions. Click Play. Then, compare the gray Total momentum vector Before and After the collision. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved Downloaded by kasra mehrinejad (kasra.mh.zh@gmail.com) lOMoARcPSD|17093303 A. How do the Before and After vectors compare? The vectors from before and after are equal in magnitude. B. What does this observation confirm? The observation confirms that the momentum and kinetic energy before and after are equal in magnitudes which conclude that the momentum and kinetic energy are conserved. Activity B: Inelastic collisions Get the Gizmo ready: ● Click Reset. ● On the CONTROLS tab, turn on Puck trails. Question: What is conserved during an inelastic collision? 1. Observe: Use the Gizmo to set up a new collision. Run the simulation first with an Elasticity of 1.0. Then, run the simulation with an Elasticity of 0.0. What was the effect of decreasing the elasticity? Upon decreasing its elasticity, the final velocity of the pucks decreases compared to the initial velocity. 2. Predict: In activity A, you found that both total kinetic energy and total momentum are conserved in a perfectly elastic collision. How do you think decreasing the elasticity of a collision will affect the total momentum and total kinetic energy after the collision? The total momentum will remain the same however; the kinetic energy will decrease after the collision. In a perfect elastic collision the kinetic energy and momentum were conserved however, in a non elastic collision, the kinetic energy will decrease and the momentum will be conserved. 3. Experiment: Move the blue puck to point (-4.0, -6.0). Set its Initial velocity to v = 3.00i + 6.00j. Set the Initial velocity of the gold puck to v = 0.00i – 6.00j. Use the Gizmo’s Elasticity slider and CALCULATION tab to complete the table. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved Downloaded by kasra mehrinejad (kasra.mh.zh@gmail.com) lOMoARcPSD|17093303 Elasticity Stage Blue puck p (kg•m/s) Gold puck Total p TKE (J) p (kg•m/s) TKE (J) (kg•m/s) Total TKE (J) Before 15.00i + 30.00j 112.50 0.00i-18.0 0j 54.00 15.00i+ 12.00j 166.50 After 9.31i + 15.69j 33.30 5.69i + 27.69j 133.20 15.00i+ 12.00j 166.50 Before 15.00i + 30.00j 112.50 0.00i-18.0 0j 54.00 15.00i + 12.00j 166.50 After 10.74i 4.27j 13.35 4.26i + 16.27j 47.15 15.00i + 12.00j 60.49 Before 15.00i + 30.00j 112.50 0.00i 18.00j 54.00 15.00i + 12.00j 166.50 After 12.16i + 7.15j 19.90 2.84i+ 4.85j 5.26 15.00i + 12.00j 25.16 1.0 0.5 0.0 4. Analyze: Study the data you collected in the table on the previous page. A. In an inelastic collision, how did the total momentum (p) of the system change? The momentum remained the same from before and after the collision. B. In an inelastic collision, how did the total kinetic energy of the system change? The total kinetic energy decreased from before and after the collision. C. How were the inelastic collisions different from the elastic collision? In an elastic collision both momentum and kinetic energy were conserved, however; in an inelastic collision only momentum is conserved and the total kinetic energy decreases. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved Downloaded by kasra mehrinejad (kasra.mh.zh@gmail.com) lOMoARcPSD|17093303 5. Make a rule: Complete the sentence: During an inelastic collision, the total momentum of the system is conserved , while kinetic energy is not conserved 6. Infer: Why do you think some of the kinetic energy is lost during an inelastic collision? After an inelastic collision the objects travel in the same direction which results in some of the kinetic energy being converted to other types of energy like thermal energy or sound energy. 7. Think about it: Suppose a meteorite collided head-on with Mars and becomes buried under Mars’s surface. What would be the elasticity of this collision? Explain your answer. The elasticity of this collision will be 0. If it was a perfectly elastic collision then the meteorite would bounce back resulting in a conservation of momentum and kinetic energy. However, in this situation, the meteorite is being buried under the surface resulting in a loss of the total kinetic energy. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved Downloaded by kasra mehrinejad (kasra.mh.zh@gmail.com) lOMoARcPSD|17093303 Get the Gizmo ready: Activity C: Center of mass ● Click Reset. Turn on Center of mass trail. ● Move the blue puck to point (-4.0, -6.0). Set its Initial velocity to v = 3.00i + 6.00j. ● Move the gold puck to point (4.0, -6.0). Set its Initial velocity to v = 3.00i + 6.00j. Introduction: Suppose you tried to balance a hammer on one finger. If you placed your finger halfway down the handle, the hammer would fall because the head is much heavier than the rest of the hammer. Instead, you would have to place your finger near the head to balance the hammer perfectly. The point where an object balances is called its center of mass. Question: How can you find the center of mass of a system? 1. Identify: The center of mass for the system of two pucks is where the balancing point would be if a weightless rod connected the pucks. Set the mass of both pucks to 5.0 kg. What do you think are the coordinates of the center of mass of the two pucks? The coordinates of the center of mass of the two pucks would be (0, -6). 2. Compare: Click Play. The teal blue trail represents the trail of the system’s center of mass. How does the center of mass trail compare to the trails of the two pucks? The centre of the mass trail is parallel to the trails of the two pucks. 3. Observe: Click Reset. Change the Blue puck mass to 10.0 kg. Click Play. How did this affect the center of mass? The centre of the mass trail is more inclined towards the blue puck because the mass is larger than that of the gold puck. 4. Calculate: Click Reset. Set the Elasticity to 1.0. Move the blue puck to (2.0, -12.0) and change its mass to 6.0 kg. Move the gold puck to (12.0, -4.0) and change its mass to 1.5 kg. A. About where do you think this system’s center of mass is? (7, -8) B. The i coordinate of the center of mass can be found using the following equation: C. What is the i coordinate of the center of mass? The i coordinate of the centre of mass is 4. The same equation can be used to find the j coordinate of the center of mass. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved Downloaded by kasra mehrinejad (kasra.mh.zh@gmail.com) lOMoARcPSD|17093303 D. What is the j coordinate for this system? -10.4 5. Run Gizmo: Click Play. The first teal dot on the center of mass trail indicates the original position of the center of mass. Were your calculations correct? If not, what were the actual coordinates? My calculations were correct. 6. Compare: Click Reset. Set the blue puck’s velocity to v = 0.00i + 6.00j and the gold puck’s velocity to v = -8.00i + 0.00j. Click Play. A. Describe the trail of the center of mass. How is it different from the pucks’ trails? The trail of the centre of the mass is linear and it does not change before or after the collision. After the collision the puck’s trail changes directions due to an elastic collision however; the centre of the mass trail stays the same. B. Based on the spacing of the trail marks, compare the velocity of the center of mass before and after the collision. The velocity of the centre of the mass trail before and after collision is constant. There is no increase or decrease and it stays constant throughout. 7. Observe: After a collision, the velocities of the pucks will almost always change, but the velocity of the center of mass remains constant. A. Why do you think the velocity of the center of mass does not change? The velocity of the centre of the mass does not change because the trail of the centre of the mass is based on the mass used. The mass used stays the same before and after the collision, therefore; the velocity remains constant. The velocity of the puck may change due to the collision however; the mass used is the same throughout. B. Select the MOMENTUM tab. How does the center of mass trail compare to the gray arrow representing the total momentum of the system? The centre of the mass trial is equivalent to the total momentum of the collision. C. Click Reset, and try out various initial settings that result in a collision. For each of these collisions, compare the center of mass trail to the arrow representing the total momentum of the system. What trend do you see? Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved Downloaded by kasra mehrinejad (kasra.mh.zh@gmail.com) lOMoARcPSD|17093303 The total momentum of the system is equivalent to the centre of the mass trail. D. What is the relationship between the velocity of a system’s center of mass and the law of conservation of momentum? The relationship between the law of conservation of momentum and velocity of the centre of the mass trail is that they both remain constant before and after the collision. No mass or momentum is lost during the collision and therefore; they both represent a linear relationship. 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