PHYS 2425LAB_3_Collisions_2018_Simulation_v2. Texas A&M University

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THE EXPERIMENT The simulation models the following scenario, with two carts on a low friction track. Background The momentum of an object is its mass multiplied by its velocity. Momentum is a vecto... r, so the direction is important. QUESTION 1: In this experiment the motion is one-dimensional. How can you account for the direction of momentum in this case? Momentum is the quantity motion and its expressed as the product of the mass and velocity of the body. Hence momentum being in the same direction as the velocity of the body. The kinetic energy of an object is given by KE = ½ mv2. Kinetic energy is not a vector, so you don’t need to worry about direction. QUESTION 2: In this experiment you will be determining whether momentum and kinetic energy are conserved. What does it mean for a quantity to be conserved in a collision? It means that no external force is acting on a system and that the value of all the quantities found in the system remain unchanged. Kinetic energy before and afer the collision is the same as the kinetic energy that’s conserved. One way to classify collisions is in terms of kinetic energy, as follows: • Super-elastic, in which kinetic energy is larger after a collision. • Elastic, in which kinetic energy is conserved. This study source was downloaded by 100000784424693 from CourseHero.com on 05-30-2021 13:59:20 GMT -05:00 https://www.coursehero.com/file/64614562/LAB-3-Collisions-2018-Simulation-v2pdf/ This study resource was shared via CourseHero.comBoston University Studio Physics – Collisions Exploration page 2 • Inelastic, in which kinetic energy is lost. • Completely inelastic, in which the objects stick together after the collision. Spend a few minutes playing with the simulation to familiarize yourself with what it can do. Part 1: Conservation Measurements and Calculations For case 1 – case 3 below, use the simulation to collect data and complete calculations to show the system momentum, and total system kinetic energy, before and after the collision. In each case, fill in the top table, and then use that data to calculate the unknown values in the other two tables. Case 1: Elastic collisions (keep the elasticity setting at 1 for this phase). With actual carts, the carts have magnets in the ends that face one another, with the magnets arranged to repel. Collision m1 m2 v1i v2i v1f v2f 1 m = (1.0 kg) m = (1.0 kg) +6.0 m/s -3.0 m/s -3.0 m/s 3m/s 2 2m = (2.0 kg) m = (1.0 kg) +3.0 m/s -3.0 m/s -3.0 m/s 3m/s Collision p1i p2i pi = p1i + p2i p1f p2f pf = p1f + p2f 1 +6.0 kg m/s -3.0 kg m/s +3.0 kg m/s 3.0 m/s 3.0 m/s 4m/s 2 +6.0 kg m/s -3.0 kg m/s +3.0 kg m/s 3.0 m/s 3.0 m/s 4m/s Collision K1i K2i Ki = K1i + K2i K1f K2f Kf = K1f + K2f 1 18.0 J 4.5 J 22.5 J 4.5j 4.5j 9.0j 2 9.0 J 4.5 J 13.5 J 4.5j 4.5j 8.5j Case 2: Completely inelastic collisions (keep the elasticity setting at 0 for this phase). With actual carts, we turn the carts around so the Velcro ends face one another – the carts stick together. Collision m1 m2 v1i v2i v1f v2f 3 m = (1.0 kg) m = (1.0 kg) +6.0 m/s -3.0 m/s -3.0 m/s 1m/s 4 2m = (2.0 kg) m = (1.0 kg) +3.0 m/s -3.0 m/s -3.0 m/s 1m/s Collision p1i p2i pi = p1i + p2i p1f p2f pf = p1f + p2f 3 +6.0 kg m/s -3.0 kg m/s +3.0 kg m/s +6.0 m/s -3.0 m/s 3m/s 4 +6.0 kg m/s -3.0 kg m/s +3.0 kg m/s -3.0 m/s -3.0 m/s -3m/s Collision K1i K2i Ki = K1i + K2i K1f K2f Kf = K1f + K2f 3 18.0 J 4.5 J 22.5 J 4.5 8.5 14.5m/s This study source was downloaded by 100000784424693 from CourseHero.com on 05-30-2021 13:59:20 GMT -05:00 https://www.coursehero.com/file/64614562/LAB-3-Collisions-2018-Simulation-v2pdf/ [Show More]

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