Understanding Conservation of Momentum Through Virtual Labs
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PHYS SCI 108
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Physics
Date
Dec 10, 2024
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9
Uploaded by JusticeHornetPerson1229
PhysicsModule 33.05 Conservation Of Momentum - Nearpod Template v.22Names:Sebastian AmayaJack BrewiGryphon PageGiovanna FuscaOlivia RiveraMateo PoujadeJacob WangMelissa Gonzalez MCameron DAlara KayaPurpose: The purpose of this lab is to determine what type of collisions follow conservation of momentum.Lab setup: This is a virtual lab. You will use PHET Collision Simulation to collect your data.Resources: equation sheet & 3.08 Nearpod Step By Step Tutorial- see resource siteStep 1. HYPOTHESISRead the lesson, then choose a, b or c to complete your hypothesis below. Highlight or bold your choice.If given elastic and inelastic collisions, the ___________ will show conservation of momentum.a.elastic collisionsb.inelastic collisionsc.elastic collisions and inelastic collisionsStep 2. PROCEDURE for DATA COLLECTIONA.Record your observations from the collision demonstration in the lab report below. (Hint: Describe what each type of collision looks like on the next page.)B.Use the PHET Collision Simulationto collect data on mass, initial velocity and final velocity.For Collisions A(Table 1) and B (Table 2)A.Begin by selecting the following settings: a.Choose the Intro option. Check the value box and keep the Elasticity at 100%B.Select mass 1 and 2 as given in the table.C.Select velocity for mass 1 and 2, by sliding the vector arrows in the simulation.D.Press play and record the final velocity for each mass after the collision occurs. ( It’s a good idea to click the pause button just shortly after the masses collide.)For Collisions C (Table 3) and D (Table 4)E.Move the Elasticity to 0%. These trials will be inelastic collisions.F.Repeat steps B-D above For Collisions E (Table 5) and F (Table 6)G.Choose your own values for mass and velocity based on the given collision type.H.Press play and record the final velocity for each mass after the collision occurs. ( It’s a good idea to click the pause button just shortly after the masses collide.)
Step 3. ANALYSIS usingEquations: p = mv and Σpi = ΣpfFor tables 1-6, calculate the initial and final momentum for each mass in each collision.For table 7, complete calculations and compare results to determine if momentum is conserved.Step 4. QUESTIONS & CONCLUSIONUse complete sentences and show all steps of your calculations!A.Collision Demonstration ObservationsB.Phet Simulation DataTable 1: Elastic Collision between equal mass:Collision AMrs CloranMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Momentum Initial(kg*m/s)Momentum Final(kg*m/s)Mass 12.003.00-3.00P = mvoP = (2.00)(3.00)P = 6.00P = mvfP =(2.00)(-3.00)P = -6.00Mass 22.00-3.003.00P = mvP = (2.00)(-3.00)P = -6.00P = mvP = (2.00)(3.00)P = 6.00Add the blue columns6 + -6 = 0-6 + 6 = 0Table 2: Elastic Collision between unequal mass:Collision BMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Momentum Initial(kg*m/s)Momentum Final(kg*m/s)
AKMass 11.003.00-5.00P = (1.00)(3.00)P = 3.00P = (1)(-5.00)P = -5.00Mass 22.00-3.001.00P = (2.00)(-3.00)P = -6.00P = (2.00)(1.00)P = 2.00Table 3: Inelastic Collision between equal mass:Collision CCDMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Momentum Initial(kg*m/s)Momentum Final(kg*m/s)Mass 12.003.000P = (2.00)(3.00)P = 6.00P = (2)(0)P = 0Mass 22.00-3.000P = (2.00)(-3.00)P = -6.00P = (2)(0)P = 0Table 4: Inelastic Collision between unequal mass:Collision DGFMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Momentum Initial(kg*m/s)Momentum Final(kg*m/s)Mass 11.003.00-1.00 m/sp=mvp= (1.00)(3.00)p= 3.00 p = (1.00)(-1.00)p = -1.00Mass 22.00-3.00-1.00 m/sp = (2.00)(-3.00)p = (2.00)(-1.00)
p = -6.00p = -2.00Table 5: Your own elastic collisionCollision EGPMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Momentum Initial(kg*m/s)Momentum Final(kg*m/s)Mass 11.502.00-2.00P = mvP = (1.50)(2.00) P = 3.00P = mvP = (1.50)(2.00)P = -3.00Mass 21.50-2.002.00P = mvP = (1.50)(2.00)P = -3.00P = mvP = (1.50)(2.00)P = 3.00Table 6: Your own inelastic collisionElastic Trial.Collision FJBMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Momentum Initial(kg*m/s)Momentum Final(kg*m/s)Mass 13.002.00-2.00P = mvP = (3.00)(2.00)P = 6.00P = mvP = (3.00)(-2.00)P = -6.00Mass 23.00-2.002.00P = mvP = (3.00)(-2.00)P = -6.00P = mvP = (3.00)(2.00)P = 6.00TYPE: InelasticCollision GMassInitial VelocityFinal VelocityMomentum InitialMomentum Final
JW(kg)(m/s)(m/s)(kg*m/s)(kg*m/s)Mass 13203 * 2 = 63 * 0 = 0Mass 23-203 * -2 = -63 * 0 = 0TYPE: ElasticCollision HMPMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Momentum Initial(kg*m/s)Momentum Final(kg*m/s)Mass 132-0.67P = mvP = (3)(2)P = 6P = mvP = (3)(-0.67)P = -2.01Mass 21.5-23.33P = mvP = (1.5)(-2)P = -3P = mvP = (1.5)(3.33)P = 5TYPE: ElasticCollision IMMGMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Momentum Initial(kg*m/s)Momentum Final(kg*m/s)Mass 131-.033P(3)(1)P = 3P = (3)(-0.33)P = -0.99Mass 21.5-11.67P(1.5)(-1)P(1.5)(1.67)
P=(-1.5)P = 2.505 P=3TYPE: inelastic CollisionJ ORMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Momentum Initial(kg*m/s)Momentum Final(kg*m/s)Mass 11.52-0.67P = mvoP= 1.5(2)P = 3 P = mvoP = 1.5 (-0.67)P = -1.0Mass 23-2-0.67P = 3(-2) P = (-6)P = 3(-0.67)P= -1.8Type: ElasticCollision KSAMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Momentum Initial(kg*m/s)Momentum Final(kg*m/s)Mass 11.52-3.33P = mvP = 1.5(2)P = 3P = mvP = 1.5(-3.33)P = -5Mass 23-20.67P = mvP = 3(-2)P = -6P = mvP = 3(0.67)P = 2.01
Table 7: Results TableCollisionTotal Momentum Initial(kg*m/s) (Hint: find sum of initial momentums)Total Momentum Final(kg*m/s)(Hint: find sum of final momentums)Is momentum conserved?(Hint: Does Total .momentum initial equal the total momentum final?)Collision ATable 1Mrs C(add the blue initial column)6.00 + -6.00 = 0-6.00 + 6.00 = 0Yes! 0 = 0Collision BTable 23.00 - 6.00 = -3.00-5.00 + 2.00 = -3.00Yes, -3.00 = -3.00 Therefore, momentum is conserved.Collision CTable 36.00 + -6.00 = 00 + 0 = 0Yes, 0 = 0 Collision DTable 43.00 + -6.00 = -3.00-2.00 + -1.00 = -3.00Yes, -3.00 = -3.00Collision ETable 53.00 + -3.00 = 0-3.00 + 3.00 = 0Yes, the initial momentumof zero is equal to the final momentum of zero.Collision FTable 66.00 + -6.00 = 0-6.00 + 6.00 = 0The total momentum initial of zero does equal the total momentum final of zero.G6 + -6 = 00 + 0 = 0Yes, 0 = 0H6 + -3 = 3-2.01 + 5 = 3Yes 3 = 3IJ3 + -6 = -3 -1.0 + - 1.8 = - 2.8KInd of yes K3+(-6)= -3-5+2.01= -2.99Yes, -3 roughly equal to -2.99 (simulation issue)
C.Questions and ConclusionAnswer the following questions in complete sentences.1.Based on your observations of the six collisions, describe the physical difference between elastic and inelastic collisions.2.For which collisions was momentum conserved? Explain how you determined this using your data. 3.The simulation limits you to a maximum of 3 kg and +/- 3 m/s for the initial velocity. Let's imagine you want to know what would happen if the masses were higher or the balls were going faster than that. Create data for a trial with higher initial velocities. Make one of the final velocities 0 m/s. What would be the final velocity of the other ball? Show your work for all calculations for individual momenta and total momenta.Create your ownMass(kg)Initial Velocity(m/s)Final Velocity(m/s)Mass 1Mass 2See the 3.06 Honors Live Lesson recording4.Application! Show your work as you answer the following conservation of momentum problems.a.A truck with mass of 3,250 kg traveling with a velocity of 25.0 m/s hits a car at rest. After the collision, the truck moves with a velocity of 19.0 m/s. The car has a mass of 2,820 kg. If the two vehicles do not stick together, how fast is the car moving after the collision?
b.Two students are sitting next to each other on chairs with wheels. They push each other and separate in opposite directions. The student with a mass of 48 kg moves to the left with a velocity of 1.8 m/s. How fast and in what direction does the other student with a 59-kg mass move?