Lab 08-Conservation of Momentum-Collision

PHYS 2101: Fall 2022 Fall 2022 COLLISIONS Investigating Momentum: one-dimensional and two-dimensional momentum Objectives: Link: https://phet.colorado.edu/sims/html/collision-lab/latest/collision- lab_all.html Choose ‘Explore 2 - D’ . Controls available to you : Vectors : If you want to see velocity and/or momentum vectors on the balls you added on the screen, you can select the “ Velocity ” and/or “ Momentum ” . It is also possible to see vectors in a separate display by selecting “ Momenta Diagram ”. You can zoom in or out using the slider on the momenta diagram to have a better look at the vectors. Elasticity : Changing the percentage of elasticity on the slide controls, the loss in kinetic energy of the system after collision can be observed. The two ends of the slide bar differentiate an elastic from the perfectly inelastic collisions. If you set the elasticity to 100%, the collision will be elastic , but it could be a perfectly Inelastic collision if you set it to 0%. All ‘ other collisions ’ that are not completely elastic/inelastic, can be represented between 0% and 100%. Additional controls : If you want to display the center of mass, check the “Center of Mass” box. You can also check the “Path” to display the footpath of the collision. You can check the “Reflecting Border ” if you want to bound the collisions inside the box. Once you pick your controls, you can click on “ Play ” button t o start the simulation. On the top right side of the ‘ simulation display ’ , you can select an up or down arrow to add or remove the balls. Select ‘ More Data ’ located below the simulation screen to get a larger box which displays the mass, location, velocity, and momenta of each ball. This is also a very important control to precisely locate and set up the initial conditions of the system. Running the simulation are obvious buttons found just at the screen. If you want to restart the simulation, just click t he “Restart” button. Introduction: When the net external force ( F net ) acting on a system of particles is zero (the system is isolated) and that no particles leave or enter the system (the system is closed), the total momentum ( P ) of the system is constant. F net = 0 🡺 Δ P/ Δ t = 0 🡺 Δ P = 0 Linear momentum is defined as, P = mv Δ P = 0 🡺 P – P 0 = 0 🡺 P = P 0 (The momentum after equals to the momentum before).

PHYS 2101: Fall 2022 Fall 2022 For example , let us consider two balls colliding with each other as shown below and rewrite the law of conservation of momentum explicitly. The momentum before a collision is always equal to the momentum after the collision. Since momentum is a vector quantity, we should pay attention to directions. If it is a one- dimensional collision, the directions are right and left or positive and negative on the horizontal axis. In two-dimensional motion, you have to resolve the momentum vectors in x- and y- directions before and after the collisions. Check out the figures below for the general case of collision in two dimensions for the two balls. After Collision m 1 v 1 + m 2 v 2 m 1 m 2 v 02 v 01 m 1 m 2 v 2 v 1 Before Collision m 1 v 01 + m 2 v 02 m 1 v 01 + m 2 v 02 = m 1 v 1 + m 2 v 2 Before Collision m 1 v 01x + m 2 v 02x m 1 v 01y + m 2 v 02y After Collision m 1 v 1x + m 2 v 2x m 1 v 1y + m 2 v 2y

PHYS 2101: Fall 2022 Fall 2022 ii. the momenta of the system before and after collisions are equal to 1 kg.m/s and 1 kg.m/s iii. the kinetic energies of the system before and after collisions are equal to .5 Joules and .5 Joules b. Now, you run the simulations with elasticity of 50% and initial velocities of ball 1 and ball 2 are 1.0 m/s and at rest respectively, [3 Points] i. the velocities of the two balls after collision are v 1 = .25 m/s and v 2 = .75 m/s ii. the momenta of the system before and after collisions are equal to .5 kgm/s and 1 kgm/s iii. the kinetic energies of the system before and after collisions are equal to .5 Joules and .31 Joules What did you observe during this exercise? [2 Points] I noticed that as when the elasticity isn’t 100 the system loses momentum and energy when objects get into contact c. Now you run the simulations with elasticity of 0% and initial velocity of ball 1 is 1 m/s while ball 2 is at rest. [3 Points] i. the velocities of the two balls after collision are v 1 = .5 m/s and v 2 = .5 m/s ii. the momenta of the system before and after the collisions are equal to 1 kgm/s and 1 kgm/s iii. the kinetic energies of the system before and after the collisions are equal to .5 Joules and .25 Joules What did you observe from the results of section 2(a), (b) and (c)? Do they all follow conservation of momentum and kinetic energy? [3 Points] When the elasticity is reduced more energy is lost thru contact. 3. Let ball 1 has a mass twice that of ball 2 and ball 1 and ball 2 are located at (0.5,0) and (1.5,0) respectively, a. If you run the simulation with elasticity of 100% and initial velocities of ball 1 and ball 2 are 1 m/s and ‘0’ m/s respectively, [3 Points] i. the velocities of the two balls after collision are v 1 = .67 m/s and v 2 = 1.33 m/s ii. the momenta of the system before and after collisions are equal to 2 kgm/s and 2 kgm/s

PHYS 2101: Fall 2022 Fall 2022 iii. the kinetic energies of the system before and after collisions are equal to 1 Joules and 1 Joules, respectively. b. If you run the simulation with elasticity of 50% and initial velocities of ball 1 and ball 2 are 1 m/s and at rest respectively, [3 Points] i. the velocities of the two balls after collision are v 1 = .5 m/s and v 2 = 1 m/s ii. the momenta of the system before and after collisions are equal to 2 kgm/s and 2 kgm/s iii. the kinetic energies of the system before and after the collisions are equal to 1 Joules and .75 Joules c. If you run the simulation with elasticity of 0% and initial velocities of ball 1 and ball 2 are 1 m/s and ‘0’ m/s, respectively, [3 Points] i. the velocities of the two balls after collision are v 1 = .67 m/s and v 2 = .67 m/s ii. the momenta of the system before and after the collisions are equal to 2 kgm/s and 2 kgm/s iii. the kinetic energies of the system before and after the collisions are equal to 1 Jules and .67 Joules What did you observe from the results of section 2(a), (b) and (c)? Do they all follow conservation of momentum and kinetic energy? Explain [3 Points] Yes because momentum and kinetic energy are preserved in the elastic collison while only momentum was preserved in the inelastic collisions. Part II: Collisions in two dimensions Reset the previous simulation and go to the “Explore 2D” Tab. Click on “More Data” and choo se ‘Path’. A. Elastic Collision 1. Apply the following settings for the simulation: a. Mass: m 1 = m 2 = 0.5 kg. b. Location: (0.5m, 0.5m) and (0.5m, 0m) for ball 1 and ball 2 respectively c. Initial velocities: v 01x = 0.4m/s v 01y = - 0.2m/s v 02x = 0.4m/s v 02y = 0.2m/s 2. Calculate the angles for initial velocity vectors. [4 Points] -26.57 and 26.57

PHYS 2101: Fall 2022 Fall 2022 .2-.2 = -.1+.1 3. How about conservation of energy ? Is that also satisfied? Show your work to prove your point. [5 Points] No it is not satisfied because a portion of the energy is lost during Collison. M1 = .5 M2 =.5 Elasticity = .5 V1ix =.4 V1iy = -.2 V2ix = .4 V2iy = .2 (.5)(.4)(.5) ≠ (.5)(.4) (.5)(-.2)(.5) ≠ (.5)(-.2) (.5)(.4)(.5) ≠ (.5)(.4) (.5)(.2)(.5) ≠ (.5)(.2) Final thoughts: What is the similarity between (A) and (B) in Part II? What is the difference? Did you observe a similar pattern for your results while working on 1-D collision? [5 Points]

PHYS 2101: Fall 2022 Fall 2022 That the Vx does not change even when the collision is inelastic. I also noticed that in a perfect elastic collision momentum and energy are conserved while in inelastic collisions only momentum is conserved. Finally, what is the big conclusion you derived for this experiment about conservation of energy and momentum? [5 Points] I derived that momentum is always conserved while energy is only conserved when the Collison is elastic