Lab 3 Collision and Momentum (Phys 2211, Friday, Oct 7, Fall 2022, Dr Xiaojun Wang) Online: https://phet.colorado.edu/sims/html/collision-lab/latest/collision-Iab all.htrnl Reports due: 11:59 pm, Wed, October 12, 2022 (Dropbox: Lab 3 Reports) Objective To observe a variety of collision types in order to determine the momentum before and after the collision as well as the change in momentum from before to after. Theory The momentum, p, of an object is the product of its mass m and its velocity v. Because it is the product of a scalar with a vector, momentum is also a vector quantity, and as such has a magnitude and direction associated with it. Momentum can be transferred in a collision from one object to another. For example, momentum can be transferred in an elastic collision when two objects bounce off each other, such as two billiard balls colliding. For another example, momentum can be transferred in an inelastic collision when two objects hit and stick together, moving as one after the collision, such as two football players during a tackle. Finally, momentum can be transferred in an explosion, when two objects push off one another. We will observe all three of these transfers of momentum in the online lab. Experiment 1 - Elastic Collision (1-dimentional collision) Parameter Ball 1 Ball 2 Mass of ball (kg) I Velocity before collision (m/s) l Momentum before collision (kg.m/s) | Velocity after collision (m/s) Momentum after collision (kg.m/s) Calculate (1) the total momenta before and after collation and (2) the total kinetic energies before and after collation. Verify the conservation for both momentum and kinetic energy. Experiment 2 - Inelastic Collision (1-dimentional collision) Parameter l Ball 1 Ball 2 Mass of ball (kg) I Velocity before collision (m/s) l Momentum before collision (kgm/s) l Velocity after collision (m/s) Momentum after collision (kgm/s) Calculate (1) the total momenta before and after collation and (2) the total kinetic energies before and after collation. Verify the conservation for momentum but NOT for kinetic energy. What is the percentage of kinetic energy loss after collision? Experiment 3 - Elastic Collision (2-dimentional collision) Collect the information (masses, angles, and velocities) before collision Balls Velocity 0.5 m 2 4 Momentum O Center of Mass C# O Kinetic Energy |vl = 0.30 mis Values Reflecting Border V = 0.68 m/s O Path Tp/ = 0.15 kg m/s Elasticity 100% HRSLIC pl = 0.34 kg mis Constant Size 0.00 s Normal G + Momenta Diagram Slow More Data Mass (kg) Position (m) Velocity (m/s) Momentum (kg m/s) 1 0.50 -0.80 -0.40 0.61 0.29 0.3 0.14 2) 0.50 0.00 0.00 0.30 0.00 0.15 0.00 Collect the information (masses, angles, and velocities) after collision Balls Velocity 0.5 m IVI = 0.52 mis 24 O Momentum O Center of Mass pl = 0.26 kg m/s O Kinetic Energy IV = 0.52 m/'s Values Reflecting Border Path Ipl = 0.26 kg mis Elasticity 100% Inclastic. Flastic Constant Size 2.88 s Normal G + Momenta Diagram Slow More Data Mass (kg) Position (m) Velocity (m/s) Momentum (kg m/s) X y Px Py 1) 0.50 0.69 0.51 0.39 0.35 0.20 0.17 N 0.50 1.13 -0.07 0.52 -0.06 0.26 -0.03 Calculate (1) the momenta before and after collation for both x- and y- components. Verify the momentum conservation, i.e., Total Pxi = Total Pxf and Total Pyi = Total Pyf. (2) the total kinetic energies before and after collation. Verify the conservation of kinetic energy.Experiment 4 - Total inelastic collision (1-dimential collision) Set mass 1 with initial velocity and mass 2 at rest before collision. 0# Velocity 0.5 m Momentum x 7 O Center of Mass Kinetic Energy IVI = 0.50 m/s IVI = 0.00 m/'s Reflecting Border O Path Ipl = 0.25 kg m/s [pl = 0.00 kg m/s Inelastic Collision Elasticity = 0% Stick . Slip _ Constant Size 0.00 s + Momenta Diagram Normal Slow More Data Mass (kg) Position (m) Velocity (m/s) Momentum (kg m/s) X VX Vy Px 0.50 -0.50 0.00 0.50 0.00 0.25 0.00 2 0.50 0.50 0.00 0.00 0.00 0.00 0.00 Custom O Velocity 0.5 m O Momentum O Center of Mass X O Kinetic Energy IVI = 0. =0:25 m/'s 1 2 Reflecting Border Path Ipl = 01pp *0/13:kg m/s Inelastic Collision Elasticity = 0% Stick O Slip _ Constant Size 3.67 s G + Momenta Diagram Normal Slow More Data Mass (kg) Position (m) Velocity (mis) Momentum (kg mis) VX Vy Py 0.50 0.77 0.00 0.25 0.00 0.13 0.00 2 0.50 1.07 0.00 0.25 0.00 0.13 0.00 Custom G- Find the velocity for the combined mass and verify the conservation of momentum. Calculate the percentage of kinetic energy loss.Experiment 5 - Let's play billiards (Optional) Set both balls at the same mass. Give ball 1 some velocity and ball 2 at rest. (1) Head on collision (before/left and after/right collision) Balls velocity 2 4 Momentum Velocity O Center of Mass Kinetic Energy Center of Mass 1 = 1.00 m/'s IM = 0.00 m/s Values O Knedl: Energy Is1 = 0.00 m's 1YI = 1.00 nIV's Reflecting Border Valuce Path Refecting Border pl = 0.50 kg m/s Ip/ = 0.00 kg m's O PEth lasticity 100%% pl = 0.00 ky m's Ipl = 0.50 ky m/'s Elasticit Constant Size Constant S ze 0.00 s + Mornemta Diagram 1.70 C + Momenta Diagram More Data More Data Mass (kg) Position (mm) Velocity (mis) Momentum (kg m/s) Mass (kj) Position [m] Velocity (mis) Momentum (kg mis) 1 0.50 -1.00 0.00 1.00 0.00 0.50 0.00 20.50 0.00 0.00 0:00 0.00 0.10 0.00 0.30 0.00 0.00 0.0 1.00 0.10 1 30 0 00 O O (2) Collision with angle (before/left and after/right collision) Balls velocity W Velocity 2 4 MomentIn 0.5 m IN1 = 0 76 mis Momentum Center o' Mass O Center of Mas Kinetic Energy Kinetic Energy 1 - 1.50 m's (2 Vales 141 = 0 37 hu Ive Voluza 121 - 0.00 kg live Reflecting Burder WI = 0.87 MIN's Reflecting Samar Path Path pl - D 50 KG m/= Elastichy 1009% Elasticity 100 3. Comeran Size Constant Size 0.00 s + Moments Diagram + Moments Diagram More Data More Data Mass (kg) Position (m ) Velocity (m's) Momentum (kg m's) Mass (ko] Position (m ] Velocity (mis) Momentum (kg mis) -1 01 1 0U 1 0.50 0.39 -0 24 1044 -0.50 0.22 -0.25 0.00 0.C0 0.00 2 0.50 0.67 0.44 0 56 0.50 0.28 0.25 O Use the momentum and energy consevation as discussed in class (the dot product term must be zero) to expain the results. momentum conservation Vli = Vif+ 12f => = Vig + Vzf + 201f.02f energy conservation va = VIf+ vzf = > Vfi = Vif+ vzf Question Use your best knowledge to discuss why conservation of momentum is effective for both elastic and inelastic collisions, while conservation of energy is only good for elastic cases. Where the energy goes for inelastic collision