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(0.5 kg . 1 m /5 ) + (1,5kg . 0 . 1. 25 kgm /s = 2 kg f Part I Elastic Collisions Use
(0.5 kg . 1 m /5 ) + (1,5kg . 0 . 1. 25 kgm /s = 2 kg f Part I Elastic Collisions Use the default mass, position and velocity, and set elasticity at 100%, press the Play button, observe the collision and see how the velocities of the objects change after the collision. Record the data in the table below. Mass vi (m/s) of (m/s) Kf (J ) Ki (J) (kg) Object 1 I m /s 10.25 5 1,25 m /s 10.445 10 .5 Object 2 1.5 0.5 m/s 10. 1875J . 25 m/s|0.445 Add the initial kinetic energy of object 1 and object 2, put your answer here 0. 4375 J this is the total kinetic energy of he system before collision. dd the final kinetic energy of object 1 and object 2, put your swer here 0. 88 J this is the total kinetic energy of e system after the collision. at is the change of the kinetic energy of the system? Is you wer making sense? Explain. ere is no change because the kine ergy before the collision is the sa ter the collision, my answer makes se because of the conservation of gy, which states that energy cannot troyed or created. In addition, th servation of energy also states that\fvi (m/s) Ki (J) Mass 10. 69m/5 0. 119 (kg) Object 1 0.44J 10.06m /5 0.00275 Object 2 0.5kg 1 m/s 15 kg 0.445 . 06 2. 1.5 0. 00 27 0.5m/'s 2 Total kinetic energy before collision _0. 445. Total kinetic energy after collision 0. 125 0 44 - 0. 12 = 0.32 Change of kinetic energy 0. 32J Compare the change of kinetic energy with the answer in part I, does it make sense? Explain. change in kinetic energy In part 2, there was The kinetic e energy has gone into collision generated heat. In an inelastic collision, some of the ene is transformed or converted to heat, It makes sense because in part 1, there was no deforma or heat generated because we had an elastic co There is reduced elasticity in part 2, which means Part Ill was an inelastic behavior between the two b meaning some of the kinetic energy was conver Use the same setting as part I and II, just change the elasticity to h to 0%. Record your data below. Mass vi (m/s) Ki (J) vf (m/s) Kf ( J ) (kg) Object 1 0.5 kg Im/5 0. 445 0, 13 m /5 . 004225J 132.. 5 = 004225 of 2 negative
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