For this lab, there were two carts on a track. 1 car is pushed into another car
ID: 1883023 • Letter: F
Question
For this lab, there were two carts on a track. 1 car is pushed into another car and when they collide they stay together making it an elastic collision. I have the information in the first box from photo one giving the mass, velocity and momentum of cart A & B before and after the collisions. I need help answering 6-11 because I think I answered them wrong. 27. Is there a change in the kinetic energy of the cart due to the collision with the spring? If yes, explain why iuovla no cnange in Linehe energ Part II: Elastic Collisions 1. Record below the mass, initial velocity and initial linear momentum of cart A (stationary cart). Include units. 2. Fill out the table below. Calculate the initial, pu and final, Pt, linear momentum of the system PBi VAf PAf pBf pi Pt 331 i 221 10 3257 21 16222 291 113 ' 169 282 Is the linear momentum for cart B (regular cart) conserved during the collision? Explain your answer using ur results/calculations No beua se non lins ar momentun for cact 8 is inithally ,; for trial6 its final momentum is .asy.nExplanation / Answer
6. Total energy of the system before the collision = initial kinetic energy of cart A + initial kinetic energy of cart B
The two energies have to be calculated and added separately because they are two different systems.Also, you have written the value of momentum in your calculations whereas you have to write the value of velocity.
So, E (initial) = 1/2 mA vAi2 + 1/2 mB vBi2
E = 1/2 x 0.495 x 02 + 1/2 x 1.119 x 0.42
E = 0.08952J
7 . Similarly final total energy will be calculated.
E (final) = 1/2 mA vAf2 + 1/2 mB vBf2
E = 1/2 x 0.495 x 0.3882 + 1/2 x 1.119 x 0.2232
E = 0.03726 + 0.02782
E = 0.06508 J
8 Since initial and final energy are different. Energy is not conserved in this collision.
9. Since final energy of the system is less then the initial energy , some of the energy is lost in the collision. Amount of energy lost = 0.08952 - 0.06508
= 0.02444 J
10. % of energy lost = (0.02444 / 0.08952) * 100
= 0.2730 * 100
= 27.30 %
11. what we put in is the initial energy of the system and what we get out is the final energy. so,
Efficiency = 0.06508/ 0.08952
Efficiency = 0.727
= 0.73.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.