Kinetic energy after the collision

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Homework Help Overview

The discussion revolves around a problem involving two carts colliding, where one cart is initially in motion and the other is at rest. The participants are tasked with calculating the velocity of the center of mass and the kinetic energy before and after the collision, specifically focusing on the differences in kinetic energy due to the nature of the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the center of mass velocity before and after the collision. There are questions about how to determine the kinetic energy after the collision and whether the collision is elastic or inelastic. Some participants express confusion regarding the differences in kinetic energy before and after the collision.

Discussion Status

The discussion is ongoing, with some participants providing calculations for kinetic energy before and after the collision. There is acknowledgment of the inelastic nature of the collision, which leads to a decrease in kinetic energy. However, confusion remains regarding the calculations and the principles involved.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information provided. There is an emphasis on understanding the differences between elastic and inelastic collisions, particularly in relation to kinetic energy.

dg_5021
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Suppose the air cart to the left has a mass of .750kg and an initial speed of .455m/s. The cart to the right is initially at rest and has a mass of .275kg. Find the velocity of the center of mass. (a) before and (b) after the carts collide and stick together (c)Find to kinetic energy of the system before and after the collision.

(a)((.750kg)(.455 m/s) +(.275kg)(0m/s))/(.750kg+ .275kg) = .332927 m/s

(b)((.750kg)(.455 m/s) +(.275kg)(0m/s))/(.750kg+ .275kg) = .332927 m/s

(c)ki= (1/2) (.750kg)(.455m/s)^2 + (1/2)(.275kg)(0)^2 = .077634 J

i don't know how to get the kinetic energy of the system after the collision? can someone help me out?
 
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dg_5021 said:
i don't know how to get the kinetic energy of the system after the collision? can someone help me out?
You found the speed of the center of mass. Now you get to use it.
 
i really can't see that the amount of kinetic energy before and after the collision are equal.. They really are not. I'm really confused with this. I need help.

I computed for the amount of kinetic energy after collision and found it to be .0568 J. It does not really tally with the initial kinetic energy which is .0776 J. Please help.
 
Last edited:
samelliz said:
i really can't see that the amount of kinetic energy before and after the collision are equal.. They really are not. I'm really confused with this. I need help.

I computed for the amount of kinetic energy after collision and found it to be .0568 J. It does not really tally with the initial kinetic energy which is .0776 J. Please help.

Is this an elastic collision, or an in-elastic collision?
 
ok i got it. the example given above is an inelastic collision so the KE after it less than the KE before. But I've encountered lots of problems involving elastic collisions but really i never computed KE as equal.
 

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