kinetic energy after the collision [Archive]

dg_5021

Oct13-04, 06:48 PM

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?

Doc Al

Oct13-04, 07:11 PM

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.

samelliz

Aug7-11, 04:04 AM

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.

PeterO

Aug7-11, 06:47 AM

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?

samelliz

Aug7-11, 08:05 AM

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.