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Kinetic energy question

  1. Nov 12, 2007 #1

    klm

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    okay i have a problem that deals with an elastic collision. so i have the mass of obj A and B and i found the initial velocity of A and B, and i have the final velocity of A and B
    so now i want to find the change in Kinetic Energy, so i am doing
    .5(mA + mB)( velocity final ^2) - .5(mA + mB)( velocity initial^2)
    but what i am confused about is what i am suppose to use for the final velocity and initial velocity. i mean do i just add if initial velocities of obj A and B and use that as Vi, and add up the final velocities of obj A and B and use that as Vf? i dont think that seems right? any help would be great!
     
  2. jcsd
  3. Nov 12, 2007 #2
    check your equations again they, as are incorrect.
     
  4. Nov 12, 2007 #3

    klm

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    the change in KE = .5mVf^2 - .5mVi^2 . is that incorrect?
     
  5. Nov 12, 2007 #4
    no but the way you combined the mass was. the masses are different objects. KE1 + KE2
     
  6. Nov 12, 2007 #5

    klm

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    i am sorry i dont understand what you mean. i cannot combine the masses to make
    .5 ( mA+mB) vf^2 ? how come?
     
  7. Nov 12, 2007 #6
    you have 2 different velocities, do you not?
     
  8. Nov 12, 2007 #7

    klm

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    yes i have 2 diff initial velocity for A and B and 2 diff final velocities for A and B
     
  9. Nov 12, 2007 #8
    what is the KE of an object?
     
  10. Nov 12, 2007 #9

    klm

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    .5mv^2
     
  11. Nov 12, 2007 #10
    so what do you get when you add the KE of 2 objects?
     
  12. Nov 12, 2007 #11

    klm

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    i dont know?
     
  13. Nov 12, 2007 #12
    why don't you try adding the KE's and see what happens?
     
  14. Nov 12, 2007 #13

    klm

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    i can't add the KE's b/c i dont know how to use the velocity in the equations. i know i need to use the final and initial, but my only problem is that i have 2 intials and 2 finals. and i dont know how to use that. i dont have a problem to solve, i am just asking because this is a concept i am getting stuck on.
     
  15. Nov 12, 2007 #14
    ...

    how do you add one plus two? now add KE1 plus KE2. the final KE equals the initial KE. does your text have examples? i can assure you if it does that there is a problem like this one.
     
  16. Nov 12, 2007 #15

    klm

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    ok thank you for your help. i will go look for an example
     
  17. Nov 12, 2007 #16
    i'm sorry but the general rules on this forum are to help people arrive at their own solutions. it is clear from your approach that you never learned the material or otherwise never remembered it. perhaps paying attention or reading your book will help. if you'd rather not do those two, i hear wikipedia has all the answers.
     
  18. Nov 12, 2007 #17
    To find the change in KE you use the formula KE = .5mv^2

    so the KEi (initial) is .5*m,a*vi,a^2 + .5*m,b*vf,b^2

    you do KE,f the same way but with the final velocities

    then you do KEf - KEi

    does this answer your question?
     
  19. Nov 12, 2007 #18

    klm

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    thank you moncheri. i did not think you were allowed to do this way because i thought you needed to keep the initial velocities separate from the final velocities. i am confused because how can you put the final velocity of b, with the initial velocity of a like you did. thank you for responding!

    and fliinghier- yes i am aware of the general rules of this board. i am not asking you to give me an answer to a problem, i am asking for help to understand an equation. thank you though
     
  20. Nov 12, 2007 #19
    i dont understand why youre confused - i kept the initial velocities separate .. you have to add the KEi of a with the KEi of b .. because before the collision the two objects a and b are seperate, correct? If they're together that's a different story - maybe I don't understand?

    Well, I thought anyway that during an elastic collision KE is conserved. Right?
     
  21. Nov 12, 2007 #20

    klm

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    no no sorry i am confusing you! you are right, the two objects are separate before and after the collision.
     
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