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Need help with a Collisions Question

  1. Jan 16, 2007 #1
    I need help with this question plz:

    A 25.0g object moving to the right at 20.0cm/s overtakes and collides elastically with a 10.0g object moving in the same direction at 15.0cm/s. Find the velocity of each object after the collision.

    The answer is: 17.1cm/s (25.0g obect), 22.1cm/s (10.0g object)

    Can someone please explain how to get these answers..im soo lost
     
  2. jcsd
  3. Jan 16, 2007 #2

    ranger

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    Before you receive help, you will have to attempt the problem. What formulas do you think need to be applied?
     
  4. Jan 16, 2007 #3
    i can only think of
    mv1(initial) + mv2(initial)= mv1(final) + mv2(final)

    i dont know what to do next
     
  5. Jan 16, 2007 #4
    umm yea is anyone going to help me..??
     
  6. Jan 16, 2007 #5
    This is posted in the wrong section. somebody move this please.

    You came up with an equation. have you plugged in the numbers yet?
     
  7. Jan 16, 2007 #6
    Yea..and it doesnt work!!!...i guess nobody knows how to do this
     
  8. Jan 16, 2007 #7

    ranger

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    That step that you did was important. You've written down the momentum conservation for the system. And if we also know that total kinetic energy remains the same before and after the collusion (write it down), can you solve for v1 final and v2 final?
     
  9. Jan 16, 2007 #8

    cristo

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    Firstly, please note that we have homework forums, where these questions should be posted. Posting there will enable your question to be answered quicker. Secondly, the people helping you on this site are volunteers, and you cannot expect an answer to your question within ten minutes! Please be patient!

    With regard to your question; you have the right equation here. Which variables do you know? Can you plug these into the equation? Do you know any other quantity that is conserved here?
     
  10. Jan 16, 2007 #9
    lol..alrite..sorry..guess im a little pushy..anyways ill try the homework forums..thanks
     
  11. Jan 16, 2007 #10

    cristo

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    That's ok. You should continue this question here now though, since you've started a thread. (This will probably be moved by a mentor at a later time anyway)

    Have you thought about my questions in the last post?
     
  12. Jan 16, 2007 #11
    1. The problem statement, all variables and given/known data
    A 25.0g object moving to the right at 20.0cm/s overtakes and collides elastically with a 10.0g object moving in the same direction at 15.0cm/s. Find the velocity of each object after the collision.


    2. Relevant equations
    mv1(initial) + mv2(initial)= mv1(final) + mv2(final)


    3. The attempt at a solution
    umm i dont even know where to begin or if the above equation is even the right one..please help
     
  13. Jan 16, 2007 #12
    umm yea and i plugged in the information i know but im not getting the right answer..
     
  14. Jan 16, 2007 #13
    Energy is also conserved, and you need a second equation to find the two unknowns
     
  15. Jan 16, 2007 #14

    cristo

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    Well, post what you get when you plug the numbers into the equation (and not just the answer.. show your working) then I'll be able to see where you're going wrong!
     
  16. Jan 16, 2007 #15
    You need another equation since you have 1 equation with two unknowns. That's why cristo was asking if there was another quantity conserved.
     
    Last edited by a moderator: Jan 20, 2007
  17. Jan 16, 2007 #16
    hmmm ummm uhhh....alrite im a blithering idiot 'cuz i cant come up with any equations..
     
  18. Jan 16, 2007 #17

    ranger

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    did you read my second reply to see what else is conserved?
     
  19. Jan 16, 2007 #18

    cristo

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    Ok, you have this equation m1v1(initial) + m2v2(initial)= m1v1(final) + m2v2(final). Now, you know the masses, and the initial speeds, so plug them in!

    Also, since we know the collision is elastic, kinetic energy is conserved. Do you know the equation for kinetic energy?
     
  20. Jan 16, 2007 #19
    umm ok how about:
    .5mv1(initial)^2 + .5mv2(initial)^2 = .5mv1(final)^2 + .5mv2(final)^2

    what now?
     
  21. Jan 16, 2007 #20
    yea i know..energy is also conserved..but how is that going to help me?
     
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