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Momentum and Kinetic Energy Equation for Elastic Collisions

  1. Jun 24, 2009 #1
    ok so, i have an assignment due tomorrow, i know, late starting, but i can't figure out how to rearange the kinetic energy and momentum equation to get the v1f and v2f

    i know that you have to find one term in the for of the other, but i dont know how to rearange my equation so that i have that...
    the equation is:

    1/2 m1v1o^2 + 1/2 m2v2o^2 = 1/2 m1v1f^2 + 1/2 m2v2f^2

    i dont know if you have to take the 1/2's from the right and multiply the LHS by 2, or if you have to take one of the 1/2mv^2 to the other side before doing that.
     
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  3. Jun 24, 2009 #2

    cepheid

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    Hi AussieBec,

    Welcome to PF!

    Remember that you can do any (mathematically allowable) thing to an equation so long as you do it to both sides of the equation (that way you ensure that the equality still holds true). In this case, since every single term is multiplied by 1/2, it seems like multiplying both sides of the equation by 2 would be a useful thing to do (because doing so would get rid of all of the factors of 1/2).
     
  4. Jun 24, 2009 #3
    :D thank you.

    ok so now i have
    2(1/2 m1v1o^2 + 1/2 m2v2o^2) = 2(1/2 m1v1f^2 + 1/2 m2v2f^2) which simplifies to
    m1v1o^2 + m2v2o^2 = m1v1f^2 + m2v2f^2

    here is where i get stuck, i dont know how to get either v1f or v2f by itself, if i take either the m1v1f^2 or m2v2f^2 to the LHS i still dont know how to solve it.
     
  5. Jun 24, 2009 #4

    cepheid

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    You have two unknowns. You cannot obtain a unique solution with only one equation. You need two equations.

    So, it suffices to solve for one unknown in terms of the other. This amounts to "getting rid of" one of the unknowns, so that you only have one left for the second equation.

    You know where the second equation comes from, right?
     
  6. Jun 24, 2009 #5
    um no... i have no idea where the second equation comes from :/
     
  7. Jun 24, 2009 #6

    cepheid

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    Hmmm...contrary to what is suggested in your thread title, conservation of kinetic energy and conservation of momentum are two separate conditions. The equation you have written only expresses conservation of kinetic energy (because it says that the total kinetic energy before the collision is equal to the total kinetic energy after the collision).

    Momentum is also conserved, so you have another equation that says:

    total momentum before collision = total momentum after collision​
     
  8. Jun 24, 2009 #7
    oh ok, so i have:

    1/2 m1v1o^2 + 1/2 m2v2o^2 = 1/2 m1v1f^2 + 1/2 m2v2f^2
    and
    m1v1o + m2v2o = m1v1f + m2v2f

    then i would find a value for say v2f in terms of v1f and then plug that into the first equation? would that work?
     
  9. Jun 24, 2009 #8

    cepheid

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    Exactly! It should work. Be advised though, that the algebra gets a bit messy here, so stick with it and try to keep everything organized.
     
  10. Jun 24, 2009 #9
    ok thatnks, i'll try that.
    and ill post again if i have any problems
     
  11. Jun 24, 2009 #10
    ok so i rearanged the momentum equation and got...

    (m1v10+m2v20-m1v1f)/m2 = v2f

    does that look kinda right? also, would the m2 in the m2v20 cancell out?
     
  12. Jun 24, 2009 #11
    wait! then you would work out the equation and then sub it into the energy one right? so youd have a number then v1f and then you could use that in the energy equation
     
  13. Jun 24, 2009 #12

    cepheid

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    Yes.

    You could distribute the 1/m2 to all three terms in the parentheses, and if you did that, it would cancel with the m2 in the middle term out of the three, but not from the leftmost or rightmost one (obviously). Doing this doesn't really gain you much of anything. What you'd be left with after would be equivalent to what you had before, and arguably not any simpler.

    Sorry, I don't really understand what you are asking here. In any case, you have solved for an expression for v2f using the momentum equation. You can now substitute this expression for v2f into the kinetic energy equation.
     
  14. Jun 24, 2009 #13
    gah. i think i did something wrong...

    i used the values... m1=1310kg, m2=1001kg, v10=22.2m/s, and v20=27.78m/s

    and i got to the last step on my equation, and i cant finish it off...

    -1381.68 = v1f^2

    and you cant square root a negative number and get a real answer
     
  15. Jun 24, 2009 #14

    cepheid

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    I can't help you figure out the problem unless if you post your steps. Start with the point where you substituted the expression for v2f into the energy equation.
     
  16. Jun 24, 2009 #15
    ok, well i figured out that (m1v10+m2v20-m1v1f)/m2 = v2f

    i then substituted m1=1310kg, m2=1001kg, v10=22.2m/s, and v20=27.78m into the equation and got

    v2f = 56.83 - 1.31v1f (i wasnt sure if you could completely eliminate the 56.83, and i couldnt figure out how to with out making the expression completely weird)

    then i put it into the 1/2 m1v1o^2 + 1/2 m2v2o^2 = 1/2 m1v1f^2 + 1/2 m2v2f^2 equation...

    1/2 m1v1o^2 + 1/2 m2v2o^2 = 1/2 m1v1f^2 + 1/2 m2v2f^
    1/2 x 1310 x 22.2^2 + 1/2 x 1001 x 27.78^2 = 1/2 x 1310 x v1f^2 + 1/2 x 1001(56.83 - 1.31v1f)^2
    322810.2 + 386250.0642 = 655v1f^2 + 500.5(56.83 - 1.31v1f)^2
    322810.2 + 386250.0642 = 655v1f^2 + 500.5 x 3229.6489 + 500.5 x 1.72 v1f^2
    709060.2642 = 655v1f^2 + 1616439.274 + 860.86v1f^2
    -907379.0103 = 1515.86v1f^2
    -598.59 = v1f^2



    when i did (56.83 - 1.31v1f)^2 was i supposed to square 1.31v1f or -1.31v1f ? cause if i was only supposed to square 1.31v1f then minus it later, that might be where it's wrong
     
  17. Jun 24, 2009 #16
    ok so i just did it with squaring 1.31v1f and it worked out as 209.91 and i think it was m/s which seems wrong. considering that they were originally travelling at 22.2m/s and 27.78m/s so i dont know if thats right
     
  18. Jun 24, 2009 #17

    cepheid

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    This is basic algebra:

    (a + b)2 = (a + b)(a + b) = a*a + a*b + b*a + b*b = a2 + 2ab + b2

    In this case, b is negative.

    a = 56.83

    b = -1.31

    So what is b*b?

    b*b = (-1.31)*(-1.31)

    Does that answer your question?
     
  19. Jun 24, 2009 #18
    ok, but then i get ...

    -907379.0103= 1515.86v1f^2
    -598.59 = v1f^2
    square root(-598.59) = v1f

    and you can't square root a negative and get a real answer. so i dont know what it is, but i must have done something wrong
     
  20. Jun 24, 2009 #19

    cepheid

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    Can I give you a word of advice? In general, you don't want to plug in the numbers too soon. There are two reasons for this:

    1. Numbers tend to add clutter and take away from the cleanness of what you're doing. When you plug in numbers it's easy to get somewhat lost in the numbers and forget what quantity is what.

    2. If you don't plug in any numbers at all and just solve for v1f, and subsequently for v2f, then you will have two algebraic relations:

    - the first tells you how v1f depends on v1i, v2i, m1, m2.

    - the second tells you how v2f depends on v1i, v2i, m1, m2.

    The beauty of these relations is that they are general results that hold true for all elastic collisions between two bodies in 1D, regardless of what the specific numbers are. So if you've solved the problem once, you've solved it for good. Although it's an oversimplification, in a sense this is how physics is done. It is the search for the relationships amongst physical quantities based on certain starting assumptions and principles.

    Granted, number 2 may entail more algebraic work than is really needed for your problem, but I stand by the statement that you should not substitute in the numbers until the latest possible stage, when you are sure you know exactly what is going on.
     
  21. Jun 24, 2009 #20
    oh ok. thank you. i did it all again and i think i got it right this time

    :D thanks for your help!
     
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