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Homework Help: Messing up simplifying formula for head on collision problem

  1. Oct 6, 2011 #1
    Ok I have been going crazy and I think I must be making a simple algebra mistake. Basically in my book I read over an example dealing with a head on collision between 2 balls and the collision is elastic so the kinetic energy intial= final KE. The momentum is also conserved so i can use momentum final=momentum initial. I tried combining these formulas but I always get quadratic. My work is in the image.
    The two formulas I am combining are the 1/2m1*vf1^2+ 1/2 m2*vf2^2=1/2m1*vo1^2
    and m1*vf1+m2*vf2=m1*vo1
    http://img825.imageshack.us/img825/2450/001copys.jpg [Broken]

    Uploaded with ImageShack.us

    the problem just gave 2 balls first one has inital velocity and second one is initially at rest. the book gave the simple formula vf1^2=v01(m1-m2/m1+m2)
    THANKS YOU FOR ANY HELP! i have done this wayy too many times and gotten wrong answers.
    The answers from book is vf1=-2.62m/s and vf2=2.38m/s
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 6, 2011 #2
    You can simplify the notation by assuming one mass is M and the other is m and the initial velocity is u and the final two velocities are v and w.
  4. Oct 6, 2011 #3
    I did that but I still couldnt get the final velocity of one to be just vf squared. I always seemed to come up with a vf^2 and a vf which led to the quadratic equation again.
  5. Oct 6, 2011 #4
    The two masses and the initial velocity of one of them are given.And the other is initially at rest.Right?
  6. Oct 6, 2011 #5
    For one thing, you are stating what the book provided as an answer incorrectly. The mass ratio is dimensionless so you have a velocity squared having the same dimensions as velocity to the first power. Something is obviously wrong there. Both sides of an equation MUST have the same dimensions.

    I have not gone through your algebra but, instead, did it myself. Use the momentum equation to get an expression for the velocity of the ball that was hit (initial velocity was zero). Plug that into the kinetic energy equation and simplify. It'll require a little factoring but I am sure you can do that.

    Keep at it.
  7. Oct 6, 2011 #6
    Last edited by a moderator: May 5, 2017
  8. Oct 6, 2011 #7
    First choose the most simple notation as possible, eg m,M,u,v,and w.
    Then do as LaurenceC adviced you and write down the conservation of momentum equation and the conservation of KE equation and then enter the known values straight away.
    Simplify the two equations to obtain a quadratic in only one unknown.
  9. Oct 6, 2011 #8
    You have a sign error in the right column of equations near the top...your last post.
  10. Oct 6, 2011 #9


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    Staff: Mentor

    If you're looking for an easy approach, it can proven that for an elastic collision the relative velocity of the two objects after collision is equal to the negative of their relative velocity before collision. This little gem can replace the conservation of KE equation as the second equation in the simultaneous equation system for solving for the velocities, and avoids the dreaded velocity-squared terms.
  11. Oct 6, 2011 #10
    Last edited by a moderator: May 5, 2017
  12. Oct 6, 2011 #11


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    Staff: Mentor

    Relative velocities: [itex](v1_o - v2_o) = -(v1_f - v2_f) [/itex]
  13. Oct 6, 2011 #12
    gneill I feel like I have found the holy grail. That so makes it easier. Thanks. would you happen to know of a website explaining the theory and all. If i use it on a test I will want to explain it.
  14. Oct 6, 2011 #13
    ahh i saw another post from you

    ok so I can just basically use the KE final=KE initial and solve for vo1-vo2 ?
  15. Oct 6, 2011 #14


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    Staff: Mentor

    Offhand I don't know of a website that deals with this little theorem directly. I did prove it myself a long ways back. It's not too hard. It's just a matter of slogging through the math using the KE (!) as one of the constraints, then comparing the relative velocities that result.

    So you only have to the the KE method once to prove the theorem, then never look back!
  16. Oct 6, 2011 #15


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    Staff: Mentor

    You can try :eek:

    There's not enough information in the KE equation alone to obtain the relative velocities.

    I just checked Wikipedia. Will wonders never cease! -- they have a derivation of the relative velocity formula.

  17. Oct 6, 2011 #16
    It's not an independent theorem. It follows from combining the energy and momentum conservation equations. It's just a matter of a few lines.
    You need to re-arrange the terms in the energy equation so that you have differences of the velocities squared.
    For example,
    where 1 and 2 mean the first and second body and i and f mean initial and respectively final.
    Once you do this you expand each parenthesis in a product of sum and difference.
    Using the momentum conservation the sum factors (and the mass) will simplify and you are left with the conservation of the relative velocity.

    Note:Sorry, I started to write before the previous message was posted.
  18. Oct 6, 2011 #17
    yes! now I will surprise my teacher with a curveball. :)
  19. Oct 6, 2011 #18
    Last edited by a moderator: May 5, 2017
  20. Oct 7, 2011 #19
    Since there seems to be so much difficulty with the algebra.....

    Subscript 11 denotes ball 1, initial velocity
    Subscript 12 denotes ball 1, final velocity

    Subscript 22 denotes ball 2, final velocity

    Attached Files:

  21. Oct 7, 2011 #20
    Thanks for the help. I usually make little mistakes so my plan is to derive the relative velocity because it is faster and applies to all elastic collisions. I can always do both to check my work though.
  22. Oct 7, 2011 #21
    It is good that the notation be simple and LaurenceC did just that.

    May I take the method of LaurenceC and simplify it further.

    Let small mass be m with initial velocy u and final velocity v and big mass be M with initial velocity 0 and final velocity w.


    (1/2)mu[itex]^{2}[/itex] = (1/2)mv[itex]^{2}[/itex] + (1/2)Mw[itex]^{2}[/itex]

    multiplying by 2: mu[itex]^{2}[/itex] = mv[itex]^{2}[/itex] + Mw[itex]^{2}[/itex]


    mu = mv + Mw
    hence m(u - v)/M = w


    mu[itex]^{2}[/itex] = mv[itex]^{2}[/itex] + M{m(u - v)/M}[itex]^{2}[/itex]

    u[itex]^{2}[/itex] = v[itex]^{2}[/itex] + (Mm){(u - v)/M}[itex]^{2}[/itex]

    (u - v)(u + v) = (m/M)(u - v)[itex]^{2}[/itex]

    (u - v)(u + v) = (m/M)(u - v)(u - v)

    But v cannot be equal to u because otherwise KE is not conserved. So one can divide by (u - v).

    therefore u + v = (m/M)(u - v)

    Mu + Mv = mu - mv
    Mv + mv = mu - Mu
    v = u(m - M)/(M + m)
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