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Conservation of Momentum 2D - 2 marbles collide - find final velocities

  1. Feb 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A 10.0 g marble slides to the left with a velocity of magnitude 0.500 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on elastic collision with a larger 30.0 g marble sliding to the right with a velocity of magnitude 0.150 m/s

    Find the velocity of each marble (magnitude and direction) after the collision, the change in momentum of each marble, and the change in kinetic energy for each marble.


    2. Relevant equations

    B=bigger marble
    S=smaller marble

    (mB)(viB) + (mS)(viS) = (mB)(vfB) + (mS)(vfS)
    .5(mB)(viB)^2 + .5(mS)(viS)^2 = .5(mB)(vfB)^2 + .5(mS)(vfS)^2

    3. The attempt at a solution

    This problem has been hard for me because I have not figured out how to substitute from one equation to the other. I have solved for the velocity of the Big marble (vfB) but when I substitute it into the kinetic energy equation I end up getting crazy numbers that are hard to simplify.

    I solved for vfB and got:
    (vfB) = .35(vfS)

    When I substitute it though, I get:

    .004625 J = .5(.01 kg)*(.05 m/s + 3vfB)^2 + .5(.03 kg)*(vfB)^2

    So, I'm getting some pretty crazy numbers that are hard for me to work with. Once I find a final velocity this is going to be easy. Does anyone have a suggestion on how to proceed?
     
  2. jcsd
  3. Feb 19, 2009 #2

    Delphi51

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    .004625 J = .5(.01 kg)*(.05 m/s + 3vfB)^2 + .5(.03 kg)*(vfB)^2
    looks good except for the .004625 J. For the total KE I get
    .5*.01*.5^2 + .5*.03*.15^2 = .0015875J

    I didn't finish it, but from your step above you would just square the binomial in the brackets (results in 3 terms), combine everything into an ax^2 + bx + c = 0 and use the quadratic general solution.
     
  4. Feb 19, 2009 #3

    AEM

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    Well, I just did a preliminary run through of your problem and you're right you will have to solve your conservation of momentum problem for one of your final velocities and substitute it into the conservation of energy equation so that you end up with a quadratic equation for the other velocity.

    Now, your expression "(vfB) = .35(vfS) isn't correct. You should have another term which will be just a number.

    Also, be aware --this is going to be a messy calculation. And, yes the numbers will look "crazy". The quadratic will give you two solutions one of which you will have to discard because it physically doesn't make sense.
     
  5. Feb 19, 2009 #4
    Oh I see! I was wondering why my units of measure were getting messed up. I'll give it another run with this instead:

    0.017 m/s + 0.333 m/s (vfS) = (vfB)

    And thanks for the quadratic equation tip. I can't believe I forgot about that.
     
  6. Feb 19, 2009 #5

    Delphi51

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    I thought you just had a typo in the first equation because I agree with what you substituted in the second for vfB. At least I did before . . . this new value doesn't look right. I get -.01*.5 + .03*.15 = -.0005 for the total momentum.
    So after the collision, -.0005 = .01Vs + .03Vb
    and Vs = (-.0005 - .03Vb)/.01 = -.05 - 3Vb
    That is what you substituted into your 2nd equation in the first post.
     
  7. Feb 19, 2009 #6
    Thanks for the save. I'm giving it another go now.
     
  8. Feb 19, 2009 #7
    Okay, I'm moving on from here and I must be doing something wrong. I'm getting a number that doesn't work:

    .004625 J = .5(.01 kg)*(-.05 m/s - 3vfB)^2 + .5(.03 kg)*(vfB)^2

    .004625 J = .0000125 kg (m^2/s^2) + .3 (m/s) + 9(vfB^2) + .015kg(vfB^2)

    0 = 9.015(vfB^2) + .3 (m/s)(vfB) - .0046125 (J?)

    so those are the A, B and C that I'm plugging into the quadratic formula. I'm getting:

    0.0114413 m/s or 0.0447192 Neither seem correct. Where did I go wrong?
     
  9. Feb 19, 2009 #8

    Delphi51

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    I mentioned in my first post that I don't agree with your .004625 for the total energy.

    EDIT
    Something else wrong here. You should have a middle term with a vfB in it from squaring the binomial: (a+b)^ = a^2 + 2ab + b^2
    Leave out the units so it is easier to read.
    Also, you can divide both sides by .005 to get numbers easier to write down.
     
  10. Feb 19, 2009 #9
    Thanks for the tip. I'm getting 12, .3 and .1025 for A, B and C. Does that look right to you?
     
  11. Feb 19, 2009 #10

    Delphi51

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    I'm getting 12, -.3, -.4869.
     
  12. Feb 19, 2009 #11
    Well, I'm about to give up. This question is worth 10% of my grade for this homework but I've been working on it for about 5 hours now. I'm getting vfB = -.189 m/s and vfS = .517 m/s.

    vfB should be closer to -.1 m/s and vfS should be closer to +.5 m/s

    I am finding those numbers at this site:
    faculty.jsd.claremont.edu/jhigdon/phys33/phys33_hw_06_solutions_07.pdf

    A marble problem with the same properties but different numbers is on page 6 of this PDF:

    faculty.jsd.claremont.edu/jhigdon/phys33/phys33_hw_06_solutions_07.pdf

    Delphi51, thanks so much for all your help. I appreciate you hanging in there with me. I hope that the above link can help anyone that has this same problem. Good luck.
     
  13. Oct 21, 2011 #12
    Last edited by a moderator: Apr 26, 2017
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