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Another Algebraic Problem Either I Just Don't Know What I'm Doing, Or I'm OutOf It

  1. May 31, 2005 #1
    Okay, so... Here I am, having another algebra problem... This is the problem:

    A glider of mass 0.141 kg is moving to the right on a frictionless, horizontal air track with a speed of 0.740 m/s. It has a head-on collision with a glider 0.310 kg that is moving to the left with a speed of 2.16 m/s. Suppose the collision is elastic.

    Find the magnitude of the final velocity of the 0.141 kg glider.
    Find the magnitude of the final velocity of the 0.310 kg glider.

    Starting with the 0.31 kg glider...
    The conservation of the x-component of momentum:
    [tex]m_{A}v_{A1} - m_{B}v_{B1} = m_{A}v_{A2x} + m_{B}v_{B2x}[/tex]
    [tex](0.141 kg)(0.74 m/s) - (0.130 kg)(2.16 m/s) = (0.141 kg)v_{A2x} + (0.310 kg)v_{B2x}[/tex]
    [tex]-.56526 = (0.141 kg)v_{A2x} + (0.310 kg)v_{B2x}

    ...I don't know how to simplify that anymore... I suppose what'd be best would be taking [tex](0.141 kg)v_{A2x}[/tex] and trying to get it to just [tex]v_{A2x}[/tex]. But when I multiply everything by 0.141's reciprocal, I still don't seem to get to the right answer.
    I get this:

    [tex]-.56526(\displaystyle{\frac{1000}{141}}) = (0.141 kg)(\displaystyle{\frac{1000}{141}})v_{A2x} + (0.310 kg)(\displaystyle{\frac{1000}{141}})v_{B2x}[/tex]
    [tex]-4.00894 = v_{A2x} + 2.19858v_{B2x}

    Which I then add to what I get for plugging in my values to the relative velocity equation:
    [tex]v_{B2x} - v_{A2x} = -(v_{B1x} - v_{A1x}) = -(-2.16 m/s - 0.74 m/s) = 2.9 m/s

    Adding the two equations:
    [tex]-4.00894 = v_{A2x} + 2.19858v_{B2x}[/tex]
    [tex]2.9 = -v_{A2x} + v_{B2x}[/tex]
    [tex]-1.10894 = 3.19858v_{B2x}[/tex]
    [tex]v_{B2x} = -.346696 m/s[/tex]

    What I'm getting for the final velocity of glider B (the larger glider) is incorrect. According to this, it'll contine on in the direction that it was initally heading... Which makes sense.

    What did I do wrong?
  2. jcsd
  3. May 31, 2005 #2


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    Where did you use the KE conservation law?

  4. May 31, 2005 #3
    I didn't use it... I know it's an elastic collision, so KE is conserved, but I was under the impression that I could use just the conservation of momentum formulas.

    But either way, I still have 2 unknowns. ...I guess I just don't know how to solve for the two unknowns.

    Any suggestions?
  5. May 31, 2005 #4


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    You were under the wrong impression. The only time conservation of momentum is sufficient to find final velocities is for totally inelastic collisions where both objects have the same final velocity. You need conservation of energy to get the additional equation you need to solve the problem.

    I suggest you write your momentum equation as

    [tex]m_{1}v_{B1} + m_{2}v_{B2} = m_{1}v_{A1} + m_{2}v_{A2} [/tex]

    You don't really need the x subscript since everything is in the x direction, but if you want to use it, be consistent. The masses are the same before and after. You only need to distinguish between object 1 and object 2. The equation now says the total momentum before the collision is the total momentum after the collision. One of the initial velocities is a positive quantity. The other one is negative.
  6. May 31, 2005 #5
    Come on verd, you're half way there. In my AP Physics classes these problems take up the most board space. There is some challenging algebraic substitution but it's totally doable.
  7. May 31, 2005 #6

    Thanks, I understand the concept, but as I said, I'm having difficulty with the algebra involved, here. As this is a physics/math help forum...

    Thanks OlderDan-- I was under the impression that I only had to use the conservation of momentum law. I also understand that the masses remain the same. Thank you.

    However, the problem I'm having here is with the algebra-- I don't see a way to find an answer without the use of variables, being that in this equation, I have two unknowns.

    pete worthington-- It's nice to know that high school students are having such an easy time with this. ...It's also nice to know that the 'challenging algebraic substitution' is totally doable. ...However, that doesn't really help me at all, because I seem to be having difficulty with it. So I don't really understand the relevance of your post. It doesn't do much more than make me feel like an idiot because I struggle with some types of algebraic substition. (I must be stupid or something if I'm having difficulty with this, right? I mean really... Physics w/ Calc in 7 weeks? Mind-numbingly easy, right?)

    I seem to still be struggling with this. How can I find these two unknowns?
    Last edited: May 31, 2005
  8. Jun 1, 2005 #7
    Theyve asked you to experiment with the kinetic energy of the system. Since it is closed, the initial and final kinetic energies will be complete, write an expression similar to the one you have above to be your second equation.
  9. Jun 1, 2005 #8
    Never said it was easy for anyone. Some textbooks refer to "totally" elastic collisions as ones where kinetic energy is conserved. Remember that this is an "ideal" condition. When this is the case you can count on the fact that you will be left with two variables after setting up the conservation of momentum equation. This should clue the student in to setting up the conservation of energy equation. When you solve for this you will have the same two unknowns. Some quick review on two equations and two unknowns is a good idea here. You'll soon see that the unknown velocities in the kenetic energy equation are squared. Therefore, one technique is to solve for the velocity (either one) in the momentum equation, square it, then plug it into the energy equation. This is where the algebra gets tricky. 7 weeks ? Sounds like a real cram course.
  10. Jun 1, 2005 #9
    by 'complete', do you mean the same? and which second equation are you referring to?

    ...Sounds to me like both velocities will be the same... But that can't be right, can it?

    ...Where can I 'review on two equations and two unknowns'... Maybe this is ridiculous, but that actually hasn't come up in any of my elementary algebra courses. So, naturally, I'd attempt to use common sense to solve whatever problem I was having.

    When I plug in the values to both the conservation of momentum and the conservation of kinetic energy equations, we experience the same two unknowns. I don't understand how I'd be able to find any value with the same two unknowns in the same two equations...

    Perhaps I'm just not understanding what you guys are telling me. I don't know.

    I checked a friend's solution manual-- the same problem was done, but with slightly different values. This is what they did:

    [tex]m_{A} = 0.15[/tex]
    [tex]m_{B} = 0.3[/tex]
    [tex]v_{A1} = 0.8[/tex]
    [tex]v_{B1} = -2.20[/tex]
    [tex]v_{A2} = ?[/tex]
    [tex]v_{B2} = ?[/tex]

    [tex]m_{A}v_{A1} + m_{B}v_{B1} = m_{A}v_{A2} + m_{B}v_{B2}[/tex]
    [tex](.015)(0.8) - (0.3)(2.20) = (0.15)v_{A2} + (0.3)v_{B2}[/tex]
    [tex]-3.6 = v_{A2} + 2v_{B2}[/tex]

    They too used the relative velocity equation:
    [tex]v_{B2} - v_{A2} = -(v_{B1} - v_{A1})[/tex]
    [tex]-(-2.2 - 0.8) = 3[/tex]
    [tex]3 = -v_{A2} + v_{B2}[/tex]

    They added the two equations and got:
    [tex]-0.6 = 3v_{B2}[/tex]
    [tex]v_{B2} = -0.2[/tex]
    [tex]v_{A2} = v_{B2} - 3 = -3.3[/tex]

    And there you have it. ...I understand most of everything going on, but when the book suddenly skips to, "Adding the two equations", I don't necessarily know how [tex]3v_{B2}[/tex] popped in there... Nor do I understand how they got [tex]2v_{B2}[/tex] up at the top...

    Anyone have any ideas?
  11. Jun 1, 2005 #10
    In looking at that, it appears that I intially did almost exactly what the book did.

    ...Atleast it looks that way... But my answer doesn't seem to be at all correct... What happened?
  12. Jun 1, 2005 #11

    Doc Al

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    I looked over your first post. Despite difficult notation, it looks good to me. What makes you think it's wrong?

    Note: By using the "relative velocity" equation you are automatically using conservation of energy. (That equation is derived by combining conservation of momentum and energy equations.)
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