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2 masses colliding (elastic)

  1. Dec 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Mass 1 has a mass of 5kg and is traveling with an initial velocity of 20 m/s at a 45 degree angle. (Starting in Quadrant II and heading toward the origin.)

    Mass 2 has a mass of 6kg and is traveling with an initial velocity of 15 m/s at a 45 degree angle. (Starting in Quadrant III and heading toward the origin.)

    The final velocity of mass 2 is known and it is 25 m/s after the collision.

    The collision is completely elastic.

    I need to find the velocity of mass 1 after collision, and the final angles of both masses.

    20161213_124419_zpsihgztx7d.jpg

    2. Relevant equations
    I don't know all of the relevant equations. But I think I need to use:
    ∑mvx(before)=∑mvx(after)
    ∑mvy(before)=∑mvy(after)


    3. The attempt at a solution
    First I broke the initial velocity of mass 1 down into its x and y components. Since the initial angle is 45°, the x and y velocities are the same.
    (20m/s)*cos(45°)≈14.14214

    Then I broke the initial velocity of mass 2 down into its x and y components.
    (15m/s)*cos(45°)≈10.60660

    Next, I thought that since I knew the final velocity of mass 2, I should be able to calculate its angle using:
    ∑mvx(before)=∑mvx(after)
    [(15m/s)*cos(45°)]/(25m/s)=(3√2)/2

    Then cos-1([3√2)/2])≈64.896°

    I'm not sure if these steps are correct up to this point ... but I went on to try to calculate the final velocity of mass 1 using:
    m1v1^2+m2v2^2=m1v'1^2+m2v'2^2

    With that equation I calculated that the final velocity of mass 1 is ≈-3.1299m/s

    I'm extremely confused on this problem and feel like everything I'm doing is wrong.
     
  2. jcsd
  3. Dec 13, 2016 #2
    Are you sure all of the values in the problem statement are correct? Maybe I messed up on the calculations, but it looks like the final kinetic energy of m2 is greater than the initial kinetic energy of m1 + m2.
     
  4. Dec 13, 2016 #3
    Well, I copied the problem down as it was written on the board. I even took a picture of the chalkboard to make sure I didn't make any transcription errors. This problem was not out of the book ... it was made up as a take-home practice problem.
     
  5. Dec 13, 2016 #4
    Yeah, the more I think about it, the more I am convinced there is a problem with the problem. I just don't think it's possible for m2's speed to increase from 15 m/s to 25 m/s due to a collision of a lesser mass object traveling at 20 m/s.
     
  6. Dec 13, 2016 #5
    Well it wouldn't surprise me ... but I still need to work the problem as much as I can in the mean time. Are you able to determine if the work I did is valid?
     
  7. Dec 13, 2016 #6
    This did not make sense to me. I don't really understand what you are doing. I agree that the sum of momentums in the x direction before = sum after, but the equation that you wrote to implement that didn't look right at all.

    Here is how I think it should look.
    ∑mvx(before)=∑mvx(after)
    m1v1xinitial + m2v2xinitial = m1v1xfinal + m2v2xfinal

    Then you should write a similar equation for the y direction.
    Then there should be a before/after equation for kinetic energy.

    But I don't think it's a good idea to try to work this problem. There is not going to be a valid solution.
     
  8. Dec 13, 2016 #7
    Looking at this equation a little more, it appears that you are trying to equate m2's initial momentum to m2's final momentum. That is not valid. It is the sum of the momentums before that are equal to the sum of the momentums after. So you have to add all of the x-direction momentums. In this case, that is m1 and m2.
     
  9. Dec 13, 2016 #8
    So basically I can't go forward because in order to calculate m1v1xinitial + m2v2xinitial = m1v1xfinal + m2v2xfinal, I need the angle of θ2, and the angle I calculated earlier isn't valid because the value of velocity 2 (after) is invalid to begin with?

    If the problem is indeed invalid by design, that is insanely confusing to someone who has practically no understanding of what's going on in the first place.
     
  10. Dec 13, 2016 #9
    Even for a valid problem, your method of calculating the angle was not valid because it was based on equating m2's initial x momentum with m2's final x momentum. If you are given the initial velocities of both objects and final velocity of one of the objects, you can find the final velocity of the other object based on conservation of energy. (This only applies to a perfectly elastic collision though.) Then you will have to use conservation of momentum equations to find the angles.
     
  11. Dec 13, 2016 #10
    Well, I emailed the professor of this class and to my surprise, he replied right away. Anyway he said to use 20 m/s as the final velocity for mass 2 instead of 25 m/s.

    So with that new information ... I now have...

    vf1=sqrt[((5kg)(20m/s)2+(6kg)(15m/s)2-(6kg)(20m/s)2)/5kg]=sqrt(190)≈13.784m/s ... (ignoring sig figs for now)

    Does that seem valid?
     
  12. Dec 13, 2016 #11
    That looks right for vf1.
     
  13. Dec 13, 2016 #12
    Great ... so to find the angles, would I break vf1 down into its x and y component and then use cosθ to find the missing angle?
     
  14. Dec 14, 2016 #13

    haruspex

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    As Tom posted, you need to write out the two momentum conservation equations (x and y). Both will involve both unknown angles. Then solve the pair of simultaneous equations.
     
  15. Dec 14, 2016 #14
    For x, I'm getting that the total momentum before is (100kgm/s)*cos(45)+(90kgm/s)*cos(45)≈134.3503, that seems reasonable. (I think.)

    But for y, I'm getting what seems like nonsense. (100kgm/s)*sin(-45)+(90kgm/s)*sin(45)≈-7.0711

    Does it make sense that the y momentum (before) would be negative?
     
  16. Dec 14, 2016 #15
    Those numbers look familiar. I think they are right. And yes, it makes sense that the y momentum could be negative.
     
  17. Dec 14, 2016 #16
    Ok ... thanks for clearing that up, but I still have a problem with too many unknowns. (I don't know the equation editor here well enough to make this clear, so please pardon the photo equation.)

    If I assume that my √190m/s (or ≈13.78m/s) velocity for mass 1 after is valid (from post #10), then I still can't solve the angles.


    2016-12-14_19-30-58_zpsmileizim.jpg
     
  18. Dec 14, 2016 #17
    Back in post #10 you solved for Vf1, so that is a known. If you look at your two equations in the form they were just before you solved them for Vf1:
    On the right hand side of the x momentum equation you have 5vf1cosθ1
    And on the right hand side of the y momentum equation you have 5vf1sinθ1
    If you square both of those equations you can then add them together and use the trig identity sin2 + cos2 = 1
    That will eliminate one of the variables.

    BE CAREFUL of your signs though. The picture shows the final velocity of m2 in the downward direction. Is that how you are defining it? That seems to be the most straightforward.
     
  19. Dec 14, 2016 #18
    Ok, I'm trying to wrap my head around this ... below is what I've done based on this new information. But I'm not seeing how this eliminates a variable. I now just have an insanely complicated equation with two unknown angles. (I'm sorry to be so dumb at this. This is my first experience with physics.)

    2016-12-14_19-58-59_zpsyd9b8sl5.jpg
     
  20. Dec 14, 2016 #19
    I did not go through your whole equations, but in your final equation (above), it looks to me like you eliminated one of the variables - namely, θ2.
    So you are left with a single equation with a single variable (θ1). Once you manipulate all of that, I think you should end up with an equation in the form:
    asinx + bcosx = c.

    http://mathforum.org/kb/servlet/JiveServlet/download/206-2697091-9755226-1202710/sin-cos-eqs.pdf
    Example IV in the link above shows how to address that.

    Check your sign in your initial y momentum equation. If you are defining the velocity of m2 in the downward direction, that term should be negative. (Edit: Replaced the word 'expression' with the word 'term'.)

    Also, do not feel dumb. Not that I am one of the experts on this forum (far from it), but I struggled quite a bit with this problem. The concept is very straightforward (equating before and after x momentum, y momentum, and kinetic energy), but the math gets quite tricky - at least for a hack like me. :)
     
  21. Dec 14, 2016 #20
    First I surely appreciate the patience and help. I will look at that link here in a moment, but before I go forward I want to make sure I am addressing the sign for m2 that you are talking about. I am assuming that when the masses collide, they will bounce off each other so mass 2 will be going in the opposite y direction from where it began. So I believe this is how the equation should look? (Now I'll continue on and get that PDF)

    2016-12-14_20-25-38_zpsccr7oteb.jpg
     
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