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Conservation of Momentum (elastic collision of masses)

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Three masses are positioned on a frictionless surface, as shown. Initially, mass m1 (1.0 kg) moves with a velocity of 2.0 m/s to the right, mass m2 (2.0 kg) is at rest, and mass m3 (3.0 kg) moves to the left with a velocity of 0.50 m/s. First, mass m1 collides elastically with mass m2 and recoils to the left. Afterwards, mass m3 collides with mass m2 and sticks. Calculate,
    1. the speeds of masses m1 and m2 after the first collision.
    2. the speeds of masses m2 and m3 after the second collision.

    2. Relevant equations

    M1V1 + M2V2 = M1V1' + M2V2'

    1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2



    3. The attempt at a solution

    Part a).

    First I put in the values and narrowed the conservation of momentum equation to:

    2 = v1' + (2) * v2' which rearranges to v1' = 2 - (2)v2' (equation 1.1)

    For the Conservation of Energy equation I did the same and got,

    2 = (0.5) v1'^2 + v2'^2 (equation 1.2)

    I then substituted the value of v1' from equation 1.1 into 1.2 and simplified to,

    v2' (3v2' -4) = 0

    v2' = 0 or 1.33

    since it is not 0 it must be 1.33 m/s

    I then substitute this answer back into equation 1.1 and get

    v1' = -0.667

    Therefore, the velocity of m1 after the collision is 0.67 m/s to the left and the velocity of m2 = 1.3 m/s to the right.

    Part b).

    for this part it is an inelastic collision and therefore only momentum is conserved.

    m2v2 + m3v3 = (m2 +m3) v2and3

    m2 = 2.0 kg
    v2 = 1.3 m/s

    m3 = 3.0 kg
    v3 = -0.50 m/s

    but these into the above equation and i get

    ((2.66 kg * m/s) + (-1.5 kg * m/s) / 5.0 kg) = v2and3

    v2and3 = 0.23 m/s (positive indicates to the right)


    Does this look correct? I made an earlier attempt but with a different answer but it didn't sit right so I retried this way.
     
  2. jcsd
  3. Apr 6, 2015 #2

    TSny

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    Your work looks correct to me.
     
  4. Apr 6, 2015 #3
    why is it in a collision that the momentum isn't divided equally between the two masses? Thats the way i thought when i first attempted the question. Or why is it that one bounces rather than simply coming to a stop and the other continuing?
     
  5. Apr 7, 2015 #4

    haruspex

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    Why should it be?
    Suppose two identical bodies moving at the same speed have a head on collision. If it's at least partly elastic, they can't both stop. By symmetry, neither stops, but they can't pass through each other.
     
  6. Apr 7, 2015 #5
    Lots of the examples used are those of billiard balls. If you shoot the ball a bit slower, then you get the cue ball bouncing back in the opposite direction of the head on collision, but what about when you increase the force and then the cue ball just stops dead after the collision, what's the difference in that collision?
     
  7. Apr 7, 2015 #6

    haruspex

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    The effect you describe is not to do with the speed of collision.
    If we pretend the balls just slide, without rolling, and collide, full-on, elastically the cue ball will stop and the object ball move at the cue ball'ss original speed.
    If you hit the cue ball gently it will have achieved pure rolling by time of impact. The cue ball will continue to roll forward after impact.
    If you hit the cue ball with a little backspin, but gently, it will lose the backspin before impact. With just the right combination, it will have stopped rotating at impact and come to a stop.
    If you hit the cue ball hard with backspin it will still have backspin on impact and will bounce back from the object ball.
    To hit the cue ball gently, no backspin, yet get it bouncing back is not within my many years of experience of playing snooker and billiards. It would require the cue ball to be lighter than the object ball.
     
  8. May 13, 2015 #7
    Every time I try to substitute the formulas of v1' = 2 - (2)v2 into 2 = 0.5v1'^2 + v2' I get what is shown on the image which doesn't give the right answer when solved through the quadratic formula, Where am i going wrong? photo.JPG
     
    Last edited: May 13, 2015
  9. May 13, 2015 #8

    TSny

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    Your equation 2 = 0.5(2-2v)(2-2v) + v2 is correct.

    However, you make one or more mistakes in going to the next equation 4 = 4 - 8v + 2v2 + v2.

    Can you spot your error(s)?
     
  10. May 13, 2015 #9
    yes I can, and did. not only was I forgetting to make it 4v^2, I was forgetting to divide the v^2 by 0.5 to give me 2v^2 for a grand total of 6v^2 and -8v giving me the exact same proportions and answer on the quadratic equation shown by v2' (3v2' -4). Wow i wasted 2 hours stuck in a mind trap.

    Thank you
     
  11. May 14, 2015 #10
    You could have also saved yourself the need to use the quadratic equation.

    If you have 6v^2-8v = 0 you can then just factor out v and have v(6v-8) = 0

    Looking at this equation you can make the relation true if v = 0 cuz then 0 times whats in the brackets will be zero. then you can also find the value of v that would make the terms in the brackets equal to zero. saves time from using the quadratic equation.
     
  12. May 14, 2015 #11
    Steve - We must be doing the exact same course through Athabasca university because every problem I have troubles with I come look at here and I often see you have posted in the thread. Although it looks like you might be done by now.

    Thanks for the help
     
  13. May 14, 2015 #12
    Yes, I am also a student with Athabasca University. I wrote my final exam on the 17th of April. I managed to get a pretty good final grade. The help here was a big asset. Tutor questions take too long and they can't really say much about assignment questions. I was amazed at the gap between the practise questions and the assignment questions. Wish they would have some lecture videos to accompany the course. I also hate the e-text.
     
  14. May 14, 2015 #13
    Oops misread response. Is the test easier in comparison to the assignment questions?
     
  15. May 14, 2015 #14
    Yes, the exam questions are easier thank goodness. They are similar but they give you a lot more figures so it's a lot easier math as it is more just about using the formulas properly.
     
  16. May 14, 2015 #15
    Good to know.

    Thanks and all the best on your next courses!
     
  17. May 14, 2015 #16
    Thanks and you too. Just finishing up organic chemistry and I'm done for the summer... 9 labs to write :(
     
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