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Analysis of collision of particles from graph

  1. Oct 12, 2013 #1
    Hi friends,
    I have an issue in Analysis of collision of particles from graph.
    Please Help me in solving this.
    Thank you all in advance.

    The problem is as:

    https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-ash4/q71/s720x720/1381762_1432382976988808_995987085_n.jpg


    Attempt:

    https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-ash3/q71/s720x720/1378626_1432380616989044_74050265_n.jpg

    Please try to help me in this.
    I will appreciate the help.
     
  2. jcsd
  3. Oct 13, 2013 #2

    Simon Bridge

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    In your answer marked "(I)" you say that "(I) is correct" - what does this mean?
    ... is it not possible for two velocities on the same axis to be in opposite directions?

    In your answer marked "(II)" you say that "it is clear from the graph" but you don't say how. What is it about the graph that tells you the velocities are the same at the instant of the collision?
    Note: if both objects are travelling with the same velocity (i.e. same speed and direction) then how can they collide?

    In your answer marked "(III)" you say there is no change in the slopes of the curves - yet don't both curves change slope twice? i.e. at t=1s and at t=3s.

    Try describing, in words, the motion of S and R as represented by the v-t graphs.
     
  4. Oct 13, 2013 #3

    haruspex

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    You are right on I, though the wording is a little unclear. What exactly do you mean by being "on the same axis"?
    In II you refer to the instant of collision. In the graph, collision occurs over a time period, and the question asks about the middle of that period. Where is that on the graph?
    For III, compare the slopes during collision. Are they the same? What does that tell you about the masses?
     
  5. Oct 14, 2013 #4

    Well,
    I though that the collision takes place at t = 2 sec. Please tell me how can we predict the instant of collision or the time limit. So far I am getting that wherever the slope of velocity - time graph is changing, these are the instants of collision. For ex. in the case of R, at points t= 1 and t = 3 the slopes are changing hence these are the instants of collision. Is it correct?

    Well, about the direction I am saying that, the velocity time graphs for the both particles lies on the same quadrant. That's why I was saying the the velocities are in the same direction for both the particles.

    Still have a big confusion in this problem.
    Thanks for the reply.
     
  6. Oct 14, 2013 #5

    Well,
    I though that the collision takes place at t = 2 sec. Please tell me how can we predict the instant of collision or the time limit. So far I am getting that wherever the slope of velocity - time graph is changing, these are the instants of collision. For ex. in the case of R, at points t= 1 and t = 3 the slopes are changing hence these are the instants of collision. Is it correct?

    Well, about the direction I am saying that, the velocity time graphs for the both particles lies on the same quadrant. That's why I was saying the the velocities are in the same direction for both the particles.

    Still have a big confusion in this problem.
    Thanks for the reply.
     
  7. Oct 15, 2013 #6

    haruspex

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    There is not an instant of collision here, the collision occurs over a period of time from t=1 to t=3.
    Yes, same quadrant makes more sense than same axis. Better still, just say they're both positive.
    Now concentrate on the period of collision. Each v-t graph has a slope there, one positive one negative. What does the slope represent? If the force between the particles at some instant is F, what equations would you write for their accelerations?
     
  8. Oct 15, 2013 #7

    ehild

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    The collision is not an instantaneous process. See video.


    When two balls collide they stay together for some time interval, and both deforming during that impact process.
    The velocity of the CM-s change with respect to the initial velocity after getting into contact. At an instant, they move together, and after that, the elastic force between them starts to accelerate both balls away from each other, till the instant when they get separated again.

    ehild
     
    Last edited by a moderator: Sep 25, 2014
  9. Oct 15, 2013 #8

    Simon Bridge

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    That is only an instant of time.
    A collision does not take an instant.
    That is the time that the velocities are equal - the velocities start changing before then.

    You read it off the graph.

    That is correct - the collision takes place over the time period from t=1 to t=2 seconds. During that time the objects experience a force which causes the acceleration.

    Note: in a real life collision, the v-t graph won't be that simple.

    You mean to say that the velocities are both positive.

    Look at the graph, and try to picture what is going on: take positive distances to the left:
    - R travels to the left at +0.8m/s for 1μs, then decelerates to +0.2m/s over 2 micro-seconds.
    - S is stationary for the first second, then it starts accelerating to the left. It reaches a velocity of +1.0m/s in 2μs and continues at a constant speed.

    So: what causes S to start moving?
     
  10. Oct 15, 2013 #9

    haruspex

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    t=1 to t=3.
     
  11. Oct 15, 2013 #10
     
  12. Oct 15, 2013 #11

    Simon Bridge

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    I'm getting tired - should take a break <makes another coffee>
     
  13. Oct 15, 2013 #12

    haruspex

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    I was thinking in terms of F = ma. If you apply that to each ball at some arbitray instant during collision, can you see a relationship between their respective accelerations and their respective masses?
     
  14. Oct 16, 2013 #13
    Taking equation for the interval t= 1 to t = 3

    For R -

    F = -0.3m1; 1≤ t ≤ 3

    For S:

    F = 0.5m2; 1≤ t ≤ 3

    During collision "F" will be same for both the objects. so, mass will be inversionally proportional to magnitude of acceleration. Hence mass of R will be grater than mass of S.

    All the doubts of this assignment are cleared with your and Simon's help. Thanks a lot to you guys. I appreciate the help.
     
  15. Oct 16, 2013 #14

    Simon Bridge

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    No worries ;)
     
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