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Collision avoidance

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Two objects with masses m1 and m2 are travelling in a frictionless surface and will collide perpendicular to each other (m1 is moving on the +x-axis, m2 is moving on the +y-axis). The distances of objects m1 and m2 from the collision point are d1 and d2 respectively.

    Q: What Δv is required for m2 to avoid collision with m1 by a distance x?

    http://dl.dropbox.com/u/12084119/image.PNG [Broken]

    2. Relevant equations

    [tex]t=\frac{d}{v}[/tex]
    [tex]x=v_0t+\frac{1}{2}at^2[/tex]

    3. The attempt at a solution

    So I'm having a bit of trouble understanding how to attack this problem. I think I'm lacking some fundamental understanding of kinematics.

    - For m2 to collide with m1,
    [tex]t_1=t_2[/tex]
    [tex]\frac{d_1}{v_1}=\frac{d_2}{v_2}[/tex]

    - For m2 to miss collision with m1 by a distance x I think that this has to be true,
    [tex]t_2new=t_1new[/tex]
    [tex]t_2new=\frac{d_1+x}{v_1}[/tex]

    In my previous attempt I'd tried to solve for a new velocity required for m2 to travel the distance d2 like so:
    [tex]v_2new=\frac{d_2v_1}{d_1+x}[/tex]
    [tex]Δv_2=v_2new-v_2[/tex]

    But this is not correct, since this would imply that there would be an immediate change in velocity when there should be an an acceleration to achieve that velocity.

    So I know I need a time value for m2 and also an acceleration, which brings me to this:

    [tex]d_2=v_2t_2+\frac{1}{2}a_2t^2_2[/tex]

    I'm stuck on how to figure out both a2 and t2 to allow for this collision to not occur. Or rather, for a new "collision" to occur where object m1 has already passed a distance of x by the time mass m2 arrives at the previous impact location. Any and all help is appreciated, I'm really stumped!
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 18, 2012 #2
    So I was thinking a bit more and thought that maybe using the formula for uniform acceleration might help. Since both times should be the same:

    [tex]d_2=v_2t+\frac{1}{2}a_2t^2[/tex]

    [tex]a_2=\frac{2(d_2-v_2t)}{t^2}[/tex]

    [tex]t=\frac{d_1+x}{v_1}[/tex]

    [tex]a=\frac{2(d_2-v_2(\frac{d_1+x}{v_1}))}{(\frac{d_1+x}{v_1})^2}[/tex]

    Is this the correct approach? It still kinda doesn't make much sense when I think it through.
     
  4. Mar 18, 2012 #3

    Redbelly98

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    Welcome to PF! Sorry your initial post didn't get a response, in spite of your showing an attempt at solving the problem.

    As I interpret the question, I think this is the correct approach -- you would just need to substitute expressions for v2new and v2 to get a final answer.

    The way I read this, there is some unexplained means be which m2 can quickly (i.e. in negligible time) brake its speed so that it goes at speed v2new over essentially the entire length d2. If they meant that m2 decelerates gradually so that it has slowed to speed v2new when it reaches the point * in the figure, they would have said this explicitly.

    p.s. you can generate longer subscripts but using curly braces: v_{2new} instead of v_2new
     
  5. Mar 19, 2012 #4
    The issue indeed is that some acceleration is required to slow the object to a slower speed so that it avoids collision. In any realistic case any change in velocity wouldn't be in a negligible time. Could I use the acceleration I found a couple posts back in another kinematics equation to get the final velocity, thus solving my problem of needing to find the change in velocity?

    I think this assumes that this constant acceleration would apply to the situation where it needs to be applied until the "fake collision" occurs, but isn't it also possible to make the object miss the collision without applying the acceleration the entire time?

    This also isn't a homework problem per se, as it's not any assignment.
     
  6. Mar 19, 2012 #5

    Redbelly98

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    Okay, let's go with the constant acceleration approach. It does look like you are on a correct approach in Post #2, and are just 1 step away from getting an expression for Δv -- plus some additional work of simplifying the expression.

    FYI, our policy on asking/helping with "homework" applies to any textbook-style problem, even if for independent study. But you are doing well with this problem so far.
     
  7. Mar 19, 2012 #6
    Ah okay, thanks for the heads up :)

    I figured that if I take the equation for acceleration that I found and use that to find a final velocity, the Δv would represent a minimum acceleration required to avoid collision, meaning any acceleration greater than that would be acceptable.

    But that leaves the question for me, how can I take what I have and be like, "what amount of time would I have to apply an acceleration for to still avoid collision by distance x?" Do I just have to make sure that the minimum acceleration times the time for this case equals a different acceleration that's applied for a shorter time? (I hope that makes sense)
     
  8. Mar 20, 2012 #7

    Redbelly98

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    Makes sense.
    So the question is, given some acceleration a, for how long must that acceleration last in order to have m2 reach point * when m1 is a distance x beyond that point. To do that, you would have to separate the motion of m2 into two parts, a constant-acceleration phase followed by a constant-velocity phase.

    You're welcome to play with equations until things work out -- that works for some people, while to others it's a confusing, convoluted mess of algebra. Or....

    ...a method that I like, but does not seem to be covered in books, is to make up a chart for the two phases of motion that looks something like this:

    Code (Text):

          _[U]a const[/U]__ _[U]v const[/U]__ __[U]Total[/U]___
     t   |_________|_________|_________|
     x   |_________|_________|_________|
     vo |_________|_________|
     v   |_________|_________|
     a   |_________|____0____|
     
    • First fill in the table with what you know from the given information.
    • Then use the usual kinematic equations to relate the variables within a single column (the "a const" and "v const" columns).
    • Also, use common sense to think about how some variables might be related across columns -- for example, the final velocity in "a const" is the initial velocity in "v const". Also the x's in the first two columns must add up to the total x (third column), which as you can see form your figure in Post #1 must be equal to ___?
     
    Last edited: Mar 20, 2012
  9. Mar 28, 2012 #8
    A week later and I'm back at it. I still think my method might be somewhat flawed.

    Just by thinking about it, the Δv should decrease as the distance of m2 from the collision point increases. Should it look exponential, approaching zero as Δv approaches infinite?

    The way I have it set up, my Δv looks like this:

    [tex]d_2=v_2t+\frac{1}{2}at^2[/tex]

    [tex]a=\frac{2(d_2-v_2t)}{t^2}[/tex]

    But,
    [tex]t=\frac{d_2}{v_2}=\frac{d_1+x}{v_1}[/tex]

    So, using uniform acceleration to speed up m2,
    [tex]a=\frac{2(d_1-\frac{v_2(d_1+x)}{v_1})}{(\frac{d_1+x}{v_1})^2}[/tex]

    [tex]v_{2new}=v_2+at[/tex]

    Meaning,
    [tex]Δv=v_{2new}-v_2=at=\frac{2(d_2-v_2t)}{t}=\frac{2v_1(d_1-\frac{v_2(d_1+x)}{v_1})}{d_1+x}[/tex]

    When I plot this Δv against increasing d_2, however, the graph is rather linear and not anything like I'd expect. It seems like something is going wrong somewhere, but I can't figure out where!
     
  10. Apr 1, 2012 #9
    The farther away the object is, the smaller the Δv should be required to make the m2 mass avoid the m1 mass. I think the way it's set up now, it's saying that the farther away m2 is from the collision point, the more acceleration required to get m2 to the collision point "on time".

    So if I were looking at impulse for instance:

    [tex]m_2Δv=F_2t[/tex]

    Where

    [tex]F_2=m_2a_2[/tex]

    So,

    [tex]Δv=a_2t[/tex]

    And when a plot is made of Δv with varying distance, as distance increases, Δv increases.

    This seems to be the case in both events where:

    [tex]t=\frac{d_1+x}{v_1}[/tex] or [tex]t=\frac{d_2}{v_2}[/tex]

    But this doesn't make sense because intuitively you'd think that at larger distances there should be a negligible Δv to avoid collision meaning it approaches 0.

    Any suggestions?
     
  11. Apr 1, 2012 #10

    Redbelly98

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    Hi,

    Not to leave you hanging, but for now I'll just say I have to run out for a while, but will be back later today and should be able to have a look at things then. (It's not a typical problem given in intro physics, so each time I come back to this thread it takes me some time to refresh myself on what is going on.)
     
  12. Apr 1, 2012 #11

    Redbelly98

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    Okay, I have three main comments. Two are about basic algebra, and the third is a pretty important conceptual observation.
    Did you literally mean exponential? There is no reason for exponential functions to come into play here. The relations between all variables involve the four basic arithmetic operations: addition, subtraction, multiplication and division. No variable or parameter appears as an exponent anywhere.
    Here is an algebra error: it should be d2, not d1, in the denominator right after the "2(".
    Carrying through with the correction I mentioned earlier, the last expression should have "2v1(d2..." in the numerator.
    Here is a key conceptual observation: you can't look at this expression, and talk about what happens if d2 increases while everything else remains constant. Why? Because v2 will also increase, because of the condition
    d2/v2 = d1/v1
    So to see what really happens, replace v2 in your expression so that d2 really is the only variable -- we'll still assume that d1 and v1 are fixed. Things will still be linear in d2, but it's not clear to me whether Δv will increase or decrease with increasing d2.
     
  13. Apr 11, 2012 #12
    That makes a lot of sense. That equality seems to be messing things up quite a bit. Another approach came to me last night, though.

    So if d_2 is increased by an amount of Δd_2 and is traveling at the same initial velocity, it will arrive at the initial collision point after m_1 has already got there, and is now a distance Δx past the collision point. However, though m_2 did miss the collision, it's still not missed by the proper amount. The mass m_1 still has to travel a distance of (d_1 + x - Δx).

    I was thinking that perhaps now, an acceleration has to be applied for the following time:

    [tex]\frac{d_1+x-Δx}{v_1}[/tex]

    This works for all Δx ≤ x and for all d_2, I think. The tricky part here for me is how am I gonna relate Δx to Δd_2? Right now I have this:

    [tex]a_2 = \frac{2[(d_2+Δd_2)- v_2(\frac{d_1+x-Δx}{v_1})]}{(\frac{d_1+x-Δx}{v_1})^2}[/tex]

    I'm not sure if this is legit (because of the issues from before), but so I don't have to worry about choosing small values for Δx and also for Δd_2, I set an equality with the time it would take m_2 to travel that Δd_2 distance and for m_1 to travel that Δx. I then solved for Δd_2 and plugged that in.

    [tex]\frac{Δd_2}{v_2} = \frac{Δx}{v_1}[/tex]

    If I choose that there's no change Δd_2, then it collapses back to what I had before, which is correct. If I choose numbers and whatnot, this does show a decrease in velocity change (as from before) and acceleration as the distance d_2 increases, but not in the right fashion. Ends up spitting something out that looks like a convex shape that hits zero instead of being something that approaches zero and is more concave. But it does have a negative direction, so I think I'm on the right track.

    I think it has something to do with what I worked out above, but I can't quite find where.
     
    Last edited: Apr 12, 2012
  14. Apr 17, 2012 #13

    Redbelly98

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    Hi,

    Sorry for not replying earlier. I'm just having trouble following your new approach, or understanding exactly what problem you are trying to figure out at this point. To me, it seems pretty straightforward to use the relation

    d2/v2 = d1/v1

    to eliminate a variable, and reduce the problem down to one variable quantity d2.
     
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