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Projectile motion collisions problem

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data
    I don't have the exact problem statement but I remember the important data.

    I have to find the value of ##x## in the figure such that the projectile reaches to its original position. The initial speed is ##20\sqrt{2}\,m/s## at an angle of ##45^{\circ}## with the horizontal. The walls are inclined at an angle of ##45^{\circ}## and the collisions with the wall are perfectly elastic.

    (Ans: 60m)

    2. Relevant equations



    3. The attempt at a solution
    Honestly, I did not where to begin with. I looked at the solution which was provided with the test paper and it only mentioned that the collision with the wall should take place at the highest point of projectile trajectory and it stated it without proof. For the overall motion, I figured that the trajectory should be something like shown in attachment 2 (I know its a poor drawing, drawing curves is difficult). But I don't see how to arrive at the result of ##60 m##. Do I have to deal with some series?

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. May 5, 2014 #2

    SammyS

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    When the projectile is at its highest position, what is the direction of its motion ?

    In what direction will it then rebound (reflect) ?
     
  4. May 5, 2014 #3
    A series solution is not required.

    I don't understand the solution provided because any point of collision with the will be the point of highest trajectory so it is a meaningless statement.

    I think the answer can be found in the same manner as if the wall were 90 degrees. What angle of incidence does the ball need to hit the wall in order for it to return to the same place?
     
  5. May 5, 2014 #4
    When it is at the highest point, just before collision it has got only horizontal component of velocity. The angle between the velocity vector and the normal vector is ##135^{\circ}## just before the collision. When it rebounds, the velocity vector is at an angle of ##45^{\circ}## with the normal (and parallel with ##\vec{g}##).

    Okay, I think I see why the collision should take place at the highest point. This is like working backwards from the answer, how am I supposed to know that it should strike the wall when it is at highest point of its path? :confused:

    Also, what if the angles were, say ##30^{\circ}## and ##45^{\circ}##, how would I analyze that case?
     
  6. May 5, 2014 #5

    SammyS

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    Not necessarily.

    Suppose the projectile has started to descend before striking the wall .


    Added in Edit:

    I suppose the writer of the solution means the highest point of trajectory the projectile would have if the were no barrier.
     
    Last edited: May 5, 2014
  7. May 5, 2014 #6
    Good point but then it seems intuitively obvious that it could never rebound back to the same position once it starts to descend.
     
  8. May 5, 2014 #7
    How can it return to it's original position if it's velocity is parallel with g?
     
  9. May 5, 2014 #8

    SammyS

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    ## \vec{g} \ ## is vertical .
     
  10. May 5, 2014 #9
    Yes, g is vertical, so the ball goes straight down and does not return to the original position. So this can't be the answer.
     
  11. May 5, 2014 #10

    SammyS

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    That solution won't work as a general solution for other launch angles / wall orientations.


    Following what paisiello2 said earlier regarding a vertical wall. If the project5ile strikes the vertical wall at a right angle, it will return to its initial position.

    In general, if the projectile strikes wall (vertical or otherwise) at a right angle, it will return to its initial position.

    Back to your question here,
    Depending on the angles involved, you may be able to find a similar path to the one in the given solution -- one with a vertical segment.
     
  12. May 5, 2014 #11

    SammyS

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    It will then bonce straight up. Right?

    Then it does return to the initial position.
     
  13. May 5, 2014 #12
    LOL, OK, you're right. I didn't think about that one. I guess there are two answers then.
     
  14. May 5, 2014 #13

    SammyS

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    What is your (the other) solution, if I might ask ?
     
  15. May 5, 2014 #14
    I was thinking the other solution would be when it hits the wall at 90 degrees, but I see now that can't be the case.
     
  16. May 5, 2014 #15
    There may be more solutions if multiple elastic bounces are allowed. Homework: Find all the solutions for a wall inclined a general angle θ
     
  17. May 5, 2014 #16

    SammyS

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    I'm not convinced that the given solution is the only one.

    BTW: The answer of 60 meters, uses the approximation, g = 10 m/s2 .
     
  18. May 6, 2014 #17
    Are you talking about the case presented in the problem statement? I am not sure, can you please explain a bit more?
     
  19. May 6, 2014 #18

    SammyS

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    Yes, the 45°, 45° problem.

    I'll get back to this when I get some time.
     
  20. May 7, 2014 #19

    haruspex

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    One infinite family of solutions for the multiple bounce case will involve the projectile retracing its path exactly. Whether there can be any other solutions I'm not sure.
    The retrace solutions arise when the projectile eventually strikes one of the surfaces at right angles, so it splits into two families. It will be easiest to consider the motion from when the retrace starts.
    The speed at a ground bounce is constant, so the sequence of bounces is characterised by θn, the angle at which the projectile rises from the ground for the nth time. It should be possible to obtain a recurrence relation for the θn then look for values of v for which the sequence includes π/4. However, the algebra gets a bit heavy, involving nested surds.
     
  21. May 7, 2014 #20

    haruspex

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    Doh! It's much simpler than that when the ceiling is at 45 degrees or more.
    It's not hard to show that the retrace path can only hit the ceiling once. It can, however, bounce along the ground any number of times. So the general solution to the OP corresponds to d = (3+4n)h where d = 3h corresponds the solution already obtained (h being the height of the bounce). There are no solutions where the retrace starts at right angles to the ceiling, since all ground bounces will be at greater than 45 degrees.

    If the ceiling is at an angle α < π/4 then it gets more complicated. E.g. if the retrace path leaves the ceiling at right angles it can reach the ceiling again if it strikes the ground at angle θ where tan(θ) > (tan(α)+cot(α))2/(4 tan(α)).
     
    Last edited: May 7, 2014
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