Collisions:


Consider a particle surrounded by a circular wall. (Top View is shown)

A particle is projected at an angle 37* with the radial direction. It returns to the point of projection after 2 collisions. Find the coeff. of restitution for the collisions between the particle and the wall.

17 comments:

  1. shaanApril 23, 2011 at 11:57 AM

    1.062?

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  2. mayankApril 23, 2011 at 1:42 PM

    e>1, are you serious?
    bhaiya, gravity is going inside the figure->as u have said that it is a top view.Then how can he reach up against gravity to reach again the point of suspension. OR is it a side view with gravity acting downward???

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  3. AnujApril 23, 2011 at 1:54 PM

    the whole thing takes place on a horizontal plane.

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  4. AnonymousApril 23, 2011 at 4:30 PM

    is the answer 0.794?

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  5. mayankApril 23, 2011 at 5:48 PM

    Hopefully, I haven't done any calculation mistake.
    e=0.811
    http://www.wolframalpha.com/input/?i=roots+of+64x^3-3x^2-36x-3

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  6. ShivankerApril 23, 2011 at 6:01 PM

    according to me.. e has to be greater than 1 here..

    m getting the cubic as
    1+e+e^2 = e^3/(tan(37)^2)
    which gives me the real root approxly 1.32

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  7. ShivankerApril 23, 2011 at 6:06 PM

    e shall be 1 for an initial angle of 30* and less than 1 only for an initial angle of less than 30*

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  8. k. rajApril 23, 2011 at 6:23 PM

    1.06161

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  9. mayankApril 23, 2011 at 6:30 PM

    yeh sab horizontal plane mein ho raha hai so, we dont have to consider the effects of gravity.
    After 2 collisions, a triangle would be formed so sum of angle=180.And finding angles is easy as angle of reflection=angle of incidence of next collision.And angle of reflection=tan inverse(tan(incident angle)/e)
    I simply used it to find e.

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  10. ShivankerApril 23, 2011 at 6:33 PM

    @mayank: ok so tan of angle of reflection is tan of incidence/e.. fine??
    now for an equilateral triangle, the angle should have been 30*.. so now, the other 2 angles have to be less than 30* =>> e>1 !!

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  11. mayankApril 23, 2011 at 6:37 PM

    seems right but, mathematically I am getting that only.may be I have done some calculation mistake.

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  12. mayankApril 23, 2011 at 6:41 PM

    ok got my mistake. Getting same as shivanker
    http://www.wolframalpha.com/input/?i=roots+of+16x^3-9x^2-9x-9

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  13. AnujApril 23, 2011 at 7:14 PM

    lets forget for some time that e<1.

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  14. shivamApril 23, 2011 at 11:21 PM

    yup same as shivankar getting same eqn but didnt solve the eqn

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  15. Tushar SinglaApril 25, 2011 at 12:43 PM

    e= 1.31
    Acc to eqn 16e^3 - 9e^2 - 9e - 9 =0

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  16. mayankApril 25, 2011 at 1:48 PM

    tushar, that's what I have got. 1.315 is approx. 1.32 and not 1.31

    ReplyDelete
  17. AnujApril 25, 2011 at 5:43 PM

    ok..we'll settle for 1.32 or 1.31.

    ReplyDelete

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