Friction, Elastic Collisions, and Projectiles

A small spherical ball undergoes an elastic collision with a rough horizontal surface. Before the collision, it is moving at an angle ß to the horizontal.



Find ß as a function of µ, such that the subsequent range is maximised.
Now determine the maximum value of ß for range to be maximum. What if ß is more than that maximum value??
Report your answer at the judge in degrees, rounded off to the nearest integer.(The judge checks only the maximum value of ß.)

19 comments:

  1. shaanMay 26, 2011 at 1:48 PM

    nice one ..

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  2. Pulkit MendirattaMay 26, 2011 at 10:46 PM

    This comment has been removed by the author.

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  3. ShivankerMay 27, 2011 at 10:05 AM

    sorry guys.. the question looked idiotic earlier..

    modified.. :)

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  4. shaanMay 27, 2011 at 11:38 AM

    B = tan^-1( 1/{ 2 mu + 1})

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  5. ShivankerMay 27, 2011 at 6:54 PM

    no u'r wrong shaan..

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  6. Pulkit MendirattaMay 27, 2011 at 8:01 PM

    I am getting Beta= cot^-1 (2u)

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  7. shouryaMay 27, 2011 at 8:06 PM

    gettin same as shaan,
    B=cot^-1(2u+1).

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  8. Pulkit MendirattaMay 27, 2011 at 8:12 PM

    Correction-
    B=2cot^-1{2u)

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  9. ShivankerMay 27, 2011 at 8:54 PM

    still.. all wrong.. but pulkit, u're quite close..

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  10. ShivankerMay 27, 2011 at 8:56 PM

    the answer is ß = {cot^-1 (2µ)}/2

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  11. ShivankerMay 27, 2011 at 8:57 PM

    by the way.. What if ß is more than that maximum value??

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  12. Pulkit MendirattaMay 27, 2011 at 9:37 PM

    OH ! sorry shivanker. i had got Cot2B=2u and wrote it as 2cot^-1(2u) !!!
    Deadly error !

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  13. shaanMay 27, 2011 at 10:01 PM

    hey, i used
    u(y)= constant = u*sinB

    and v(x)= usinB so that R is max..

    I(f)= mu*I(n)......
    so mu*(2u sinB)= u( cosB- sinB).....

    so B = tan^-1( 1/{ 2 mu + 1}).......
    pls tell d mistake..!

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  14. ShivankerMay 28, 2011 at 10:22 AM

    @shaan: ur "assumption" that vx = u.sinß for max R is baseless..

    try to do without it and u'll get it

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  15. AnonymousMay 29, 2011 at 5:09 PM

    shivanker tune khud likha hai collision is elastic. i agree with shaan & shourya.

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  16. tushar guptaMay 29, 2011 at 5:15 PM

    agree with shivanker

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  17. ShivankerMay 29, 2011 at 6:46 PM

    yes.. colision is elastic..
    and yes vy = u.sinß

    BUT, the fact(or rather "assumption") that Range wud b max when vx = vy is wrong..

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  18. ShivankerMay 29, 2011 at 6:50 PM

    ok.. here's the thing why.. vx = vy is wrong because when that happens, both of them might be really small! (magnitudes matter here.. dont they?)

    sure enough, vy = u.sinß
    using the impulse comments as stated above by shaan, vx = u(cosß - 2µ.sinß)
    now maximise vx.vy... ie, maximise: sinß(cosß-2µ.sinß) to get ß = 1/2 cot^(-1) 2µ

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  19. shaanMay 30, 2011 at 12:45 AM

    ohh, thanks shivanker, nice one!!

    ReplyDelete

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