Mirrored

A particle moving with initial velocity u = ( 3i + 5j ) units collides with a smooth plane wall placed at some orientation to the particle's trajectory such that the resulting velocity of the particle is v = ( -2i - j ) units.

Determine the orientation of the reflecting plane.

21 comments:

  1. shaanMay 10, 2011 at 10:59 AM

    5x+6y= k?

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  2. ShivankerMay 10, 2011 at 3:45 PM

    ok.. u cn check ur answers at the judge in the format "5,6,0" if the dr's of the normal are (5,6,0)..

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  3. shivamMay 10, 2011 at 6:17 PM

    ya shaan's solution seems correct but the magnitude of the velocity vectors are not equal in magnitude so basically we cant calculate the normal vector unless the magnitudes of vectors are equal or e is given

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  4. Sambhav GuptaMay 10, 2011 at 7:05 PM

    the given information is enough to calculate 'e' also, if needed

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  5. shaanMay 10, 2011 at 7:34 PM

    well this is what i did..

    let the vector on the reflecting plane {in plane of u, v}
    be X= ai+bJ.

    now, cons. of momentum along the plane{ ie. along X) ---> u.X/mod X= v.X/modX
    => 3a+ 5b= -2a-b

    so, 5a=-6b.......
    so we'll take a=-6, b=5..

    NOW, NORMAL(N)= Xx(UxV) (x= cross prod)

    so N= 5i+6j

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  6. shaanMay 10, 2011 at 7:35 PM

    This comment has been removed by the author.

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  7. shaanMay 10, 2011 at 7:55 PM

    yes, and acc to my soln, e=16/45

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  8. Shivam TiwariMay 11, 2011 at 5:58 AM

    5i + 6j ?? & m also getting e = 16/45, is that correct ?

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  9. ShivankerMay 11, 2011 at 9:34 AM

    well.. the original question WAS to find e only.. :D

    yes the answers are 5i+6j and e=16/45..

    though.. Can anybody suggest a shorter way to calculate e..?? maybe bypassing the calculation of the normal vector??

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  10. Shivam TiwariMay 11, 2011 at 8:24 PM

    u + (-v) :D :D I used this btw :D

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  11. shaanMay 11, 2011 at 9:48 PM

    "u + (-v)"

    it is working!! even for general values of u,v :O

    can someone think of a reason why it is working..
    (cos' i can't)

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  12. Sambhav GuptaMay 12, 2011 at 12:34 AM

    haha, alas, this question was so easy shivanker!...(remember the time we spent hours over it?)

    we have to conserve momentum along the wall, so components of velocities along the wall are equal, so components of velocity perpendicular to the normal are equal, if n` is the normal vector, (uxn)=(vxn), means, (u-v)xn=0, or n is along u-v!
    simple mathematics! :-)
    thanks to Shivam And Shaan

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  13. Sambhav GuptaMay 12, 2011 at 12:41 AM

    proceeding, coefficient of restitution comes out to be the absolute value of [v.(u-v)]/[u.(u-v)]=16/45

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  14. AnonymousMay 12, 2011 at 12:49 AM

    Shaan, alternately you can think of angle bisector of u and -v (why? )

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  15. Sambhav GuptaMay 12, 2011 at 1:20 AM

    no anonymous, angle bisect nhi ho rha, angle is bisected only when e=1, also angle bisector simple addn ya subtraction se nhi aata, we have to make sure we add equal magnitudes of the 2 vectors, (3,5) and (-2,-1) are not of equal magnitude

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  16. shaanMay 12, 2011 at 3:10 AM

    please suggest the mistake in my method in this LIGHT question
    http://anujkalia.blogspot.com/2011/04/light.html?Question=Light&Answer=1.500#comments

    i'm still getting mew = (root 3)/2

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  17. ShivankerMay 12, 2011 at 10:28 AM

    awesome shivam! :)

    i wonder why it was so difficult to strike..? :D

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  18. AnujMay 14, 2011 at 4:49 AM

    i remember this post having 17 comments...did someone delete them?? or is it a problem with blogger (i've been having some) ??

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  19. Sambhav GuptaMay 14, 2011 at 5:39 AM

    yup, the kinematics one had 20, they also got deleted, probably some problem with blogger..(it also didnt allow me to log in and post something for a day)

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  20. AnujMay 14, 2011 at 5:25 PM

    hmm...same with me.

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  21. ShivankerMay 14, 2011 at 8:50 PM

    yep..
    some problem..
    here's the detail: http://twitter.com/#!/blogger

    ReplyDelete

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