This might be tough!

The figure is self explanatory, you just have to find the final velocity the particle attains.

(Pardon my drawing.. The figure shows an inclined plane of inclination 30 degrees, on which a particle is given a velocity 60m/s)

The mathematics might be tougher, so u MAY just report the DE you get.

14 comments:

  1. guntashMay 29, 2011 at 5:57 PM

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  2. ishanMay 29, 2011 at 8:31 PM

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  3. ishanMay 30, 2011 at 12:12 AM

    btw...nice name to the pic bro...:D

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  4. Sambhav GuptaMay 30, 2011 at 4:26 AM

    wooo, that was good!

    and sry for the name of the pic, i had something else in mind while saving the pic :P

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  5. MayankMay 30, 2011 at 5:25 PM

    Is it 30m/s ?

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  6. Sambhav GuptaMay 31, 2011 at 9:59 PM

    :-|
    i didn't expect this many answers :P

    yes, 30 it is.
    solution, anyone?

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  7. MayankMay 31, 2011 at 10:42 PM

    Very much similar to the 'Cat-Dog' problem.
    let the v be the vel. at any time instant t & vel. vector makes an angle Q with the base of the plane in the same plane as of the inclined plane at any time t.
    dv(y) = (gsin30-µgcos30.sinQ)dt....(i)
    integrate LHS from 0 to v
    dv = -(µgcos30-gsin30.sinQ)dt....(ii)
    integrate LHS from u to v
    v = 30m/s [substituting the value of integral from (i) in (ii)]

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  8. AnonymousJune 1, 2011 at 2:04 AM

    nice soln mayank..

    these DE.s weren't that difficult...
    sambhav, u had a different solution in mind?

    PS: i cudn't solve this...

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  9. Sambhav GuptaJune 1, 2011 at 8:48 PM

    very well done Mayank, i had a really different (AND difficult) solution in mind!

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  10. Ishan GuptaJune 1, 2011 at 9:58 PM

    check this out.............

    acc along inst velocity= acc down the incline= mu*g*sin30...

    so INcrease in vel "parallel to incline"= DEcrease in vel "along instantaneous velocity"....

    let vel "parallel to incline"= u
    instantaneous velocity = v

    so u + v= constant.
    initially...u=0, v=Vo....so constt= Vo

    finally....u=v
    hence v(final )= Vo/2

    suppose u= vel

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  11. Ishan GuptaJune 1, 2011 at 9:59 PM

    ignore last line...

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  12. guntashJune 1, 2011 at 10:20 PM

    a much better question will be to express v(x) and v(y) as function of theta(angle made by instantaneous velocity with horizontal or v(x)) and angular velocity too...to solve this one,u have take a more difficult route(which sambhav have in his mind)

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  13. Sambhav GuptaJune 2, 2011 at 2:22 AM

    Ishan, i 'doubt' your solution
    you forgot that both u and v, as well accelerations, are vectors

    you can't add them the way we add scalars
    also, both gsin30 and friction affect velocity "along instantaneous velocity", you considered that only friction is responsible.

    I might be wrong though

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  14. Sambhav GuptaJune 2, 2011 at 7:45 AM

    ok im sry, Ishan, you were correct!
    i had to sit for an hour in the morning wind to prove this mathematically :P

    good job! :-)

    ReplyDelete

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