IIT JEE , AIEEE Forum - Physics , Mechanics

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A PARTICLE OF MASS m IS MADE TO MOVE WITH UNIFORM SPEED u
ALONG THE PERIMETER OF A REGULAR POLYGON OF 2n SIDE. FIND THE MAGNITUDE OF IMPULSE ON THE PARTICLE AT EACH CORNER OF THE
POLYGON.

computer001

mu[root of{ 2+ 2cos(pi*(2n-2)/2n)}]

karthik2007

Consider one of the sides as the x axis.

Velocity along this line = vi

Now, velocity along the next side, which makes an angle of (

-

/n) with the positive x axis is v(-cos(pi/n) + v(sin(pi/n))

required impulse = mass x difference between the two velocity vectors. To get magnitude, take sum of square root fo components

computer001

kartik ..it simple method...
along 1 side stop body with mu..along next side give it mu...
angle bet them is @=pi(2n-2)/2n
then vector addn

hash_include

impulse = change in momentum
consider the angle of the polygon = @.

i am considering the moment in which the particle goes from the top most edge to the subsequent edge.
let the initial velocity be u i
=> initial momentum = mu i
final momentum = mu cos (180 - @) i + mu sin(180 - @) (-j)
= -mu cos@ i - mu sin @ j

=> change in momentum = -(mucos @ + mu) i - mu sin @ j
=> magnitude = mu root(2(1+cos@) )
= mu sin(@/2)

sunstitute
@=pi(2n-2)/2n

computer001

i dun understand y u have to resolve n all...pl look @ my method...its a bit more direct i think

hash_include

yeah yea.. anyway, answer is same..
and my method'll takes a bit longer to explain thts all..

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