Discuss with colleagues and IITians> problem on collision...
2 AnswersLast Activity: 12 Years ago
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Last Activity: 12 Years ago
I guess u meant coefficient of restitution, by coefficient of friction!!
First we should know that coefficient of restitution, e= relative speed of seperation = RSS/RSA
relative speed of approach
(Wish i could help you with the visuals but i don''t know how to)
Never mind, lets continue this way..
before collision
let the ball fall at angle of 37degree, with the normal, then dividing the velocity into components we have-
vsin37 along verticall downward, direction is -j^, and
vcos37 along horizontal towards the wall, i.e direction along -i^
after collision
let the velocity be u, alog vertically downward it is usin37 -j^
and along horizontal it is ucos37 +i^
Now vertical velocity is not affect during collision, because only horizontal velocity takes part in collision.
therefore, usin37=vsin37
or, u=v
therefore horizontal velocity after collision is, vcos37
applying very first formulae, we have e= RSS/RSA
or, e=vcos37/vco37=1
therefore, we can say collision is perfectly elastic.
Now we have a shortcut also, when we have angles of collision and refelection, that might help (:D)-
e=tanΘ1/tanΘ2
here tanΘ1 is intial angle and tanΘ2 isfinal angle..
hope this helps :D
AJ
(Hope you liked this, if yes then please vote this answer ;))
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