a ball with mass m strikes a rough vertical wall with an angle of 37 degress ( along the common normal)and rebounds with the same angle. Find the coefficient of friction

0.25

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

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