a small conceptual one

Let the wall be inclined at an angle θ to x-axis
then a unit vector along the wall W=cosθ i +sinθ j
and a unit vector perpendicular to the wall will be Wp=-sinθ i+cosθ j

let U be the approaching velocity, U=5 i+3 j
and V=-3 i+4 j
Now, we know that component of velocities along the wall would not change and the component of velocities perpendicular to the wall will have a factor of 'e'

that is,

V.W=U.W .... (i) and
V.Wp=-e (U.Wp) ... (ii)

from (i)
-3 cosθ+4sinθ = 5 cosθ+3sinθ
or tanθ=8 ...(iii)

from (ii)
3 sinθ +4cosθ=-e(-5sinθ+3cosθ)

or e=(3 sinθ +4cosθ)/(5sinθ-3cosθ)

=(3tanθ+4)/(5tanθ-3)
=28/37 .. using (iii)

yes prashant bhaiyya u are right good work :)

answer is 28/37 :)

but there is an easier method (as this is an objective one)

u see the concept is that
the direction of the change in velocity is the direction of the normal reaction
so you just have to find the component of velocity using dot product alog the normal two times
now divide them you will get the answer

sorry bhaiyya earlier i called u by the name (saw ur status just now )

but sir the solution rohan gave is really easier

all we have to do is subtract the vectors and get -8i^{^} + j^{^}
then we get the ans as
-(-3i^{^}+4j^{^}).(-8i^{^} + j^{^}) divided by
(5i^{^}+3j^{^}).(-8i^{^} + j^{^})

actually we should take the unit vector but since the magnitude would cancel out anyway on division we need not bother