A smooth sphere falls vertically and strikes a fixed smooth plane inclined at an angle of ∅ to the horizontal.If the coefficient of restitution is (2/3) and the sphere rebounds horizontally,
Its speed before impact is u and after impact v
calculate the fraction of kinetic energy lost during impact.
The Attempt at a Solution
Diagram of the question is attached to this post.
i = horizontal component vector and j = vertical component vector
Taking the inclined plane as the x axis
v, vcos∅i + vsin∅j ..... alternate angle to inclined angle
u, vcos(90 - ∅ )i - v sin(90 - ∅)j = usin∅i - ucos∅j
the I component remains the same
vcos∅ = usin∅
v = utan∅
vsin∅/(ucos∅) = 2/3 (v/u)tan∅ = 2/3 (utan∅/u)tan∅ =2/3
(tan∅)^2 = 2/3 tan∅ = (√6)/3
v = u(√6)/3
.5mu^2 - .5mv^2 = energy loss
.5mu^2 -.5m(6/9)u^2 = .5mu^2 - m(6/18)u^2 = .5mu^2 - (1/3)mu^2 = (1/6)mu^2
fraction of kinetic energy lost is 1/6
my book says the answer is 1/3 but I got 1/6. Any help would be appreciated.