Direct collisions on an inclined plane

[Woolyabyss]

Homework Statement

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.

Homework Equations

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∅jthe 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)/3v = 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 1/6my book says the answer is 1/3 but I got 1/6. Any help would be appreciated.

 

[Andrew Mason]

vcos∅ = usin∅

v = utan∅

Correct so far

vsin∅/(ucos∅) = 2/3 (v/u)tan∅ = 2/3 (utan∅/u)tan∅ =2/3

It is a bit convoluted there. I can't follow what you have done. Just use:

vsin∅/(ucos∅) = 2/3

so: tan∅ = 2/3(u/v)

Since you also have tan∅ = v/u

∴ v^2/u^2 = 2/3

AM

 

[Woolyabyss]

I used the equation

v^2/u^2 = 2/3

v^2 = (2/3)u^2

so .5mu^2 - .5m(2/3)u^2 = 1/6mu^2

fraction of kinetic energy lost during impact = 1/6

I got the same answer do you think my book could be wrong?

 

[haruspex]

so .5mu^2 - .5m(2/3)u^2 = 1/6mu^2
fraction of kinetic energy lost during impact = 1/6
I got the same answer do you think my book could be wrong?

What fraction is 1/6mu^2 of .5mu^2?

 

[Woolyabyss]

What fraction is 1/6mu^2 of .5mu^2?

1/3! thanks