Question

"A ball falls on an inclined plane of inclination θ from a height h above the point of impact and makes a perfectly elastic collision. Where will it hit the plane again?"

Solve the previous problem if the coefficient of restitution is e. Use θ = 45°, e = ¾, and h = 5 m.

Answer 1

Thanks for asking the question!

ANSWER::

Ф = 45°, e = ¾, and h = 5 m

Velocity with which it should strike = v = √(2g x 5) = √100 = 10 m/s

Let is make an angle β with horizontal .

The horizontal component of velocity 10 cos 45° will remain unchanged and velocity in perpendicular direction to plane after will be

V₁ (in y or perpendicular direction) = e x 10 sin 45°

(3/4) x 10 x 1/√2 = (3.75)√2 m/s

V₂(in x or horizontal direction) = 10 cos 45° = 5√2 m/s

So ,

u = √(V₁² + V₂²) = √(50 + 28.125) = √78.125 = 8.83 m/s

Angle of reflection from wall β = tan⁻¹ (3.75√2 / 5√2) = tan⁻¹(3/4) = 37°

Angle of projection = α = 90° - (Ф + β) = 90° - (45° + 37°) = 8°

Let the distance where it falls = L

x = L cos Ф

y = - L sin Ф

Angle of projection = α = - 8°

Using equation of trajectory ,

y = x tan α - (gx² sec²α / 2u²)

- L sin Ф = L cos Ф x tan 8° - (g x Lcos²Фsec²8°) / 2u²

- sin 45° = cos 45° - tan 8° - (10 cos²45°sec 8° L) / (8.83)²

- 1/√2 = - 1/√2 - 0.14 - (10 cos²45° x 1 x L) / (8.83)²

Solving above equation we get ,

L = 18.5 m

Hope it helps!