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? 

Answer 1

Given:
The angle of inclination of the inclined plane is θ.
A ball falls on the inclined plane from height h.
The coefficient of restitution is e.

Let the ball strikes the inclined plane at origin with velocity \[v_0 = \sqrt{2gh}\] .

As the ball elastically rebounds, it recalls with the same velocity v₀ at the same angle θ from the normal or y-axis.
Let the ball strikes the incline second time at any point P, which is at a distance l from the origin along the incline.
From the equation,

\[y = v_{oy} t + \frac{1}{2} a_y t^2 \]

\[\text{ we may write: }\]

\[0 = v_0 \cos \theta t - \frac{1}{2}g \cos \theta t^2 \]

where t is time of motion of the ball in air, as the ball moves from origin to point P.
\[As t \neq 0, \text{ the value of t is }\frac{2 v_0}{g} .\]

Now from the equation,

\[x = v_0 t + \frac{1}{2} a_x t^2 \]

\[\text{ we can write }, \]

\[l = v_0 \sin \theta t + \frac{1}{2}g \sin \theta t^2 \]

\[ \therefore l = v_0 \sin \theta\left( \frac{2 v_0}{g} \right) + \frac{1}{2}g \sin \theta \left( \frac{2 v_0}{g} \right)^2 \]

\[ \Rightarrow l = \frac{4 v_0^2 \sin \theta}{g}\]

Substituting the value of v₀ in the above equation, we get:
l = 8h sin  θ

Therefore, the ball again hits the plain at a distance, l = 8h sin θ