Solve the previous problem if the coefficient of restitution is e. \[\text{ Use } \theta = 45^\circ\, e = \frac{3}{4} \text{ and h = 5 m } .\]
Solution
Given:
The angle of inclination of the plane, θ = 45°
A ball falls on the inclined plane from a height (h) of 5 m.
The coefficient of restitution,\[e = \left( \frac{3}{4} \right)\]
The velocity with which ball strikes the inclined plane is given as,
\[v = \sqrt{2g \times 5} = 10 m/s\]
The ball makes an angle β with the horizontal, after the collision.
The horizontal component of velocity, 10cos45° remains unchanged.
However, the velocity in perpendicular direction to the plane after the collision will now be:
\[v_1 = e \times 10 \sin45^\circ\]
\[ = \left( \frac{3}{4} \right) \times 10 \times \frac{1}{\sqrt{2}}\]
\[ = (3 . 75) \sqrt{2} m/s\]
=\[\text{ Similarly,} v_2 = 5\sqrt{2}\text{m/s} \]
\[\text{ Now,} \]
\[u = \sqrt{v_2^2 + v_1^2}\]
\[ = \sqrt{50 + 28 . 125}\]
\[ = \sqrt{78 . 125}\]
\[ = 8 . 83 \text{m/s}\]
Angle of reflection from the wall is given as,
\[\beta = \tan^{- 1} \frac{(3 . 75\sqrt{2})}{5\sqrt{2}}\]
\[ = \tan^{- 1} \left( \frac{3}{4} \right) = 37^\circ\]
Angle of projection α = 90 − (θ + β)
\[\Rightarrow\] α = 90° − (45° + 37°) = 8°
Let the distance where the ball falls after the collision be L.
⇒ x = L cos θ
Angle of projection (α) = −8°
Now,
Y = \[x \tan \alpha - \frac{g x^2 se c^2 \alpha}{2 u^2}\]
\[\Rightarrow\] −L sin θ = L cos θ × tan 8° \[- \frac{g}{2}\frac{L^2 \cos^2 \theta \sec^2 80^\circ\ }{(u )^2}\]
\[\Rightarrow - \sin 45^\circ = \cos 45^\circ \times \tan 8^\circ - \frac{10 \cos^2 45 \sec^28^\circ }{(8 . 83 )^2}\]
On solving the above equation, we get:
L = 18.5 m
