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Question
A projectile is fired with a speed u at an angle θ above a horizontal field. The coefficient of restitution of collision between the projectile and the field is e. How far from the starting point, does the projectile makes its second collision with the field?
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Solution
Given:
Initial velocity of the projectile = u
Angle of projection of the projectile with respect to ground = θ
When the projectile hits the ground for the first time, the velocity remains same i.e. u.
The component of velocity parallel to ground, u cos θ should remain constant.
However, the vertical component of the projectile undergoes a change after the collision.
If the coefficient of restitution of collision between the projectile and the field is e,
The velocity of separation is given by,
⇒ v = eu sin θ
Therefore, for the second projectile motion,
Velocity of projection (u') will be,
\[u' = \sqrt{(u \cos \theta )^2 + (eu \sin \theta )^2}\]
\[\text{ Angle of projection,} \alpha = \tan^{- 1} \left( \frac{eu \sin \theta}{u \cos \theta} \right)\]
\[ \Rightarrow \alpha = \tan^{- 1} (e \tan \theta)\]
\[\text{or}, \tan \alpha = e \tan \theta . . . (2)\]
\[\text{Also, y} = x \tan \alpha - \left( \frac{g x^2 \sec^2 \alpha}{2u '^2} \right) . . . (3)\]
\[\text{Here, y} = 0 \]
\[ \therefore \tan \alpha = e \tan \theta\]
\[ \Rightarrow \sec^2 \alpha = 1 + e^2 \tan^2 \theta\]
\[\text{and }u '^2 = u^2 \cos^2 \theta + e^2 u^2 \sin^2 \theta\]
Putting the above calculated values in equation (3), we get:
\[\text{xe } tan\theta = \frac{g x^2 (1 + e^2 \tan^2 \theta)}{2 u^2 ( \cos^2 \theta + e^2 \sin^2 \theta)}\]
\[\text{ or, x } = \frac{2e u^2 \tan \theta ( \cos^2 \theta + e^2 \sin^2 \theta)}{g(1 + e^2 \tan^2 \theta)}\]
\[ \Rightarrow x = \frac{2e u^2 \tan \theta . \cos^2 \theta}{g}\]
\[ \Rightarrow x = \frac{e u^2 \sin 2\theta}{g}\]
Thus, from the starting point the projectile makes its second collision with the field at a distance,
\[x' = \frac{u^2 \sin 2\theta}{g} + \frac{e u^2 \sin 2\theta}{g}\]
\[ \Rightarrow x' = \frac{u^2 \sin 2\theta}{g}(1 + e)\]
