Question

A GIRL THROWS A BALL WITH INITIAL VELOCITY V AT AN INCLINATION OF 45. THE BALL STRIKES THE SMOOTH VERTICAL WALL AT A HORIZONTAL DISTANCE D FROM THE GIRL AND AFTER REBOUNCING RETURNS TO HER HAND. WHAT IS THE COEFFICIENT OF RESTITUTION BETWEEN WALL AND THE BALL ?

Answer 1

Answer on Question #40898, Physics, Mechanics

A GIRL THROWS A BALL WITH INITIAL VELOCITY V AT AN INCLINATION OF 45. THE BALL STRIKES THE SMOOTH VERTICAL WALL AT A HORIZONTAL DISTANCE D FROM THE GIRL AND AFTER REBOUNCING RETURNS TO HER HAND. WHAT IS THE COEFFICIENT OF RESTITUTION BETWEEN WALL AND THE BALL?

Solution

Initial x-component of velocity is $v_{0x} = v \cos 45{}^\circ = \frac{v}{\sqrt{2}}$, initial y-component of velocity is

$v_{0y} = v \sin 45{}^\circ = \frac{v}{\sqrt{2}}$. Time before the impact is $t_1 = \frac{d}{v_{0x}} = \frac{\sqrt{2}d}{v}$. The y-component of velocity before the impact is $v_y = v_{0y} - gt_1 = \frac{v}{\sqrt{2}} - g\frac{\sqrt{2}d}{v} = \frac{v^2 - 2gd}{\sqrt{2}v}$. The height of the ball immediately before the impact is

$$h_1 = v_{0y} t_1 - \frac{g t_1^2}{2} = \frac{v}{\sqrt{2}} \cdot \frac{\sqrt{2}d}{v} - \frac{g \left(\frac{\sqrt{2}d}{v}\right)^2}{2} = d - g \frac{d^2}{v^2}.$$

The y-component of velocity didn't change after the impact:

$$u_{0y} = v_y = \frac{v^2 - 2gd}{\sqrt{2}v}.$$

The x-component of velocity after the impact is $u_x = ev_{0x} = e \frac{v}{\sqrt{2}}$, where $e$ is the coefficient of restitution.

Time before the ball returns to girl's hand after the impact is

$$t_2 = \frac{d}{u_x} = \frac{\sqrt{2}d}{ev}.$$

The height of the ball when it returns to girl's hand is

$$h_2 = 0 = h_1 + u_{0y} t_2 - \frac{g t_2^2}{2} = \left(d - g \frac{d^2}{v^2}\right) + \frac{v^2 - 2gd}{\sqrt{2}v} \cdot \frac{\sqrt{2}d}{ev} - \frac{g}{2} \cdot \left(\frac{\sqrt{2}d}{ev}\right)^2.$$

We have a quadratic equation on $e$:

$$\left(d - g \frac{d^2}{v^2}\right) e^2 + \frac{d}{v^2} (v^2 - 2gd) e - \frac{g d^2}{v^2} = 0.$$

Discriminant is

$$D = d^2.$$

The solutions of this quadratic equation are

$$e_1 = \frac{gd}{v^2 - gd}, \quad e_2 = -1.$$

Second solution is unphysical, because the coefficient of restitution can be $0 \leq e \leq 1$.

Answer: $\frac{gd}{v^2 - gd}$