Bouncy Bouncing Ball!

Let $e = 0.5$

Now, just before $n^{th}$ bounce, the vertical velocity of ball will be $e^{n-1}v$ and just after $n^{th}$ bounce, vertical velocity of ball would be $e^nv$ .

Also, let just after $n^{th}$ bounce, let angular velocity be $\omega_n$ and horizontal velocity be $v_n$

Applying Impulse - momentum theorem, considering normal reaction $N$

$\int Ndt = m(e^nv + e^{n-1}v)$ $\mu\int Ndt = m(v_n - v_{n-1})$

By above two equations

$\mu v(e+1)e^{n-1} = v_n - v_{n-1}$

Solving above series, we get

$\boxed{\mu v\frac{1+e}{1-e}(1-e^n) = v_n}$

Apply Angular impulse - angular momentum theorem

$\mu R \int Ndt = \frac{mR^2}{2}(\omega_{n-1} - \omega_n)$

Putting value of $\int Ndt$ from first equation

$\frac{2\mu v}{R}(1+e)e^{n-1} = \omega_{n-1} - \omega_n$

Solving above series, we get

$\frac{2\mu v}{R}\frac{1+e}{1-e}(1-e^n) = \omega - \omega_n$

Now, checking $\omega_n R \leq v_n$ , we see it not true for any value of $n$ for given values , so friction will act at every bounce in motion.

Putting $n = \infty$ and other values in question, we get

$\boxed{\omega' = 1000}$

Now, let range after $n^{th}$ and before $(n+1)^{th}$ bounce be $r_n$

$r_n = \frac{2v_xv_y}{g}$ $r_n = \frac{2\mu v\frac{1+e}{1-e}(1-e^n)e^nv}{g}$ $r_n = \frac{2\mu v^2}{g}\frac{1+e}{1-e}(e^n - e^{2n})$ $\sum_{n=0}^{\infty}r_n = \sum_{n=0}^{\infty}\frac{2\mu v^2}{g}\frac{1+e}{1-e}(e^n - e^{2n})$

$r = \frac{2\mu v^2}{g}\frac{e}{(1-e)^2}$

Putting values, we get

$\boxed{r = 4000}$

$\color{#3D99F6}{\boxed{\omega' + r = 1000 + 4000 = 5000}}$

It should be stated in the problem that the gravitational acceleration is $10$ .

A few basic principles behind this problem:

1) When the ball bounces, there is a vertical impulse and associated upward normal force. This causes the vertical speed after the bounce to be half of its value before the bounce. In order for this to happen, the magnitude of the impulse must be $1.5$ times the vertical momentum right before the bounce (because the force has to get rid of all the downward speed before imparting half as much speed upward).

2) Because the friction coefficient is unity, the horizontal impulse is equal to the vertical impulse. This means that for each bounce, the horizontal speed increases by $1.5$ times the vertical speed right before the bounce.

3) The horizontal friction force also reduces the angular speed. The relationships between translational and rotational quantities are given below ( $I$ is the disk moment of inertia):

$N \Delta t = m \Delta v_x \\ \tau \Delta t = -I \Delta \omega = N R \Delta t = m R \Delta v_x \\ \Delta \omega = - \frac{m R \Delta v_x}{I}$

Thus, the change in angular momentum for the bounce can be calculated once the change in horizontal speed is known. The attached code calculates the parameters right after the first bounce. It then simulates a large number of bounces to get limiting values for the horizontal distance traveled and the angular speed. These are:

$x_f = 4000 \\ \omega_f = 1000$

I'm sure it would also be possible to evaluate some geometric series analytically, as an alternative to the numerical approach.

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import math

N = 10**5     # Number of bounces

# Constants

h = 500.0         # initial height
R = 0.9           # disk radius
w = 5000.0/3.0    # initial angular speed
u = 1.0           # friction coefficient
g = 10.0           # gravitational acceleration

m = 1.0             # mass of disk
I = 0.5*m*(R**2.0)  # disk moment of inertia

#####################################

# Initialize parameters to their values right after first bounce

# m*g*h = 0.5*m*(vpre**2.0)   

vpre = math.sqrt(2.0*g*h)  # speed just before first bounce

x = 0.0        # initial horizontal distance to zero
vy = 0.5*vpre  # vertical speed to to 50% of initial speed
vx = 1.5*vpre  # horizontal speed increased to 150% of initial speed
w = w - m*R*(1.5*vpre)/I  # update angular speed

#####################################

# Simulate a large number of additional bounces

for j in range(0,N):

    tf = 2.0*vy/g              # flight time from bounce to bounce
    x = x + vx*tf              # update horizontal distance

    vyold = vy                 # store old vertical speed

    vy = 0.5*vyold             # update vertical speed
    vx = vx + 1.5*vyold        # update horizontal speed
    w = w - m*R*(1.5*vyold)/I  # update angular speed

#####################################

print N
print vy
print ""
print x
print w
print ""
print (x+w)

#>>> 
#1000
#4.66631809252e-300

#4000.0
#1000.0

#5000.0
#>>> ================================ RESTART ================================
#>>> 
#100000
#0.0

#4000.0
#1000.0

#5000.0
#>>>