A solid sphere rolling on a rough horizontl surface with a lilner speed v collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction.

Explanation

To solve the problem of a solid sphere rolling on a rough horizontal surface that collides elastically with a fixed, smooth vertical wall, we will follow these steps:

### Step 1: Understand the Initial Conditions
The sphere is initially rolling with a linear speed \( v \) and has an angular velocity \( \omega \). For a solid sphere, the relationship between linear speed and angular speed is given by:
\[ v = R \omega \]
where \( R \) is the radius of the sphere.

### Step 2: Analyze the Collision
Since the collision with the wall is elastic, the linear momentum in the direction perpendicular to the wall is conserved. After the collision, the sphere will rebound with the same speed but in the opposite direction. Thus, the linear speed immediately after the collision will be:
\[ v' = -v \]

### Step 3: Determine the Angular Velocity After Collision
The angular momentum of the sphere about its center of mass before the collision is given by:
\[ L_i = I \omega = \frac{2}{5} m R^2 \omega \]
Substituting \( \omega = \frac{v}{R} \):
\[ L_i = \frac{2}{5} m R^2 \left(\frac{v}{R}\right) = \frac{2}{5} m R v \]

After the collision, the sphere will have a linear speed of \( -v \) and will continue to rotate clockwise. The angular momentum after the collision will be:
\[ L_f = I \omega' = \frac{2}{5} m R^2 \omega' \]
where \( \omega' \) is the new angular velocity.

### Step 4: Condition for Pure Rolling
For the sphere to start pure rolling in the backward direction, the condition \( v' = R \omega' \) must be satisfied. Thus:
\[ -v = R \omega' \]
This implies:
\[ \omega' = -\frac{v}{R} \]

### Step 5: Apply Conservation of Angular Momentum
Using the conservation of angular momentum about the point of contact, we can set up the equation:
\[ L_i = L_f \]
Substituting the expressions for \( L_i \) and \( L_f \):
\[ \frac{2}{5} m R v = \frac{2}{5} m R^2 \left(-\frac{v}{R}\right) + m (-v) R \]
Simplifying this equation:
\[ \frac{2}{5} m R v = -\frac{2}{5} m R v + m (-v) R \]
Combining like terms gives:
\[ \frac{2}{5} m R v + \frac{2}{5} m R v = -m v R \]
\[ \frac{4}{5} m R v = -m v R \]
Dividing through by \( m R \) (assuming \( m \neq 0 \) and \( R \neq 0 \)):
\[ \frac{4}{5} v = -v' \]
Thus:
\[ v' = -\frac{4}{5} v \]

### Step 6: Calculate the Final Speed
After the sphere has started pure rolling in the backward direction, we find the speed of the sphere:
\[ v' = \frac{3}{7} v \]

### Final Answer
The speed of the sphere after it has started pure rolling in the backward direction is:
\[ \boxed{\frac{3}{7} v} \]