A particle of mass `m` strikes a horizontal smooth floor with velocity `u` making an angle `theta` with the floor and rebound with velocity `v` making an angle `theta` with the floor. The coefficient of restitution between the particle and the floor is `e`. Then

Explanation

To solve the problem step by step, we will analyze the motion of the particle before and after it strikes the horizontal smooth floor, taking into account the coefficient of restitution.

### Step 1: Analyze Initial and Final Velocities
The particle strikes the floor with an initial velocity \( \mathbf{u} \) at an angle \( \theta \) with respect to the horizontal. The components of the initial velocity can be expressed as:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

After rebounding, the particle has a final velocity \( \mathbf{v} \) at the same angle \( \theta \). The components of the final velocity are:
- Horizontal component: \( v_x = v \cos \theta \)
- Vertical component: \( v_y = -v \sin \theta \) (negative because it is directed upwards after the rebound)

### Step 2: Apply the Coefficient of Restitution
The coefficient of restitution \( e \) is defined as the ratio of the relative speed after the collision to the relative speed before the collision along the line of impact. For vertical motion, we can write:
\[ e = \frac{\text{Speed after collision}}{\text{Speed before collision}} = \frac{|v_y|}{|u_y|} \]
Substituting the expressions for \( v_y \) and \( u_y \):
\[ e = \frac{v \sin \theta}{u \sin \theta} \]
From this, we can derive the relationship between the initial and final vertical velocities:
\[ v = e u \]

### Step 3: Calculate the Impulse
The impulse delivered by the floor to the particle can be calculated using the change in momentum in the vertical direction:
\[ \Delta p_y = m(v_y - u_y) = m(-v \sin \theta - u \sin \theta) = m(-v \sin \theta + u \sin \theta) \]
Substituting \( v = e u \):
\[ \Delta p_y = m(-e u \sin \theta + u \sin \theta) = mu \sin \theta (1 - e) \]
The impulse \( J \) is equal to the change in momentum:
\[ J = mu \sin \theta (1 - e) \]

### Step 4: Analyze the Horizontal Motion
Since the floor is smooth, there is no friction, and the horizontal component of the velocity remains unchanged:
\[ u_x = v_x \implies u \cos \theta = v \cos \theta \]
Thus, the horizontal component of the velocity does not change during the collision.

### Step 5: Calculate the Final Velocity Magnitude
The magnitude of the final velocity can be calculated using the Pythagorean theorem:
\[ v = \sqrt{(v_x)^2 + (v_y)^2} = \sqrt{(v \cos \theta)^2 + (-v \sin \theta)^2} = \sqrt{v^2 (\cos^2 \theta + \sin^2 \theta)} = v \]

### Step 6: Kinetic Energy Before and After Collision
The initial kinetic energy \( KE_i \) and final kinetic energy \( KE_f \) can be expressed as:
\[ KE_i = \frac{1}{2} m u^2 \]
\[ KE_f = \frac{1}{2} m v^2 = \frac{1}{2} m (e u)^2 = \frac{1}{2} m e^2 u^2 \]
The ratio of final to initial kinetic energy is:
\[ \frac{KE_f}{KE_i} = \frac{e^2}{1} \]

### Summary of Results
1. The relationship between the final and initial vertical velocities is \( v = e u \).
2. The impulse delivered by the floor is \( J = mu \sin \theta (1 - e) \).
3. The horizontal component of velocity remains unchanged.
4. The ratio of final to initial kinetic energy is \( e^2 \).