A ball moving with velocity u strikes a smooth horizontal floor at an angle `alpha ` with the normal to the surface at the point of contact . After collision , ball moves with velocity v at an angle `beta ` with the normal . Assume e as the coefficient of restitution .

Options

A. `tan alpha = e tan beta `
B. Impulse applied by the surface on the ball is mu(1+e)cos `
C. `v = usqrt((sin^(2)alpha+e^(2) cos^(2)alpha))`
D. v=` tan alpha `

Explanation

To solve the problem, we will analyze the collision of the ball with the smooth horizontal floor, applying the principles of momentum and the coefficient of restitution.

### Step-by-Step Solution:

1. **Identify the Components of Velocity:**
- The ball strikes the floor with an initial velocity \( u \) at an angle \( \alpha \) with the normal.
- The components of the initial velocity \( u \) can be expressed as:
- \( u_x = u \cos \alpha \) (component along the normal)
- \( u_y = u \sin \alpha \) (component parallel to the surface)

2. **Post-Collision Velocity Components:**
- After the collision, the ball moves with a velocity \( v \) at an angle \( \beta \) with the normal.
- The components of the final velocity \( v \) are:
- \( v_x = v \cos \beta \) (component along the normal)
- \( v_y = v \sin \beta \) (component parallel to the surface)

3. **Apply the Coefficient of Restitution:**
- The coefficient of restitution \( e \) is defined as the ratio of the relative speed after collision to the relative speed before collision along the normal.
- Therefore, we can write:
\[ e = \frac{v_x}{u_x} = \frac{v \cos \beta}{u \cos \alpha} \]
- Rearranging gives us:
\[ v \cos \beta = e u \cos \alpha \quad \text{(Equation 1)} \]

4. **Conservation of Momentum in the Horizontal Direction:**
- Since no external forces act along the horizontal direction (the surface is smooth), the horizontal component of momentum is conserved:
\[ u_y = v_y \]
- This gives us:
\[ u \sin \alpha = v \sin \beta \quad \text{(Equation 2)} \]

5. **Square and Add the Equations:**
- Square both equations:
\[ (u \sin \alpha)^2 = (v \sin \beta)^2 \quad \text{(from Equation 2)} \]
\[ (v \cos \beta)^2 = (e u \cos \alpha)^2 \quad \text{(from Equation 1)} \]
- Adding these two equations:
\[ u^2 \sin^2 \alpha + e^2 u^2 \cos^2 \alpha = v^2 \sin^2 \beta + v^2 \cos^2 \beta \]
- Since \( \sin^2 \beta + \cos^2 \beta = 1 \), we can simplify:
\[ u^2 (\sin^2 \alpha + e^2 \cos^2 \alpha) = v^2 \]

6. **Solve for \( v \):**
- Rearranging gives us:
\[ v^2 = u^2 (\sin^2 \alpha + e^2 \cos^2 \alpha) \]
- Taking the square root:
\[ v = u \sqrt{\sin^2 \alpha + e^2 \cos^2 \alpha} \]

### Final Result:
The final velocity \( v \) of the ball after the collision is given by:
\[ v = u \sqrt{\sin^2 \alpha + e^2 \cos^2 \alpha} \]