A ball is projected form a point `A` on a smooth inclined plane which makes an angle a to the horizontal. The velocity of projection makes an angle `theta` with the plane upwards. If on the second bounce the ball is moving perpendicular to the plane, find `e` in terms of `alpha` and `theta`. Here `e` is the coefficient of restitution between the ball and the plane.

Explanation

To solve the problem step by step, we need to analyze the motion of the ball on the inclined plane, taking into account the angles and the coefficient of restitution.

### Step 1: Understanding the Projection
The ball is projected from point A on a smooth inclined plane at an angle \( \alpha \) to the horizontal. The velocity of projection makes an angle \( \theta \) with the plane upwards.

### Step 2: Components of Initial Velocity
Let the initial velocity of projection be \( u \). The components of the initial velocity can be expressed as:
- The component along the incline (x-direction):
\[ u_x = u \cos \theta \]
- The component perpendicular to the incline (y-direction):
\[ u_y = u \sin \theta \]

### Step 3: Time of Flight to the First Bounce
The time taken for the ball to reach the first bounce (point B) can be calculated using the vertical motion equations. The effective acceleration acting on the ball in the direction perpendicular to the incline is \( g \cos \alpha \). The time of flight \( T_1 \) can be calculated as:
\[ T_1 = \frac{2u_y}{g \cos \alpha} = \frac{2u \sin \theta}{g \cos \alpha} \]

### Step 4: Velocity Just Before the Second Bounce
When the ball strikes the inclined plane for the second time, its velocity will change due to the coefficient of restitution \( e \). The velocity just before the second bounce can be expressed as:
\[ v = 2e u \sin \theta \]
This is because the velocity component perpendicular to the incline will be reduced by the coefficient of restitution.

### Step 5: Time of Flight After the First Bounce
After the first bounce, the ball will again travel upwards and then downwards. The time of flight \( T_2 \) for this motion can be calculated similarly:
\[ T_2 = \frac{2(2e u \sin \theta)}{g \cos \alpha} = \frac{4e u \sin \theta}{g \cos \alpha} \]

### Step 6: Total Time of Flight
The total time of flight from point A to point C (after the second bounce) is:
\[ T_{total} = T_1 + T_2 = \frac{2u \sin \theta}{g \cos \alpha} + \frac{4e u \sin \theta}{g \cos \alpha} \]
This simplifies to:
\[ T_{total} = \frac{2u \sin \theta (1 + 2e)}{g \cos \alpha} \]

### Step 7: Condition for Perpendicular Motion
At the second bounce, the ball is moving perpendicular to the inclined plane, which means the horizontal component of the velocity must be zero. The horizontal component of the velocity after the first bounce can be expressed as:
\[ u_x - g \sin \alpha \cdot T_{total} = 0 \]
Substituting \( u_x = u \cos \theta \) and \( T_{total} \):
\[ u \cos \theta - g \sin \alpha \cdot \frac{2u \sin \theta (1 + 2e)}{g \cos \alpha} = 0 \]

### Step 8: Solving for Coefficient of Restitution \( e \)
From the equation above, we can simplify and solve for \( e \):
\[ u \cos \theta = 2u \sin \theta (1 + 2e) \frac{\sin \alpha}{\cos \alpha} \]
Dividing through by \( u \) and rearranging gives:
\[ \cos \theta = 2 \sin \theta \tan \alpha (1 + 2e) \]
Solving for \( e \):
\[ 1 + 2e = \frac{\cos \theta}{2 \sin \theta \tan \alpha} \]
\[ 2e = \frac{\cos \theta}{2 \sin \theta \tan \alpha} - 1 \]
\[ e = \frac{1}{2} \left( \frac{\cos \theta}{2 \sin \theta \tan \alpha} - 1 \right) \]

### Final Expression for Coefficient of Restitution
Thus, the coefficient of restitution \( e \) in terms of \( \alpha \) and \( \theta \) is:
\[ e = \frac{1}{2} \left( \frac{\cos \theta}{2 \sin \theta \tan \alpha} - 1 \right) \]