A particle of mass m is projected from the ground with an initial speed `u_0` at an angle `alpha` with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed `u_0.` The angle that the composite system makes with the horizontal immediately after the collision is

Options

A. (a)`pi/4`
B. (b)`pi/4 +alpha`
C. (c)`pi/4 - alpha`
D. (d)`pi/2`

Explanation

To solve the problem step by step, we will analyze the motion of the two particles and apply the principles of conservation of momentum.

### Step 1: Analyze the motion of the first particle
The first particle is projected from the ground with an initial speed \( u_0 \) at an angle \( \alpha \) with the horizontal. At the highest point of its trajectory, the vertical component of its velocity becomes zero, and only the horizontal component remains.

**Horizontal component of velocity at the highest point:**
\[ v_{1x} = u_0 \cos \alpha \]
\[ v_{1y} = 0 \]

### Step 2: Analyze the motion of the second particle
The second particle is thrown vertically upward with the same initial speed \( u_0 \). We need to find its velocity at the moment of collision, which occurs at the highest point of the first particle's trajectory.

Using the equations of motion, we can find the velocity of the second particle just before the collision. The time taken to reach the highest point can be calculated using:
\[ t = \frac{u_0}{g} \]
where \( g \) is the acceleration due to gravity.

**Velocity of the second particle at the moment of collision:**
At the highest point, its velocity will be:
\[ v_{2y} = u_0 - g t = u_0 - g \left(\frac{u_0}{g}\right) = 0 \]
Thus, the vertical component of the second particle's velocity just before the collision is:
\[ v_{2y} = 0 \]
The horizontal component is also zero since it is moving vertically.

### Step 3: Apply conservation of momentum
Since the collision is completely inelastic, the two particles stick together after the collision. We can apply the conservation of momentum in the horizontal and vertical directions.

**Horizontal momentum before collision:**
\[ p_{x, \text{initial}} = m v_{1x} + m v_{2x} = m (u_0 \cos \alpha) + m (0) = m u_0 \cos \alpha \]

**Vertical momentum before collision:**
\[ p_{y, \text{initial}} = m v_{1y} + m v_{2y} = m (0) + m (u_0) = m u_0 \]

After the collision, let \( V_x \) and \( V_y \) be the horizontal and vertical components of the velocity of the combined mass \( 2m \).

**Horizontal momentum after collision:**
\[ p_{x, \text{final}} = (2m) V_x \]

Setting the horizontal momentum before and after the collision equal:
\[ m u_0 \cos \alpha = 2m V_x \implies V_x = \frac{u_0 \cos \alpha}{2} \]

**Vertical momentum after collision:**
\[ p_{y, \text{final}} = (2m) V_y \]

Setting the vertical momentum before and after the collision equal:
\[ m u_0 = 2m V_y \implies V_y = \frac{u_0}{2} \]

### Step 4: Calculate the angle after the collision
The angle \( \theta \) that the composite system makes with the horizontal can be found using the tangent function:
\[ \tan \theta = \frac{V_y}{V_x} = \frac{\frac{u_0}{2}}{\frac{u_0 \cos \alpha}{2}} = \frac{1}{\cos \alpha} \]

Thus, we can write:
\[ \theta = \tan^{-1} \left( \frac{1}{\cos \alpha} \right) \]

### Conclusion
The angle \( \theta \) that the composite system makes with the horizontal immediately after the collision is:
\[ \theta = 45^\circ \quad \text{(when } \alpha = 45^\circ\text{)} \]