Inelastic Collision of Projectile at Trajectory Apex with Vertically Thrown Body

To find the angle that the trajectory of the combined system makes with the horizontal immediately after the collision, we analyze the motion and velocity of each particle right before the impact.

Step 1: Velocity of the first particle at its trajectory apex

The first particle of mass $m$ is projected with an initial speed $u_0$ at an angle $\alpha$ above the horizontal. Its velocity components at any time $t$ are given by: $u_{1x} = u_0 \cos\alpha$ $u_{1y} = u_0 \sin\alpha - gt$

At the highest point (apex) of its trajectory, the vertical component of velocity becomes zero ($v_{1y} = 0$). Thus, the velocity vector of the first particle right before collision is purely horizontal: $\vec{v}_1 = (u_0 \cos\alpha) \hat{i}$

The maximum height $H$ reached by this particle is: $H = \frac{u_0^2 \sin^2\alpha}{2g}$

---

Step 2: Velocity of the second particle at height $H$

The second identical particle of mass $m$ is launched vertically upward from the ground with speed $u_0$. It collides with the first particle at height $H$.

Using the third equation of motion (or conservation of mechanical energy) for the second particle at height $H$: $v_{2y}^2 = u_0^2 - 2gH$

Substituting $H = \frac{u_0^2 \sin^2\alpha}{2g}$ into the equation: $v_{2y}^2 = u_0^2 - 2g \left(\frac{u_0^2 \sin^2\alpha}{2g}\right) = u_0^2 (1 - \sin^2\alpha) = u_0^2 \cos^2\alpha$

Taking the square root for the upward motion: $v_{2y} = u_0 \cos\alpha$

Thus, the velocity vector of the second particle right before collision is purely vertical: $\vec{v}_2 = (u_0 \cos\alpha) \hat{j}$

---

Step 3: Completely inelastic collision

Since the collision is completely inelastic, the two particles stick together to form a composite mass of $2m$. By the law of conservation of linear momentum: $(2m) \vec{V} = m \vec{v}_1 + m \vec{v}_2$

Dividing by $2m$: $\vec{V} = \frac{\vec{v}_1 + \vec{v}_2}{2} = \left(\frac{u_0 \cos\alpha}{2}\right) \hat{i} + \left(\frac{u_0 \cos\alpha}{2}\right) \hat{j}$

---

Step 4: Angle of the combined velocity vector

Let $\theta$ be the angle that the velocity vector $\vec{V}$ makes with the horizontal right after impact. Then: $\tan\theta = \frac{V_y}{V_x} = \frac{\frac{u_0 \cos\alpha}{2}}{\frac{u_0 \cos\alpha}{2}} = 1$

Solving for $\theta$: $\theta = \arctan(1) = \frac{\pi}{4}$

Thus, the correct option is A.