Particles $P$ and $Q$ of mass $20\,\text{gm}$ and $40\,\text{gm}$ respectively are simultaneously projected from points $A$ and $B$ on the ground. The initial velocities of $P$ and $Q$ make $45^\circ$ and $135^\circ$ angles respectively with the horizontal $AB$ as shown in the figure. Each particle has an initial speed of $49\,\text{m/s}$. The separation $AB$ is $245\,\text{m}$. Both particle travel in the same vertical plane and undergo a collision After the collision, $P$ retraces its path, Determine the position of $Q$ when it hits the ground. How much time after the collision does the particle $Q$ take to reach the ground? Take $g = 9.8\,\text{m/s}^{2}$.

Step 1: Calculate the components of initial velocity and the time of collision.

The initial horizontal and vertical velocity components for both particles are equal in magnitude.

$$u_x = u_y = 49 \cos 45^\circ = \frac{49}{\sqrt{2}}\text{ m/s}$$

The relative horizontal velocity of approach is the sum of their horizontal speeds.

$$v_{\text{rel}} = 2 u_x = 49\sqrt{2}\text{ m/s}$$

The time $T$ from projection to collision is determined by the initial separation.

$$T = \frac{AB}{v_{\text{rel}}} = \frac{245}{49\sqrt{2}} = 2.5\sqrt{2}\text{ s} \approx 3.54\text{ s}$$

Step 2: Determine the position and velocity of the particles just before collision.

The time to reach the maximum height for each particle is equal to the collision time.

$$t_{\text{apex}} = \frac{u_y}{g} = \frac{49/\sqrt{2}}{9.8} = 2.5\sqrt{2}\text{ s}$$

Since $T = t_{\text{apex}}$, the collision occurs at the apex of their trajectories where vertical velocities are zero.

$$h = u_y T - \frac{1}{2}gT^2$$ $$h = \left(\frac{49}{\sqrt{2}}\right)(2.5\sqrt{2}) - \frac{1}{2}(9.8)(2.5\sqrt{2})^2$$ $$h = 61.25\text{ m}$$

Step 3: Apply conservation of momentum to find the velocity of $Q$ after collision.

Since $P$ retraces its path, its horizontal velocity is reversed to $-u_x$, while the vertical velocities remain zero.

$$m_P u_x - m_Q u_x = m_P (-u_x) + m_Q v'_Q$$ $$v'_Q = \frac{2m_P - m_Q}{m_Q} u_x$$

Substitute the masses $m_P = 20\text{ g}$ and $m_Q = 40\text{ g}$.

$$v'_Q = \frac{2(20) - 40}{40} u_x = 0\text{ m/s}$$

Step 4: Calculate the landing position and time of flight for $Q$ after collision.

Since $Q$ has zero velocity immediately after the collision, it falls vertically from rest from height $h$.

$$t' = \sqrt{\frac{2h}{g}}$$ $$t' = \sqrt{\frac{2 \times 61.25}{9.8}} = 2.5\sqrt{2}\text{ s} \approx 3.54\text{ s}$$

The landing position is directly below the collision point, which is the midpoint of $AB$.

$$x = \frac{AB}{2} = \frac{245}{2} = 122.5\text{ m}$$

Answer: Particle $Q$ hits the ground at a distance of $122.5\text{ m}$ from $A$ (the midpoint of $AB$), and it takes $3.54\text{ s}$ (or $2.5\sqrt{2}\text{ s}$) after the collision to reach the ground.