A particle is projected from the ground with speed u at angle `60^@` from horizontal. It collides with a second particle of same mass moving with horizontal speed u in same direction at highest point of its trajectory. If collision is perfectly inelastic then find horizontal distance travelled by them after collision when they reached at ground

Options

A. `5/8(u^(2))/(g)`
B. `(3sqrt3)/(8)(u^(2))/(g)`
C. `2sqrt2(u^(2))/(g)`
D. `(3sqrt2)/(4)(u^(2))/(g)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the components of the initial velocity
The particle is projected with speed \( u \) at an angle of \( 60^\circ \) from the horizontal. We can find the horizontal and vertical components of the velocity:

- Horizontal component:
\[ u_x = u \cos(60^\circ) = u \cdot \frac{1}{2} = \frac{u}{2} \]

- Vertical component:
\[ u_y = u \sin(60^\circ) = u \cdot \frac{\sqrt{3}}{2} = \frac{u\sqrt{3}}{2} \]

### Step 2: Calculate the time to reach the maximum height
The time taken to reach the maximum height can be calculated using the vertical component of the velocity and the acceleration due to gravity \( g \):

At maximum height, the vertical velocity becomes zero. Using the formula:
\[ v_y = u_y - gt \]
Setting \( v_y = 0 \):
\[ 0 = \frac{u\sqrt{3}}{2} - gt \implies gt = \frac{u\sqrt{3}}{2} \implies t = \frac{u\sqrt{3}}{2g} \]

### Step 3: Calculate the maximum height
The maximum height \( h \) can be calculated using the formula:
\[ h = u_y t - \frac{1}{2} g t^2 \]
Substituting \( t \):
\[ h = \frac{u\sqrt{3}}{2} \cdot \frac{u\sqrt{3}}{2g} - \frac{1}{2} g \left(\frac{u\sqrt{3}}{2g}\right)^2 \]
Calculating the first term:
\[ h = \frac{u^2 \cdot 3}{4g} - \frac{1}{2} g \cdot \frac{3u^2}{4g^2} = \frac{3u^2}{4g} - \frac{3u^2}{8g} = \frac{6u^2 - 3u^2}{8g} = \frac{3u^2}{8g} \]

### Step 4: Find the horizontal distance traveled before collision
At the maximum height, the horizontal distance \( x \) traveled by the first particle is:
\[ x = u_x \cdot t = \frac{u}{2} \cdot \frac{u\sqrt{3}}{2g} = \frac{u^2\sqrt{3}}{4g} \]

### Step 5: Determine the velocity after collision
Since the collision is perfectly inelastic, the two particles stick together. The total momentum before the collision is:
\[ \text{Total momentum} = m \cdot u + m \cdot \frac{u}{2} = mu + \frac{mu}{2} = \frac{3mu}{2} \]
The combined mass after the collision is \( 2m \). Thus, the velocity \( v \) after the collision is:
\[ v = \frac{\text{Total momentum}}{\text{Total mass}} = \frac{\frac{3mu}{2}}{2m} = \frac{3u}{4} \]

### Step 6: Calculate the time to fall from maximum height to the ground
The time \( t_f \) taken to fall from the maximum height \( h \) to the ground can be calculated using:
\[ h = \frac{1}{2} g t_f^2 \implies t_f^2 = \frac{2h}{g} \implies t_f = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot \frac{3u^2}{8g}}{g}} = \sqrt{\frac{3u^2}{4g^2}} = \frac{u\sqrt{3}}{2g} \]

### Step 7: Calculate the horizontal distance traveled after collision
The horizontal distance \( d \) traveled after the collision until they reach the ground is:
\[ d = v \cdot t_f = \frac{3u}{4} \cdot \frac{u\sqrt{3}}{2g} = \frac{3u^2\sqrt{3}}{8g} \]

### Final Answer
The total horizontal distance traveled by the two particles after the collision until they reach the ground is:
\[ \text{Total distance} = x + d = \frac{u^2\sqrt{3}}{4g} + \frac{3u^2\sqrt{3}}{8g} = \frac{2u^2\sqrt{3}}{8g} + \frac{3u^2\sqrt{3}}{8g} = \frac{5u^2\sqrt{3}}{8g} \]