A particle of mass m is dropped from a height h above the ground. At the same time another particle of mass 2m is thrown vertically upwards from the ground with a speed of `sqrt(gh)` . If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of `sqrt(h/g)` is:

Options

A. `(sqrt40 -2)/6`
B. `(sqrt(30)-3)/7`
C. `(sqrt(20)-2)/5`
D. `sqrt(3/2)`

Explanation

To solve the problem, we need to analyze the motion of both particles and their collision. Here's a step-by-step breakdown of the solution:

### Step 1: Analyze the motion of the first particle (mass m)
The first particle is dropped from a height \( h \). The time taken for it to reach the ground can be found using the equation of motion:
\[ h = \frac{1}{2} g t_1^2 \]
From this, we can solve for \( t_1 \):
\[ t_1 = \sqrt{\frac{2h}{g}} \]

### Step 2: Analyze the motion of the second particle (mass 2m)
The second particle is thrown upwards with an initial velocity of \( \sqrt{gh} \). The maximum height it reaches can be calculated using:
\[ h_2 = u t - \frac{1}{2} g t^2 \]
where \( u = \sqrt{gh} \) and \( t \) is the time taken to reach the maximum height. At maximum height, its velocity becomes zero:
\[ 0 = \sqrt{gh} - g t \implies t = \frac{\sqrt{gh}}{g} = \sqrt{\frac{h}{g}} \]
Now, the total height reached by the second particle is:
\[ h_2 = \sqrt{gh} \cdot \sqrt{\frac{h}{g}} - \frac{1}{2} g \left(\sqrt{\frac{h}{g}}\right)^2 = h - \frac{h}{2} = \frac{h}{2} \]

### Step 3: Determine the time of collision
The two particles will collide when the distance fallen by the first particle equals the distance risen by the second particle. Let \( t_0 \) be the time taken for the collision:
\[ h - \frac{1}{2} g t_0^2 + \sqrt{gh} t_0 = h \]
This simplifies to:
\[ \frac{1}{2} g t_0^2 = \sqrt{gh} t_0 \]
Dividing both sides by \( t_0 \) (assuming \( t_0 \neq 0 \)):
\[ \frac{1}{2} g t_0 = \sqrt{gh} \implies t_0 = \frac{2\sqrt{h}}{g} \]

### Step 4: Calculate the velocities at the time of collision
The velocity of the first particle just before collision:
\[ v_1 = g t_0 = g \cdot \frac{2\sqrt{h}}{g} = 2\sqrt{gh} \]
The velocity of the second particle just before collision (upward):
\[ v_2 = \sqrt{gh} - g t_0 = \sqrt{gh} - g \cdot \frac{2\sqrt{h}}{g} = -\sqrt{gh} \]

### Step 5: Use conservation of momentum for the inelastic collision
The total momentum before collision:
\[ m v_1 + 2m v_2 = m(2\sqrt{gh}) + 2m(-\sqrt{gh}) = m(2\sqrt{gh} - 2\sqrt{gh}) = 0 \]
After the collision, the combined mass \( 3m \) has zero velocity.

### Step 6: Determine the time taken for the combined mass to reach the ground
Since the combined mass has zero velocity at the moment of collision, it will fall from the height where the collision occurred. The height at the moment of collision can be calculated as:
\[ h' = h - \frac{1}{2} g t_0^2 = h - \frac{1}{2} g \left(\frac{2\sqrt{h}}{g}\right)^2 = h - \frac{2h}{g} = \frac{h}{2} \]
Now, using the equation of motion to find the time \( t_f \) to fall from height \( h' \):
\[ h' = \frac{1}{2} g t_f^2 \implies \frac{h}{2} = \frac{1}{2} g t_f^2 \implies t_f^2 = \frac{h}{g} \implies t_f = \sqrt{\frac{h}{g}} \]

### Step 7: Total time to reach the ground
The total time taken for the combined mass to reach the ground is:
\[ t_{total} = t_0 + t_f = \frac{2\sqrt{h}}{g} + \sqrt{\frac{h}{g}} = \frac{2\sqrt{h}}{g} + \frac{\sqrt{h}}{g} = \frac{3\sqrt{h}}{g} \]

### Final Answer
Expressing the total time in units of \( \sqrt{\frac{h}{g}} \):
\[ t_{total} = 3 \sqrt{\frac{h}{g}} \]