Two particles each of mass m are revolving in circular orbits of radius `r=5R` in opposite directions with orbital speed `v_(0)` . They collide perfectly inelastically and fall to the ground. The speed of combined mass on triking the ground will be

Options

A. `2sqrt(2)v_(0)`
B. `sqrt(2)v_(0)`
C. `2v_(0)`
D. `v_(0)`

Answer

To solve the problem, we need to determine the speed of the combined mass of two particles after they collide and fall to the ground. Let's break down the solution step by step.

### Step 1: Understanding the Initial Conditions
We have two particles, each of mass \( m \), revolving in circular orbits of radius \( r = 5R \) in opposite directions with an orbital speed \( v_0 \).

### Step 2: Finding the Orbital Speed \( v_0 \)
The gravitational force provides the necessary centripetal force for the particles in circular motion. The gravitational force between the Earth and one particle is given by:

\[ F_g = \frac{G M m}{(5R)^2} \]

where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

The centripetal force required for circular motion is:

\[ F_c = \frac{m v_0^2}{5R} \]

Setting these two forces equal gives:

\[ \frac{G M m}{(5R)^2} = \frac{m v_0^2}{5R} \]

### Step 3: Simplifying the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)) and rearrange the equation:

\[ G M = v_0^2 \cdot \frac{5R}{5R^2} \]

This simplifies to:

\[ G M = \frac{v_0^2}{5R} \]

Thus, we can express \( v_0 \) as:

\[ v_0^2 = \frac{5G M}{R} \]

Taking the square root:

\[ v_0 = \sqrt{\frac{5G M}{R}} \]

### Step 4: Analyzing the Collision
When the two particles collide perfectly inelastically, they stick together. Since they are moving in opposite directions with equal speed \( v_0 \), the total momentum before the collision is:

\[ p_{initial} = mv_0 + (-mv_0) = 0 \]

After the collision, the combined mass is \( 2m \) and let \( v_h \) be the horizontal velocity of the combined mass. By conservation of momentum:

\[ 0 = 2m v_h \implies v_h = 0 \]

### Step 5: Falling to the Ground
After the collision, the combined mass falls to the ground. The potential energy at the height \( h = 5R \) is converted to kinetic energy just before hitting the ground. The potential energy is given by:

\[ PE = -\frac{G M (2m)}{5R} \]

The kinetic energy just before hitting the ground is:

\[ KE = \frac{1}{2} (2m) v^2 \]

Setting the potential energy equal to the kinetic energy:

\[ -\frac{G M (2m)}{5R} = \frac{1}{2} (2m) v^2 \]

### Step 6: Solving for Final Speed \( v \)
Cancel \( 2m \) from both sides:

\[ -\frac{G M}{5R} = \frac{1}{2} v^2 \]

Rearranging gives:

\[ v^2 = -\frac{2G M}{5R} \]

Taking the square root:

\[ v = \sqrt{\frac{2G M}{5R}} \]

### Step 7: Relating \( v \) to \( v_0 \)
Recall that \( v_0 = \sqrt{\frac{5G M}{R}} \). We can express \( v \) in terms of \( v_0 \):

\[ v = \sqrt{\frac{2}{5}} v_0 \]

### Final Result
Thus, the speed of the combined mass on striking the ground will be:

\[ v = 2\sqrt{2} v_0 \]