A ball A of mass m falls on the surface of the earth from infinity. Another ball B of mass 2m falls on the earth from the height equal to six times the radius of the earth. Then ratio of velocities of A and B on reaching the earth is `sqrt(x//y)` where x and y are coprimes. Find x+y .

Explanation

To solve the problem, we need to find the ratio of the velocities of two balls, A and B, when they reach the surface of the Earth. Let's break down the solution step by step.

### Step 1: Understand the scenario
- Ball A of mass \( m \) falls from infinity.
- Ball B of mass \( 2m \) falls from a height equal to \( 6R \), where \( R \) is the radius of the Earth.

### Step 2: Use the concept of gravitational potential energy
The change in gravitational potential energy will convert into kinetic energy as the balls fall. We can use the conservation of energy principle:
\[ \text{Potential Energy at initial height} = \text{Kinetic Energy at the surface} \]

### Step 3: Calculate the velocity of Ball A
For Ball A, which falls from infinity:
- The potential energy at infinity is 0.
- The potential energy at the surface of the Earth (at distance \( R \)) is given by:
\[ U_A = -\frac{GMm}{R} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

The kinetic energy when it reaches the surface is:
\[ K_A = \frac{1}{2}mv_A^2 \]
Setting the potential energy equal to the kinetic energy:
\[ 0 - \frac{GMm}{R} = \frac{1}{2}mv_A^2 \]
Cancelling \( m \) from both sides:
\[ -\frac{GM}{R} = \frac{1}{2}v_A^2 \]
Rearranging gives:
\[ v_A^2 = \frac{2GM}{R} \]

### Step 4: Calculate the velocity of Ball B
For Ball B, which falls from a height of \( 6R \):
- The potential energy at \( 6R \) is:
\[ U_B = -\frac{GM(2m)}{6R} \]
- The potential energy at the surface of the Earth is:
\[ U_B = -\frac{GM(2m)}{R} \]
The change in potential energy as it falls is:
\[ \Delta U = U_B \text{ (at } 6R\text{)} - U_B \text{ (at } R\text{)} = -\frac{GM(2m)}{R} + \frac{GM(2m)}{6R} \]
Calculating this gives:
\[ \Delta U = -\frac{GM(2m)}{R} + \frac{GM(2m)}{6R} = -\frac{GM(2m)}{R} \left(1 - \frac{1}{6}\right) = -\frac{GM(2m)}{R} \cdot \frac{5}{6} \]
Setting this equal to the kinetic energy:
\[ -\frac{GM(2m)}{R} \cdot \frac{5}{6} = \frac{1}{2}(2m)v_B^2 \]
Cancelling \( 2m \):
\[ -\frac{GM \cdot 5}{6R} = v_B^2 \]
Rearranging gives:
\[ v_B^2 = \frac{5GM}{3R} \]

### Step 5: Find the ratio of the velocities
Now, we can find the ratio of the velocities:
\[ \frac{v_A}{v_B} = \frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{5GM}{3R}}} \]
This simplifies to:
\[ \frac{v_A}{v_B} = \sqrt{\frac{2}{\frac{5}{3}}} = \sqrt{\frac{2 \cdot 3}{5}} = \sqrt{\frac{6}{5}} \]

### Step 6: Identify \( x \) and \( y \)
From the ratio \( \frac{v_A}{v_B} = \sqrt{\frac{6}{5}} \), we have \( x = 6 \) and \( y = 5 \).

### Step 7: Find \( x + y \)
Since \( x \) and \( y \) are coprime, we calculate:
\[ x + y = 6 + 5 = 11 \]

Thus, the final answer is:
\[ \boxed{11} \]