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 velocities of two balls A and B just before they reach the surface of the Earth and then determine the ratio of these velocities.

### Step 1: Calculate the velocity of ball A
Ball A of mass \( m \) falls from infinity. We can use the conservation of mechanical energy to find its velocity just before it hits the Earth.

- **Initial Energy (at infinity)**:
- Potential Energy (PE) = 0 (since it is at infinity)
- Kinetic Energy (KE) = 0 (since it starts from rest)

- **Final Energy (just before hitting the Earth)**:
- Potential Energy (PE) = \( -\frac{GMm}{R} \) (where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth)
- Kinetic Energy (KE) = \( \frac{1}{2} mv_A^2 \)

Using conservation of energy:
\[ \text{Initial Energy} = \text{Final Energy} \]
\[ 0 = -\frac{GMm}{R} + \frac{1}{2} mv_A^2 \]
Rearranging gives:
\[ \frac{1}{2} mv_A^2 = \frac{GMm}{R} \]
Dividing by \( m \) and multiplying by 2:
\[ v_A^2 = \frac{2GM}{R} \]
Thus,
\[ v_A = \sqrt{\frac{2GM}{R}} \]

### Step 2: Calculate the velocity of ball B
Ball B of mass \( 2m \) falls from a height of \( 6R \) (which is 6 times the radius of the Earth).

- **Initial Energy (at height \( 6R \))**:
- Potential Energy (PE) = \( -\frac{GM(2m)}{7R} \) (the distance from the center of the Earth is \( 7R \))
- Kinetic Energy (KE) = 0 (since it starts from rest)

- **Final Energy (just before hitting the Earth)**:
- Potential Energy (PE) = \( -\frac{GM(2m)}{R} \)
- Kinetic Energy (KE) = \( \frac{1}{2} (2m) v_B^2 \)

Using conservation of energy:
\[ \text{Initial Energy} = \text{Final Energy} \]
\[ -\frac{GM(2m)}{7R} = -\frac{GM(2m)}{R} + \frac{1}{2} (2m) v_B^2 \]
Rearranging gives:
\[ -\frac{GM(2m)}{7R} + \frac{GM(2m)}{R} = mv_B^2 \]
Finding a common denominator:
\[ \frac{-2GM}{7R} + \frac{14GM}{7R} = mv_B^2 \]
This simplifies to:
\[ \frac{12GM}{7R} = mv_B^2 \]
Dividing by \( m \):
\[ v_B^2 = \frac{12GM}{7R} \]
Thus,
\[ v_B = \sqrt{\frac{12GM}{7R}} \]

### Step 3: Calculate the ratio of velocities
Now we can find the ratio of the velocities \( \frac{v_A}{v_B} \):
\[ \frac{v_A}{v_B} = \frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{12GM}{7R}}} \]
This simplifies to:
\[ \frac{v_A}{v_B} = \sqrt{\frac{2GM}{R} \cdot \frac{7R}{12GM}} = \sqrt{\frac{14}{12}} = \sqrt{\frac{7}{6}} \]

### Step 4: Identify \( x \) and \( y \)
Here, \( x = 7 \) and \( y = 6 \). Since 7 and 6 are coprime, we can find \( x + y \):
\[ x + y = 7 + 6 = 13 \]

### Final Answer
Thus, the final answer is \( 13 \).