Two small bodies of masses 10 kg and 20 kg are kept a distance 1.0 m apart and released. Assuming that only mutual gravitational force are acting, find the speeds of the particles when the separation decreases to 0.5 m.

Explanation

To solve the problem of finding the speeds of two small bodies of masses 10 kg and 20 kg when the separation decreases from 1.0 m to 0.5 m, we can follow these steps:

### Step 1: Understand the Initial Conditions
We have two masses:
- \( m_1 = 10 \, \text{kg} \)
- \( m_2 = 20 \, \text{kg} \)
The initial distance between them is \( r_1 = 1.0 \, \text{m} \) and the final distance is \( r_2 = 0.5 \, \text{m} \).

### Step 2: Apply Conservation of Momentum
Since the only forces acting are gravitational and internal, the total linear momentum of the system remains constant. Initially, both bodies are at rest, so:
\[ m_1 v_1 + m_2 v_2 = 0 \]
This implies:
\[ m_1 v_1 = -m_2 v_2 \]
Let \( v_1 \) be the velocity of mass \( m_1 \) and \( v_2 \) be the velocity of mass \( m_2 \). We can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = -\frac{m_2}{m_1} v_2 \]
Substituting the values:
\[ v_1 = -\frac{20}{10} v_2 = -2 v_2 \]

### Step 3: Calculate the Change in Gravitational Potential Energy
The gravitational potential energy (U) is given by:
\[ U = -\frac{G m_1 m_2}{r} \]
Where \( G \) is the gravitational constant \( (6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \).

1. Calculate the initial potential energy at \( r_1 = 1.0 \, \text{m} \):
\[ U_1 = -\frac{(6.67 \times 10^{-11}) \cdot (10) \cdot (20)}{1.0} = -1.334 \times 10^{-9} \, \text{J} \]

2. Calculate the potential energy at \( r_2 = 0.5 \, \text{m} \):
\[ U_2 = -\frac{(6.67 \times 10^{-11}) \cdot (10) \cdot (20)}{0.5} = -2.668 \times 10^{-9} \, \text{J} \]

### Step 4: Apply Conservation of Energy
The change in potential energy will be converted into kinetic energy:
\[ \Delta U = K.E. \]
The change in potential energy is:
\[ \Delta U = U_2 - U_1 = -2.668 \times 10^{-9} - (-1.334 \times 10^{-9}) = -1.334 \times 10^{-9} \, \text{J} \]

The total kinetic energy when the separation is 0.5 m is:
\[ K.E. = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

Substituting \( v_1 = 2 v_2 \):
\[ K.E. = \frac{1}{2} (10) (2v_2)^2 + \frac{1}{2} (20) v_2^2 \]
\[ = 20 v_2^2 + 10 v_2^2 = 30 v_2^2 \]

### Step 5: Set Up the Equation
Now we can set the kinetic energy equal to the change in potential energy:
\[ 30 v_2^2 = 1.334 \times 10^{-9} \]

### Step 6: Solve for \( v_2 \)
\[ v_2^2 = \frac{1.334 \times 10^{-9}}{30} \]
\[ v_2^2 = 4.44667 \times 10^{-11} \]
\[ v_2 = \sqrt{4.44667 \times 10^{-11}} \approx 2.11 \times 10^{-5} \, \text{m/s} \]

### Step 7: Calculate \( v_1 \)
Using \( v_1 = 2 v_2 \):
\[ v_1 = 2 \times 2.11 \times 10^{-5} \approx 4.22 \times 10^{-5} \, \text{m/s} \]

### Final Answers
- \( v_1 \approx 4.22 \times 10^{-5} \, \text{m/s} \)
- \( v_2 \approx 2.11 \times 10^{-5} \, \text{m/s} \)