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

Explanation

To solve the problem of two small bodies of masses 10 kg and 20 kg that are initially 1.0 m apart and then come to a separation of 0.5 m, we will use the principles of conservation of momentum and conservation of energy.

### Step 1: Conservation of Momentum
The momentum of the system before the bodies are released is zero since both bodies are at rest. When they start moving towards each other due to gravitational attraction, we can express the conservation of momentum as:

\[ m_1 v_1 + m_2 v_2 = 0 \]

Where:
- \( m_1 = 10 \, \text{kg} \)
- \( m_2 = 20 \, \text{kg} \)
- \( v_1 \) is the speed of the 10 kg mass
- \( v_2 \) is the speed of the 20 kg mass

From this equation, we can express \( v_1 \) in terms of \( v_2 \):

\[ 10 v_1 + 20 v_2 = 0 \]
\[ v_1 = -2 v_2 \]

Since we are interested in speeds (magnitudes), we can ignore the negative sign:

\[ v_1 = 2 v_2 \]

### Step 2: Conservation of Energy
Next, we apply the conservation of energy. The initial potential energy when the masses are 1.0 m apart is given by:

\[ U_i = -\frac{G m_1 m_2}{r_1} \]

Where:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( r_1 = 1.0 \, \text{m} \)

Substituting the values:

\[ U_i = -\frac{(6.67 \times 10^{-11}) \cdot (10) \cdot (20)}{1} = -1.334 \times 10^{-9} \, \text{J} \]

When the bodies are at a separation of 0.5 m, the potential energy is:

\[ U_f = -\frac{G m_1 m_2}{r_2} \]

Where \( r_2 = 0.5 \, \text{m} \):

\[ U_f = -\frac{(6.67 \times 10^{-11}) \cdot (10) \cdot (20)}{0.5} = -2.668 \times 10^{-9} \, \text{J} \]

The total mechanical energy at the initial state (potential energy only) is equal to the total mechanical energy at the final state (potential energy + kinetic energy):

\[ U_i + 0 = U_f + K \]

Where \( K \) is the total kinetic energy at that moment:

\[ K = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

Substituting the values we have:

\[ -1.334 \times 10^{-9} = -2.668 \times 10^{-9} + \frac{1}{2} (10) v_1^2 + \frac{1}{2} (20) v_2^2 \]

### Step 3: Substitute \( v_1 \) in terms of \( v_2 \)
Using \( v_1 = 2 v_2 \):

\[ -1.334 \times 10^{-9} = -2.668 \times 10^{-9} + \frac{1}{2} (10) (2 v_2)^2 + \frac{1}{2} (20) v_2^2 \]

This simplifies to:

\[ -1.334 \times 10^{-9} = -2.668 \times 10^{-9} + 20 v_2^2 + 10 v_2^2 \]
\[ -1.334 \times 10^{-9} = -2.668 \times 10^{-9} + 30 v_2^2 \]

### Step 4: Solve for \( v_2^2 \)
Rearranging gives:

\[ 30 v_2^2 = -1.334 \times 10^{-9} + 2.668 \times 10^{-9} \]
\[ 30 v_2^2 = 1.334 \times 10^{-9} \]
\[ v_2^2 = \frac{1.334 \times 10^{-9}}{30} \]
\[ v_2^2 = 4.447 \times 10^{-11} \]
\[ v_2 = \sqrt{4.447 \times 10^{-11}} \]
\[ v_2 \approx 2.11 \times 10^{-5} \, \text{m/s} \]

### Step 5: Calculate \( v_1 \)
Now substituting back to find \( v_1 \):

\[ v_1 = 2 v_2 = 2 \times 2.11 \times 10^{-5} \]
\[ v_1 \approx 4.22 \times 10^{-5} \, \text{m/s} \]

### Final Speeds
Thus, the speeds of the particles when the separation decreases to 0.5 m are:
- \( v_1 \approx 4.22 \times 10^{-5} \, \text{m/s} \) (for the 10 kg mass)
- \( v_2 \approx 2.11 \times 10^{-5} \, \text{m/s} \) (for the 20 kg mass)