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 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 will use the principles of conservation of momentum and conservation of energy.

### Step-by-Step Solution:

**Step 1: Define the system and initial conditions.**
- Let mass \( m_1 = 10 \, \text{kg} \) and \( m_2 = 20 \, \text{kg} \).
- The initial separation \( R_i = 1.0 \, \text{m} \).
- The final separation \( R_f = 0.5 \, \text{m} \).
- Initially, both masses are at rest, so their initial velocities \( v_1 = 0 \) and \( v_2 = 0 \).

**Step 2: Apply conservation of momentum.**
- The total initial momentum \( P_i = m_1 \cdot 0 + m_2 \cdot 0 = 0 \).
- Let the final velocities be \( v_1 \) for mass 1 and \( v_2 \) for mass 2.
- According to the conservation of momentum:
\[ P_f = m_1 v_1 + m_2 v_2 = 0 \]
This implies:
\[ 10 v_1 + 20 v_2 = 0 \quad \Rightarrow \quad v_1 = -2 v_2 \]

**Step 3: Apply conservation of energy.**
- The initial potential energy \( U_i \) when the masses are 1 m apart is given by:
\[ U_i = -\frac{G m_1 m_2}{R_i} = -\frac{(6.67 \times 10^{-11}) \cdot (10) \cdot (20)}{1} = -1.334 \times 10^{-9} \, \text{J} \]
- The final potential energy \( U_f \) when the masses are 0.5 m apart is:
\[ U_f = -\frac{G m_1 m_2}{R_f} = -\frac{(6.67 \times 10^{-11}) \cdot (10) \cdot (20)}{0.5} = -2.668 \times 10^{-9} \, \text{J} \]
- The change in potential energy \( \Delta U = U_f - U_i \):
\[ \Delta U = -2.668 \times 10^{-9} - (-1.334 \times 10^{-9}) = -1.334 \times 10^{-9} \, \text{J} \]
- The kinetic energy \( K \) of the two masses when they are at 0.5 m apart is:
\[ K = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
- By conservation of energy:
\[ \Delta U + K = 0 \quad \Rightarrow \quad K = -\Delta U = 1.334 \times 10^{-9} \, \text{J} \]

**Step 4: Substitute \( v_1 \) in terms of \( v_2 \) into the kinetic energy equation.**
- Substitute \( v_1 = -2 v_2 \):
\[ K = \frac{1}{2} (10)(-2 v_2)^2 + \frac{1}{2} (20)(v_2)^2 \]
\[ K = \frac{1}{2} (10)(4 v_2^2) + \frac{1}{2} (20)(v_2^2) = 20 v_2^2 + 10 v_2^2 = 30 v_2^2 \]
- Set this equal to the kinetic energy we found:
\[ 30 v_2^2 = 1.334 \times 10^{-9} \]
\[ v_2^2 = \frac{1.334 \times 10^{-9}}{30} = 4.44667 \times 10^{-11} \]
\[ v_2 = \sqrt{4.44667 \times 10^{-11}} \approx 2.11 \times 10^{-5} \, \text{m/s} \]

**Step 5: 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 Results:
- Speed of mass 1 (\( v_1 \)): \( 4.22 \times 10^{-5} \, \text{m/s} \)
- Speed of mass 2 (\( v_2 \)): \( 2.11 \times 10^{-5} \, \text{m/s} \)