To solve the problem, we will break it down into two parts as stated in the question.
### Part (a): Finding the distance of the center of mass below the release point at \( t = 300 \, \text{ms} \)
1. **Identify the masses and times**:
- Let the mass of the first stone (Stone A) be \( m \).
- The mass of the second stone (Stone B) is \( 2m \).
- Stone A is dropped at \( t = 0 \, \text{ms} \).
- Stone B is dropped at \( t = 100 \, \text{ms} \).
2. **Calculate the time each stone has been falling at \( t = 300 \, \text{ms} \)**:
- For Stone A:
\[
t_1 = 300 \, \text{ms} - 0 \, \text{ms} = 300 \, \text{ms} = 0.3 \, \text{s}
\]
- For Stone B:
\[
t_2 = 300 \, \text{ms} - 100 \, \text{ms} = 200 \, \text{ms} = 0.2 \, \text{s}
\]
3. **Calculate the distance fallen by each stone**:
- Using the formula for distance fallen under gravity:
\[
d = \frac{1}{2} g t^2
\]
- For Stone A:
\[
d_1 = \frac{1}{2} \cdot 10 \, \text{m/s}^2 \cdot (0.3 \, \text{s})^2 = \frac{1}{2} \cdot 10 \cdot 0.09 = 0.45 \, \text{m}
\]
- For Stone B:
\[
d_2 = \frac{1}{2} \cdot 10 \, \text{m/s}^2 \cdot (0.2 \, \text{s})^2 = \frac{1}{2} \cdot 10 \cdot 0.04 = 0.2 \, \text{m}
\]
4. **Calculate the center of mass (CM) position**:
- The formula for the center of mass \( d_{cm} \) of two masses is:
\[
d_{cm} = \frac{m_1 d_1 + m_2 d_2}{m_1 + m_2}
\]
- Substituting the values:
\[
d_{cm} = \frac{m \cdot 0.45 + 2m \cdot 0.2}{m + 2m} = \frac{0.45m + 0.4m}{3m} = \frac{0.85m}{3m} = \frac{0.85}{3} \approx 0.2833 \, \text{m} = 28.3 \, \text{cm}
\]
### Part (b): Finding the velocity of the center of mass at \( t = 300 \, \text{ms} \)
1. **Calculate the velocity of each stone**:
- The velocity of an object falling under gravity is given by:
\[
v = gt
\]
- For Stone A:
\[
v_1 = g \cdot t_1 = 10 \, \text{m/s}^2 \cdot 0.3 \, \text{s} = 3 \, \text{m/s}
\]
- For Stone B:
\[
v_2 = g \cdot t_2 = 10 \, \text{m/s}^2 \cdot 0.2 \, \text{s} = 2 \, \text{m/s}
\]
2. **Calculate the velocity of the center of mass**:
- Using the formula for the velocity of the center of mass:
\[
v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}
\]
- Substituting the values:
\[
v_{cm} = \frac{m \cdot 3 + 2m \cdot 2}{m + 2m} = \frac{3m + 4m}{3m} = \frac{7m}{3m} = \frac{7}{3} \approx 2.33 \, \text{m/s}
\]
### Final Answers:
(a) The center of mass is approximately \( 28.3 \, \text{cm} \) below the release point.
(b) The velocity of the center of mass is approximately \( 2.33 \, \text{m/s} \).
