To solve the problem step by step, we will break it down into parts (a) and (b) as described.
### Part (a): Finding the position of the center of mass at `t = 300 ms`
1. **Identify the time intervals for each stone:**
- Stone 1 is dropped at `t = 0 ms` and has been falling for `300 ms`.
- Stone 2 is dropped at `t = 100 ms` and has been falling for `200 ms` by `t = 300 ms`.
2. **Calculate the displacement of Stone 1 (y1):**
- The formula for displacement under constant acceleration is:
\[
y = ut + \frac{1}{2} a t^2
\]
- For Stone 1:
- Initial velocity \( u = 0 \)
- Acceleration \( a = g = 10 \, \text{m/s}^2 \)
- Time \( t = 0.3 \, \text{s} = 300 \, \text{ms} \)
- Plugging in the values:
\[
y_1 = 0 + \frac{1}{2} \cdot 10 \cdot (0.3)^2 = 0.5 \cdot 10 \cdot 0.09 = 0.45 \, \text{m}
\]
- Since it is downward, we take it as:
\[
y_1 = -0.45 \, \text{m}
\]
3. **Calculate the displacement of Stone 2 (y2):**
- For Stone 2:
- It has fallen for \( 200 \, \text{ms} = 0.2 \, \text{s} \).
- Using the same formula:
\[
y_2 = 0 + \frac{1}{2} \cdot 10 \cdot (0.2)^2 = 0.5 \cdot 10 \cdot 0.04 = 0.20 \, \text{m}
\]
- Again, since it is downward:
\[
y_2 = -0.20 \, \text{m}
\]
4. **Calculate the center of mass (SCM):**
- The formula for the center of mass of two objects is:
\[
SCM = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}
\]
- Let \( m_1 = m \) and \( m_2 = 2m \):
\[
SCM = \frac{m(-0.45) + 2m(-0.20)}{m + 2m} = \frac{-0.45m - 0.40m}{3m} = \frac{-0.85m}{3m} = -\frac{0.85}{3} \, \text{m}
\]
- Therefore:
\[
SCM = -0.283 \, \text{m}
\]
### Part (b): Finding the velocity of the center of mass at `t = 300 ms`
1. **Calculate the velocity of Stone 1 (v1):**
- Using the formula:
\[
v = u + at
\]
- For Stone 1:
\[
v_1 = 0 + 10 \cdot 0.3 = 3 \, \text{m/s}
\]
- Direction is downward, so:
\[
v_1 = -3 \, \text{m/s}
\]
2. **Calculate the velocity of Stone 2 (v2):**
- For Stone 2, which has fallen for \( 200 \, \text{ms} \):
\[
v_2 = 0 + 10 \cdot 0.2 = 2 \, \text{m/s}
\]
- Direction is downward, so:
\[
v_2 = -2 \, \text{m/s}
\]
3. **Calculate the velocity of the center of mass (Vcm):**
- Using the formula:
\[
V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}
\]
- Substituting the values:
\[
V_{cm} = \frac{m(-3) + 2m(-2)}{m + 2m} = \frac{-3m - 4m}{3m} = \frac{-7m}{3m} = -\frac{7}{3} \, \text{m/s}
\]
- Therefore:
\[
V_{cm} = -2.33 \, \text{m/s}
\]
### Final Answers:
(a) The center of mass is at \( -0.283 \, \text{m} \) below the release point.
(b) The velocity of the center of mass is \( -2.33 \, \text{m/s} \).
