Answer:
Step-by-step explanation:
To solve this problem, we can use the principles of conservation of momentum and gravitational potential energy. The center of mass of a system of particles will behave as if all the mass is concentrated at that point.
Let's break down the problem step by step:
1. **Initial Conditions:**
- Ball 1 (0.50 kg) is dropped from 25 m above Earth's surface, so its initial height is 25 m.
- Ball 2 (0.25 kg) is thrown upward with an initial speed of 15 m/s, so its initial height is 0 m (Earth's surface).
2. **Time = 2.0 seconds:**
- We want to find the height of the center of mass of the two-ball system above Earth's surface.
3. **Center of Mass Position:**
- The center of mass position at any time is given by:
\[x_{\text{cm}} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2}\]
- Where \(m_1\) and \(m_2\) are the masses of the two balls, and \(x_1\) and \(x_2\) are their respective heights above Earth's surface.
4. **Calculations:**
- Ball 1's height \(x_1\) after 2.0 seconds:
\(x_1 = 25 - \frac{1}{2} g t^2\)
- Ball 2's height \(x_2\) after 2.0 seconds:
\(x_2 = v_0 t - \frac{1}{2} g t^2\), where \(v_0 = 15 \, \text{m/s}\) is the initial velocity of Ball 2.
- Substituting the values:
\(x_1 = 25 - \frac{1}{2} \cdot 9.81 \cdot (2.0)^2\) (assuming \(g = 9.81 \, \text{m/s}^2\))
\(x_2 = 15 \cdot 2.0 - \frac{1}{2} \cdot 9.81 \cdot (2.0)^2\)
- Now plug in the values into the center of mass formula:
\(x_{\text{cm}} = \frac{0.50 \cdot x_1 + 0.25 \cdot x_2}{0.50 + 0.25}\)
Calculate using the calculated values of \(x_1\) and \(x_2\). This will give you the height above Earth's surface of the center of mass of the two-ball system after 2.0 seconds.