During hockey practice, two pucks are sliding across the ice in the same direction. At one instant, a 0.18-kg puck is moving at 16 m/s while the other puck has a mass of 0.14 kg and a speed of 3.8 m/s. What is the velocity of the center of mass of the two pucks?

Options

A. 5.0 m/s
B. 9.0 m/s
C. 7.0 m/s
D. 13 m/s

Answer

To find the velocity of the center of mass of the two pucks, we can use the formula for the velocity of the center of mass (V_cm) of a system of particles:

\[ V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]

Where:
- \(m_1\) and \(m_2\) are the masses of the two pucks.
- \(v_1\) and \(v_2\) are the velocities of the two pucks.

### Step-by-step Solution:

1. **Identify the masses and velocities:**
- Mass of the first puck, \(m_1 = 0.18 \, \text{kg}\)
- Velocity of the first puck, \(v_1 = 16 \, \text{m/s}\)
- Mass of the second puck, \(m_2 = 0.14 \, \text{kg}\)
- Velocity of the second puck, \(v_2 = 3.8 \, \text{m/s}\)

2. **Substitute the values into the formula:**
\[ V_{cm} = \frac{(0.18 \, \text{kg} \times 16 \, \text{m/s}) + (0.14 \, \text{kg} \times 3.8 \, \text{m/s})}{0.18 \, \text{kg} + 0.14 \, \text{kg}} \]

3. **Calculate the numerator:**
- For the first puck:
\[ 0.18 \, \text{kg} \times 16 \, \text{m/s} = 2.88 \, \text{kg m/s} \]
- For the second puck:
\[ 0.14 \, \text{kg} \times 3.8 \, \text{m/s} = 0.532 \, \text{kg m/s} \]
- Adding these results:
\[ 2.88 \, \text{kg m/s} + 0.532 \, \text{kg m/s} = 3.412 \, \text{kg m/s} \]

4. **Calculate the denominator:**
\[ m_1 + m_2 = 0.18 \, \text{kg} + 0.14 \, \text{kg} = 0.32 \, \text{kg} \]

5. **Calculate the velocity of the center of mass:**
\[ V_{cm} = \frac{3.412 \, \text{kg m/s}}{0.32 \, \text{kg}} = 10.66 \, \text{m/s} \]

### Final Answer:
The velocity of the center of mass of the two pucks is \(10.66 \, \text{m/s}\).