by:

$$ v_{\text{CM}} = \frac{m_1 v_1 + m_2 v_2 + m_3 v_3}{m_1 + m_2 + m_3} $$

Given:

Masses: $ m_1 = 5 \, \text{kg}, \, m_2 = 4 \, \text{kg}, \, m_3 = 2 \, \text{kg} $

Velocities: $ v_1 = 5 \, \text{m/s}, \, v_2 = 4 \, \text{m/s}, \, v_3 = 2 \, \text{m/s} $

Substitute these values into the formula:

$ v_{\text{CM}} = \frac{(5 \, \text{kg} \times 5 \, \text{m/s}) + (4 \, \text{kg} \times 4 \, \text{m/s}) + (2 \, \text{kg} \times 2 \, \text{m/s})}{5 \, \text{kg} + 4 \, \text{kg} + 2 \, \text{kg}} $

Calculate the numerator:

$ 5 \, \text{kg} \times 5 \, \text{m/s} = 25 \, \text{kg} \cdot \text{m/s} $

$ 4 \, \text{kg} \times 4 \, \text{m/s} = 16 \, \text{kg} \cdot \text{m/s} $

$ 2 \, \text{kg} \times 2 \, \text{m/s} = 4 \, \text{kg} \cdot \text{m/s} $

Thus, the total momentum is:

$ 25 \, \text{kg} \cdot \text{m/s} + 16 \, \text{kg} \cdot \text{m/s} + 4 \, \text{kg} \cdot \text{m/s} = 45 \, \text{kg} \cdot \text{m/s} $

The total mass is:

$ 5 \, \text{kg} + 4 \, \text{kg} + 2 \, \text{kg} = 11 \, \text{kg} $

Now, calculate $ v_{\text{CM}} $:

$ v_{\text{CM}} = \frac{45 \, \text{kg} \cdot \text{m/s}}{11 \, \text{kg}} = 4.09 \, \text{m/s} $

Thus, the magnitude of the velocity of the center of mass is approximately 4.1 m/s. Choice not matching entirely the given options, however, closest is Option B 4 m/s following rounding approximations commonly in physics problem settings.

Center Of Mass And Collision

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