Question

Three particles A, B, and C of masses \( m \), \( 2m \), and \( 3m \) are moving towards north, south, and east, respectively. If the velocities of the particles A, B, and C are \( 6 \) m/s, \( 12 \) m/s, and \( 8 \) m/s respectively, then the velocity of the center of mass of the system of particles is:

• A. \( 7 \) m/s
• B. \( 5 \) m/s
• C. \( 26 \) m/s
• D. \( 8 \) m/s

Hint:

The velocity of the center of mass is computed separately for x and y components. - The net velocity is obtained using the Pythagorean theorem.

Answer 1

The Correct Option is B

Step 1: Compute velocity of center of mass
The velocity of the center of mass is given by: \[ V_{cm} = \frac{\sum m_i v_i}{\sum m_i}. \] The x-component is: \[ V_{cm,x} = \frac{3m \times 8}{m + 2m + 3m} = \frac{24m}{6m} = 4 \text{ m/s}. \] The y-component is: \[ V_{cm,y} = \frac{m (6) + 2m (-12)}{6m} = \frac{6m - 24m}{6m} = -3 \text{ m/s}. \] Step 2: Compute magnitude of \( V_{cm} \)
\[ V_{cm} = \sqrt{(V_{cm,x})^2 + (V_{cm,y})^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ m/s}. \] Thus, the correct answer is \( \boxed{5} \) m/s.