Question

A body of mass 2 kg is moving towards north with a velocity of 20 m/s and another body of mass 3 kg is moving towards east with a velocity of 10 m/s. The magnitude of the velocity of the centre of mass of the system of the two bodies is

• A. 20 m/s
• B. 10 m/s
• C. 15 m/s
• D. \(2 \sqrt{5}\) m/s

Hint:

The velocity of the centre of mass of a system is the mass-weighted average of the velocities of the individual bodies. Use vector addition to find the resultant velocity magnitude.

Answer 1

The Correct Option is B

Given: Mass of body 1, \(m_1 = 2\, kg\), velocity \(v_1 = 20\, m/s\) (north)
Mass of body 2, \(m_2 = 3\, kg\), velocity \(v_2 = 10\, m/s\) (east)
Velocity of centre of mass \(\vec{V}_{cm}\) components: \[ V_{cm,x} = \frac{m_1 v_{1x} + m_2 v_{2x}}{m_1 + m_2}, \quad V_{cm,y} = \frac{m_1 v_{1y} + m_2 v_{2y}}{m_1 + m_2} \] Assigning coordinate axes: north as y-axis, east as x-axis, we get: \[ v_{1x} = 0, \quad v_{1y} = 20 \] \[ v_{2x} = 10, \quad v_{2y} = 0 \] Calculate components: \[ V_{cm,x} = \frac{0 + 3 \times 10}{5} = \frac{30}{5} = 6\, m/s \] \[ V_{cm,y} = \frac{2 \times 20 + 0}{5} = \frac{40}{5} = 8\, m/s \] Magnitude of \(\vec{V}_{cm}\): \[ |\vec{V}_{cm}| = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\, m/s \]