Question:

A particle of mass 10 g is moving towards east with a velocity of $10 \text{ ms}^{-1}$ and another particle of mass 15 g is moving towards north with a velocity of $5 \text{ ms}^{-1}$. The magnitude of the velocity of the ntre of mass of the system of the two particles is

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Tip: Use vector components to calculate velocity of centre of mass. Final result is a vector, so apply Pythagoras to find magnitude.
Updated On: May 20, 2025
  • $5 \text{ ms}^{-1}$
  • $10 \text{ ms}^{-1}$
  • $15 \text{ ms}^{-1}$
  • $7.5 \text{ ms}^{-1}$
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The Correct Option is A

Approach Solution - 1

Let $m_1 = 10 \text{ g} = 0.01 \text{ kg}, \vec{v}_1 = 10\hat{i}$, and $m_2 = 15 \text{ g} = 0.015 \text{ kg}, \vec{v}_2 = 5\hat{j}$. \[ \vec{v}_{CM} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2} = \frac{0.01 \cdot 10\hat{i} + 0.015 \cdot 5\hat{j}}{0.025} = \frac{0.1\hat{i} + 0.075\hat{j}}{0.025} = 4\hat{i} + 3\hat{j} \] Now calculate the magnitude: \[ |\vec{v}_{CM}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ ms}^{-1} \]
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Approach Solution -2

Step 1: Identify given data
Mass of first particle, \(m_1 = 10 \text{ g} = 0.01 \text{ kg}\)
Velocity of first particle, \(\vec{v}_1 = 10 \text{ ms}^{-1}\) towards east
Mass of second particle, \(m_2 = 15 \text{ g} = 0.015 \text{ kg}\)
Velocity of second particle, \(\vec{v}_2 = 5 \text{ ms}^{-1}\) towards north

Step 2: Calculate total mass of the system
\[ M = m_1 + m_2 = 0.01 + 0.015 = 0.025 \text{ kg} \]

Step 3: Calculate momentum components
Momentum of first particle:
\[ \vec{p}_1 = m_1 \vec{v}_1 = 0.01 \times 10 = 0.1 \text{ kg m/s (east)} \]
Momentum of second particle:
\[ \vec{p}_2 = m_2 \vec{v}_2 = 0.015 \times 5 = 0.075 \text{ kg m/s (north)} \]

Step 4: Calculate velocity of center of mass (CM)
Velocity components of CM:
\[ v_{x} = \frac{p_1}{M} = \frac{0.1}{0.025} = 4 \text{ ms}^{-1} \]
\[ v_{y} = \frac{p_2}{M} = \frac{0.075}{0.025} = 3 \text{ ms}^{-1} \]

Step 5: Calculate magnitude of CM velocity
\[ v_{CM} = \sqrt{v_x^2 + v_y^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ ms}^{-1} \]

Step 6: Conclusion
The magnitude of the velocity of the center of mass of the system is \(5 \text{ ms}^{-1}\).
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