Question:
If three particles of masses $2m$, $m$ and $4m$ are moving in three mutually perpendicular directions with velocities $3\text{ ms}^{-1}$, $4\text{ ms}^{-1}$ and $3\text{ ms}^{-1}$ respectively, then the magnitude of the velocity of the center of mass of the system of three particles is
If three particles of masses $2m$, $m$ and $4m$ are moving in three mutually perpendicular directions with velocities $3\text{ ms}^{-1}$, $4\text{ ms}^{-1}$ and $3\text{ ms}^{-1}$ respectively, then the magnitude of the velocity of the center of mass of the system of three particles is
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Assigning mutually perpendicular directions to orthogonal unit vectors ($\hat{i}, \hat{j}, \hat{k}$) converts a multi-dimensional physics problem into a simple vector magnitude calculation.
Updated On: Jun 3, 2026
- $3.5\text{ ms}^{-1}$
- $2\text{ ms}^{-1}$
- $2.5\text{ ms}^{-1}$
- $3\text{ ms}^{-1}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The velocity vector of the center of mass for a system of multiple particles is given by the weighted vector sum formula: $\vec{v}_{\text{cm}} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2 + m_3\vec{v}_3}{m_1 + m_2 + m_3}$.
Step 2: Meaning
Since the three directions are mutually perpendicular, we can align them along the standard Cartesian coordinate axes $\hat{i}$, $\hat{j}$, and $\hat{k}$: $\vec{v}_1 = 3\hat{i}$, $\vec{v}_2 = 4\hat{j}$, and $\vec{v}_3 = 3\hat{k}$. The total mass is $M = 2m + m + 4m = 7m$.
Step 3: Analysis
Now construct the center of mass velocity vector: $\vec{v}_{\text{cm}} = \frac{(2m)(3\hat{i}) + (m)(4\hat{j}) + (4m)(3\hat{k})}{7m} = \frac{6\hat{i} + 4\hat{j} + 12\hat{k}}{7}$. Now calculate the magnitude of this vector: $|\vec{v}_{\text{cm}}| = \frac{\sqrt{6^2 + 4^2 + 12^2}}{7} = \frac{\sqrt{36 + 16 + 144}}{7} = \frac{\sqrt{196}}{7} = \frac{14}{7} = 2\text{ ms}^{-1}$.
Step 4: Conclusion
The calculated value is exactly $2\text{ ms}^{-1}$, which matches option (B).
Final Answer: (B)
The velocity vector of the center of mass for a system of multiple particles is given by the weighted vector sum formula: $\vec{v}_{\text{cm}} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2 + m_3\vec{v}_3}{m_1 + m_2 + m_3}$.
Step 2: Meaning
Since the three directions are mutually perpendicular, we can align them along the standard Cartesian coordinate axes $\hat{i}$, $\hat{j}$, and $\hat{k}$: $\vec{v}_1 = 3\hat{i}$, $\vec{v}_2 = 4\hat{j}$, and $\vec{v}_3 = 3\hat{k}$. The total mass is $M = 2m + m + 4m = 7m$.
Step 3: Analysis
Now construct the center of mass velocity vector: $\vec{v}_{\text{cm}} = \frac{(2m)(3\hat{i}) + (m)(4\hat{j}) + (4m)(3\hat{k})}{7m} = \frac{6\hat{i} + 4\hat{j} + 12\hat{k}}{7}$. Now calculate the magnitude of this vector: $|\vec{v}_{\text{cm}}| = \frac{\sqrt{6^2 + 4^2 + 12^2}}{7} = \frac{\sqrt{36 + 16 + 144}}{7} = \frac{\sqrt{196}}{7} = \frac{14}{7} = 2\text{ ms}^{-1}$.
Step 4: Conclusion
The calculated value is exactly $2\text{ ms}^{-1}$, which matches option (B).
Final Answer: (B)
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