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

• A. $3.5\text{ ms}^{-1}$
• B. $2\text{ ms}^{-1}$
• C. $2.5\text{ ms}^{-1}$
• D. $3\text{ ms}^{-1}$

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Write the centre of mass velocity rule.
The velocity of the centre of mass is the mass-weighted average of all the velocities: \[ \vec{v}_{cm} = \frac{m_{1}\vec{v}_{1} + m_{2}\vec{v}_{2} + m_{3}\vec{v}_{3}}{m_{1}+m_{2}+m_{3}} \]
Step 2: Use the perpendicular directions as axes.
The three motions are at right angles, so we can call them $\hat{i}$, $\hat{j}$, $\hat{k}$. So $\vec{v}_{1}=3\hat{i}$, $\vec{v}_{2}=4\hat{j}$, $\vec{v}_{3}=3\hat{k}$.

Step 3: Add up the masses.
Total mass $= 2m + m + 4m = 7m$.

Step 4: Build the numerator.
\[ (2m)(3\hat{i}) + (m)(4\hat{j}) + (4m)(3\hat{k}) = 6m\hat{i} + 4m\hat{j} + 12m\hat{k} \]
Step 5: Divide by total mass.
\[ \vec{v}_{cm} = \frac{6\hat{i} + 4\hat{j} + 12\hat{k}}{7} \]
Step 6: Take the magnitude.
\[ |\vec{v}_{cm}| = \frac{\sqrt{6^{2}+4^{2}+12^{2}}}{7} = \frac{\sqrt{196}}{7} = \frac{14}{7} = 2 \text{ m s}^{-1} \]This is option 2.
\[ \boxed{2 \text{ m s}^{-1}} \]