Particles of masses `1kg` and `3kg` are moving with the velocities `(10i-7j-3k)ms^(-1)` and `(7i-9j+6k)ms^(-1)` then the instantaneous velocity of their centre of mass is

Options

A. `(1)/(4)(31i-34j+15k)ms^(-1)`
B. `(1)/(4)(30i-32j+15k)ms^(-1)`
C. `(1)/(4)(25i-28j+15k)ms^(-1)`
D. `(1)/(4)(31i-34j+20k)ms^(-1)`

Answer

To find the instantaneous velocity of the center of mass of two particles with given masses and velocities, we can follow these steps:

### Step 1: Identify the masses and velocities
- Mass of particle 1, \( m_1 = 1 \, \text{kg} \)
- Velocity of particle 1, \( \mathbf{v_1} = (10\mathbf{i} - 7\mathbf{j} - 3\mathbf{k}) \, \text{m/s} \)
- Mass of particle 2, \( m_2 = 3 \, \text{kg} \)
- Velocity of particle 2, \( \mathbf{v_2} = (7\mathbf{i} - 9\mathbf{j} + 6\mathbf{k}) \, \text{m/s} \)

### Step 2: Use the formula for the velocity of the center of mass
The formula for the velocity of the center of mass (\( \mathbf{V_{cm}} \)) is given by:

\[ \mathbf{V_{cm}} = \frac{m_1 \mathbf{v_1} + m_2 \mathbf{v_2}}{m_1 + m_2} \]

### Step 3: Substitute the values into the formula
Substituting the values we have:

\[ \mathbf{V_{cm}} = \frac{1 \cdot (10\mathbf{i} - 7\mathbf{j} - 3\mathbf{k}) + 3 \cdot (7\mathbf{i} - 9\mathbf{j} + 6\mathbf{k})}{1 + 3} \]

### Step 4: Calculate the numerator
Calculating the individual components:

1. For \( m_1 \mathbf{v_1} \):
\[ 1 \cdot (10\mathbf{i} - 7\mathbf{j} - 3\mathbf{k}) = 10\mathbf{i} - 7\mathbf{j} - 3\mathbf{k} \]

2. For \( m_2 \mathbf{v_2} \):
\[ 3 \cdot (7\mathbf{i} - 9\mathbf{j} + 6\mathbf{k}) = 21\mathbf{i} - 27\mathbf{j} + 18\mathbf{k} \]

Now, adding these two results together:

\[ (10\mathbf{i} - 7\mathbf{j} - 3\mathbf{k}) + (21\mathbf{i} - 27\mathbf{j} + 18\mathbf{k}) = (10 + 21)\mathbf{i} + (-7 - 27)\mathbf{j} + (-3 + 18)\mathbf{k} \]

This simplifies to:

\[ 31\mathbf{i} - 34\mathbf{j} + 15\mathbf{k} \]

### Step 5: Divide by the total mass
Now, divide by the total mass:

\[ \mathbf{V_{cm}} = \frac{31\mathbf{i} - 34\mathbf{j} + 15\mathbf{k}}{4} \]

### Step 6: Final result
This gives us:

\[ \mathbf{V_{cm}} = \frac{31}{4}\mathbf{i} - \frac{34}{4}\mathbf{j} + \frac{15}{4}\mathbf{k} \]

Thus, the instantaneous velocity of the center of mass is:

\[ \mathbf{V_{cm}} = 7.75\mathbf{i} - 8.5\mathbf{j} + 3.75\mathbf{k} \, \text{m/s} \]