Particles of masses 1 kg and 3 kg at `(2i+5j+13 k)` m and `(-6i+4j-2k)m` then instantaneous position of their centre of mass is

Options

A. `(1)/(4)(-16i+17j+7k)m`
B. `(1)/(4)(-8i+17j+7k)m`
C. `(1)/(4)(-6i+17j+7k)m`
D. `(1)/(4)(-6i+17j+5k)m`

Answer

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

### Step 1: Identify the given data
- Mass of particle 1, \( m_1 = 1 \, \text{kg} \)
- Position vector of particle 1, \( \vec{r_1} = 2\hat{i} + 5\hat{j} + 13\hat{k} \, \text{m} \)
- Mass of particle 2, \( m_2 = 3 \, \text{kg} \)
- Position vector of particle 2, \( \vec{r_2} = -6\hat{i} + 4\hat{j} - 2\hat{k} \, \text{m} \)

### Step 2: Calculate the total mass
The total mass \( M \) is given by:
\[ M = m_1 + m_2 = 1 \, \text{kg} + 3 \, \text{kg} = 4 \, \text{kg} \]

### Step 3: Use the formula for the center of mass
The position vector of the center of mass \( \vec{R} \) is given by:
\[ \vec{R} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2}}{M} \]

### Step 4: Substitute the values into the formula
Substituting the values we have:
\[ \vec{R} = \frac{1 \cdot (2\hat{i} + 5\hat{j} + 13\hat{k}) + 3 \cdot (-6\hat{i} + 4\hat{j} - 2\hat{k})}{4} \]

### Step 5: Calculate the numerator
Calculating the individual components:
- For the \( \hat{i} \) component:
\[ 1 \cdot 2 + 3 \cdot (-6) = 2 - 18 = -16 \]
- For the \( \hat{j} \) component:
\[ 1 \cdot 5 + 3 \cdot 4 = 5 + 12 = 17 \]
- For the \( \hat{k} \) component:
\[ 1 \cdot 13 + 3 \cdot (-2) = 13 - 6 = 7 \]

### Step 6: Combine the components
Now we can combine these results into the numerator:
\[ \vec{R} = \frac{-16\hat{i} + 17\hat{j} + 7\hat{k}}{4} \]

### Step 7: Divide by the total mass
Now we divide each component by the total mass:
\[ \vec{R} = -4\hat{i} + \frac{17}{4}\hat{j} + \frac{7}{4}\hat{k} \]

### Final Result
Thus, the instantaneous position of the center of mass is:
\[ \vec{R} = -4\hat{i} + \frac{17}{4}\hat{j} + \frac{7}{4}\hat{k} \]