Two bodies of mass 2 kg and 5 kg have position (1 m, 2 m, 1 m) and (3 m, 2 m, -1 m) respectively. The position vector of centre of mass is

Options

A. `(frac{17}{7}hati + 2hatj - frac{3}{7}hatk)m`
B. `(17hati + 14hatj - 3hatk)m`
C. `(2hati + frac{17}{7}hatj - frac{3}{7}hatk)m`
D. `(14hati + 17hatj - 3hatk)m`

Answer

To find the position vector of the center of mass for the two bodies, we can follow these steps:

### Step 1: Identify the Masses and Position Vectors
We have two bodies with the following properties:
- Mass \( m_1 = 2 \, \text{kg} \) at position \( \mathbf{r_1} = (1 \, \text{m}, 2 \, \text{m}, 1 \, \text{m}) \)
- Mass \( m_2 = 5 \, \text{kg} \) at position \( \mathbf{r_2} = (3 \, \text{m}, 2 \, \text{m}, -1 \, \text{m}) \)

### Step 2: Write the Position Vectors
The position vectors can be expressed in unit vector notation:
- \( \mathbf{r_1} = 1 \hat{i} + 2 \hat{j} + 1 \hat{k} \)
- \( \mathbf{r_2} = 3 \hat{i} + 2 \hat{j} - 1 \hat{k} \)

### Step 3: Use the Formula for Center of Mass
The formula for the position vector of the center of mass \( \mathbf{r_{cm}} \) is given by:
\[ \mathbf{r_{cm}} = \frac{m_1 \mathbf{r_1} + m_2 \mathbf{r_2}}{m_1 + m_2} \]

### Step 4: Calculate the Numerator
Substituting the values into the formula:
\[ \mathbf{r_{cm}} = \frac{2 \cdot (1 \hat{i} + 2 \hat{j} + 1 \hat{k}) + 5 \cdot (3 \hat{i} + 2 \hat{j} - 1 \hat{k})}{2 + 5} \]

Calculating each term:
- \( 2 \cdot \mathbf{r_1} = 2 \cdot (1 \hat{i} + 2 \hat{j} + 1 \hat{k}) = 2 \hat{i} + 4 \hat{j} + 2 \hat{k} \)
- \( 5 \cdot \mathbf{r_2} = 5 \cdot (3 \hat{i} + 2 \hat{j} - 1 \hat{k}) = 15 \hat{i} + 10 \hat{j} - 5 \hat{k} \)

Now, combine these:
\[ 2 \hat{i} + 4 \hat{j} + 2 \hat{k} + 15 \hat{i} + 10 \hat{j} - 5 \hat{k} = (2 + 15) \hat{i} + (4 + 10) \hat{j} + (2 - 5) \hat{k} \]
This simplifies to:
\[ 17 \hat{i} + 14 \hat{j} - 3 \hat{k} \]

### Step 5: Calculate the Denominator
The total mass is:
\[ m_1 + m_2 = 2 + 5 = 7 \]

### Step 6: Final Calculation of Center of Mass
Now substitute back into the formula:
\[ \mathbf{r_{cm}} = \frac{17 \hat{i} + 14 \hat{j} - 3 \hat{k}}{7} \]
This gives:
\[ \mathbf{r_{cm}} = \frac{17}{7} \hat{i} + \frac{14}{7} \hat{j} - \frac{3}{7} \hat{k} \]

### Final Result
The position vector of the center of mass is:
\[ \mathbf{r_{cm}} = \frac{17}{7} \hat{i} + 2 \hat{j} - \frac{3}{7} \hat{k} \]

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