The centres of three spherical masses of 1 kg, 2 kg and 3 kg have co-orinates (4,0) m,(0,3)m and (-2,5)m respectively. What is the position vector of its centre of mass is terms of its x and y co-ordinates?

Options

A. `vecR_(cm)=2hati+3hatj`
B. `vecR_(cM)=1/3hati+5/2hatj`
C. `veR_(cm)=-=-1/3hati+3.5hatj`
D. `vecr_(cm)=2/3hati-5/3hatj`

Explanation

To find the position vector of the center of mass of the three spherical masses, we will follow these steps:

### Step 1: Identify the masses and their coordinates
- Mass \( m_1 = 1 \, \text{kg} \) at coordinates \( (x_1, y_1) = (4, 0) \)
- Mass \( m_2 = 2 \, \text{kg} \) at coordinates \( (x_2, y_2) = (0, 3) \)
- Mass \( m_3 = 3 \, \text{kg} \) at coordinates \( (x_3, y_3) = (-2, 5) \)

### Step 2: Calculate the total mass
The total mass \( M \) is given by:
\[ M = m_1 + m_2 + m_3 = 1 + 2 + 3 = 6 \, \text{kg} \]

### Step 3: Calculate the x-coordinate of the center of mass
The x-coordinate of the center of mass \( x_{cm} \) is calculated using the formula:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M} \]
Substituting the values:
\[ x_{cm} = \frac{(1 \cdot 4) + (2 \cdot 0) + (3 \cdot -2)}{6} = \frac{4 + 0 - 6}{6} = \frac{-2}{6} = -\frac{1}{3} \]

### Step 4: Calculate the y-coordinate of the center of mass
The y-coordinate of the center of mass \( y_{cm} \) is calculated using the formula:
\[ y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} \]
Substituting the values:
\[ y_{cm} = \frac{(1 \cdot 0) + (2 \cdot 3) + (3 \cdot 5)}{6} = \frac{0 + 6 + 15}{6} = \frac{21}{6} = 3.5 \]

### Step 5: Write the position vector of the center of mass
The position vector of the center of mass \( \vec{R}_{cm} \) can be expressed as:
\[ \vec{R}_{cm} = x_{cm} \hat{i} + y_{cm} \hat{j} = -\frac{1}{3} \hat{i} + 3.5 \hat{j} \]

### Final Answer
The position vector of the center of mass is:
\[ \vec{R}_{cm} = -\frac{1}{3} \hat{i} + 3.5 \hat{j} \]

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