Two bodies of masses 10 kg and 20 kg are located in x-y plane at (0, 1) and (1, 0). The position of their centre of mass is

Options

A. `(2//3,1//3)`
B. `(1//3,2//3)`
C. `(2,1)`
D. `(1//3,4//3)`

Answer

To find the center of mass of two bodies located in the x-y plane, we can use the formula for the center of mass (RCM) given by:

\[ R_{CM} = \frac{m_1 \cdot r_1 + m_2 \cdot r_2}{m_1 + m_2} \]

Where:
- \(m_1\) and \(m_2\) are the masses of the two bodies,
- \(r_1\) and \(r_2\) are the position vectors of the two bodies.

### Step 1: Identify the masses and their positions
- Mass \(m_1 = 10 \, \text{kg}\) located at \(r_1 = (0, 1)\)
- Mass \(m_2 = 20 \, \text{kg}\) located at \(r_2 = (1, 0)\)

### Step 2: Write down the formula for the center of mass
The center of mass coordinates \( (x_{CM}, y_{CM}) \) can be calculated using the following equations:

\[ x_{CM} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2} \]

\[ y_{CM} = \frac{m_1 \cdot y_1 + m_2 \cdot y_2}{m_1 + m_2} \]

### Step 3: Substitute the values into the equations
For the x-coordinate:

\[ x_{CM} = \frac{10 \cdot 0 + 20 \cdot 1}{10 + 20} = \frac{0 + 20}{30} = \frac{20}{30} = \frac{2}{3} \]

For the y-coordinate:

\[ y_{CM} = \frac{10 \cdot 1 + 20 \cdot 0}{10 + 20} = \frac{10 + 0}{30} = \frac{10}{30} = \frac{1}{3} \]

### Step 4: Combine the results
The position of the center of mass is:

\[ R_{CM} = \left( \frac{2}{3}, \frac{1}{3} \right) \]

### Final Answer
The position of the center of mass is \(\left( \frac{2}{3}, \frac{1}{3} \right)\).

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