In a certain experiments to measure the ratio of charge to mass of elementry particles, a surprising result was obtained in which two particle, a surprising result was obtained in which two particles moved in such a way that the distance between them always remained constant. It was also noticed that this two-particle system was isolated from all other particles and no force was acting on this system except the force between these two mases. After careful observation followed bu intensive calculation, it was deduced that velocity of these two particles was always opposite in direction and magnitude of velocity was `10^(3) ms^(-1) and 2 xx 10^(3) ms^(-1)` for first and second particle, respectively, and mass of these particles were `2 xx 10^(-30) kg and 10^(-30)kg`, respectively. Distance between them were 12Å(1Å = 10^(-`10)m).`
if the first particle is stopped for a moment and then released, the velocity of center of mass of the system just after the release will be

Options

A. `1/3 xx 10^(-30)ms^(-1)`
B. `1/3 xx 10^(3) ms^(-1)`
C. `2/3 xx 10^(3) ms^(-1)`
D. none of these

Explanation

To find the velocity of the center of mass of the two-particle system just after the first particle is stopped and then released, we can use the formula for the velocity of the center of mass (V_cm):

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

Where:
- \(m_1\) is the mass of the first particle,
- \(v_1\) is the velocity of the first particle,
- \(m_2\) is the mass of the second particle,
- \(v_2\) is the velocity of the second particle.

### Step 1: Identify the given values
- Mass of the first particle, \(m_1 = 2 \times 10^{-30} \, \text{kg}\)
- Velocity of the first particle before stopping, \(v_1 = 10^3 \, \text{m/s}\)
- Mass of the second particle, \(m_2 = 10^{-30} \, \text{kg}\)
- Velocity of the second particle, \(v_2 = 2 \times 10^3 \, \text{m/s}\)

### Step 2: Set the velocity of the first particle to zero
Since the first particle is stopped for a moment, we set:
- \(v_1 = 0 \, \text{m/s}\)

### Step 3: Substitute the values into the center of mass formula
Now we can substitute the values into the center of mass formula:

\[ V_{cm} = \frac{(2 \times 10^{-30} \, \text{kg} \cdot 0 \, \text{m/s}) + (10^{-30} \, \text{kg} \cdot 2 \times 10^3 \, \text{m/s})}{(2 \times 10^{-30} \, \text{kg} + 10^{-30} \, \text{kg})} \]

### Step 4: Simplify the equation
Calculating the numerator:
\[ = 0 + (10^{-30} \cdot 2 \times 10^3) = 2 \times 10^{-27} \, \text{kg m/s} \]

Calculating the denominator:
\[ = 2 \times 10^{-30} + 10^{-30} = 3 \times 10^{-30} \, \text{kg} \]

Now substituting back into the equation:
\[ V_{cm} = \frac{2 \times 10^{-27}}{3 \times 10^{-30}} \]

### Step 5: Final calculation
\[ V_{cm} = \frac{2}{3} \times 10^{3} \, \text{m/s} = \frac{2}{3} \times 1000 \, \text{m/s} \approx 666.67 \, \text{m/s} \]

### Final Answer
The velocity of the center of mass of the system just after the release will be approximately \(666.67 \, \text{m/s}\).

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