As observed in the laboratory system , a 6 MeV proton is incident on a stationary 12 C target. The velocity of centre of mass of the system is (Take mass of proton to be 1 amu)

Options

A. `2.6xx10^(6) ms^(-1)`
B. `6.2xx10^(6) ms ^(-1)`
C. `10xx10^(6) ms ^(-1)`
D. `10 ms ^(-1)`

Answer

To find the velocity of the center of mass of the system when a 6 MeV proton is incident on a stationary 12 C target, we can follow these steps:

### Step 1: Identify the masses and energies
- Mass of the proton, \( m_1 = 1 \, \text{amu} \)
- Mass of the carbon target, \( m_2 = 12 \, \text{amu} \)
- Kinetic energy of the proton, \( KE = 6 \, \text{MeV} \)

### Step 2: Convert kinetic energy to joules
1 MeV = \( 1.6 \times 10^{-13} \, \text{J} \)

So,
\[ KE = 6 \, \text{MeV} = 6 \times 1.6 \times 10^{-13} \, \text{J} = 9.6 \times 10^{-13} \, \text{J} \]

### Step 3: Calculate the velocity of the proton
Using the kinetic energy formula:
\[ KE = \frac{1}{2} m v^2 \]
We can rearrange this to find the velocity \( v \):
\[ v = \sqrt{\frac{2 \times KE}{m_1}} = \sqrt{\frac{2 \times 9.6 \times 10^{-13}}{1.66 \times 10^{-27}}} \]
(Note: \( 1 \, \text{amu} = 1.66 \times 10^{-27} \, \text{kg} \))

Calculating this gives:
\[ v \approx \sqrt{\frac{1.92 \times 10^{-12}}{1.66 \times 10^{-27}}} \approx \sqrt{1.157 \times 10^{15}} \approx 1.075 \times 10^7 \, \text{m/s} \]

### Step 4: Calculate the velocity of the center of mass
The velocity of the center of mass \( v_{cm} \) is given by:
\[ v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]
Since the carbon target is stationary, \( v_2 = 0 \):
\[ v_{cm} = \frac{m_1 v_1 + 0}{m_1 + m_2} = \frac{m_1 v_1}{m_1 + m_2} \]

Substituting the values:
\[ v_{cm} = \frac{(1 \, \text{amu}) \times (1.075 \times 10^7 \, \text{m/s})}{1 \, \text{amu} + 12 \, \text{amu}} = \frac{1.075 \times 10^7}{13} \approx 8.27 \times 10^5 \, \text{m/s} \]

### Final Answer
The velocity of the center of mass of the system is approximately \( 8.27 \times 10^5 \, \text{m/s} \).

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