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 (V_com) of a system consisting of a proton and a stationary carbon-12 target, we can follow these steps:

### Step 1: Identify the masses and kinetic energy
- Mass of the proton (m₁) = 1 amu
- Mass of the carbon-12 nucleus (m₂) = 12 amu
- Kinetic energy (KE) of the proton = 6 MeV

### Step 2: Convert kinetic energy from MeV to Joules
1 MeV = \(1.6 \times 10^{-13}\) Joules. Therefore,
\[ KE = 6 \, \text{MeV} = 6 \times 1.6 \times 10^{-13} \, \text{J} = 9.6 \times 10^{-13} \, \text{J} \]

### Step 3: Relate kinetic energy to momentum
The relationship between kinetic energy (KE) and momentum (p) is given by:
\[ KE = \frac{p^2}{2m} \]
From this, we can express momentum as:
\[ p = \sqrt{2m \cdot KE} \]

### Step 4: Calculate the momentum of the proton
Using the mass of the proton (m₁ = 1 amu), we can calculate the momentum:
\[ p = \sqrt{2 \cdot 1 \, \text{amu} \cdot 9.6 \times 10^{-13} \, \text{J}} \]

### Step 5: Substitute the values and simplify
Convert 1 amu to kg:
\[ 1 \, \text{amu} = 1.67 \times 10^{-27} \, \text{kg} \]
Thus, the momentum becomes:
\[ p = \sqrt{2 \cdot (1.67 \times 10^{-27} \, \text{kg}) \cdot (9.6 \times 10^{-13} \, \text{J})} \]

### Step 6: Calculate the center of mass velocity
The velocity of the center of mass (V_com) is given by:
\[ V_{com} = \frac{p}{m_1 + m_2} \]
Substituting the values:
\[ V_{com} = \frac{p}{1 \, \text{amu} + 12 \, \text{amu}} = \frac{p}{13 \, \text{amu}} = \frac{p}{13 \cdot 1.67 \times 10^{-27} \, \text{kg}} \]

### Step 7: Final calculation
Now we can substitute the value of momentum (p) into the equation for V_com and calculate it:
\[ V_{com} = \frac{\sqrt{2 \cdot (1.67 \times 10^{-27}) \cdot (9.6 \times 10^{-13})}}{13 \cdot 1.67 \times 10^{-27}} \]

### Step 8: Solve for V_com
After performing the calculations, we find:
\[ V_{com} \approx 2.6 \times 10^6 \, \text{m/s} \]

### Conclusion
The velocity of the center of mass of the system is approximately \(2.6 \times 10^6 \, \text{m/s}\).

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