A proton has a mass of `1.673xx10^(-27)kg`. If the proton is accelerated to a speed of 0.93c , what is the magnitude of the relativistic momentum of the proton?

Options

A. `6.2xx10^(-17)kg*m//s`
B. `4.7xx10^(-19)kg*m//s`
C. `1.3xx10^(-18)kg*m//s`
D. `5.9xx10^(-24)kg*m//s`

Answer

To find the magnitude of the relativistic momentum of a proton moving at a speed of \(0.93c\), we can use the formula for relativistic momentum:

\[ p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} \]

where:
- \(p\) is the relativistic momentum,
- \(m_0\) is the rest mass of the proton,
- \(v\) is the speed of the proton,
- \(c\) is the speed of light.

### Step 1: Identify the values
- The rest mass of the proton, \(m_0 = 1.673 \times 10^{-27} \, \text{kg}\)
- The speed of the proton, \(v = 0.93c\)

### Step 2: Substitute the values into the formula
We need to calculate \(p\):

\[ p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} \]

Substituting \(v = 0.93c\):

\[ p = \frac{1.673 \times 10^{-27} \, \text{kg} \cdot (0.93c)}{\sqrt{1 - \frac{(0.93c)^2}{c^2}}} \]

### Step 3: Simplify the equation
The term \(\frac{(0.93c)^2}{c^2}\) simplifies to \(0.93^2\):

\[ p = \frac{1.673 \times 10^{-27} \, \text{kg} \cdot (0.93c)}{\sqrt{1 - 0.93^2}} \]

Calculating \(0.93^2\):

\[ 0.93^2 = 0.8649 \]

Thus:

\[ 1 - 0.93^2 = 1 - 0.8649 = 0.1351 \]

### Step 4: Calculate the square root
Now we can calculate the square root:

\[ \sqrt{0.1351} \approx 0.3678 \]

### Step 5: Substitute back into the momentum formula
Now we substitute back into the momentum formula:

\[ p = \frac{1.673 \times 10^{-27} \, \text{kg} \cdot (0.93 \cdot 3 \times 10^8 \, \text{m/s})}{0.3678} \]

Calculating \(0.93 \cdot 3 \times 10^8\):

\[ 0.93 \cdot 3 \times 10^8 \approx 2.79 \times 10^8 \, \text{m/s} \]

### Step 6: Calculate the momentum
Now substituting this value into the equation:

\[ p = \frac{1.673 \times 10^{-27} \cdot 2.79 \times 10^8}{0.3678} \]

Calculating the numerator:

\[ 1.673 \times 10^{-27} \cdot 2.79 \times 10^8 \approx 4.67 \times 10^{-19} \, \text{kg m/s} \]

Now divide by \(0.3678\):

\[ p \approx \frac{4.67 \times 10^{-19}}{0.3678} \approx 1.27 \times 10^{-18} \, \text{kg m/s} \]

### Final Result
Thus, the magnitude of the relativistic momentum of the proton is approximately:

\[ p \approx 1.27 \times 10^{-18} \, \text{kg m/s} \]