If the speed of a particle moving at a relativistic speed is doubled, its linear momentum will

Options

A. Become double
B. bocome more than double
C. remain equal
D. become less than double.

Answer

To solve the problem of how the linear momentum of a particle changes when its speed is doubled at relativistic speeds, we will follow these steps:

### Step 1: Understand the formula for relativistic momentum
The linear momentum \( P \) of a particle moving at relativistic speeds is given by the formula:
\[ P = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} \]
where:
- \( m_0 \) is the rest mass of the particle,
- \( v \) is the velocity of the particle,
- \( c \) is the speed of light.

### Step 2: Define the new velocity
If the original speed of the particle is \( v \), and it is doubled, the new speed \( v' \) will be:
\[ v' = 2v \]

### Step 3: Calculate the new momentum
Substituting \( v' \) into the momentum formula, we get the new momentum \( P' \):
\[ P' = \frac{m_0 (2v)}{\sqrt{1 - \frac{(2v)^2}{c^2}}} \]
This simplifies to:
\[ P' = \frac{2m_0 v}{\sqrt{1 - \frac{4v^2}{c^2}}} \]

### Step 4: Compare the new momentum with the original momentum
Now, we can express the new momentum in terms of the original momentum \( P \):
\[ P = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} \]
Thus, we can relate \( P' \) to \( P \):
\[ P' = 2P \cdot \frac{\sqrt{1 - \frac{v^2}{c^2}}}{\sqrt{1 - \frac{4v^2}{c^2}}} \]

### Step 5: Analyze the ratio
Since \( \frac{1 - \frac{4v^2}{c^2}}{1 - \frac{v^2}{c^2}} < 1 \) when \( v \) is a relativistic speed (i.e., \( v \) is close to \( c \)), it implies that:
\[ P' > 2P \]
This indicates that the new momentum \( P' \) is more than double the original momentum \( P \).

### Conclusion
Therefore, if the speed of a particle moving at a relativistic speed is doubled, its linear momentum will become more than double.