For a body moving with relativistic speed, if the velocity is doubled, then

Options

A. Its linear momentum is doubled
B. Its linear momentum will be less than double
C. Its linear momentum will be more than double
D. Its linear momentum remains unchanged

Answer

To solve the problem, we need to analyze how the momentum of a body moving at relativistic speeds changes when its velocity is doubled.

### Step-by-Step Solution:

1. **Understanding Relativistic Momentum**:
The relativistic momentum \( p \) of a body is given by the formula:
\[ p = m' v \]
where \( m' \) is the relativistic mass and \( v \) is the velocity of the body.

2. **Relativistic Mass**:
The relativistic mass \( m' \) is defined as:
\[ m' = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \]
where \( m_0 \) is the rest mass of the body and \( c \) is the speed of light.

3. **Initial Momentum**:
The initial momentum \( p_1 \) when the body is moving with velocity \( v \) is:
\[ p_1 = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} \]

4. **Doubling the Velocity**:
If we double the velocity, the new velocity becomes \( 2v \). We need to find the new momentum \( p_2 \):
\[ p_2 = m' (2v) = \frac{m_0 (2v)}{\sqrt{1 - \frac{(2v)^2}{c^2}}} \]
Simplifying this, we get:
\[ p_2 = \frac{2m_0 v}{\sqrt{1 - \frac{4v^2}{c^2}}} \]

5. **Comparing the Momenta**:
To compare \( p_2 \) with \( p_1 \), we can express \( p_2 \) in terms of \( p_1 \):
\[ p_2 = 2p_1 \cdot \frac{\sqrt{1 - \frac{v^2}{c^2}}}{\sqrt{1 - \frac{4v^2}{c^2}}} \]
The factor \( \frac{\sqrt{1 - \frac{v^2}{c^2}}}{\sqrt{1 - \frac{4v^2}{c^2}}} \) is less than 1 since \( 1 - \frac{4v^2}{c^2} < 1 - \frac{v^2}{c^2} \).

6. **Conclusion**:
Since \( p_2 \) is equal to \( 2p_1 \) multiplied by a factor that is less than 1, we conclude that the new momentum \( p_2 \) is more than double the initial momentum \( p_1 \). Thus, the momentum increases by more than double when the velocity is doubled.

### Final Answer:
The linear momentum will be more than double when the velocity is doubled.