Understanding the Problem
* Relativistic Momentum: At high speeds, we need to use the relativistic momentum formula:
* p = γmv where:
* p is momentum
* γ (gamma) is the Lorentz factor: γ = 1 / √(1 - (v²/c²))
* m is mass
* v is velocity
* c is the speed of light
* Doubling Momentum: The problem states the momentum doubles after the acceleration. This means the final momentum (p₂) is twice the initial momentum (p₁): p₂ = 2p₁.
Setting up the Equations
1. Initial Momentum (p₁):
* p₁ = γ₁mv₁
* where γ₁ is the Lorentz factor at the initial velocity (v₁)
2. Final Momentum (p₂):
* p₂ = γ₂mv₂
* where γ₂ is the Lorentz factor at the final velocity (v₂)
3. Doubling Momentum:
* p₂ = 2p₁
* γ₂mv₂ = 2γ₁mv₁
Solving for the Final Velocity (v₂)
1. Cancel Common Terms: The mass (m) and the speed of light (c) are constants in this problem, so they cancel out:
* γ₂v₂ = 2γ₁v₁
2. Substitute Lorentz Factors:
* (1 / √(1 - (v₂²/c²))) * v₂ = 2 * (1 / √(1 - (v₁²/c²))) * v₁
3. Solve for v₂: This equation is a bit tricky to solve directly. You'll likely need to use numerical methods (like a calculator or computer program) to solve for v₂. However, we can simplify the equation further:
* √(1 - (v₁²/c²)) * v₂ = 2√(1 - (v₂²/c²)) * v₁
* Square both sides to get rid of the square roots.
* (1 - (v₁²/c²)) * v₂² = 4(1 - (v₂²/c²)) * v₁²
4. Rearrange and Solve: Rearrange the equation to solve for v₂. You'll end up with a quadratic equation. Use the quadratic formula to find the solutions for v₂.
Important Note: Keep in mind that the initial velocity (8 E8 meters per second) is already a significant fraction of the speed of light. The final velocity will be even closer to the speed of light.
Let me know if you'd like to try solving the quadratic equation to find a numerical value for the final velocity.
