When the speed of a particle travelling at a relativistic speed is doubled, its linear momentum will

Step-by-step solution

The relativistic momentum of a particle is given by p = mv 1 - v^2 c^2 When the speed is v , the initial momentum is p₁ = mv 1 - v^2 c^2 When the speed is doubled to 2v , the new momentum is p₂ = m(2v) 1 - 4v^2 c^2 Comparing p₂ with 2p₁ : 2p₁ = 2mv 1 - v^2 c^2 Since 1 - 4v^2 c^2 Therefore, 1 1 - 4v^2 c^2 > 1 1 - v^2 c^2 This implies p₂ > 2p₁ Thus, the linear momentum becomes more than double.