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If a high-speed spaceship appears shrunken to half its normal length, how does its momentum compare with the classical formula p=mvp=m v ?

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The idea of the problem is to conclude the factor γ\gamma which will enable us to conclude the relativistic momentum compared with classical momentum, since the relativistic momentum is given by the simple formula

prelativistic=γmvp_{relativistic} = \gamma mv

We are given the fact that the length contracted and is now twice its classical length:

L=L02L = \frac{L_{0}}{2}

The length contraction is given by the formula

L=L0γL = \frac{L_{0}}{\gamma}

so if we have that conclude that γ=0.5\gamma = 0.5. Thus we have:

prelativistic=0.5mvp_{relativistic} = 0.5 mv

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