## Related questions with answers

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

Solutions

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

$p_{relativistic} = \gamma mv$

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

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

The length contraction is given by the formula

$L = \frac{L_{0}}{\gamma}$

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

$p_{relativistic} = 0.5 mv$

**Explanation:** It is said that the high speed spaceship shrunk to half of its length.

And the momentum of the spaceship is given by,

$\begin{aligned} P&=mv\\ \textbf{where,}\ m&=\text{the mass of the moving spaceship}\\ &=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}} }} \\ m_{0}&=\text{the rest mass of the spaceship}\\ v&=\text{the speed of the moving stick}\\ \textbf{after putting the value,}\\ P&=\frac{m_{0}v}{\sqrt{1-\frac{v^{2}}{c^{2}} }}----\left( 1 \right) \end{aligned}$

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