Relativistic Momentum

Imagine a car that drives with the speed $w,$ which is much smaller than $c$, in $x$-direction against a wall and creates a certain amount of damage. An observer may now move parallel to the wall in $y$-directon with the speed $v$. The value of $v$ should be very large so that relativistic effects can be ignored any longer. For the observer, the car travels at the speed $$w' = \frac{\Delta x'}{\Delta t'}$$ Because the observer moves perpendicular to the $x$-axis, $\Delta x'$ is equal to $\Delta x$. Only the time difference has to be transformed with the help of the time dilation which leads to $$w' = \frac{\Delta x}{\gamma\Delta t} = \frac{w}{\gamma}$$ The momentum of the car in the system of the observer can then be written as $$p' = m'w' = m'\frac{w}{\gamma}$$ Because the damage in the wall has to be independent of the observer, the momentum of the car in both systems has to equal: $$p' = p$$ Hence, it follows that the mass has to be transformed in the following way: $$\boxed{m' = \gamma m}$$ The quantity $m'$ is called relativistic mass or dynamic mass and is always larger than the rest mass $m$. The relativistic momentum can therefore be written as $$\boxed{p' = \gamma mv}$$