Can relativistic mass be treated as rest mass?

Question

In what real sense does the mass of an object increase with its speed? When we learn that the mass of an object increases according to the equation,

$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

We can think of the mass of an object as its resistance to being moved (inertia). But suppose we devised a contraption that increased its mass by the motion of its internal parts. Would the inertia of the contraption be greater? In the following, I devise the simplest possible contraption that takes advantage of this fact and ask what happens when an object collides with it.

Consider a box with a negligibly light frame in which two balls each of mass $\frac{m_0}{2}$ start in the middle and move in opposite directions at velocity $v$. Call this box the "balls box".

The balls box has mass $m = \frac{m_{0}}{\sqrt{1 - \frac{v^2}{c^2}}}$.

Consider a box with rest mass $m$ when another box of rest mass $m$ hits it moving at velocity $v$. In an elastic collision, the first mass with move with velocity $v$ and the second will come to rest.

Now replace the box with rest mass $m$ with the balls box. Because the balls inside it are moving, it has rest mass $m_0$ but relativistic mass $m$. The balls box is at rest when a box with mass $m$ hits it at velocity $v$ in the direction along which the balls move. What happens?

According to a naive analysis, after the collision there will be some jiggling as information is transmitted down the length of the balls box. Eventually, the moving box will come to rest and the balls box will move with velocity $v$ to the right. Is this right?

What happens when we turn the balls box perpendicular to the motion of the moving ball?

Edit:

The rationale behind the naive analysis is that when the balls box is treated as a self-contained system—and when it's at rest—it has mass $m$. By that reasoning, it should behave the same as a box with rest mass $m$. The problem arises when we look inside the box after the collision. After information has propagated, one ball moves at $\frac{v - v}{\sqrt{1 + \frac{v^2}{c^2}}} = 0$. The other moves at $\frac{2v}{\sqrt{1 + \frac{v^2}{c^2}}}$. The box previously in motion is at rest. This leads to a momentum and energy different from the energy and momentum before the collision.

Answer 1

This answer written in the "modern" perspective.

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The balls box has rest mass $m_0$

But it doesn't. The (rest or invariant) mass of the box is found in the usual way from the square or it's energy-momentum four vector $(mc^2)^2 = \mathbf{p}^2 = E^2 - (\vec{p}c)^2$.

The energy-momentum four vector of the box is found by adding the four vectors of it's three parts together, so (noting that all the momenta are co-linear) \begin{align*} (mc^2)^2 &= \mathbf{p}^2 \\ &= \left[\mathbf{p}_{b1} + \mathbf{p}_{b2} + \mathbf{p}_{frame}\right]^2 \\ &= \left[ (\gamma m_0 c^2/2,+\gamma m_0 \vec{v}c/2) + (\gamma m_0 c^2/2,-\gamma m_0 \vec{v}c/2) + (0,\vec{0}) \right]^2 \\ &= \left[ (\gamma m_0 c^2, \vec{0}) \right]^2 \\ &= \gamma m_0 c^2 \end{align*}

So the invariant mass of the box is $\gamma m_0$ in the first place.

In relativity, the mass of a system is not necessarily equal to the sum of masses of it's component parts. Indeed, that is where the energy of nuclear reactions comes from/goes to: changes in the mass of the system by putting it together or taking it apart.

Answer 2

The naive analysis cannot be correct because then the velocity of one ball would be $0$ and the other would be $\frac{2v}{1 + \frac{v^2}{c^2}}$, which gives an energy and momentum different from the initial energy and momentum of the system. As to what actually happens, I'll leave that to other posters...