Solution

Let's take a qualitative look at the picture. To do this, it is convenient to switch to the center of mass system, because in it the projectiles move towards each other at a speed of $\upsilon$.

The induction has a different value for the cross section of each projectile in both tubes, which is determined by the law of conservation of magnetic flux:

$$B_1 = \frac{2BS}{2S - s}$$

In this case, the energy of the magnetic field in the two sections of the tubes is equal to:

$$E_1 = \frac{B_1^2}{2\mu_0} = \frac{4B^2S^2l}{\mu_0(2S-s)}$$

If the first projectile can align with the second, then they are pushed apart by magnetic forces, so we want to consider the boundary case when all the total energy in the center of mass system has shifted into a change in the energy of the magnetic field and the projectiles have aligned, then the boundary energy of the magnetic field:

$$E_2 = \frac{B^2lS^2}{\mu_0 (S - s)}$$

We find the difference between $E_2$ and $_E_1$ and get the desired solutions.:

For $$m \upsilon^2 < \frac{B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$$

{\upsilon_1}' = \upsilon

{\upsilon_2}' = 3\upsilon

For $$m \upsilon^2 > \frac{B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$$

{\upsilon_1}' = 3 \upsilon

{\upsilon_2}' = \upsilon

I don't know why it doesn't match the answer in Savchenko, I think it's a typo, if you see a flaw in the solution, please write to me.