$11.5.20.$ The two-channel magnetic energy redistributor of projectiles has the following design. Two metal pipes with a slot are connected by metal bridges as shown in the figure. The homogeneous magnetic field of induction $B$ is directed along the axis of the tubes. The same long superconducting projectiles move along the axis of each tube. One of the projectiles, which has a velocity of $3v$, catches up with the second projectile, which has a velocity of $v$. The length of each projectile is $l$, the cross section is $s$, and the mass is $m$. The cross section of each pipe is $S$. Determine the velocity of the projectiles after their interaction. Pipe resistance should be ignored.
+
### 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:
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:
@@ -18,22 +18,32 @@Solution
$$E_2 = \frac{B^2lS^2}{\mu_0 (S - s)}$$
−
We find the difference between $E_2$ and $_E_1$ and get the desired solutions.:
+
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
+
$${\upsilon_1}' = \upsilon$$
−
{\upsilon_2}' = 3\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
+
$${\upsilon_1}' = 3 \upsilon$$
−
{\upsilon_2}' = \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.
#### Answer
−
[Insert a concise answer or boxed result]
+
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 $$m \upsilon^2 > \frac{B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$$
+
+
$${\upsilon_1}' = 3 \upsilon$$
+
+
$${\upsilon_2}' = \upsilon$$
@@ -1,7 +1,7 @@
### Statement
### Statement
$11.5.20.$ [Insert the problem statement]
$11.5.20.$ The two-channel magnetic energy redistributor of projectiles has the following design. Two metal pipes with a slot are connected by metal bridges as shown in the figure. The homogeneous magnetic field of induction $B$ is directed along the axis of the tubes. The same long superconducting projectiles move along the axis of each tube. One of the projectiles, which has a velocity of $3v$, catches up with the second projectile, which has a velocity of $v$. The length of each projectile is $l$, the cross section is $s$, and the mass is $m$. The cross section of each pipe is $S$. Determine the velocity of the projectiles after their interaction. Pipe resistance should be ignored.
### Solution
### 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 $\upsilon$.
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:
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:
$$B_1 = \frac{2BS}{2S - s}$$
$$B_1 = \frac{2BS}{2S - s}$$
In this case, the energy of the magnetic field in the two sections of the tubes is equal to:
In this case, the energy of the magnetic field in the two sections of the tubes is equal to:
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:
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:
@@ -18,22 +18,32 @@Solution
$$E_2 = \frac{B^2lS^2}{\mu_0 (S - s)}$$
$$E_2 = \frac{B^2lS^2}{\mu_0 (S - s)}$$
We find the difference between $E_2$ and $_E_1$ and get the desired solutions.:
We find the difference between $E_2$ and $E_1$ and get the desired solutions.:
For $$m \upsilon^2 < \frac{B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$$
For $$m \upsilon^2 < \frac{B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$$
{\upsilon_1}' = \upsilon
$${\upsilon_1}' = \upsilon$$
{\upsilon_2}' = 3\upsilon
$${\upsilon_2}' = 3\upsilon$$
For $$m \upsilon^2 > \frac{B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$$
For $$m \upsilon^2 > \frac{B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$$
{\upsilon_1}' = 3 \upsilon
$${\upsilon_1}' = 3 \upsilon$$
{\upsilon_2}' = \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.
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.
#### Answer
#### Answer
[Insert a concise answer or boxed result]
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 $$m \upsilon^2 > \frac{B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$$
