$11.5.20.$ 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.
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### Solution
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Let's take a qualitative look at the picture. To do this, it is convenient to switch to the centerofmass system, because in it the projectiles move towards each other at a speed of $\upsilon$.
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The induction has a different value for the crosssection of each projectile in both tubes, which is determined by the law of conservation of magnetic flux:
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+
Let's take a closer look at the picture. To do this, it is convenient to switch to the center-of-mass system, as the projectiles are moving towards each other with a velocity of $\upsilon$.
+
In the part of the pipe that is free from projectiles, the energy per unit length is equal to:
+
$$\omega_0 = \frac{B^2 S}{\mu_0}$$
+
In the cross-section of each projectile in both pipes, the induction has a different value, 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:
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$$E_2 =\frac{B^2lS^2}{\mu_0 (S - s)}$$
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We find the difference between $E_2$ and $E_1$ and get the desired solutions.:
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For $$m \upsilon^2 < \frac{B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$$
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Hence the energy per unit length in a cross-section with one projectile:
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$$\omega_1 = \frac{2B^2 S^2}{\mu_0 (2S - s)}$$
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And finally, if there are two projectiles in the cross-section, the induction is:
+
$$B_2 = \frac{BS}{S-s}$$
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The energy per unit length is:
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$$\omega_2 = \frac{B^2 S^2}{\mu_0 (S - s)}$$
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While the projectiles are moving without overlapping, the energy is equal to:
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$$U_1 = 2\omega_1l + \omega_0L$$
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When the projectiles slightly overlap by an amount x, the energy is equal to:
$$\Delta U = -2\omega_1 x + \omega_2 x + \omega_0 x$$
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Substitute $x=l$ and find:
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$$\Delta U = \frac{B^2 S s^2l}{\mu_0(2S - s)(S-s)}$$
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If there is enough speed in this system to prevent a complete overlap, then this is an equivalent of an absolutely elastic collision, but if not, then the speeds remain the same:
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When $$m \upsilon^2 <\frac{B^2 l S s^2}{\mu_0 (S - s)(2S - s)}$$
When $$m \upsilon^2 > \frac{B^2 l S s^2}{\mu_0(S - s)(2S - s)}$$
$${\upsilon_1}' = 3 \upsilon$$
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$${\upsilon_2}' = \upsilon$$
### Statement
### 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.
$11.5.20.$ 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.
@@ -4,46 +4,36 @@Statement
### Solution
### Solution
Let's take a qualitative look at the picture. To do this, it is convenient to switch to the centerofmass system, because in it the projectiles move towards each other at a speed of $\upsilon$.
Let's take a closer look at the picture. To do this, it is convenient to switch to the center-of-mass system, as the projectiles are moving towards each other with a velocity of $\upsilon$.
In the part of the pipe that is free from projectiles, the energy per unit length is equal to:
The induction has a different value for the crosssection of each projectile in both tubes, which is determined by the law of conservation of magnetic flux:
$$\omega_0 = \frac{B^2 S}{\mu_0}$$
In the cross-section of each projectile in both pipes, the induction has a different value, which is determined by the law of conservation of magnetic flux:
$$B_1 = \frac{2BS}{2S - s}$$
$$B_1 = \frac{2BS}{2S - s}$$
Hence the energy per unit length in a cross-section with one projectile:
In this case, the energy of the magnetic field in the two sections of the tubes is equal to:
$$\omega_1 = \frac{2B^2 S^2}{\mu_0 (2S - s)}$$
And finally, if there are two projectiles in the cross-section, the induction is:
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:
$$\omega_2 = \frac{B^2 S^2}{\mu_0 (S - s)}$$
While the projectiles are moving without overlapping, the energy is equal to:
$$E_2 =\frac{B^2lS^2}{\mu_0 (S - s)}$$
$$U_1 = 2\omega_1l + \omega_0L$$
When the projectiles slightly overlap by an amount x, the energy is equal to:
We find the difference between $E_2$ and $E_1$ and get the desired solutions.:
For $$m \upsilon^2 < \frac{B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$$
$$\Delta U = -2\omega_1 x + \omega_2 x + \omega_0 x$$
Substitute $x=l$ and find:
$$\Delta U = \frac{B^2 S s^2l}{\mu_0(2S - s)(S-s)}$$
If there is enough speed in this system to prevent a complete overlap, then this is an equivalent of an absolutely elastic collision, but if not, then the speeds remain the same:
When $$m \upsilon^2 <\frac{B^2 l S s^2}{\mu_0 (S - s)(2S - s)}$$
$${\upsilon_1}' = \upsilon$$
$${\upsilon_1}' = \upsilon$$
$${\upsilon_2}' = 3\upsilon$$
$${\upsilon_2}' = 3\upsilon$$
When $$m \upsilon^2 > \frac{B^2 l S s^2}{\mu_0(S - s)(2S - s)}$$
