Edit to “Statement”

naz edited Jun 29, 2026, 6:46 PM

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			### Statement

		−	$11.5.21.$ Решите задачу $11.5.20$ в случае, если масса первого снаряда $m_1$, а второго $m_2$, а скорость снарядов равна соответственно $v_1$ и $v_2$ ($v_1 > v_2$).
		+	$11.5.21.$Solve the problem $11.5.20$ if the mass of the first projectile is $m_1$ and the second is $m_2$, and the velocity of the projectiles is $v_1$ and $v_2$, respectively ($v_1 > v_2$).

			### Solution

			The boundary condition will be the same as in the previous problem, but the left side of the inequality will change.

			The velocity of the center of mass of the system:

			$$\upsilon_C = \frac{m_1\upsilon_1 + m_2\upsilon_2}{m_1 + m_2}$$

			Hence the velocity of our projectiles:

			$$\upsilon_{1_C} = \frac{m_2(\upsilon_1 - \upsilon_2)}{m_1 + m_2}$$
			$$\upsilon_{2_C} = -\frac{m_1(\upsilon_1 - \upsilon_2)}{m_1 + m_2}$$

			Now let's calculate the total kinetic energy in the SCM, it will go to the left side of the inequality:

			$$E_K = \frac{(\upsilon_1 - \upsilon_2)^2}{2(\frac{1}{m_1}+\frac{1}{m_2})}$$

			Finite velocities in the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S^2(3s - 2S)}{\mu_0 (S - s)(2S - s)}$：

			They are obtained by changing the sign of the speeds in the CM system and adding its speed:

			$${\upsilon_1}' = \frac{2m_2\upsilon_2 + (m_1 - m_2)\upsilon_1}{m_1 + m_2}$$
			$${\upsilon_2}' = \frac{2m_1\upsilon_1 + (m_2 - m_1)\upsilon_2}{m_1 + m_2}$$

			For the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:

			$${\upsilon_1}' = \upsilon_1$$

			$${\upsilon_2}' = \upsilon_2$$

			#### Answer
			For:

			$$\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S^2(3s - 2S)}{\mu_0 (S - s)(2S - s)}$$



			$${\upsilon_1}' = \frac{2m_2\upsilon_2 + (m_1 - m_2)\upsilon_1}{m_1 + m_2}$$
			$${\upsilon_2}' = \frac{2m_1\upsilon_1 + (m_2 - m_1)\upsilon_2}{m_1 + m_2}$$

			For the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:

			$${\upsilon_1}' = \upsilon_1$$

			$${\upsilon_2}' = \upsilon_2$$
		unchanged lines 43

	@@ -1,6 +1,6 @@
	### Statement		### Statement

	$11.5.21.$ Решите задачу $11.5.20$ в случае, если масса первого снаряда $m_1$, а второго $m_2$, а скорость снарядов равна соответственно $v_1$ и $v_2$ ($v_1 > v_2$).		$11.5.21.$Solve the problem $11.5.20$ if the mass of the first projectile is $m_1$ and the second is $m_2$, and the velocity of the projectiles is $v_1$ and $v_2$, respectively ($v_1 > v_2$).

	### Solution		### Solution

	The boundary condition will be the same as in the previous problem, but the left side of the inequality will change.		The boundary condition will be the same as in the previous problem, but the left side of the inequality will change.

	The velocity of the center of mass of the system:		The velocity of the center of mass of the system:

	$$\upsilon_C = \frac{m_1\upsilon_1 + m_2\upsilon_2}{m_1 + m_2}$$		$$\upsilon_C = \frac{m_1\upsilon_1 + m_2\upsilon_2}{m_1 + m_2}$$

	Hence the velocity of our projectiles:		Hence the velocity of our projectiles:

	$$\upsilon_{1_C} = \frac{m_2(\upsilon_1 - \upsilon_2)}{m_1 + m_2}$$		$$\upsilon_{1_C} = \frac{m_2(\upsilon_1 - \upsilon_2)}{m_1 + m_2}$$
	$$\upsilon_{2_C} = -\frac{m_1(\upsilon_1 - \upsilon_2)}{m_1 + m_2}$$		$$\upsilon_{2_C} = -\frac{m_1(\upsilon_1 - \upsilon_2)}{m_1 + m_2}$$

	Now let's calculate the total kinetic energy in the SCM, it will go to the left side of the inequality:		Now let's calculate the total kinetic energy in the SCM, it will go to the left side of the inequality:

	$$E_K = \frac{(\upsilon_1 - \upsilon_2)^2}{2(\frac{1}{m_1}+\frac{1}{m_2})}$$		$$E_K = \frac{(\upsilon_1 - \upsilon_2)^2}{2(\frac{1}{m_1}+\frac{1}{m_2})}$$

	Finite velocities in the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S^2(3s - 2S)}{\mu_0 (S - s)(2S - s)}$：		Finite velocities in the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S^2(3s - 2S)}{\mu_0 (S - s)(2S - s)}$：

	They are obtained by changing the sign of the speeds in the CM system and adding its speed:		They are obtained by changing the sign of the speeds in the CM system and adding its speed:

	$${\upsilon_1}' = \frac{2m_2\upsilon_2 + (m_1 - m_2)\upsilon_1}{m_1 + m_2}$$		$${\upsilon_1}' = \frac{2m_2\upsilon_2 + (m_1 - m_2)\upsilon_1}{m_1 + m_2}$$
	$${\upsilon_2}' = \frac{2m_1\upsilon_1 + (m_2 - m_1)\upsilon_2}{m_1 + m_2}$$		$${\upsilon_2}' = \frac{2m_1\upsilon_1 + (m_2 - m_1)\upsilon_2}{m_1 + m_2}$$

	For the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:		For the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:

	$${\upsilon_1}' = \upsilon_1$$		$${\upsilon_1}' = \upsilon_1$$

	$${\upsilon_2}' = \upsilon_2$$		$${\upsilon_2}' = \upsilon_2$$

	#### Answer		#### Answer
	For:		For:

	$$\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S^2(3s - 2S)}{\mu_0 (S - s)(2S - s)}$$		$$\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S^2(3s - 2S)}{\mu_0 (S - s)(2S - s)}$$



	$${\upsilon_1}' = \frac{2m_2\upsilon_2 + (m_1 - m_2)\upsilon_1}{m_1 + m_2}$$		$${\upsilon_1}' = \frac{2m_2\upsilon_2 + (m_1 - m_2)\upsilon_1}{m_1 + m_2}$$
	$${\upsilon_2}' = \frac{2m_1\upsilon_1 + (m_2 - m_1)\upsilon_2}{m_1 + m_2}$$		$${\upsilon_2}' = \frac{2m_1\upsilon_1 + (m_2 - m_1)\upsilon_2}{m_1 + m_2}$$

	For the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:		For the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:

	$${\upsilon_1}' = \upsilon_1$$		$${\upsilon_1}' = \upsilon_1$$

	$${\upsilon_2}' = \upsilon_2$$		$${\upsilon_2}' = \upsilon_2$$
	unchanged lines 43