$11.5.21.$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.
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 $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S s^2}{\mu_0 (S - s)(2S - s)}$：
They are obtained by changing the sign of the speeds in the CM system and adding its speed:
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 $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S s^2}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:
$${\upsilon_1}' = \upsilon_1$$
$${\upsilon_2}' = \upsilon_2$$
#### Answer
For:
−
$$\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S^2(3s - 2S)}{\mu_0 (S - s)(2S - s)}$$
+
$$\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S s^2}{\mu_0 (S - s)(2S - s)}$$
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 $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S s^2}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:
$${\upsilon_1}' = \upsilon_1$$
$${\upsilon_2}' = \upsilon_2$$
### Statement
### Statement
$11.5.21.$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 ($v_1 > v_2$).
$11.5.21.$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.
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 $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S s^2}{\mu_0 (S - s)(2S - s)}$：
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:
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 $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S s^2}{\mu_0(S - s)(2S - s)}$ 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:
$$\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S^2(3s - 2S)}{\mu_0 (S - s)(2S - s)}$$
$$\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S s^2}{\mu_0 (S - s)(2S - s)}$$
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 $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S s^2}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:
