A1  ^?? Using the mass conservation law, show that $\rho_1 v_1 = \rho_2 v_2$.

Let's choose a plot of area $S$ on the front surface. Mass flow through this surface can be calculated twice: \[\frac{dm}{dt} = \rho_1 v_ 1 S = \rho_2 v_2 S\]
This gives us: \[\rho_1 v_ 1 S = \rho_2 v_2 \]

А2  ^?? By writing down the law of conservation of momentum, get that $P_1 - P_2 = \rho_1 v_1 (v_2 - v_1)$

Lets select a cylinder with a base area $S$, mentioned later. Force, acting on it's left face equals $P_1S$. Taking into account the force on it's right face, we derive that the whole force \[F = (P_1 - P_2)S\]
This force causes momentum to change: \[F = \frac{dp}{dt} = \frac{dm}{dt} \cdot \Delta V = S\rho_1v_1(v_2 - v_1)\]
Reducing by S we get: \[P_1 - P_2 = \rho_1 v_1 (v_2 - v_1)\]

А3  ^?? Show that the relation between speeds is as follows:
\[ k = \frac{v_2}{v_1} = \frac{M_1^2(\gamma - 1) + 2 }{M_1^2(\gamma + 1)},\]
where $M_1 = \frac{v_1}{\sqrt{\gamma P_1/\rho_1}}$ - Mach number of the shock wave.

\[\frac{\gamma}{\gamma - 1}\frac{P_1}{\rho_1} + \frac{v_1^2}{2} = \frac{\gamma}{\gamma - 1}\frac{P_1 - \rho_1v_1(v_2 - v_1)}{\rho_1 v_1 / v_2} + \frac{v_2^2}{2}\]
\[\frac{\gamma}{\gamma - 1}\frac{P_1}{\rho_1 v_1^2} + \frac{1}{2} = \frac{\gamma}{\gamma - 1}\frac{P_1/\rho_1v_1^2 - k+1}{1/k} + \frac{k^2}{2}\]
\[\frac{\gamma}{\gamma - 1}\frac{1}{\gamma M_1^{2}} + \frac{1}{2} = \frac{\gamma}{\gamma - 1}k(\frac{1}{\gamma M_1^{2}} + 1 - k) + \frac{k^2}{2}\]
\[\frac{1}{\gamma - 1} M_1^{-2}(1 - k) + \frac{1}{2} (1 - k^2) = \frac{\gamma}{\gamma - 1}k(1 - k) \]
Reducing by $(1-k)$:
(It is obvious that such a reduce was supposed to happen due to Bezout's theorem)
\[\frac{1}{\gamma - 1}M_1^{-2} + \frac{1}{2} (1+k) = \frac{\gamma}{\gamma - 1}k\]
\[\frac{1}{\gamma - 1}M_1^{-2} + \frac{1}{2} = k(\frac{\gamma}{\gamma - 1} - \frac{1}{2})\]
Finally:
\[k = \frac{\frac{1}{\gamma - 1}M_1^{-2} + \frac{1}{2}}{\frac{\gamma}{\gamma - 1} - \frac{1}{2}}\]
\[ k = \frac{v_2}{v_1} = \frac{M_1^2(\gamma - 1) + 2 }{M_1^2(\gamma + 1)},\]

A4  ^?? Show that the ratio of pressures on different sides of the front is equal to:\[\frac{P_2}{P_1} = \frac{2\gamma M_1^2 - (\gamma - 1)}{\gamma + 1}\]

\[\frac{P_2}{P_1} = \frac{P_1 - \rho_1 v_1 ^2 (k - 1)}{P_1} = 1 - \gamma M_1^2\cdot(k - 1) = 1 + \gamma M_1^2- \gamma M_1^2 \cdot \frac{M_1^2(\gamma - 1) + 2 }{M_1^2(\gamma + 1)} = \]
\[ = 1 - \frac{2\gamma}{\gamma + 1} + \gamma M_1^2(1 - \frac{\gamma - 1}{\gamma + 1}) = \frac{2 \gamma M_1^2 +(\gamma - 1)}{\gamma + 1}\]

B1  ^?? Prove that the tangential component of the air velocity is preserved when crossing the border.

Consider the forces acting on a small volume of gas crossing the front. The area of its section by the plane of the front is equal to $S$. The gas to the left of the front presses on our piece with a force of $F_1 = P_1 S$ in a direction perpendicular to the boundary. Similarly, there is an oppositely directed force $F_2 = P_2 S$. These are all forces that act on the allocated volume. We get that there are no forces operating along the front. It follows from this that the tangential component of momentum, and hence velocity, is preserved.

B2  ^?? Using the above results, get the so-called $\theta -\beta - M$ equation:
\[\tan{\theta} = 2 \cot{\beta} \frac{M_1^2 \sin^2{\beta} - 1}{M_1^2(\gamma + \cos{2\beta} )+ 2} \tag{1}\]

The condition for preserving the tangential component:

\[ v_2 \cos{(\beta - \theta)} = v_1 \cos{\beta} \]
The equation from Part A:
\[v_2 \sin{(\beta - \theta)} = v_1 \sin{\beta} \cdot \frac{(\gamma - 1)M_{1}^2 \sin^2{\beta} + 2 }{(\gamma + 1)M_{1}^2 \sin^2{\beta}}\]
Let's divide one by the other:
\[\tan{(\beta - \theta)} = \tan {\beta} \cdot \frac{(\gamma - 1)M_{1}^2 \sin^2{\beta} + 2 }{(\gamma + 1)M_{1}^2 \sin^2{\beta}}\]
Convert:
\[(\tan{\beta} - \tan{\theta})(\gamma + 1)M_{1}^2 \sin^2{\beta} = (1 + \tan{\beta}\tan{\theta})\tan {\beta} \cdot ((\gamma - 1)M_{1}^2 \sin^2{\beta} + 2 )\]
Grouping:
\[\tan{\beta} \cdot \big((\gamma + 1)M_{1}^2 \sin^2{\beta} - (\gamma - 1)M_{1}^2 \sin^2{\beta} - 2\big) = \tan{\theta} \cdot \big((\gamma + 1)M_{1}^2 \sin^2{\beta} + \tan^2{\beta}((\gamma - 1)M_{1}^2 \sin^2{\beta} + 2 ) \big)\]
\[\tan{\theta} = \cot{\beta} \frac{2M_1^2\sin^2{\beta} - 2}{M_1^2\big((\gamma + 1)\cos^2{\beta}) + (\gamma - 1)\sin^2{\beta}\big) + 2} = 2 \cot{\beta} \frac{M_1^2 \sin^2{\beta} - 1}{M_1^2(\gamma + \cos{2\beta} )+ 2} \]

B3  ^?? Use the program to plot $\beta$ graphs from $\theta$ at various $M_1$. For $\theta = 10^\circ$ find $M_{min}$, at which the solution of equation (1) for the angle $\beta$ still exists.
The values of $M$ are set in the array $M_{values}$. For convenience, the vertical line $\theta = 10^{\circ}$ is also drawn on the chart. Program output for the values of $M_{values} = [1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0]$ is shown below.

С1  ^?? What is the mach number of the aircraft $M_1$?

The speed of sound at this altitude is $c = \sqrt{\gamma P/\rho} = 295 m/s$. Then the Mach number is $M = 2.03$

С2  ^?? Graphically find two solutions to the $\theta -\beta - M$ equation - $\beta_1$ and $\beta_2$. Use a smaller root in next questions.

С3  ^?? What is the air pressure under the wing of the aircraft $P_2$?

Using the expression from the part A with the normal component of the Mach number we have: \[\frac{P_2}{P_1} = \frac{2\gamma M_1^2 \sin^2{\beta}- (\gamma - 1)}{\gamma + 1} = 1.72\]
From where: $P_2 = 9509~\text{Pa}$

С4  ^?? Find the vertical force $F$ acting on the wing. With what additional cargo $\Delta M$ can the aircraft fly at the given altitude?

The force acting on the right end has no vertical component. Then, considering the angle $\theta$ small, \[F = (P_2 - P_1)HL = 59700~\text{N}\] The mass of the additional cargo is found by the formula: \[(M + \Delta M )g = F\]

From where \[\Delta M =\frac{F}{g} - M =1160~\text{kg} \]

The proposed model is rather inaccurate, but, as we can see, it gives force values very close to real ones.