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+### Statement
+
+$2.2.3.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$2.2.3.$ A spaceship must, having changed course, move with the previous momentum magnitude $p$ at an angle $\alpha$ to the original direction. For what minimum time is it necessary to turn on the engine with a thrust force $F$, and how should the axis of the engine be oriented in this case?
+
+### Solution
+
+When turning on the engine with a thrust force $F$ for a time $\Delta t$, the ship's momentum will be:
+$$\vec{p}_1 = \vec{p} + \Delta\vec{p} \quad (1)$$
+where
+$$\Delta\vec{p} = \vec{F} \cdot \Delta t \quad (2)$$
+
+Let's choose the direction of the x-axis and project equation (2):
+$$\Delta p = F \cdot \Delta t \quad (3)$$
+
+On the other hand, from equation (1) in projection onto the x-axis:
+$$\Delta p = 2p \cdot \sin\frac{\alpha}{2} \quad (4)$$
+
+Equating the right sides of equations (3) and (4):
+$$F \cdot \Delta t = 2p \cdot \sin\frac{\alpha}{2}$$
+
+From where the required engine burn time is:
+$$\Delta t = \frac{2p \cdot \sin\frac{\alpha}{2}}{F}$$
+
+The axis of the engine must be oriented to the original direction at an angle:
+$$\beta = \frac{\pi}{2} + \frac{\alpha}{2}$$
+
+<b>Alternative solution:</b>
+
+It is necessary to add momentum:
+$$\Delta p = p\sqrt{2 - 2\cos\alpha} = 2p \cdot \sin\left(\frac{\alpha}{2}\right)$$
+Then the time:
+$$\Delta t = \frac{\Delta p}{F} = \frac{2p}{F} \cdot \sin\left(\frac{\alpha}{2}\right)$$
+
+The angle between the initial velocity and the force $F$ will be equal to the base angle of an isosceles triangle with the vertex angle $\alpha$:
+$$\frac{1}{2}(\pi - \alpha)$$
+The engine is oriented in the opposite direction to the force, meaning:
+$$\beta = \pi - \frac{1}{2}(\pi - \alpha) = \frac{1}{2}(\pi + \alpha)$$
+
+#### Answer
+$$\Delta t = \frac{2p \cdot \sin\frac{\alpha}{2}}{F}; \quad \beta = \frac{\pi}{2} + \frac{\alpha}{2}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]