$2.2.3.$ 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: $$\beta = \frac{\pi}{2} + \frac{\alpha}{2}$$
Alternative solution:
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: $$\beta = \pi - \frac{1}{2}(\pi - \alpha) = \frac{1}{2}(\pi + \alpha)$$