Edits to “Statement”, “Answer”

Valter edited Sep 3, 2026, 6:33 PM

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revision #20518 parent #20517 ← older

en/2.2.3.md +0 −10

		@@ -1,11 +1,5 @@
			### 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:
			$$\beta = \pi - \frac{1}{2}(\pi - \alpha) = \frac{1}{2}(\pi + \alpha)$$
		@@ -44,7 +38,3 @@Solution

			#### 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]

	@@ -1,11 +1,5 @@
	### Statement		### 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?		$2.2.3.$ 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		### Solution

	When turning on the engine with a thrust force $F$ for a time $\Delta t$, the ship's momentum will be:		When turning on the engine with a thrust force $F$ for a time $\Delta t$, the ship's momentum will be:
	$$\vec{p}_1 = \vec{p} + \Delta\vec{p} \quad (1)$$		$$\vec{p}_1 = \vec{p} + \Delta\vec{p} \quad (1)$$
	where		where
	$$\Delta\vec{p} = \vec{F} \cdot \Delta t \quad (2)$$		$$\Delta\vec{p} = \vec{F} \cdot \Delta t \quad (2)$$

	Let's choose the direction of the x-axis and project equation (2):		Let's choose the direction of the x-axis and project equation (2):
	$$\Delta p = F \cdot \Delta t \quad (3)$$		$$\Delta p = F \cdot \Delta t \quad (3)$$

	On the other hand, from equation (1) in projection onto the x-axis:		On the other hand, from equation (1) in projection onto the x-axis:
	$$\Delta p = 2p \cdot \sin\frac{\alpha}{2} \quad (4)$$		$$\Delta p = 2p \cdot \sin\frac{\alpha}{2} \quad (4)$$

	Equating the right sides of equations (3) and (4):		Equating the right sides of equations (3) and (4):
	$$F \cdot \Delta t = 2p \cdot \sin\frac{\alpha}{2}$$		$$F \cdot \Delta t = 2p \cdot \sin\frac{\alpha}{2}$$

	From where the required engine burn time is:		From where the required engine burn time is:
	$$\Delta t = \frac{2p \cdot \sin\frac{\alpha}{2}}{F}$$		$$\Delta t = \frac{2p \cdot \sin\frac{\alpha}{2}}{F}$$

	The axis of the engine must be oriented to the original direction at an angle:		The axis of the engine must be oriented to the original direction at an angle:
	$$\beta = \frac{\pi}{2} + \frac{\alpha}{2}$$		$$\beta = \frac{\pi}{2} + \frac{\alpha}{2}$$

	<b>Alternative solution:</b>		<b>Alternative solution:</b>

	It is necessary to add momentum:		It is necessary to add momentum:
	$$\Delta p = p\sqrt{2 - 2\cos\alpha} = 2p \cdot \sin\left(\frac{\alpha}{2}\right)$$		$$\Delta p = p\sqrt{2 - 2\cos\alpha} = 2p \cdot \sin\left(\frac{\alpha}{2}\right)$$
	Then the time:		Then the time:
	$$\Delta t = \frac{\Delta p}{F} = \frac{2p}{F} \cdot \sin\left(\frac{\alpha}{2}\right)$$		$$\Delta t = \frac{\Delta p}{F} = \frac{2p}{F} \cdot \sin\left(\frac{\alpha}{2}\right)$$

	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 $\alpha$:		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 $\alpha$:
	$$\frac{1}{2}(\pi - \alpha)$$		$$\frac{1}{2}(\pi - \alpha)$$
	The engine is oriented in the opposite direction to the force, meaning:		The engine is oriented in the opposite direction to the force, meaning:
	$$\beta = \pi - \frac{1}{2}(\pi - \alpha) = \frac{1}{2}(\pi + \alpha)$$		$$\beta = \pi - \frac{1}{2}(\pi - \alpha) = \frac{1}{2}(\pi + \alpha)$$
	@@ -44,7 +38,3 @@Solution

	#### Answer		#### Answer
	$$\Delta t = \frac{2p \cdot \sin\frac{\alpha}{2}}{F}; \quad \beta = \frac{\pi}{2} + \frac{\alpha}{2}$$		$$\Delta t = \frac{2p \cdot \sin\frac{\alpha}{2}}{F}; \quad \beta = \frac{\pi}{2} + \frac{\alpha}{2}$$

	#### Answer

	[Insert a concise answer or boxed result]