Новое решение

Valter правка от 30 авг. 2026 г., 19:12

правка #20422 позже →

en/2.3.1.md +28 −0

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		+	### Statement
		+
		+	$2.3.1.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	### Statement
		+
		+	$2.3.1.$ A beam of charged particles of various masses, having the same velocity $v$, is directed along the normal to two grid electrodes, between which the same force $F$ acts on each particle. At what minimum particle mass in the beam will all of them reach the second grid, if the gap width between the electrodes is $\ell$?
		+
		+	### Solution
		+
		+	According to the problem statement, it is required to find the minimum mass of the particles in the beam at which all of them will reach the second grid. In this case, the work of the force between the grid electrodes over the distance $\ell$ is equal to the change in the kinetic energy of the particles:
		+	$$F\ell = \Delta E_k = \frac{mv^2}{2},$$
		+	here it is taken into account that the work of the force $F$ is negative, and the change in kinetic energy is:
		+	$$\Delta E_k = 0 - \frac{mv^2}{2}$$
		+
		+	Then the minimum mass of the particles is:
		+	$$m = \frac{2F\ell}{v^2}$$
		+
		+	<b>Note:</b> the problem can also be solved dynamically, using the impulse of force. Try to reach the answer independently using this method.
		+
		+	#### Answer
		+	The minimum mass of the particles for them to reach the second grid is $m = \frac{2F\ell}{v^2}$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$2.3.1.$ [Insert the problem statement]

			### Solution

			### Statement

			$2.3.1.$ A beam of charged particles of various masses, having the same velocity $v$, is directed along the normal to two grid electrodes, between which the same force $F$ acts on each particle. At what minimum particle mass in the beam will all of them reach the second grid, if the gap width between the electrodes is $\ell$?

			### Solution

			According to the problem statement, it is required to find the minimum mass of the particles in the beam at which all of them will reach the second grid. In this case, the work of the force between the grid electrodes over the distance $\ell$ is equal to the change in the kinetic energy of the particles:
			$$F\ell = \Delta E_k = \frac{mv^2}{2},$$
			here it is taken into account that the work of the force $F$ is negative, and the change in kinetic energy is:
			$$\Delta E_k = 0 - \frac{mv^2}{2}$$

			Then the minimum mass of the particles is:
			$$m = \frac{2F\ell}{v^2}$$

			<b>Note:</b> the problem can also be solved dynamically, using the impulse of force. Try to reach the answer independently using this method.

			#### Answer
			The minimum mass of the particles for them to reach the second grid is $m = \frac{2F\ell}{v^2}$

			#### Answer

			[Insert a concise answer or boxed result]