Правка разделов «Statement», «Answer»
en/2.3.2.md
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| ### Statement | |||
| − | $2.3.2.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
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| $2.3.2.$ Determine the force acting on a particle of mass $m$ in a gap of width $l$ between grid electrodes, if its velocity changed from the value $v_1$ at the first electrode to the value $v_2$ at the second electrode. How can one determine the direction of the force acting on the particle from its velocity values? | |||
| ### Solution | |||
| From the velocity values of the particle, we can determine its kinetic energy in the two states: | |||
| $$E_{k1} = \frac{mv_1^2}{2}, \quad E_{k2} = \frac{mv_2^2}{2}$$ | |||
| Two situations are possible: the particle increases its speed, which is possible when the direction of the force coincides with the velocity vector, or the particle decreases its speed, when the velocity vector and the direction of the force are opposite. | |||
| In either case: | |||
| $$\Delta E_k = Fl \implies F = \frac{\Delta E_k}{l} = \frac{m}{2l}(v_2^2 - v_1^2)$$ | |||
| If $v_2 > v_1$, the particle accelerates and the projection of the force on the direction of motion is $F > 0$. If $v_2 < v_1$, the particle decelerates and, correspondingly, $F < 0$. | |||
| @@ -22,7 +16,3 @@Solution | |||
| #### Answer | |||
| $F = \frac{m}{2l}(v_2^2 - v_1^2)$. If $v_2 > v_1$, the particle accelerates and the projection of the force on the direction of motion is $F > 0$. If $v_2 < v_1$, the particle decelerates and $F < 0$. | |||
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| − | #### Answer | ||
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| − | [Insert a concise answer or boxed result] | ||
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| ### Statement | ### Statement | ||
| $2.3.2.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.3.2.$ Determine the force acting on a particle of mass $m$ in a gap of width $l$ between grid electrodes, if its velocity changed from the value $v_1$ at the first electrode to the value $v_2$ at the second electrode. How can one determine the direction of the force acting on the particle from its velocity values? | $2.3.2.$ Determine the force acting on a particle of mass $m$ in a gap of width $l$ between grid electrodes, if its velocity changed from the value $v_1$ at the first electrode to the value $v_2$ at the second electrode. How can one determine the direction of the force acting on the particle from its velocity values? | ||
| ### Solution | ### Solution | ||
| From the velocity values of the particle, we can determine its kinetic energy in the two states: | From the velocity values of the particle, we can determine its kinetic energy in the two states: | ||
| $$E_{k1} = \frac{mv_1^2}{2}, \quad E_{k2} = \frac{mv_2^2}{2}$$ | $$E_{k1} = \frac{mv_1^2}{2}, \quad E_{k2} = \frac{mv_2^2}{2}$$ | ||
| Two situations are possible: the particle increases its speed, which is possible when the direction of the force coincides with the velocity vector, or the particle decreases its speed, when the velocity vector and the direction of the force are opposite. | Two situations are possible: the particle increases its speed, which is possible when the direction of the force coincides with the velocity vector, or the particle decreases its speed, when the velocity vector and the direction of the force are opposite. | ||
| In either case: | In either case: | ||
| $$\Delta E_k = Fl \implies F = \frac{\Delta E_k}{l} = \frac{m}{2l}(v_2^2 - v_1^2)$$ | $$\Delta E_k = Fl \implies F = \frac{\Delta E_k}{l} = \frac{m}{2l}(v_2^2 - v_1^2)$$ | ||
| If $v_2 > v_1$, the particle accelerates and the projection of the force on the direction of motion is $F > 0$. If $v_2 < v_1$, the particle decelerates and, correspondingly, $F < 0$. | If $v_2 > v_1$, the particle accelerates and the projection of the force on the direction of motion is $F > 0$. If $v_2 < v_1$, the particle decelerates and, correspondingly, $F < 0$. | ||
| @@ -22,7 +16,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $F = \frac{m}{2l}(v_2^2 - v_1^2)$. If $v_2 > v_1$, the particle accelerates and the projection of the force on the direction of motion is $F > 0$. If $v_2 < v_1$, the particle decelerates and $F < 0$. | $F = \frac{m}{2l}(v_2^2 - v_1^2)$. If $v_2 > v_1$, the particle accelerates and the projection of the force on the direction of motion is $F > 0$. If $v_2 < v_1$, the particle decelerates and $F < 0$. | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||