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131 lines
4.7 KiB
Markdown
131 lines
4.7 KiB
Markdown
# Elastic Collisions (One Dimension)
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## 1. Introduction
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## 2. Relevance
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There are numerous reasons why this project is relevant, as well as a wide range of potential applications. The following will outline the primary reasons and applications:
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- It can be used in the educational process. For instance, it can be used to illustrate some principles of mechanics to students and to visualize some physics problems.
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- It can be used to simulate some physical experiments. For instance, it can be used to simulate the one described in _G. A. Galperin's work_[^1].
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- It can be used as a basis for future projects.
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## 3. The consequences of collisions
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Firstly, it is essential to comprehend how the velocity of the particles after collision is calculated. We will solve the problem in the elastic collision model (one-dimensional Newtonian).
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### 3.1 With particle
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Consider two particles, designated as particles `1` and `2`, with respective masses $m_1$ and $m_2$. Before the collision, the velocities of particles 1 and 2 are $v_1$ and $v_2$, respectively. After the collision, the velocities of particles 1 and 2 are $v_1'$ and $v_2'$, respectively.
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In any collision, momentum is conserved. A collision between to particles is said to be elastic if it involves no change in their internal state. Accordingly, when the law of conservation of energy is applied to such a collision, the internal energy of the particles may be neglected.[^2]
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The conservation of momentum before and after the collision is expressed by the following equation:
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$$
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m_{1}\vec{v}_{1} + m_{2}\vec{v}_{2} = m_{1}\vec{v'}_{1} + m_{2}\vec{v'}_{2}
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$$
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And the conservation of the total kinetic energy before and after the collision is expressed by the following equation:
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$$
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\frac{m_{1}{v_{1}}^2}{2} + \frac{m_{2}{v_{2}}^2}{2} = \frac{m_{1}{v'_{1}}^2}{2} + \frac{m_{2}{v'_{2}}^2}{2}
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$$
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If we consider the law of conservation of momentum in projection on the X-axis, we can transform the previous equations into the following system:
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$$
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\begin{cases}
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m_{1}({v_{1}} - {v'_{1}}) = m_{2}({v'_{2}} - {v_{2}}) \\
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m_{1}({v_{1}}^2 - {v'_{1}}^2) = m_{2}({v'_{2}}^2 - {v_{2}}^2)
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\end{cases}
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$$
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The division of each side of the top equation by each side of the bottom equation yields:
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$$
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\frac{v_{1} - v'_{1}}{{v_{1}^2} - {v'_{1}}^2} = \frac{v'_{2} - v_{2}}{{{v'_{2}}^2} - {v_{2}}^2}
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$$
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The preceding equation may be transformed using the difference of two squares formula to obtain the following:
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$$
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\frac{v_{1} - v'_{1}}{(v_1 + v'_1)(v_1 - v'_1)} = \frac{v'_{2} - v_{2}}{(v'_2 - v_2)(v'_2 + v_2)}
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$$
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This expression can be transformed as follows:
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$$
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\frac{1}{v_1 + v'_1} = \frac{1}{v'_2 + v_2}
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$$
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From this expression follows:
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$$
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v_1 + v'_1 = v'_2 + v_2
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$$
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From this equation, the following system can be derived:
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$$
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\begin{cases}
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v'_1 = v_2 + v'_2 - v_1 \\
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v'_2 = v_1 + v'_1 - v_2
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\end{cases}
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$$
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The law of conservation of momentum then yields the following:
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$$
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\begin{cases}
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m_1v_1 + m_2v_2 = m_1{v'}_1 + m_2(v_1 + {v'}_1 - v_2) \\
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m_1v_1 + m_2v_2 = m_1(v_2 + v'_2 - v_1) + m_2v'_2
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\end{cases}
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$$
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From this, we can derive the velocity of the particles following the collision as follows:
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$$
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\begin{cases}
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v'_1 = \frac{2m_2v_2 + v_1(m_1-m_2)}{m_1+m_2} \\
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v'_2 = \frac{2m_1v_1 + v_2(m_2-m_1)}{m_1+m_2}
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\end{cases}
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$$
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### 3.2 With wall
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Consider particle and a wall with respective masses $m$ and $m_w \rightarrow \infty$. Before the collision, the velocities of the particle and the wall are $v$ and $v_w = 0$, respectively. After the collision, the velocities of the particle and the wall are $v'$ and $v_w'$, respectively.
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In order to proceed, we will utilize the formulas derived in section **3.1**.
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The velocity of the particle will take the following form:
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$$
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v' = \lim\limits_{m_w \to \infty} \frac{2m_wv_w + v(m-m_w)}{m+m_w} = \lim\limits_{m_w \to \infty} \frac{v(m-m_w)}{m+m_w} = -v
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$$
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The velocity of the wall will take the following form:
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$$
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v_w' = \lim\limits_{m_w \to \infty} \frac{2mv + v_w(m_w-m)}{m+m_w} = \lim\limits_{m_w \to \infty} \frac{2mv}{m+m_w} = 0
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$$
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This leads to the conclusion that the wall will not change its position, but the particle will change its velocity value to the opposite.
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---
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## Note
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* If you decide to use this project as the basis of your project, it would be advisable to [contact me](https://arbuz.icu/mail) first.
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* If you have any interesting projects that I could be involved in, or if you would like to contact me, you can do so [here](https://arbuz.icu/mail).
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## Reference
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[^1]: Galperin, G. A., _[PLAYING POOL WITH π (THE NUMBER π FROM A BILLIARD POINT OF VIEW)](http://rcd.ics.org.ru/RD2003v008n04ABEH000252/)_, _Regular and Chaotic Dynamics_, 2003, vol. 8, no. 4, pp. 375-394.
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[^2]: Landau, L.D. and Lifshitz, E.M., _Course of Theoretical Physics_, vol. 1: _Mechanics_, Elsevier Science, 1982.
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