4.5 KiB
2. Relevance
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 work1.
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It can be used as a basis for future projects.
3. The consequences of collisions
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).
3.1 With particle
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.
In any collision, momentum is conserved. An elastic collision is an collision, as a result of which the total kinetic energy of the colliding particles is conserved.
The conservation of momentum before and after the collision is expressed by the following equation:
$$
m_{1}\vec{v}{1} + m{2}\vec{v}{2} = m{1}\vec{v'}{1} + m{2}\vec{v'}_{2}
And the conservation of the total kinetic energy before and after the collision is expressed by the following equation:
$$
\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}
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:
$$
\begin{cases}
m_{1}({v_{1}} - {v'{1}}) = m{2}({v'{2}} - {v{2}}) \
m_{1}({v_{1}}^2 - {v'{1}}^2) = m{2}({v'{2}}^2 - {v{2}}^2)
\end{cases}
The division of each side of the top equation by each side of the bottom equation yields:
$$
\frac{v_{1} - v'{1}}{{v{1}^2} - {v'{1}}^2} = \frac{v'{2} - v_{2}}{{{v'{2}}^2} - {v{2}}^2}
The preceding equation may be transformed using the difference of two squares formula to obtain the following:
$$
\frac{v_{1} - v'_{1}}{(v_1 + v'1)(v_1 - v'1)} = \frac{v'{2} - v{2}}{(v'_2 - v_2)(v'_2 + v_2)}
This expression can be transformed as follows:
$$
\frac{1}{v_1 + v'_1} = \frac{1}{v'_2 + v_2}
From this expression follows:
$$
v_1 + v'_1 = v'_2 + v_2
From this equation, the following system can be derived:
$$
\begin{cases}
v'_1 = v_2 + v'_2 - v_1 \
v'_2 = v_1 + v'_1 - v_2
\end{cases}
The law of conservation of momentum then yields the following:
$$
\begin{cases}
m_1v_1 + m_2v_2 = m_1{v'}_1 + m_2(v_1 + {v'}_1 - v_2) \
m_1v_1 + m_2v_2 = m_1(v_2 + v'_2 - v_1) + m_2v'_2
\end{cases}
From this, we can derive the velocity of the particles following the collision as follows:
$$
\begin{cases}
v'_1 = \frac{2m_2v_2 + v_1(m_1-m_2)}{m_1+m_2} \
v'_2 = \frac{2m_1v_1 + v_2(m_2-m_1)}{m_1+m_2}
\end{cases}
3.2 With wall
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.
In order to proceed, we will utilize the formulas derived in section 3.1.
The velocity of the particle will take the following form:
$$
v' = \lim\limits_{m_w \to \infty} \frac{2m_wv_w + v(m-m_w)}{m+m_w} = \lim\limits_{m_w \to \infty} \frac{v(m-m_w)}{m+m_w} = -v
The velocity of the wall will take the following form:
$$
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
This leads to the conclusion that the wall will not change its position, but the particle will change its velocity value to the opposite.
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 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.
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Galperin G. A., "PLAYING POOL WITH π (THE NUMBER π FROM A BILLIARD POINT OF VIEW)", Regular and Chaotic Dynamics, 2003, Volume 8, Number 4, pp. 375-394 DOI: 10.1070/RD2003v008n04ABEH000252 ↩︎