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@ -16,7 +16,7 @@ There are numerous reasons why this project is relevant, as well as a wide range
## 3. The consequences of collisions ## 3. The consequences of collisions
Firstly, it is essential to comprehend how the velocity of the bodies after collision is calculated. We will solve the problem in the elastic collision model (one-dimensional Newtonian). Firstly, it is essential to comprehend how the velocity of the body after collision is calculated. We will solve the problem in the elastic collision model (one-dimensional Newtonian).
### 3.1 With body ### 3.1 With body
@ -98,11 +98,11 @@ $$
### 3.2 With wall ### 3.2 With wall
Consider bodie and a wall with respective masses $m$ and $m_w \rightarrow \infty$. Before the collision, the velocities of the bodie and the wall are $v$ and $v_w = 0$, respectively. After the collision, the velocities of the bodie and the wall are $v'$ and $v_w'$, respectively. Consider body and a wall with respective masses $m$ and $m_w \rightarrow \infty$. Before the collision, the velocities of the body and the wall are $v$ and $v_w = 0$, respectively. After the collision, the velocities of the body and the wall are $v'$ and $v_w'$, respectively.
In order to proceed, we will utilize the formulas derived in section **3.1**. In order to proceed, we will utilize the formulas derived in section **3.1**.
The velocity of the bodie will take the following form: The velocity of the body will take the following form:
$$ $$
v' = \lim\limits_{m_w \to \infty} \frac{2m_wv_w + v(m-m_w)}{m+m_w} = \lim\limits_{m_w \to \infty} \frac{v(m-m_w)}{m+m_w} = -v v' = \lim\limits_{m_w \to \infty} \frac{2m_wv_w + v(m-m_w)}{m+m_w} = \lim\limits_{m_w \to \infty} \frac{v(m-m_w)}{m+m_w} = -v
@ -114,7 +114,7 @@ $$
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 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 bodie will change its velocity value to the opposite. This leads to the conclusion that the wall will not change its position, but the body will change its velocity value to the opposite.
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