diff --git a/content/blog/1d-collisions/index.md b/content/blog/1d-collisions/index.md
index 88d7a7a..c5310af 100644
--- a/content/blog/1d-collisions/index.md
+++ b/content/blog/1d-collisions/index.md
@@ -9,15 +9,20 @@ Consider two particles, designated as particles `1` and `2`, with respective mas
 
 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:
+
 $$
 \frac{m_{1}{v_{1}}^2}{2} + \frac{m_{2}{v_{2}}^2}{2} = \frac{m_{1}{v'_{1}}^2}{2} + \frac{m_{2}{v'_{2}}^2}{2}
 $$
 
+The aforementioned equations can be transformed into the following system:
+
 $$
 \begin{cases}
 m_{1}({v_{1}} - {v'_{1}}) = m_{2}({v'_{2}} - {v_{2}}) \\
@@ -25,22 +30,32 @@ m_{1}({v_{1}}^2 - {v'_{1}}^2) = m_{2}({v'_{2}}^2 - {v_{2}}^2)
 \end{cases}
 $$
 
+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:
+
 $$
 \begin{cases}
 v'_1 = v_2 + v'_2 - v_1 \\
@@ -48,6 +63,8 @@ v'_2 = v_1 + v'_1 - v_2
 \end{cases}
 $$
 
+The law of conservation of momentum then yields the following:
+
 $$
 \begin{cases}
 m_1v_1 + m_2v_2 = m_1{v'}_1 + m_2(v_1 + {v'}_1 - v_2) \\
@@ -55,6 +72,8 @@ m_1v_1 + m_2v_2 = m_1(v_2 + v'_2 - v_1) + m_2v'_2
 \end{cases}
 $$
 
+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} \\