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+---
+title: "Elastic collision"
+date: 2021-10-04
+categories:
+- Physics
+- Classical mechanics
+layout: "concept"
+---
+
+In an **elastic collision**,
+the sum of the colliding objects' kinetic energies
+is the same before and after the collision.
+In contrast, in an **inelastic collision**,
+some of that energy is converted into another form,
+for example heat.
+
+
+## One dimension
+
+In 1D, not only the kinetic energy is conserved, but also the total momentum.
+Let $v_1$ and $v_2$ be the initial velocities of objects 1 and 2,
+and $v_1'$ and $v_2'$ their velocities afterwards:
+
+$$\begin{aligned}
+    \begin{cases}
+        \quad\! m_1 v_1 +\:\:\: m_2 v_2
+        = \quad m_1 v_1' +\:\:\: m_2 v_2'
+        \\
+        \displaystyle\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2
+        = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2
+    \end{cases}
+\end{aligned}$$
+
+After some rearranging,
+these two equations can be written as follows:
+
+$$\begin{aligned}
+    \begin{cases}
+        m_1 (v_1 - v_1')
+        \qquad\quad\:\;\; = m_2 (v_2' - v_2)
+        \\
+        m_1 (v_1 - v_1') (v_1 + v_1')
+        = m_2 (v_2' - v_2) (v_2 + v_2')
+    \end{cases}
+\end{aligned}$$
+
+Using the first equation to replace $m_1 (v_1 \!-\! v_1')$
+with $m_2 (v_2 \!-\! v_2')$ in the second:
+
+$$\begin{aligned}
+    m_2 (v_1 + v_1') (v_2' - v_2)
+    = m_2 (v_2 + v_2') (v_2' - v_2)
+\end{aligned}$$
+
+Dividing out the common factors
+then leads us to a simplified system of equations:
+
+$$\begin{aligned}
+    \begin{cases}
+        \qquad\;\; v_1 + v_1'
+        = v_2 + v_2'
+        \\
+        m_1 v_1 + m_2 v_2
+        = m_1 v_1' + m_2 v_2'
+    \end{cases}
+\end{aligned}$$
+
+Note that the first relation is equivalent to $v_1 - v_2 = v_2' - v_1'$,
+meaning that the objects' relative velocity
+is reversed by the collision.
+Moving on, we replace $v_1'$ in the second equation:
+
+$$\begin{aligned}
+    m_1 v_1 + m_2 v_2
+    &= m_1 (v_2 + v_2' - v_1) + m_2 v_2'
+    \\
+    (m_1 + m_2) v_2'
+    &= 2 m_1 v_1 + (m_2 - m_1) v_2
+\end{aligned}$$
+
+Dividing by $m_1 + m_2$,
+and going through the same process for $v_1'$,
+we arrive at:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            v_1'
+            &= \frac{(m_1 - m_2) v_1 + 2 m_2 v_2}{m_1 + m_2}
+            \\
+            v_2'
+            &= \frac{2 m_1 v_1 + (m_2 - m_1) v_2}{m_1 + m_2}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+To analyze this result,
+for practicality, we simplify it by setting $v_2 = 0$.
+In that case:
+
+$$\begin{aligned}
+    v_1'
+    = \frac{(m_1 - m_2) v_1}{m_1 + m_2}
+    \qquad \quad
+    v_2'
+    = \frac{2 m_1 v_1}{m_1 + m_2}
+\end{aligned}$$
+
+How much of its energy and momentum does object 1 transfer to object 2?
+The following ratios compare $v_1$ and $v_2'$ to quantify the transfer:
+
+$$\begin{aligned}
+    \frac{m_2 v_2'}{m_1 v_1}
+    = \frac{2 m_2}{m_1 + m_2}
+    \qquad \quad
+    \frac{m_2 v_2'^2}{m_1 v_1^2}
+    = \frac{4 m_1 m_2}{(m_1 + m_2)^2}
+\end{aligned}$$
+
+If $m_1 = m_2$, both ratios reduce to $1$,
+meaning that all energy and momentum is transferred,
+and object 1 is at rest after the collision.
+Newton's cradle is an example of this.
+
+If $m_1 \ll m_2$, object 1 simply bounces off object 2,
+barely transferring any energy.
+Object 2 ends up with twice object 1's momentum,
+but $v_2'$ is very small and thus negligible:
+
+$$\begin{aligned}
+    \frac{m_2 v_2'}{m_1 v_1}
+    \approx 2
+    \qquad \quad
+    \frac{m_2 v_2'^2}{m_1 v_1^2}
+    \approx \frac{4 m_1}{m_2}
+\end{aligned}$$
+
+If $m_1 \gg m_2$, object 1 barely notices the collision,
+so not much is transferred to object 2:
+
+$$\begin{aligned}
+    \frac{m_2 v_2'}{m_1 v_1}
+    \approx \frac{2 m_2}{m_1}
+    \qquad \quad
+    \frac{m_2 v_2'^2}{m_1 v_1^2}
+    \approx \frac{4 m_2}{m_1}
+\end{aligned}$$
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.