AcademyMomentum Systems
Academy
Elastic Collision Models
Level 1 - Physics topic page in Momentum Systems.
Principle
An elastic collision conserves both total momentum and total kinetic energy.
Notation
\(m_1,m_2\)
masses
\(\mathrm{kg}\)
\(u_1,u_2\)
initial velocities
\(\mathrm{m\,s^{-1}}\)
\(v_1,v_2\)
final velocities
\(\mathrm{m\,s^{-1}}\)
\(K\)
kinetic energy
\(\mathrm{J}\)
\(e\)
coefficient of restitution
1
Method
The useful one-dimensional elastic shortcut comes from combining momentum conservation with kinetic-energy conservation.
Momentum
\[m_1(u_1-v_1)=m_2(v_2-u_2)\]
Energy
\[m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)\]
Divide equations
\[u_1+v_1=v_2+u_2\]
This step assumes the relative velocity changes during the collision.
Relative speed
\[v_2-v_1=u_1-u_2\]
The graph shows the equal-mass limiting case: one cart gives its velocity to the other.
For unequal masses, use the same two equations; the velocity-exchange picture is only the symmetric case.
Rules
These are the compact one-dimensional elastic collision results.
Momentum conserved
\[m_1u_1+m_2u_2=m_1v_1+m_2v_2\]
Kinetic conserved
\[\frac12m_1u_1^2+\frac12m_2u_2^2=\frac12m_1v_1^2+\frac12m_2v_2^2\]
Relative speed
\[v_2-v_1=u_1-u_2\]
Rest target
\[v_1=\frac{m_1-m_2}{m_1+m_2}u_1,\quad v_2=\frac{2m_1}{m_1+m_2}u_1\]
Examples
Question
A
\[2.0\,\mathrm{kg}\]
cart moving at \[3.0\,\mathrm{m\,s^{-1}}\]
elastically hits a \[4.0\,\mathrm{kg}\]
cart at rest. Find final velocities.Answer
Use the rest-target formulas.
\[v_1=\frac{2.0-4.0}{6.0}(3.0)=-1.0\,\mathrm{m\,s^{-1}}\]
\[v_2=\frac{2(2.0)}{6.0}(3.0)=2.0\,\mathrm{m\,s^{-1}}\]
Checks
- Momentum and kinetic energy must both match before and after.
- Identical masses exchange velocities in a 1D elastic collision.
- Relative speed of separation equals relative speed of approach.
- Signs are more reliable than direction words.